💡Alright, so I've got this problem about Brenda and Sally running on a circular track. They start at diametrically opposite points, which means they're half a track length apart initially. They run in opposite directions and meet twice: first after Brenda has run 80 meters, and then again after Sally has run an additional 180 meters past their first meeting point. I need to find the length of the track.Okay, let's break this down step by step. First, let's denote the length of the track as ( x ) meters. Since they start diametrically opposite, the initial distance between them is ( frac{x}{2} ) meters.When they first meet, Brenda has run 80 meters. Since they're moving towards each other (because they're running in opposite directions), the sum of the distances they've covered should equal the initial distance between them, which is ( frac{x}{2} ). So, if Brenda has run 80 meters, Sally must have run ( frac{x}{2} - 80 ) meters by the time they meet the first time.Now, after their first meeting, they continue running in their respective directions. They meet again after Sally has run an additional 180 meters past their first meeting point. So, from the first meeting point, Sally runs 180 meters to reach the second meeting point. At this second meeting, the total distance Sally has run since the start is ( frac{x}{2} - 80 + 180 ) meters, which simplifies to ( frac{x}{2} + 100 ) meters. Similarly, Brenda, from the first meeting point, would have run the remaining distance around the track to meet Sally again. Since the track is circular, the distance Brenda runs from the first meeting point to the second meeting point is ( x - left( frac{x}{2} + 100 right) ) meters, which simplifies to ( frac{x}{2} - 100 ) meters.Now, here's the key part: since both Brenda and Sally are running at constant speeds, the ratio of the distances they cover should remain constant. That is, the ratio of Brenda's distance to Sally's distance in the first meeting should be the same as the ratio in the second meeting.From the first meeting:- Brenda ran 80 meters.- Sally ran ( frac{x}{2} - 80 ) meters.So, the ratio is ( frac{80}{frac{x}{2} - 80} ).From the second meeting:- Brenda ran ( frac{x}{2} - 100 ) meters.- Sally ran ( frac{x}{2} + 100 ) meters.So, the ratio is ( frac{frac{x}{2} - 100}{frac{x}{2} + 100} ).Setting these ratios equal to each other:[frac{80}{frac{x}{2} - 80} = frac{frac{x}{2} - 100}{frac{x}{2} + 100}]Now, let's solve this equation for ( x ). First, cross-multiply to eliminate the fractions:[80 left( frac{x}{2} + 100 right) = left( frac{x}{2} - 80 right) left( frac{x}{2} - 100 right)]Expanding both sides:Left side:[80 times frac{x}{2} + 80 times 100 = 40x + 8000]Right side:[left( frac{x}{2} times frac{x}{2} right) - left( frac{x}{2} times 100 right) - left( 80 times frac{x}{2} right) + (80 times 100)]Simplifying:[frac{x^2}{4} - 50x - 40x + 8000 = frac{x^2}{4} - 90x + 8000]So, putting it all together:[40x + 8000 = frac{x^2}{4} - 90x + 8000]Subtracting ( 40x + 8000 ) from both sides:[0 = frac{x^2}{4} - 130x]Multiplying both sides by 4 to eliminate the fraction:[0 = x^2 - 520x]Factoring out an ( x ):[0 = x(x - 520)]So, the solutions are ( x = 0 ) or ( x = 520 ). Since a track can't have a length of 0 meters, the valid solution is ( x = 520 ) meters.Therefore, the length of the track is 520 meters.

💡Alright, so I have this problem where I need to find the point on a given line that's closest to the point (1, 2, 3). The line is defined by the vector equation:[begin{pmatrix} 3 1 2 end{pmatrix} + s begin{pmatrix} -1 4 -2 end{pmatrix}]Okay, let me break this down. First, I know that any point on the line can be expressed as:[begin{pmatrix} x y z end{pmatrix} = begin{pmatrix} 3 - s 1 + 4s 2 - 2s end{pmatrix}]So, for some parameter ( s ), this gives me the coordinates of any point on the line. Now, I need to find the specific value of ( s ) such that the point on the line is closest to (1, 2, 3).I remember that the closest point on a line to a given point is where the vector connecting the given point to the line is perpendicular to the direction vector of the line. That makes sense because the shortest distance from a point to a line is along the perpendicular.So, let's define the vector from (1, 2, 3) to a general point on the line. If the point on the line is ( (3 - s, 1 + 4s, 2 - 2s) ), then the vector from (1, 2, 3) to this point is:[begin{pmatrix} (3 - s) - 1 (1 + 4s) - 2 (2 - 2s) - 3 end{pmatrix} = begin{pmatrix} 2 - s -1 + 4s -1 - 2s end{pmatrix}]This vector should be orthogonal to the direction vector of the line, which is ( begin{pmatrix} -1 4 -2 end{pmatrix} ). To find orthogonality, their dot product should be zero.So, let's compute the dot product:[begin{pmatrix} 2 - s -1 + 4s -1 - 2s end{pmatrix} cdot begin{pmatrix} -1 4 -2 end{pmatrix} = 0]Calculating each component:1. First component: ( (2 - s)(-1) = -2 + s )2. Second component: ( (-1 + 4s)(4) = -4 + 16s )3. Third component: ( (-1 - 2s)(-2) = 2 + 4s )Adding them all up:[(-2 + s) + (-4 + 16s) + (2 + 4s) = 0]Simplify the equation:Combine like terms:- Constants: ( -2 - 4 + 2 = -4 )- ( s ) terms: ( s + 16s + 4s = 21s )So, the equation becomes:[21s - 4 = 0]Solving for ( s ):[21s = 4 s = frac{4}{21}]Okay, so ( s = frac{4}{21} ). Now, I need to substitute this back into the parametric equations of the line to find the coordinates of the closest point.Let's compute each coordinate:1. ( x = 3 - s = 3 - frac{4}{21} = frac{63}{21} - frac{4}{21} = frac{59}{21} )2. ( y = 1 + 4s = 1 + 4 times frac{4}{21} = 1 + frac{16}{21} = frac{21}{21} + frac{16}{21} = frac{37}{21} )3. ( z = 2 - 2s = 2 - 2 times frac{4}{21} = 2 - frac{8}{21} = frac{42}{21} - frac{8}{21} = frac{34}{21} )Wait, hold on. When I calculated ( y ), I think I made a mistake. Let me check that again.( y = 1 + 4s = 1 + 4 times frac{4}{21} = 1 + frac{16}{21} ). Hmm, 1 is ( frac{21}{21} ), so adding ( frac{16}{21} ) gives ( frac{37}{21} ). Wait, but in the initial solution, it was ( frac{25}{21} ). Did I do something wrong?Wait, no. Let me double-check the substitution. The parametric equation for ( y ) is ( 1 + 4s ). So, substituting ( s = frac{4}{21} ):( y = 1 + 4 times frac{4}{21} = 1 + frac{16}{21} ). Converting 1 to ( frac{21}{21} ), so ( frac{21}{21} + frac{16}{21} = frac{37}{21} ). Hmm, that's different from the initial solution's ( frac{25}{21} ). Did I make a mistake?Wait, let me go back. The vector from (1,2,3) to the point on the line was:[begin{pmatrix} 2 - s -1 + 4s -1 - 2s end{pmatrix}]Then, the dot product with the direction vector:[(2 - s)(-1) + (-1 + 4s)(4) + (-1 - 2s)(-2) = 0]Calculating each term:1. ( (2 - s)(-1) = -2 + s )2. ( (-1 + 4s)(4) = -4 + 16s )3. ( (-1 - 2s)(-2) = 2 + 4s )Adding them up:( (-2 + s) + (-4 + 16s) + (2 + 4s) = (-2 - 4 + 2) + (s + 16s + 4s) = (-4) + (21s) = 21s - 4 = 0 )So, ( 21s = 4 ), so ( s = frac{4}{21} ). That seems correct.Then, substituting back into ( y = 1 + 4s ):( 1 + 4 times frac{4}{21} = 1 + frac{16}{21} = frac{37}{21} ). So, why did the initial solution say ( frac{25}{21} )? That must be a mistake.Wait, let me check the initial solution again. It had:[begin{pmatrix} 3 - frac{4}{21} 1 + 4 cdot frac{4}{21} 2 - 2 cdot frac{4}{21} end{pmatrix} = begin{pmatrix} frac{59}{21} frac{25}{21} frac{34}{21} end{pmatrix}]Wait, in the initial solution, ( y = 1 + 4 times frac{4}{21} = 1 + frac{16}{21} = frac{37}{21} ), but it was written as ( frac{25}{21} ). That's incorrect. So, the initial solution had an error in the calculation of ( y ).Therefore, the correct ( y )-coordinate should be ( frac{37}{21} ), not ( frac{25}{21} ). Let me verify that.Wait, no, hold on. Let me re-examine the vector from (1,2,3) to the point on the line. The vector is:[begin{pmatrix} (3 - s) - 1 (1 + 4s) - 2 (2 - 2s) - 3 end{pmatrix} = begin{pmatrix} 2 - s -1 + 4s -1 - 2s end{pmatrix}]So, the vector components are correct. Then, the dot product with the direction vector:[(2 - s)(-1) + (-1 + 4s)(4) + (-1 - 2s)(-2) = 0]Calculations:1. ( (2 - s)(-1) = -2 + s )2. ( (-1 + 4s)(4) = -4 + 16s )3. ( (-1 - 2s)(-2) = 2 + 4s )Adding them:( (-2 + s) + (-4 + 16s) + (2 + 4s) = (-2 - 4 + 2) + (s + 16s + 4s) = (-4) + (21s) = 21s - 4 = 0 )So, ( s = frac{4}{21} ). That seems correct.Substituting back into the parametric equations:1. ( x = 3 - s = 3 - frac{4}{21} = frac{63}{21} - frac{4}{21} = frac{59}{21} )2. ( y = 1 + 4s = 1 + frac{16}{21} = frac{21}{21} + frac{16}{21} = frac{37}{21} )3. ( z = 2 - 2s = 2 - frac{8}{21} = frac{42}{21} - frac{8}{21} = frac{34}{21} )So, the correct point should be ( left( frac{59}{21}, frac{37}{21}, frac{34}{21} right) ). Therefore, the initial solution had an error in calculating ( y ). It incorrectly wrote ( frac{25}{21} ) instead of ( frac{37}{21} ).Wait, but let me double-check the calculation for ( y ). Maybe I made a mistake in the initial substitution.Given ( y = 1 + 4s ), and ( s = frac{4}{21} ):( y = 1 + 4 times frac{4}{21} = 1 + frac{16}{21} ). Converting 1 to ( frac{21}{21} ), so ( frac{21}{21} + frac{16}{21} = frac{37}{21} ). Yes, that's correct.Therefore, the correct closest point is ( left( frac{59}{21}, frac{37}{21}, frac{34}{21} right) ).Wait, but in the initial solution, the answer was ( left( frac{59}{21}, frac{25}{21}, frac{34}{21} right) ). So, there must be an error in the initial solution's calculation of ( y ). It seems like they might have miscalculated ( 4 times frac{4}{21} ) as ( frac{16}{21} ) but then added it incorrectly to 1.Alternatively, perhaps the initial solution had a different approach or made a different mistake. Let me try another method to verify.Another way to find the closest point is to minimize the distance squared from (1,2,3) to a general point on the line. The distance squared is:[D^2 = (3 - s - 1)^2 + (1 + 4s - 2)^2 + (2 - 2s - 3)^2][= (2 - s)^2 + (-1 + 4s)^2 + (-1 - 2s)^2]Expanding each term:1. ( (2 - s)^2 = 4 - 4s + s^2 )2. ( (-1 + 4s)^2 = 1 - 8s + 16s^2 )3. ( (-1 - 2s)^2 = 1 + 4s + 4s^2 )Adding them up:[4 - 4s + s^2 + 1 - 8s + 16s^2 + 1 + 4s + 4s^2][= (4 + 1 + 1) + (-4s - 8s + 4s) + (s^2 + 16s^2 + 4s^2)][= 6 - 8s + 21s^2]To minimize ( D^2 ), take the derivative with respect to ( s ) and set it to zero:[frac{d}{ds}(6 - 8s + 21s^2) = -8 + 42s = 0][42s = 8 s = frac{8}{42} = frac{4}{21}]So, ( s = frac{4}{21} ), which matches the previous result. Therefore, substituting back into the parametric equations:1. ( x = 3 - frac{4}{21} = frac{59}{21} )2. ( y = 1 + 4 times frac{4}{21} = 1 + frac{16}{21} = frac{37}{21} )3. ( z = 2 - 2 times frac{4}{21} = 2 - frac{8}{21} = frac{34}{21} )So, the correct closest point is ( left( frac{59}{21}, frac{37}{21}, frac{34}{21} right) ). Therefore, the initial solution had an error in the ( y )-coordinate calculation.Alternatively, maybe the initial solution used a different method or made a different mistake. Let me check if I can find another way to approach this problem.Another approach is to use the formula for the projection of a vector onto a line. The closest point on the line to the given point can be found by projecting the vector from a point on the line to the given point onto the direction vector of the line.Let me denote:- Point on the line: ( mathbf{a} = begin{pmatrix} 3 1 2 end{pmatrix} )- Direction vector: ( mathbf{v} = begin{pmatrix} -1 4 -2 end{pmatrix} )- Given point: ( mathbf{p} = begin{pmatrix} 1 2 3 end{pmatrix} )The vector from ( mathbf{a} ) to ( mathbf{p} ) is:[mathbf{p} - mathbf{a} = begin{pmatrix} 1 - 3 2 - 1 3 - 2 end{pmatrix} = begin{pmatrix} -2 1 1 end{pmatrix}]The projection of ( mathbf{p} - mathbf{a} ) onto ( mathbf{v} ) is:[text{proj}_{mathbf{v}} (mathbf{p} - mathbf{a}) = left( frac{(mathbf{p} - mathbf{a}) cdot mathbf{v}}{mathbf{v} cdot mathbf{v}} right) mathbf{v}]Calculating the dot products:1. ( (mathbf{p} - mathbf{a}) cdot mathbf{v} = (-2)(-1) + (1)(4) + (1)(-2) = 2 + 4 - 2 = 4 )2. ( mathbf{v} cdot mathbf{v} = (-1)^2 + 4^2 + (-2)^2 = 1 + 16 + 4 = 21 )So, the projection is:[left( frac{4}{21} right) begin{pmatrix} -1 4 -2 end{pmatrix} = begin{pmatrix} -frac{4}{21} frac{16}{21} -frac{8}{21} end{pmatrix}]Therefore, the closest point ( mathbf{q} ) on the line is:[mathbf{a} + text{proj}_{mathbf{v}} (mathbf{p} - mathbf{a}) = begin{pmatrix} 3 1 2 end{pmatrix} + begin{pmatrix} -frac{4}{21} frac{16}{21} -frac{8}{21} end{pmatrix} = begin{pmatrix} 3 - frac{4}{21} 1 + frac{16}{21} 2 - frac{8}{21} end{pmatrix} = begin{pmatrix} frac{59}{21} frac{37}{21} frac{34}{21} end{pmatrix}]This confirms the earlier result. Therefore, the correct closest point is ( left( frac{59}{21}, frac{37}{21}, frac{34}{21} right) ).So, in conclusion, the initial solution had an error in the calculation of the ( y )-coordinate, incorrectly stating ( frac{25}{21} ) instead of the correct ( frac{37}{21} ). The correct closest point is ( left( frac{59}{21}, frac{37}{21}, frac{34}{21} right) ).

💡Okay, so I have this problem where Xiao Ming is attending the live recording of "Chinese Poetry Conference" with his parents and grandparents, making a total of five people. They are sitting in a row of five seats. The condition is that at least one of Xiao Ming's parents must sit next to him. I need to find the total number of different seating arrangements possible under this condition.First, I need to understand the problem clearly. There are five people: Xiao Ming, his father, his mother, his grandfather, and his grandmother. They are sitting in a row of five seats. The requirement is that at least one parent (either father or mother) must be sitting next to Xiao Ming. So, I need to count all possible seating arrangements where this condition is satisfied.To approach this, I think it's helpful to use the principle of complementary counting. That is, instead of directly counting the number of arrangements where at least one parent is next to Xiao Ming, I can subtract the number of arrangements where neither parent is next to Xiao Ming from the total number of possible arrangements.Let me start by calculating the total number of possible seating arrangements without any restrictions. Since there are five people, the total number of ways they can arrange themselves in a row is 5 factorial, which is:5! = 5 × 4 × 3 × 2 × 1 = 120So, there are 120 possible seating arrangements in total.Now, I need to calculate the number of arrangements where neither of Xiao Ming's parents is sitting next to him. Once I have that, I can subtract it from the total to find the number of valid arrangements.To find the number of arrangements where neither parent is next to Xiao Ming, I can think of it as arranging the five people such that Xiao Ming is not adjacent to either his father or his mother.One way to approach this is to first fix Xiao Ming's position and then count the number of ways to arrange the other four people such that his parents are not sitting next to him. However, since Xiao Ming can be in any position, it might be more efficient to use the inclusion-exclusion principle or consider the problem as arranging the non-parents first and then placing Xiao Ming and his parents in the remaining seats.Alternatively, I can model this problem by considering the seats as positions 1 to 5. Xiao Ming can be in any of these positions, and we need to ensure that his parents are not in the adjacent seats.Let me try to break it down step by step.1. **Total number of arrangements without restrictions:** 120.2. **Number of arrangements where neither parent is next to Xiao Ming:** To calculate this, I can consider the following approach: a. First, place Xiao Ming in one of the five seats. b. Then, determine the number of ways to place his parents such that they are not adjacent to Xiao Ming. c. Finally, arrange the remaining two grandparents in the remaining seats.Let's go through each step.**Step 2a: Placing Xiao Ming**Xiao Ming can be placed in any of the five seats. However, depending on where Xiao Ming is placed, the number of adjacent seats varies. For example, if Xiao Ming is at one end (seat 1 or seat 5), there is only one adjacent seat. If he is in the middle seats (2, 3, or 4), there are two adjacent seats.Therefore, I need to consider two cases:- Case 1: Xiao Ming is at one of the ends (seat 1 or seat 5).- Case 2: Xiao Ming is in one of the middle seats (seat 2, 3, or 4).Let me calculate each case separately.**Case 1: Xiao Ming is at one end (seat 1 or seat 5).**There are 2 choices for Xiao Ming's seat (seat 1 or seat 5).Once Xiao Ming is placed at an end, there is only one seat adjacent to him (seat 2 if he is at seat 1, or seat 4 if he is at seat 5). We need to ensure that neither of his parents is in this adjacent seat.So, the adjacent seat can be occupied by either of the two grandparents. There are 2 grandparents, so 2 choices for the adjacent seat.After placing Xiao Ming and one grandparent in the adjacent seat, we have 3 remaining seats and 3 people left (the two parents and the other grandparent). These 3 people can be arranged in the remaining 3 seats in 3! = 6 ways.Therefore, the number of arrangements for this case is:Number of choices for Xiao Ming's seat × number of choices for the adjacent seat × number of arrangements for the remaining people= 2 × 2 × 6 = 24**Case 2: Xiao Ming is in one of the middle seats (seat 2, 3, or 4).**There are 3 choices for Xiao Ming's seat.Each middle seat has two adjacent seats. For example, if Xiao Ming is in seat 2, the adjacent seats are 1 and 3. Similarly, seat 3 has adjacent seats 2 and 4, and seat 4 has adjacent seats 3 and 5.We need to ensure that neither of Xiao Ming's parents is in either of these two adjacent seats.So, the two adjacent seats must be occupied by the two grandparents. There are 2 grandparents, so the number of ways to arrange them in the two adjacent seats is 2! = 2.After placing Xiao Ming and the two grandparents in the adjacent seats, we have 2 remaining seats and 2 people left (the two parents). These 2 people can be arranged in the remaining 2 seats in 2! = 2 ways.Therefore, the number of arrangements for this case is:Number of choices for Xiao Ming's seat × number of arrangements for the adjacent seats × number of arrangements for the remaining people= 3 × 2 × 2 = 12**Total number of arrangements where neither parent is next to Xiao Ming:**= Case 1 + Case 2 = 24 + 12 = 36**Number of arrangements where at least one parent is next to Xiao Ming:**= Total arrangements - arrangements where neither parent is next to Xiao Ming= 120 - 36 = 84Therefore, the total number of different seating arrangements where at least one of Xiao Ming's parents sits next to him is 84.Wait, let me double-check my calculations to ensure I haven't missed anything.In Case 1, Xiao Ming is at the end. There are 2 choices for his seat. The adjacent seat can be occupied by either grandparent (2 choices). Then, the remaining 3 seats are occupied by the two parents and the other grandparent, which can be arranged in 3! = 6 ways. So, 2 × 2 × 6 = 24. That seems correct.In Case 2, Xiao Ming is in the middle. There are 3 choices. The two adjacent seats must be occupied by the two grandparents, which can be arranged in 2! = 2 ways. The remaining two seats are occupied by the two parents, which can be arranged in 2! = 2 ways. So, 3 × 2 × 2 = 12. That also seems correct.Adding them up gives 24 + 12 = 36 arrangements where neither parent is next to Xiao Ming. Subtracting this from the total 120 gives 84 valid arrangements. That seems consistent.Alternatively, I can approach this problem using the principle of inclusion-exclusion.Let me try that method to verify.**Using Inclusion-Exclusion Principle:**We need to count the number of arrangements where at least one parent is next to Xiao Ming. Let's define:- A: The set of arrangements where the father is next to Xiao Ming.- B: The set of arrangements where the mother is next to Xiao Ming.We need to find |A ∪ B|, which is the number of arrangements where either the father or the mother is next to Xiao Ming.By the inclusion-exclusion principle:|A ∪ B| = |A| + |B| - |A ∩ B|So, I need to calculate |A|, |B|, and |A ∩ B|.**Calculating |A|: Number of arrangements where the father is next to Xiao Ming.**Treat Xiao Ming and his father as a single unit. This unit can be arranged in 2 ways (Xiao Ming first or father first). Now, we have four units to arrange: [Xiao Ming-Father], mother, grandfather, grandmother.These four units can be arranged in 4! = 24 ways. However, we need to consider the positions of this unit.But wait, actually, when treating Xiao Ming and his father as a single unit, we have to consider the number of ways this unit can be placed in the row of five seats.The number of ways to arrange the unit is equal to the number of positions the unit can occupy. Since the unit takes up two adjacent seats, there are 4 possible positions: seats 1-2, 2-3, 3-4, or 4-5.For each position, the unit can be arranged in 2 ways (Xiao Ming first or father first). Then, the remaining three people (mother, grandfather, grandmother) can be arranged in the remaining three seats in 3! = 6 ways.Therefore, |A| = 4 × 2 × 6 = 48Similarly, |B| will be the same as |A| because the situation is symmetric for the mother. So, |B| = 48**Calculating |A ∩ B|: Number of arrangements where both the father and the mother are next to Xiao Ming.**This means that Xiao Ming is between his father and mother. So, the three of them form a single unit. The unit can be arranged in 2 ways: father-Xiao Ming-mother or mother-Xiao Ming-father.Now, we have three units to arrange: [Father-Mother-Xiao Ming], grandfather, grandmother.These three units can be arranged in 3! = 6 ways. However, similar to before, we need to consider the number of positions this unit can occupy.The unit takes up three adjacent seats. In a row of five seats, the possible positions for this unit are:- Seats 1-2-3- Seats 2-3-4- Seats 3-4-5So, there are 3 possible positions for the unit.For each position, the unit can be arranged in 2 ways (father first or mother first). Then, the remaining two people (grandfather and grandmother) can be arranged in the remaining two seats in 2! = 2 ways.Therefore, |A ∩ B| = 3 × 2 × 2 = 12**Applying Inclusion-Exclusion:**|A ∪ B| = |A| + |B| - |A ∩ B| = 48 + 48 - 12 = 84This confirms the previous result. Therefore, the number of valid seating arrangements is 84.I think this method also makes sense. By treating the father and Xiao Ming as a unit and calculating the arrangements, then doing the same for the mother, and subtracting the overlap where both parents are next to Xiao Ming, we arrive at the same answer.Just to ensure thoroughness, let me consider another approach: fixing Xiao Ming's position and counting the valid arrangements for each position.**Approach 3: Fixing Xiao Ming's Position**Xiao Ming can be in any of the five seats. For each seat, calculate the number of valid arrangements where at least one parent is adjacent.1. **Xiao Ming in seat 1:** Adjacent seat: seat 2. We need at least one parent in seat 2. Total arrangements without restriction: 4! = 24. Arrangements where neither parent is in seat 2: seat 2 must be occupied by a grandparent (2 choices). The remaining three seats (3,4,5) are occupied by the two parents and the other grandparent, which can be arranged in 3! = 6 ways. So, invalid arrangements: 2 × 6 = 12. Valid arrangements: 24 - 12 = 12.2. **Xiao Ming in seat 2:** Adjacent seats: seat 1 and seat 3. We need at least one parent in seat 1 or seat 3. Total arrangements: 4! = 24. Invalid arrangements: neither parent is in seat 1 or seat 3. So, seats 1 and 3 must be occupied by grandparents. There are 2 grandparents, so arranging them in seats 1 and 3: 2! = 2 ways. The remaining two seats (4 and 5) are occupied by the two parents: 2! = 2 ways. So, invalid arrangements: 2 × 2 = 4. Valid arrangements: 24 - 4 = 20.3. **Xiao Ming in seat 3:** Adjacent seats: seat 2 and seat 4. Similar to seat 2. Total arrangements: 4! = 24. Invalid arrangements: neither parent is in seat 2 or seat 4. Seats 2 and 4 must be occupied by grandparents: 2! = 2 ways. Remaining seats (1 and 5) are occupied by parents: 2! = 2 ways. Invalid arrangements: 2 × 2 = 4. Valid arrangements: 24 - 4 = 20.4. **Xiao Ming in seat 4:** Adjacent seats: seat 3 and seat 5. Similar to seat 2. Total arrangements: 4! = 24. Invalid arrangements: neither parent is in seat 3 or seat 5. Seats 3 and 5 must be occupied by grandparents: 2! = 2 ways. Remaining seats (1 and 2) are occupied by parents: 2! = 2 ways. Invalid arrangements: 2 × 2 = 4. Valid arrangements: 24 - 4 = 20.5. **Xiao Ming in seat 5:** Adjacent seat: seat 4. Similar to seat 1. Total arrangements: 4! = 24. Invalid arrangements: neither parent is in seat 4. Seat 4 must be occupied by a grandparent (2 choices). Remaining seats (1,2,3) are occupied by parents and the other grandparent: 3! = 6 ways. Invalid arrangements: 2 × 6 = 12. Valid arrangements: 24 - 12 = 12.**Total valid arrangements:**= (12) + (20) + (20) + (20) + (12) = 84This method also gives the same result, 84. It seems consistent across different approaches.Therefore, I can confidently conclude that the total number of different seating arrangements where at least one of Xiao Ming's parents sits next to him is 84.**Final Answer**The total number of different seating arrangements is boxed{84}.

💡Okay, so I have this function ( g(x) = 3x^2 - 6x - 8 ), and I need to figure out the largest interval that includes the point ( x = 2 ) where the function can be restricted to make it invertible. Hmm, invertible means that the function has to pass the horizontal line test, right? So, it needs to be one-to-one, meaning each y-value corresponds to exactly one x-value.First, I remember that quadratic functions are parabolas, and they aren't one-to-one over their entire domains because they fail the horizontal line test. They have a vertex, and they either open upwards or downwards. In this case, the coefficient of ( x^2 ) is positive (3), so the parabola opens upwards. That means the vertex is the minimum point of the parabola.I think I need to find the vertex of this parabola to understand where the function starts increasing or decreasing. The general form of a quadratic function is ( ax^2 + bx + c ), and the x-coordinate of the vertex is given by ( -frac{b}{2a} ). Let me calculate that.For ( g(x) = 3x^2 - 6x - 8 ), ( a = 3 ) and ( b = -6 ). So, the x-coordinate of the vertex is:[x = -frac{-6}{2 times 3} = frac{6}{6} = 1]So, the vertex is at ( x = 1 ). Since the parabola opens upwards, the function decreases to the left of the vertex and increases to the right of the vertex.Now, I need to include ( x = 2 ) in the interval. Since ( x = 2 ) is to the right of the vertex at ( x = 1 ), and the function is increasing to the right of the vertex, I can consider the interval from the vertex to the right, which would be ( [1, infty) ).Wait, but I should verify if this interval actually makes the function invertible. If I restrict ( g(x) ) to ( [1, infty) ), then the function is strictly increasing, which means it will pass the horizontal line test. Therefore, it should be invertible on this interval.Let me think if there's a larger interval that includes ( x = 2 ) where the function is still invertible. If I try to include any values less than 1, the function starts decreasing, and that would mean it's not one-to-one anymore because it would fail the horizontal line test. So, ( [1, infty) ) seems to be the largest possible interval containing ( x = 2 ) where ( g(x) ) is invertible.Just to be thorough, maybe I should also consider the function's behavior. The function ( g(x) = 3x^2 - 6x - 8 ) can be rewritten in vertex form to make it clearer. Let me complete the square.Starting with ( g(x) = 3x^2 - 6x - 8 ), factor out the 3 from the first two terms:[g(x) = 3(x^2 - 2x) - 8]Now, to complete the square inside the parentheses, take half of the coefficient of x, which is -2, so half of that is -1, and square it to get 1. Add and subtract this inside the parentheses:[g(x) = 3(x^2 - 2x + 1 - 1) - 8 = 3((x - 1)^2 - 1) - 8]Distribute the 3:[g(x) = 3(x - 1)^2 - 3 - 8 = 3(x - 1)^2 - 11]So, in vertex form, the function is ( g(x) = 3(x - 1)^2 - 11 ). This confirms that the vertex is at ( (1, -11) ), and since the coefficient is positive, it opens upwards.This means that to the right of ( x = 1 ), the function increases without bound, and to the left, it decreases. Therefore, the function is one-to-one on ( [1, infty) ) because it's strictly increasing there. Including any point left of 1 would cause the function to decrease, making it not one-to-one.Therefore, the largest interval containing ( x = 2 ) where ( g(x) ) is invertible is ( [1, infty) ).I think that's it. I don't see any mistakes in my reasoning. The key was to find the vertex and determine the direction the parabola opens, then choose the interval where the function is strictly increasing or decreasing to ensure it's one-to-one. Since ( x = 2 ) is to the right of the vertex, the interval starts at the vertex and goes to infinity.

💡Alright, so I have this problem here: I need to calculate the value of 0.overline{246} minus 0.overline{135} minus 0.overline{579}, and then express the answer as a fraction in its lowest terms. Hmm, okay, repeating decimals can sometimes be tricky, but I remember that there's a method to convert them into fractions. Let me try to recall how that works.First off, I think the general approach is to set the repeating decimal equal to a variable, multiply both sides by a power of 10 that shifts the decimal point to the right past the repeating part, and then subtract the original equation from this new one to eliminate the repeating part. That should give me an equation that I can solve for the variable, which will be the fraction representation of the decimal.Let me start with the first repeating decimal: 0.overline{246}. I'll call this x. So, x equals 0.246246246..., right? Since the repeating part is three digits long, I'll multiply both sides by 1000 to shift the decimal three places to the right. That gives me 1000x equals 246.246246246..., which is the same as 246.overline{246}.Now, if I subtract the original x from this equation, the repeating parts should cancel out. So, 1000x minus x equals 246.overline{246} minus 0.overline{246}. That simplifies to 999x equals 246. To solve for x, I divide both sides by 999, so x equals 246 over 999. Okay, that seems right. I'll write that down: 0.overline{246} is equal to 246/999.Next, I'll tackle the second repeating decimal: 0.overline{135}. Let's call this y. So, y equals 0.135135135..., and again, the repeating part is three digits. I'll multiply both sides by 1000 to get 1000y equals 135.135135135..., or 135.overline{135}.Subtracting the original y from this equation: 1000y minus y equals 135.overline{135} minus 0.overline{135}, which simplifies to 999y equals 135. Solving for y gives me y equals 135 over 999. So, 0.overline{135} is 135/999.Now, onto the third repeating decimal: 0.overline{579}. I'll call this z. So, z equals 0.579579579..., and the repeating part is three digits again. Multiplying both sides by 1000 gives me 1000z equals 579.579579579..., or 579.overline{579}.Subtracting the original z from this equation: 1000z minus z equals 579.overline{579} minus 0.overline{579}, which simplifies to 999z equals 579. Solving for z, I get z equals 579 over 999. So, 0.overline{579} is 579/999.Alright, so now I have all three repeating decimals converted into fractions:- 0.overline{246} = 246/999- 0.overline{135} = 135/999- 0.overline{579} = 579/999The problem asks for 0.overline{246} minus 0.overline{135} minus 0.overline{579}, so substituting the fractions in, that becomes:246/999 - 135/999 - 579/999Since all the denominators are the same, I can combine the numerators:(246 - 135 - 579)/999Let me calculate the numerator step by step. First, 246 minus 135. Let's see, 246 minus 100 is 146, minus another 35 is 111. So, 246 - 135 equals 111.Now, subtracting 579 from that result: 111 minus 579. Hmm, that's going to be negative. 579 minus 111 is 468, so 111 minus 579 is -468. Therefore, the numerator is -468, and the denominator remains 999.So, the expression simplifies to -468/999. Now, I need to express this fraction in its lowest terms. To do that, I have to find the greatest common divisor (GCD) of 468 and 999 and then divide both the numerator and the denominator by that GCD.Let me recall how to find the GCD using the Euclidean algorithm. The Euclidean algorithm involves dividing the larger number by the smaller one and then replacing the larger number with the smaller number and the smaller number with the remainder from the division, repeating this process until the remainder is zero. The last non-zero remainder is the GCD.So, let's apply that to 999 and 468.First, divide 999 by 468. 468 times 2 is 936, which is less than 999. 468 times 3 would be 1404, which is too much. So, 999 divided by 468 is 2 with a remainder. Let's calculate the remainder: 999 minus (468 times 2) equals 999 minus 936, which is 63. So, the remainder is 63.Now, replace 999 with 468 and 468 with 63, and repeat the process.Divide 468 by 63. 63 times 7 is 441, which is less than 468. 63 times 8 is 504, which is too much. So, 468 divided by 63 is 7 with a remainder. The remainder is 468 minus (63 times 7) equals 468 minus 441, which is 27.Now, replace 468 with 63 and 63 with 27.Divide 63 by 27. 27 times 2 is 54, which is less than 63. 27 times 3 is 81, which is too much. So, 63 divided by 27 is 2 with a remainder of 63 minus 54, which is 9.Replace 63 with 27 and 27 with 9.Divide 27 by 9. 9 times 3 is exactly 27, so the remainder is 0.Since the remainder is now 0, the last non-zero remainder is 9. Therefore, the GCD of 468 and 999 is 9.Wait, hold on, earlier I thought the GCD was 21, but according to this calculation, it's 9. Did I make a mistake somewhere?Let me double-check my steps.Starting with 999 and 468:999 divided by 468 is 2 with a remainder of 63 (since 468*2=936; 999-936=63).Then, 468 divided by 63: 63*7=441; 468-441=27.Then, 63 divided by 27: 27*2=54; 63-54=9.Then, 27 divided by 9: 9*3=27; remainder 0.So, the GCD is indeed 9, not 21. Hmm, I must have made a mistake earlier when I thought it was 21. Okay, so the GCD is 9.Therefore, to simplify -468/999, I divide both numerator and denominator by 9.First, let's divide 468 by 9. 9*50=450, so 468-450=18. 18 divided by 9 is 2. So, 468 divided by 9 is 52.Similarly, 999 divided by 9: 9*111=999, so 999/9=111.So, simplifying -468/999 by dividing numerator and denominator by 9 gives us -52/111.Wait, but earlier I thought the GCD was 21, leading to -24/51, but that seems incorrect now. Let me check that again.If I consider 468 and 999, and I thought the GCD was 21, but according to the Euclidean algorithm, it's 9. So, perhaps my initial thought was wrong, and the correct GCD is 9.Let me verify by checking if 52 and 111 have any common divisors. 52 factors are 2*2*13, and 111 factors are 3*37. There are no common factors between 52 and 111, so -52/111 is indeed in its simplest form.Wait, but in my initial thought process, I thought the GCD was 21, leading to -24/51, which simplifies further to -8/17, but that's not necessary because -24/51 can be simplified by dividing numerator and denominator by 3, giving -8/17. However, based on the Euclidean algorithm, the GCD is 9, not 21, so perhaps I made a mistake in my initial calculation.Let me double-check the Euclidean algorithm steps again to be sure.Starting with 999 and 468:999 ÷ 468 = 2 with remainder 63.468 ÷ 63 = 7 with remainder 27.63 ÷ 27 = 2 with remainder 9.27 ÷ 9 = 3 with remainder 0.So, GCD is 9.Therefore, the correct simplified fraction is -52/111.Wait, but earlier I thought the GCD was 21, which would have given a different result. Let me see where that came from.Perhaps I confused 468 with another number. Let me check 468 and 999 again.468 divided by 3 is 156, 999 divided by 3 is 333.156 and 333: 156 divided by 3 is 52, 333 divided by 3 is 111.So, 468/999 simplifies to 52/111 when divided by 9, which is the GCD.Alternatively, if I thought the GCD was 21, I would have divided both by 21, but 468 divided by 21 is 22.2857..., which is not an integer, so that can't be right. Therefore, my initial thought of GCD being 21 was incorrect. The correct GCD is 9.So, the simplified fraction is -52/111.Wait, but 52 and 111: 52 is 4*13, and 111 is 3*37. No common factors, so -52/111 is indeed in its simplest form.Therefore, the final answer should be -52/111.But in my initial thought process, I thought it was -24/51, which simplifies to -8/17, but that was based on an incorrect GCD of 21. So, I need to correct that.So, to summarize:0.overline{246} = 246/9990.overline{135} = 135/9990.overline{579} = 579/999Then, 246/999 - 135/999 - 579/999 = (246 - 135 - 579)/999 = (-468)/999Simplify -468/999 by dividing numerator and denominator by GCD 9: -52/111So, the final answer is -52/111.Wait, but in the initial problem, the user had written:"Calculate the value of 0.overline{246} - 0.overline{135} - 0.overline{579} and express your answer as a fraction in lowest terms."And in their initial thought process, they had arrived at -24/51, which simplifies to -8/17, but that was based on an incorrect GCD.So, to ensure accuracy, I need to confirm the GCD again.Let me calculate GCD of 468 and 999.Using prime factorization:468: Let's factorize 468.468 ÷ 2 = 234234 ÷ 2 = 117117 ÷ 3 = 3939 ÷ 3 = 13So, prime factors of 468 are 2^2 * 3^2 * 13Now, 999:999 ÷ 3 = 333333 ÷ 3 = 111111 ÷ 3 = 37So, prime factors of 999 are 3^3 * 37Now, the common prime factors are 3^2, since 468 has 3^2 and 999 has 3^3.Therefore, GCD is 3^2 = 9.So, indeed, the GCD is 9, not 21. Therefore, the simplified fraction is -52/111.Wait, but in the initial thought process, the user had written:"Calculate the value of 0.overline{246} - 0.overline{135} - 0.overline{579} and express your answer as a fraction in lowest terms."And in their initial thought process, they had arrived at -24/51, which simplifies to -8/17, but that was based on an incorrect GCD of 21.So, to correct that, the accurate simplified fraction is -52/111.Therefore, the final answer is -52/111.

💡Alright, so I'm trying to prove this lemma that says if I have two numbers, a and b, where 0 < a < b, and then I pick two other numbers x and y between a and b such that their product is equal to a times b, then the sum of g(a) and g(b) is greater than or equal to the sum of g(x) and g(y). First off, I need to understand what the function g(x) is doing here. The problem gives me an inequality involving g(x1) and g(x2) with some exponents and a parameter t. It looks like it's some kind of convexity condition, but I'm not entirely sure. Maybe I can use that inequality to get some properties of g(x).So, the lemma is about rearranging or pairing numbers in a certain way to get an inequality involving g. I remember something about the rearrangement inequality, which deals with ordering of sequences and how their products behave. Maybe that's related here.Given that a < x < b and a < y < b, and xy = ab, I wonder if there's a way to express x and y in terms of a and b. Maybe using exponents or logarithms? Let me think about that.If I take logarithms, maybe I can turn the product into a sum, which might make it easier to handle. So, if I let log(x) = something and log(y) = something else, then log(xy) = log(a) + log(b). That could be useful.Alternatively, maybe I can express x and y as a combination of a and b. For example, x could be a weighted average of a and b, and y could be the complementary weighted average. That might help me apply the given inequality for g.Wait, the given inequality has g(x1 squared times x2 to the power of 1-t) on the left side and a combination of g(x1) and g(x2) on the right side. It seems like it's a kind of convexity or concavity condition. If I can figure out whether g is convex or concave, that might help me apply Jensen's inequality or something similar.But I'm not sure if g is convex or concave. Maybe I can assume it's convex and see if that leads me somewhere. If not, maybe I can adjust my approach.Let me try to express x and y in terms of a and b. Suppose x = a^α b^{1-α} and y = a^β b^{1-β}, where α and β are between 0 and 1. Then, since xy = ab, I can set up the equation:a^α b^{1-α} * a^β b^{1-β} = abSimplifying the left side, I get a^{α + β} b^{2 - α - β} = abSo, matching the exponents, I have:α + β = 1And 2 - α - β = 1, which also gives α + β = 1.So, that means β = 1 - α.Therefore, x = a^α b^{1-α} and y = a^{1-α} b^{α}That's interesting. So, x and y are related through the parameter α.Now, I can try to express g(x) and g(y) in terms of g(a) and g(b). Maybe using the given inequality.Given that g(x1^2 x2^{1-t}) ≤ t g(x1) + (1-t) g(x2)I need to see if I can choose x1 and x2 such that x1^2 x2^{1-t} equals x or y.Let me try to set x1 = a and x2 = b, and choose t such that x1^2 x2^{1-t} equals x.So, x = a^α b^{1-α} = a^2 b^{1-t}Wait, that would mean:a^α b^{1-α} = a^2 b^{1-t}Which implies:α = 2 and 1 - α = 1 - tBut α is between 0 and 1, so this doesn't make sense. Maybe I need to adjust my approach.Alternatively, maybe I can set x1 = x and x2 = y, but I'm not sure.Wait, maybe I can use the given inequality in reverse. Since I have g(x) ≤ something involving g(a) and g(b), and similarly for g(y), then adding them up might give me the desired inequality.Let me try that.From the given inequality, for x = a^α b^{1-α}, I can set x1 = a, x2 = b, and t = α.Then, g(x) = g(a^α b^{1-α}) ≤ α g(a) + (1 - α) g(b)Similarly, for y = a^{1-α} b^{α}, setting x1 = a, x2 = b, and t = 1 - α, we get:g(y) = g(a^{1-α} b^{α}) ≤ (1 - α) g(a) + α g(b)Now, if I add these two inequalities together:g(x) + g(y) ≤ [α g(a) + (1 - α) g(b)] + [(1 - α) g(a) + α g(b)] = g(a) + g(b)So, that's exactly what I needed to prove! Therefore, g(a) + g(b) ≥ g(x) + g(y)Wait, that seems too straightforward. Did I miss something? Let me double-check.I expressed x and y in terms of a and b with exponents α and 1 - α, used the given inequality for g with appropriate choices of x1, x2, and t, and then added the resulting inequalities. It seems correct.But I'm a bit confused because I thought it was going to be more complicated. Maybe I should consider if there are any edge cases or if the function g has any specific properties that I need to be aware of.Also, I should make sure that the choices of x1, x2, and t are valid within the given conditions. Since α is between 0 and 1, t = α and t = 1 - α are both in (0,1), which is required by the given inequality.Okay, I think this approach works. I'll write it up formally now.