💡Okay, so I have this problem with five logarithmic equations, and I need to figure out which one is incorrect. The equations are:① ( lg 3 = 2a - b )② ( lg 5 = a + c )③ ( lg 8 = 3 - 3a - 3c )④ ( lg 9 = 4a - 2b )⑤ ( lg 15 = 3a - b + c + 1 )The problem states that exactly one of these equations doesn't hold, so I need to determine which one is incorrect.First, I should recall some logarithmic identities and properties that might help me here. I know that:- ( lg 10 = 1 ) because ( 10^1 = 10 ).- ( lg(ab) = lg a + lg b )- ( lg(a^n) = n lg a )- ( lg frac{a}{b} = lg a - lg b )Also, since all these equations involve ( lg ) (which is base 10 logarithm), I can express other logarithms in terms of known values or in terms of each other.Looking at the equations, I notice that they involve variables ( a ), ( b ), and ( c ). So, it seems like these variables are defined in terms of logarithms. Maybe I can express ( a ), ( b ), and ( c ) in terms of known logarithms or relate them to each other.Let me try to express each equation in terms of ( lg 2 ), ( lg 3 ), and ( lg 5 ) because these are prime numbers and their logarithms are independent.I know that:- ( lg 2 + lg 5 = lg 10 = 1 ), so ( lg 2 = 1 - lg 5 )- ( lg 3 ) is a separate value, which can't be simplified further without knowing its exact value.Given that, let me see if I can express all the equations in terms of ( lg 2 ), ( lg 3 ), and ( lg 5 ).Starting with equation ①: ( lg 3 = 2a - b )Equation ②: ( lg 5 = a + c )Equation ③: ( lg 8 = 3 - 3a - 3c )Equation ④: ( lg 9 = 4a - 2b )Equation ⑤: ( lg 15 = 3a - b + c + 1 )Let me analyze each equation step by step.Starting with equation ①: ( lg 3 = 2a - b )This gives me a relationship between ( a ) and ( b ). If I can express ( a ) or ( b ) in terms of the other, it might help.Equation ②: ( lg 5 = a + c )Similarly, this relates ( a ) and ( c ).Equation ③: ( lg 8 = 3 - 3a - 3c )I know that ( lg 8 = lg 2^3 = 3 lg 2 ). So, ( 3 lg 2 = 3 - 3a - 3c ). Dividing both sides by 3, I get ( lg 2 = 1 - a - c ).But from equation ②, ( lg 5 = a + c ). So, ( lg 2 = 1 - (a + c) = 1 - lg 5 ). That makes sense because ( lg 2 + lg 5 = 1 ). So, equation ③ seems consistent with equation ②.Moving on to equation ④: ( lg 9 = 4a - 2b )I know that ( lg 9 = lg 3^2 = 2 lg 3 ). From equation ①, ( lg 3 = 2a - b ), so ( 2 lg 3 = 4a - 2b ). Therefore, equation ④ is just twice equation ①, which is consistent. So, equation ④ seems correct.Now, equation ⑤: ( lg 15 = 3a - b + c + 1 )Let me express ( lg 15 ) in terms of known logarithms. ( 15 = 3 times 5 ), so ( lg 15 = lg 3 + lg 5 ).From equation ①, ( lg 3 = 2a - b ).From equation ②, ( lg 5 = a + c ).Therefore, ( lg 15 = (2a - b) + (a + c) = 3a - b + c ).But equation ⑤ says ( lg 15 = 3a - b + c + 1 ). So, according to my calculation, it should be ( 3a - b + c ), but equation ⑤ adds an extra 1. That seems inconsistent.Wait, maybe I made a mistake. Let me double-check.( lg 15 = lg 3 + lg 5 ) is correct.From equation ①: ( lg 3 = 2a - b )From equation ②: ( lg 5 = a + c )Adding them: ( lg 15 = (2a - b) + (a + c) = 3a - b + c ). So, yes, that's correct.But equation ⑤ says ( 3a - b + c + 1 ). So, it's adding 1 more than what I calculated. That suggests that equation ⑤ is incorrect.But before I conclude, let me check if there's another way equation ⑤ could be correct. Maybe I need to consider another relationship.Alternatively, maybe I can express ( lg 15 ) in terms of other logarithms.Wait, ( 15 ) is also ( 10 times 1.5 ), but that might not help directly. Alternatively, ( 15 = frac{30}{2} ), but again, not sure.Alternatively, maybe I can express ( lg 15 ) in terms of ( lg 2 ) and ( lg 3 ) and ( lg 5 ), but I already did that.Alternatively, maybe I can use equation ③ to express ( lg 2 ) in terms of ( a ) and ( c ), and then use that to find another relationship.From equation ③, ( lg 8 = 3 - 3a - 3c ), which is ( 3 lg 2 = 3 - 3a - 3c ), so ( lg 2 = 1 - a - c ).From equation ②, ( lg 5 = a + c ), so ( lg 2 = 1 - lg 5 ), which is consistent because ( lg 2 + lg 5 = 1 ).So, that's consistent.Now, let me see if I can find the values of ( a ), ( b ), and ( c ) from the equations to see if equation ⑤ holds.From equation ①: ( lg 3 = 2a - b ) => ( b = 2a - lg 3 )From equation ②: ( lg 5 = a + c ) => ( c = lg 5 - a )From equation ③: ( lg 8 = 3 - 3a - 3c ). But we already used that to get ( lg 2 = 1 - a - c ), which is consistent.From equation ④: ( lg 9 = 4a - 2b ). But ( lg 9 = 2 lg 3 ), and from equation ①, ( 2 lg 3 = 4a - 2b ), which is consistent.So, all equations except ⑤ are consistent with each other.Now, let's compute ( lg 15 ) using the expressions for ( a ), ( b ), and ( c ).From equation ① and ②, we have:( b = 2a - lg 3 )( c = lg 5 - a )So, ( 3a - b + c = 3a - (2a - lg 3) + (lg 5 - a) = 3a - 2a + lg 3 + lg 5 - a = (3a - 2a - a) + (lg 3 + lg 5) = 0 + lg 15 = lg 15 )Therefore, ( 3a - b + c = lg 15 ), but equation ⑤ says ( 3a - b + c + 1 = lg 15 ). So, that would imply ( lg 15 = lg 15 + 1 ), which is impossible because 1 is not equal to 0. Therefore, equation ⑤ is incorrect.Wait, but let me think again. Maybe I made a mistake in substitution.Let me compute ( 3a - b + c ) step by step.Given:( b = 2a - lg 3 )( c = lg 5 - a )So,( 3a - b + c = 3a - (2a - lg 3) + (lg 5 - a) )Simplify term by term:- ( 3a )- ( - (2a - lg 3) = -2a + lg 3 )- ( + (lg 5 - a) = + lg 5 - a )Combine all terms:( 3a - 2a + lg 3 + lg 5 - a )Simplify:( (3a - 2a - a) + (lg 3 + lg 5) = 0 + lg 15 = lg 15 )So, ( 3a - b + c = lg 15 ), but equation ⑤ says ( 3a - b + c + 1 = lg 15 ). Therefore, equation ⑤ is adding an extra 1, which is incorrect.Therefore, equation ⑤ is the incorrect one.Wait, but let me check if maybe equation ③ is incorrect. Because sometimes, when equations are interdependent, an error in one can affect others.From equation ③: ( lg 8 = 3 - 3a - 3c )But ( lg 8 = 3 lg 2 ), and from equation ②, ( lg 5 = a + c ), so ( lg 2 = 1 - a - c ). Therefore, ( 3 lg 2 = 3(1 - a - c) = 3 - 3a - 3c ), which matches equation ③. So, equation ③ is correct.Similarly, equation ④ is correct because it's just twice equation ①.Equation ① and ② are given, and equation ③, ④ are derived from them, so they are correct.Therefore, the only incorrect equation is ⑤.

💡Alright, so I have this problem where a motorist and a cyclist start from points A and B, respectively, heading towards each other but not at the same time. When they meet at point C, they both turn around and go back to their starting points. After reaching A or B, they turn around again and continue this pattern. They meet again at point D, and this continues. The question is asking where their 2015th meeting will occur on segment AB.Hmm, okay, let's break this down step by step. First, I need to visualize the scenario. Points A and B are fixed, and the motorist starts from A, the cyclist from B. They don't start at the same time, so one has a head start. When they meet at point C, they both turn around immediately and go back to their respective starting points. Then, upon reaching A or B, they turn around again and head back towards each other, meeting again at point D, and so on.I think the key here is to figure out the pattern of their meeting points. If I can determine whether the meeting points alternate between C and D or follow some other sequence, I can predict where the 2015th meeting will be.Let me denote the speeds of the motorist and cyclist as ( v_m ) and ( v_c ), respectively. Since they start at different times, there's a time difference between their departures. Let's assume the motorist starts first, so by the time the cyclist starts, the motorist has already covered some distance towards B.When they meet at point C for the first time, the total distance they've covered together is equal to the distance between A and B. Let's denote the distance between A and B as ( D ). So, the time taken for them to meet at C can be calculated based on their speeds and the head start of the motorist.After meeting at C, they both turn around and head back to their starting points. The motorist goes back to A, and the cyclist goes back to B. Upon reaching their respective starting points, they turn around again and start heading towards each other once more. This time, when they meet at point D, it's their second meeting.I think the pattern here is that each time they meet, they're effectively covering the distance between A and B twice—once going towards each other and once going away from each other. So, each meeting after the first one might be happening at alternating points between C and D.Wait, let me think about that again. If they meet at C, then turn around and go back, then meet again at D, and then turn around again, would the third meeting be back at C? Or would it be somewhere else?Actually, I think it's more likely that the meeting points alternate between C and D because each time they complete a round trip, their meeting point shifts. So, the first meeting is at C, the second at D, the third back at C, the fourth at D, and so on.If that's the case, then the odd-numbered meetings (1st, 3rd, 5th, etc.) occur at point C, and the even-numbered meetings (2nd, 4th, 6th, etc.) occur at point D.Given that, the 2015th meeting would be an odd-numbered meeting, which would occur at point C.But wait, I should verify this pattern to make sure it holds. Let's consider the first few meetings:1. First meeting: C2. Second meeting: D3. Third meeting: C4. Fourth meeting: D5. Fifth meeting: CYes, it seems like the pattern alternates between C and D with each meeting. Therefore, the odd-numbered meetings are at C, and the even-numbered ones are at D.Since 2015 is an odd number, the 2015th meeting should occur at point C.I think that makes sense. Each time they meet, they're essentially swapping places in terms of starting points, leading to alternating meeting points. So, the pattern continues consistently, and the parity of the meeting number determines the location.Therefore, after analyzing the pattern and verifying it with the first few meetings, I conclude that the 2015th meeting will occur at point C.

💡Okay, so I have this system of equations to solve:1. ( x(3y^2 + 1) = y(y^2 + 3) )2. ( y(3z^2 + 1) = z(z^2 + 3) )3. ( z(3x^2 + 1) = x(x^2 + 3) )Hmm, all three equations look similar. Maybe there's some symmetry here that I can exploit. Let me see if I can find a pattern or maybe assume that all variables are equal. That often helps in symmetric systems.So, let's suppose that ( x = y = z ). If that's the case, then each equation becomes the same. Let's substitute ( y ) and ( z ) with ( x ) in the first equation:( x(3x^2 + 1) = x(x^2 + 3) )Let me simplify this:Left side: ( x(3x^2 + 1) = 3x^3 + x )Right side: ( x(x^2 + 3) = x^3 + 3x )So, setting them equal:( 3x^3 + x = x^3 + 3x )Subtract ( x^3 + 3x ) from both sides:( 2x^3 - 2x = 0 )Factor out a 2x:( 2x(x^2 - 1) = 0 )So, ( 2x = 0 ) or ( x^2 - 1 = 0 )Which gives ( x = 0 ) or ( x = pm 1 )Therefore, if ( x = y = z ), the possible solutions are ( (0, 0, 0) ), ( (1, 1, 1) ), and ( (-1, -1, -1) ).But wait, is that all? Maybe there are solutions where ( x ), ( y ), and ( z ) are not all equal. Let me check.Looking back at the first equation:( x(3y^2 + 1) = y(y^2 + 3) )Let me rearrange this:( x(3y^2 + 1) - y(y^2 + 3) = 0 )Expanding:( 3x y^2 + x - y^3 - 3y = 0 )Hmm, not sure if that helps. Maybe I can solve for ( x ) in terms of ( y ):( x = frac{y(y^2 + 3)}{3y^2 + 1} )Similarly, from the second equation:( y = frac{z(z^2 + 3)}{3z^2 + 1} )And from the third equation:( z = frac{x(x^2 + 3)}{3x^2 + 1} )So, substituting each into the next, we can write:( x = frac{y(y^2 + 3)}{3y^2 + 1} )But ( y = frac{z(z^2 + 3)}{3z^2 + 1} ), so:( x = frac{frac{z(z^2 + 3)}{3z^2 + 1} left( left( frac{z(z^2 + 3)}{3z^2 + 1} right)^2 + 3 right)}{3 left( frac{z(z^2 + 3)}{3z^2 + 1} right)^2 + 1} )This seems complicated. Maybe there's a better approach.Alternatively, let's consider the function ( f(t) = frac{t(t^2 + 3)}{3t^2 + 1} ). Then, from the equations, we have:( x = f(y) ), ( y = f(z) ), and ( z = f(x) ).So, ( x = f(f(f(x))) ). That is, ( x ) is a fixed point of the function ( f ) composed three times.If I can find the fixed points of ( f ), that might help. A fixed point satisfies ( f(t) = t ).So, set ( frac{t(t^2 + 3)}{3t^2 + 1} = t )Multiply both sides by ( 3t^2 + 1 ):( t(t^2 + 3) = t(3t^2 + 1) )Simplify:( t^3 + 3t = 3t^3 + t )Bring all terms to one side:( t^3 + 3t - 3t^3 - t = 0 )Combine like terms:( -2t^3 + 2t = 0 )Factor:( -2t(t^2 - 1) = 0 )So, ( t = 0 ) or ( t = pm 1 )Therefore, the fixed points are ( t = 0, 1, -1 ). So, if ( x = y = z ), we get these solutions.But could there be other solutions where ( x ), ( y ), ( z ) are not equal? Let's see.Suppose ( x neq y ). Then, from ( x = f(y) ), ( y = f(z) ), ( z = f(x) ), we have a cyclic dependency.Let me check if there are cycles of length 3, meaning ( x neq y neq z neq x ).But this might be complicated. Alternatively, let's assume that two variables are equal and see if that leads us anywhere.Suppose ( x = y ). Then, from the first equation:( x(3x^2 + 1) = x(x^2 + 3) )Which simplifies to:( 3x^3 + x = x^3 + 3x )( 2x^3 - 2x = 0 )( 2x(x^2 - 1) = 0 )So, ( x = 0 ) or ( x = pm 1 ). So, again, same solutions.Similarly, if ( y = z ), we get the same result.Alternatively, suppose ( x = -y ). Let's see:From the first equation:( x(3y^2 + 1) = y(y^2 + 3) )If ( x = -y ), then:( -y(3y^2 + 1) = y(y^2 + 3) )Multiply both sides by -1:( y(3y^2 + 1) = -y(y^2 + 3) )If ( y neq 0 ), we can divide both sides by ( y ):( 3y^2 + 1 = -(y^2 + 3) )Simplify:( 3y^2 + 1 = -y^2 - 3 )Bring all terms to one side:( 4y^2 + 4 = 0 )Which gives ( y^2 = -1 ), which has no real solutions. So, ( x = -y ) is not possible unless ( y = 0 ), but then ( x = 0 ) as well, which is already covered in our previous solutions.So, it seems that the only real solutions are when ( x = y = z ), giving us ( (0, 0, 0) ), ( (1, 1, 1) ), and ( (-1, -1, -1) ).Let me double-check these solutions in all three equations.For ( (0, 0, 0) ):1. ( 0(3*0^2 + 1) = 0 = 0(0^2 + 3) ) → True2. Similarly for the other equations.For ( (1, 1, 1) ):1. ( 1(3*1^2 + 1) = 4 = 1(1^2 + 3) = 4 ) → True2. Similarly for the other equations.For ( (-1, -1, -1) ):1. ( -1(3*(-1)^2 + 1) = -4 = -1((-1)^2 + 3) = -4 ) → True2. Similarly for the other equations.So, all three solutions satisfy the system.I don't see any other possible solutions, especially since assuming any two variables equal leads back to these solutions, and assuming a cycle doesn't seem to yield new solutions without leading to contradictions or non-real results.Therefore, the real solutions to the system are ( (0, 0, 0) ), ( (1, 1, 1) ), and ( (-1, -1, -1) ).

💡Okay, so I have this problem with two parts. First, I need to find the ordinary equations for a line ( l ) and a curve ( C ) given their parametric equations. Then, I have to find the value of ( |PA| cdot |PB| ) where ( P ) is the point ( (2, 0) ) and ( A ) and ( B ) are the intersection points of line ( l ) and curve ( C ).Starting with part (1): finding the ordinary equations.For the line ( l ), the parametric equations are:[begin{cases}x = 2 + frac{sqrt{2}}{2} t y = frac{sqrt{2}}{2} tend{cases}]where ( t ) is the parameter.I remember that to convert parametric equations to an ordinary equation, I need to eliminate the parameter. Here, both ( x ) and ( y ) are expressed in terms of ( t ). Let me see if I can express ( t ) from one equation and substitute into the other.Looking at the equation for ( y ):[y = frac{sqrt{2}}{2} t]I can solve for ( t ):[t = frac{2}{sqrt{2}} y = sqrt{2} y]Now, substitute this expression for ( t ) into the equation for ( x ):[x = 2 + frac{sqrt{2}}{2} cdot sqrt{2} y]Simplify the terms:[frac{sqrt{2}}{2} cdot sqrt{2} = frac{2}{2} = 1]So, the equation becomes:[x = 2 + y]Which can be rearranged to:[x - y = 2]So, the ordinary equation of line ( l ) is ( x - y - 2 = 0 ).Now, for the curve ( C ), the parametric equations are:[begin{cases}x = 4 cos theta y = 2 sqrt{3} sin thetaend{cases}]where ( theta ) is the parameter.I recall that parametric equations involving sine and cosine often represent conic sections, like ellipses. To find the ordinary equation, I need to eliminate ( theta ).I know that ( cos theta = frac{x}{4} ) and ( sin theta = frac{y}{2 sqrt{3}} ). Since ( cos^2 theta + sin^2 theta = 1 ), I can substitute these expressions into that identity.So, substituting:[left( frac{x}{4} right)^2 + left( frac{y}{2 sqrt{3}} right)^2 = 1]Simplify each term:[frac{x^2}{16} + frac{y^2}{12} = 1]So, the ordinary equation of curve ( C ) is ( frac{x^2}{16} + frac{y^2}{12} = 1 ). That looks like an ellipse centered at the origin with semi-major axis 4 and semi-minor axis ( 2 sqrt{3} ).Alright, part (1) is done. Now, moving on to part (2): finding ( |PA| cdot |PB| ) where ( P(2, 0) ).First, I need to find the points ( A ) and ( B ) where line ( l ) intersects curve ( C ). So, I should solve the system of equations given by the ordinary equations of ( l ) and ( C ).The equations are:1. ( x - y - 2 = 0 ) (from line ( l ))2. ( frac{x^2}{16} + frac{y^2}{12} = 1 ) (from curve ( C ))Let me solve equation 1 for ( x ) in terms of ( y ):[x = y + 2]Now, substitute this into equation 2:[frac{(y + 2)^2}{16} + frac{y^2}{12} = 1]I need to solve this equation for ( y ). Let me expand ( (y + 2)^2 ):[(y + 2)^2 = y^2 + 4y + 4]So, substituting back:[frac{y^2 + 4y + 4}{16} + frac{y^2}{12} = 1]To combine these fractions, I should find a common denominator. The denominators are 16 and 12. The least common multiple of 16 and 12 is 48. So, I'll multiply each term by 48 to eliminate the denominators:[48 cdot frac{y^2 + 4y + 4}{16} + 48 cdot frac{y^2}{12} = 48 cdot 1]Simplify each term:- ( 48 / 16 = 3 ), so the first term becomes ( 3(y^2 + 4y + 4) )- ( 48 / 12 = 4 ), so the second term becomes ( 4y^2 )- The right side is 48So, the equation becomes:[3(y^2 + 4y + 4) + 4y^2 = 48]Expanding the first term:[3y^2 + 12y + 12 + 4y^2 = 48]Combine like terms:- ( 3y^2 + 4y^2 = 7y^2 )- ( 12y ) remains as is- ( 12 ) remains as isSo, the equation is:[7y^2 + 12y + 12 = 48]Subtract 48 from both sides:[7y^2 + 12y + 12 - 48 = 0]Simplify:[7y^2 + 12y - 36 = 0]So, we have a quadratic equation in terms of ( y ):[7y^2 + 12y - 36 = 0]I need to solve this quadratic equation for ( y ). Let me use the quadratic formula:[y = frac{-b pm sqrt{b^2 - 4ac}}{2a}]Where ( a = 7 ), ( b = 12 ), and ( c = -36 ).Calculating the discriminant:[b^2 - 4ac = 12^2 - 4 cdot 7 cdot (-36) = 144 + 1008 = 1152]So, the solutions are:[y = frac{-12 pm sqrt{1152}}{14}]Simplify ( sqrt{1152} ). Let's factor 1152:[1152 = 1024 + 128 = 1024 + 128 = 1024 is 32^2, but 1152 is 1024 + 128 = 1024 + 128 = 1024 + 128 = 1024 + 128 = 1024 + 128 = 1024 + 128 = 1024 + 128 = 1024 + 128 = 1024 + 128 = 1024 + 128 = 1024 + 128 = 1024 + 128 = 1024 + 128.Wait, that's not helpful. Let me factor 1152 properly.1152 divided by 16 is 72. So, 1152 = 16 * 72.72 can be factored into 36 * 2, so 1152 = 16 * 36 * 2 = 16 * 36 * 2.So, ( sqrt{1152} = sqrt{16 times 36 times 2} = sqrt{16} times sqrt{36} times sqrt{2} = 4 times 6 times sqrt{2} = 24 sqrt{2} ).So, ( sqrt{1152} = 24 sqrt{2} ).Therefore, the solutions are:[y = frac{-12 pm 24 sqrt{2}}{14}]Simplify the fractions by dividing numerator and denominator by 2:[y = frac{-6 pm 12 sqrt{2}}{7}]So, the two y-coordinates of points ( A ) and ( B ) are:[y_1 = frac{-6 + 12 sqrt{2}}{7}, quad y_2 = frac{-6 - 12 sqrt{2}}{7}]Now, recall that ( x = y + 2 ) from the equation of line ( l ). So, let's find the corresponding ( x )-coordinates.For ( y_1 ):[x_1 = y_1 + 2 = frac{-6 + 12 sqrt{2}}{7} + 2 = frac{-6 + 12 sqrt{2} + 14}{7} = frac{8 + 12 sqrt{2}}{7}]For ( y_2 ):[x_2 = y_2 + 2 = frac{-6 - 12 sqrt{2}}{7} + 2 = frac{-6 - 12 sqrt{2} + 14}{7} = frac{8 - 12 sqrt{2}}{7}]So, the coordinates of points ( A ) and ( B ) are:- ( A left( frac{8 + 12 sqrt{2}}{7}, frac{-6 + 12 sqrt{2}}{7} right) )- ( B left( frac{8 - 12 sqrt{2}}{7}, frac{-6 - 12 sqrt{2}}{7} right) )Now, I need to find ( |PA| ) and ( |PB| ), where ( P(2, 0) ), and then compute their product.Let me recall the distance formula between two points ( (x_1, y_1) ) and ( (x_2, y_2) ):[|PQ| = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}]So, let's compute ( |PA| ) first.Point ( A ) is ( left( frac{8 + 12 sqrt{2}}{7}, frac{-6 + 12 sqrt{2}}{7} right) ), and point ( P ) is ( (2, 0) ).Compute the differences:- ( x )-difference: ( frac{8 + 12 sqrt{2}}{7} - 2 = frac{8 + 12 sqrt{2} - 14}{7} = frac{-6 + 12 sqrt{2}}{7} )- ( y )-difference: ( frac{-6 + 12 sqrt{2}}{7} - 0 = frac{-6 + 12 sqrt{2}}{7} )So, ( |PA| = sqrt{ left( frac{-6 + 12 sqrt{2}}{7} right)^2 + left( frac{-6 + 12 sqrt{2}}{7} right)^2 } )Since both differences are the same, this simplifies to:[|PA| = sqrt{ 2 left( frac{-6 + 12 sqrt{2}}{7} right)^2 } = sqrt{2} cdot left| frac{-6 + 12 sqrt{2}}{7} right|]Similarly, let's compute ( |PB| ).Point ( B ) is ( left( frac{8 - 12 sqrt{2}}{7}, frac{-6 - 12 sqrt{2}}{7} right) ), and point ( P ) is ( (2, 0) ).Compute the differences:- ( x )-difference: ( frac{8 - 12 sqrt{2}}{7} - 2 = frac{8 - 12 sqrt{2} - 14}{7} = frac{-6 - 12 sqrt{2}}{7} )- ( y )-difference: ( frac{-6 - 12 sqrt{2}}{7} - 0 = frac{-6 - 12 sqrt{2}}{7} )So, ( |PB| = sqrt{ left( frac{-6 - 12 sqrt{2}}{7} right)^2 + left( frac{-6 - 12 sqrt{2}}{7} right)^2 } )Again, both differences are the same, so:[|PB| = sqrt{ 2 left( frac{-6 - 12 sqrt{2}}{7} right)^2 } = sqrt{2} cdot left| frac{-6 - 12 sqrt{2}}{7} right|]Now, let's compute ( |PA| cdot |PB| ):[|PA| cdot |PB| = sqrt{2} cdot left| frac{-6 + 12 sqrt{2}}{7} right| cdot sqrt{2} cdot left| frac{-6 - 12 sqrt{2}}{7} right|]Simplify the constants:[sqrt{2} cdot sqrt{2} = 2]So, the expression becomes:[2 cdot left| frac{(-6 + 12 sqrt{2})(-6 - 12 sqrt{2})}{49} right|]First, compute the numerator:[(-6 + 12 sqrt{2})(-6 - 12 sqrt{2}) = (-6)^2 - (12 sqrt{2})^2 = 36 - (144 cdot 2) = 36 - 288 = -252]So, the expression inside the absolute value is:[frac{-252}{49} = -5.142857...]But since we have an absolute value, it becomes:[left| frac{-252}{49} right| = frac{252}{49} = 5.142857...]But let me compute it exactly:[frac{252}{49} = frac{252 div 7}{49 div 7} = frac{36}{7}]So, putting it all together:[|PA| cdot |PB| = 2 cdot frac{36}{7} = frac{72}{7}]Wait, hold on, that doesn't match the initial thought. Let me double-check my calculations.Wait, earlier I had:[|PA| cdot |PB| = 2 cdot left| frac{(-6 + 12 sqrt{2})(-6 - 12 sqrt{2})}{49} right| = 2 cdot left| frac{(-6)^2 - (12 sqrt{2})^2}{49} right| = 2 cdot left| frac{36 - 288}{49} right| = 2 cdot left| frac{-252}{49} right| = 2 cdot frac{252}{49}]Wait, 252 divided by 49 is 5.142857, but 252 divided by 49 is actually 5.142857, but let me compute 49 times 5 is 245, so 252 - 245 is 7, so 252/49 is 5 + 7/49 = 5 + 1/7 = 5.142857, which is 36/7? Wait, 36 divided by 7 is approximately 5.142857. Wait, 36/7 is 5.142857.Wait, 252 divided by 49 is 5.142857, which is 36/7 because 36 divided by 7 is 5.142857.Wait, 36/7 is 5.142857, and 252/49 is also 5.142857 because 252 divided by 49 is 5.142857.But 252 divided by 49 is 5.142857, which is 36/7 because 36/7 is 5.142857.Wait, 252 is 36 * 7, so 252/49 = (36 * 7)/49 = 36/7.Yes, that's correct.So, 252/49 = 36/7.Therefore, ( |PA| cdot |PB| = 2 * (36/7) = 72/7 ).Wait, but in my initial thought, I thought it was 48/7. Hmm, maybe I made a mistake somewhere.Wait, let me go back.When I computed ( (-6 + 12 sqrt{2})(-6 - 12 sqrt{2}) ), I got ( (-6)^2 - (12 sqrt{2})^2 = 36 - 288 = -252 ). That's correct.So, the numerator is -252, and the denominator is 49, so the absolute value is 252/49 = 36/7.Then, multiplying by 2, we get 72/7.Wait, but in the initial thought, the user had 48/7. So, perhaps I made a mistake in the calculation.Wait, let me check the parametric equations again.Wait, in the parametric equation of line ( l ), it's ( x = 2 + (sqrt{2}/2) t ), ( y = (sqrt{2}/2) t ). So, when I converted to the ordinary equation, I got ( x - y = 2 ). That seems correct.For the curve ( C ), parametric equations are ( x = 4 cos theta ), ( y = 2 sqrt{3} sin theta ). So, converting to ordinary equation, ( x^2 / 16 + y^2 / 12 = 1 ). That seems correct.Then, substituting ( x = y + 2 ) into the ellipse equation, I got to ( 7y^2 + 12y - 36 = 0 ). Let me check that step again.Starting from substituting ( x = y + 2 ) into ( x^2 / 16 + y^2 / 12 = 1 ):[frac{(y + 2)^2}{16} + frac{y^2}{12} = 1]Expanding ( (y + 2)^2 = y^2 + 4y + 4 ), so:[frac{y^2 + 4y + 4}{16} + frac{y^2}{12} = 1]Multiplying both sides by 48:[3(y^2 + 4y + 4) + 4y^2 = 48]Expanding:[3y^2 + 12y + 12 + 4y^2 = 48]Combine like terms:[7y^2 + 12y + 12 = 48]Subtract 48:[7y^2 + 12y - 36 = 0]Yes, that's correct.Then, solving ( 7y^2 + 12y - 36 = 0 ) using quadratic formula:Discriminant ( D = 12^2 - 4 * 7 * (-36) = 144 + 1008 = 1152 )So, ( y = [-12 pm sqrt{1152}]/14 )Simplify ( sqrt{1152} = 24 sqrt{2} ), so:( y = [-12 pm 24 sqrt{2}]/14 = [-6 pm 12 sqrt{2}]/7 )So, that's correct.Then, computing ( |PA| ) and ( |PB| ):For point ( A ), ( x = (8 + 12 sqrt{2})/7 ), ( y = (-6 + 12 sqrt{2})/7 )Distance from ( P(2, 0) ):( x )-difference: ( (8 + 12 sqrt{2})/7 - 14/7 = (-6 + 12 sqrt{2})/7 )( y )-difference: ( (-6 + 12 sqrt{2})/7 - 0 = (-6 + 12 sqrt{2})/7 )So, distance ( |PA| = sqrt{ [(-6 + 12 sqrt{2})/7]^2 + [(-6 + 12 sqrt{2})/7]^2 } = sqrt{2 * [(-6 + 12 sqrt{2})/7]^2 } = sqrt{2} * |(-6 + 12 sqrt{2})/7| )Similarly for ( |PB| ), it's ( sqrt{2} * |(-6 - 12 sqrt{2})/7| )Multiplying them together:( |PA| * |PB| = 2 * | [(-6 + 12 sqrt{2})(-6 - 12 sqrt{2})]/49 | )Compute numerator:( (-6 + 12 sqrt{2})(-6 - 12 sqrt{2}) = (-6)^2 - (12 sqrt{2})^2 = 36 - 288 = -252 )So, absolute value is 252, so:( |PA| * |PB| = 2 * (252/49) = 2 * (36/7) = 72/7 )Wait, but in the initial thought, the user had 48/7. So, perhaps I made a mistake in the calculation of the distances.Wait, let me check the distances again.Wait, for point ( A ), ( x = (8 + 12 sqrt{2})/7 ), ( y = (-6 + 12 sqrt{2})/7 )Distance from ( P(2, 0) ):( x )-difference: ( (8 + 12 sqrt{2})/7 - 14/7 = (-6 + 12 sqrt{2})/7 )( y )-difference: ( (-6 + 12 sqrt{2})/7 - 0 = (-6 + 12 sqrt{2})/7 )So, the distance is ( sqrt{ [(-6 + 12 sqrt{2})/7]^2 + [(-6 + 12 sqrt{2})/7]^2 } = sqrt{2 * [(-6 + 12 sqrt{2})/7]^2 } = sqrt{2} * |(-6 + 12 sqrt{2})/7| )Similarly, for ( |PB| ), it's ( sqrt{2} * |(-6 - 12 sqrt{2})/7| )Multiplying them:( |PA| * |PB| = 2 * | [(-6 + 12 sqrt{2})(-6 - 12 sqrt{2})]/49 | )Compute numerator:( (-6 + 12 sqrt{2})(-6 - 12 sqrt{2}) = (-6)^2 - (12 sqrt{2})^2 = 36 - 288 = -252 )So, absolute value is 252, so:( |PA| * |PB| = 2 * (252/49) = 2 * (36/7) = 72/7 )Wait, so according to my calculations, it's 72/7, but in the initial thought, it was 48/7. So, perhaps I made a mistake in the initial thought.Wait, let me check the initial thought again.In the initial thought, the user had:"Let the ( x )-coordinates of points ( A ) and ( B ) be ( x_1 ) and ( x_2 ), respectively. Then, we have ( x_{1}+x_{2}= frac {16}{7} ) and ( x_{1}x_{2}=- frac {32}{7} ).Now, we need to find ( |PA|⋅|PB| ). Using the distance formula between two points, we get:( |PA|⋅|PB|= sqrt {2}|x_{1}-2|+ sqrt {2}|x_{2}-2|= sqrt {2}|x_{1}-x_{2}|= sqrt {2} sqrt {(x_{1}+x_{2})^{2}-4x_{1}x_{2}}= sqrt {2} sqrt { frac {1152}{7^{2}}}= boxed{frac {48}{7}} )."Wait, in the initial thought, the user made a mistake in the distance formula. They wrote ( |PA|⋅|PB|= sqrt {2}|x_{1}-2|+ sqrt {2}|x_{2}-2| ), which is incorrect because the product of distances is not the sum of distances multiplied by sqrt(2). Instead, the product should be computed as sqrt(2) * |x1 - 2| multiplied by sqrt(2) * |x2 - 2|, which is 2 * |x1 - 2| * |x2 - 2|.But in the initial thought, the user incorrectly wrote it as a sum instead of a product, which led to an incorrect result.So, in reality, the correct approach is:( |PA| = sqrt{2} |x1 - 2| )( |PB| = sqrt{2} |x2 - 2| )Thus, ( |PA| * |PB| = 2 |x1 - 2| |x2 - 2| )Now, ( |x1 - 2| |x2 - 2| ) can be expressed as |(x1 - 2)(x2 - 2)|Let me compute ( (x1 - 2)(x2 - 2) )From the quadratic equation ( 7y^2 + 12y - 36 = 0 ), we found the roots ( y1 ) and ( y2 ), and from ( x = y + 2 ), we have ( x1 = y1 + 2 ), ( x2 = y2 + 2 )But perhaps it's easier to express ( x1 ) and ( x2 ) in terms of the quadratic equation.Wait, in the quadratic equation in ( y ), we have ( 7y^2 + 12y - 36 = 0 ), so the roots are ( y1 ) and ( y2 ). Then, ( x1 = y1 + 2 ), ( x2 = y2 + 2 )So, ( x1 - 2 = y1 ), ( x2 - 2 = y2 )Thus, ( (x1 - 2)(x2 - 2) = y1 y2 )From the quadratic equation, ( y1 + y2 = -12/7 ), and ( y1 y2 = -36/7 )So, ( (x1 - 2)(x2 - 2) = y1 y2 = -36/7 )Therefore, ( |x1 - 2| |x2 - 2| = |y1 y2| = | -36/7 | = 36/7 )Thus, ( |PA| * |PB| = 2 * (36/7) = 72/7 )So, the correct answer is 72/7.But in the initial thought, the user incorrectly wrote ( |PA|⋅|PB|= sqrt {2}|x_{1}-2|+ sqrt {2}|x_{2}-2| ), which is wrong because it's a sum instead of a product. The correct approach is to compute the product as 2 * |x1 - 2| * |x2 - 2|, which is 2 * |y1 y2| = 2 * (36/7) = 72/7.Therefore, the correct value of ( |PA| cdot |PB| ) is 72/7.Wait, but in my earlier detailed calculation, I also got 72/7, so that's consistent.But in the initial thought, the user had 48/7, which is incorrect. So, the correct answer should be 72/7.But wait, let me check again.Wait, in the quadratic equation, the roots are ( y1 ) and ( y2 ), and ( x1 = y1 + 2 ), ( x2 = y2 + 2 )So, ( x1 - 2 = y1 ), ( x2 - 2 = y2 )Thus, ( (x1 - 2)(x2 - 2) = y1 y2 = c/a = (-36)/7 )So, ( |x1 - 2| |x2 - 2| = |y1 y2| = 36/7 )Thus, ( |PA| * |PB| = 2 * (36/7) = 72/7 )Yes, that's correct.So, the correct answer is 72/7.But in the initial thought, the user had 48/7, which is incorrect. So, the correct answer is 72/7.Wait, but in my detailed calculation earlier, I also got 72/7, so that's consistent.Therefore, the correct value of ( |PA| cdot |PB| ) is ( boxed{dfrac{72}{7}} )

💡First, I recognize that receiving money on WeChat is represented as a positive number, such as +5 yuan for receiving 5 yuan.Expenditure, on the other hand, involves spending money, which should be represented as a negative number to indicate a decrease in funds.Therefore, Aunt Zhang's expenditure of 3 yuan should be represented as -3 yuan.This corresponds to option A.

💡Okay, so I have this problem where every student in a class attends exactly two extracurricular clubs, and any group of three students shares at least one common club. I need to prove that there is at least one club that every child attends. Hmm, let me try to wrap my head around this.First, let me restate the problem in my own words to make sure I understand it. We have a class of students, each of whom is a member of two clubs. Now, if I pick any three students from this class, there must be at least one club that all three of them are members of. The goal is to show that under these conditions, there must be a single club that every student in the class is part of.Alright, so maybe I can start by considering a simple case. Let's say there are only a few students. Maybe three students. If each attends two clubs, and any three share at least one club, does that mean there's a common club? Well, with three students, each attending two clubs, and all three sharing at least one club, it seems like that common club must be the one that all three are in. So in this case, yes, there is a club that all three attend. But this is a very small case. I need to think about a more general scenario.Let me try to think about it with more students. Suppose there are four students. Each attends two clubs. Any three of them share at least one club. Does that mean there's a club that all four attend? Hmm. Let me try to visualize this.Let's denote the students as A, B, C, D. Each of them is in two clubs. Let's say A is in clubs 1 and 2, B is in clubs 1 and 3, C is in clubs 1 and 4, and D is in clubs 1 and 5. In this case, every group of three students would share club 1. So, in this setup, club 1 is the common club that everyone is in. So, that satisfies the condition.But what if the clubs are arranged differently? Suppose A is in clubs 1 and 2, B is in clubs 1 and 3, C is in clubs 2 and 3, and D is in clubs 2 and 3. Now, let's see if any three students share a common club. Take A, B, C: A is in 1 and 2, B is in 1 and 3, C is in 2 and 3. The common club here is... hmm, actually, there is no common club. A is in 1 and 2, B is in 1 and 3, C is in 2 and 3. So, A and B share club 1, A and C share club 2, B and C share club 3, but all three together don't share a common club. That violates the condition given in the problem. So, this arrangement wouldn't satisfy the problem's condition.Therefore, such an arrangement where not all students are in a common club might not satisfy the condition. So, maybe the only way to satisfy the condition is if there is a common club that everyone is in.Wait, but in the previous example, I tried to make an arrangement without a common club, but it didn't satisfy the condition. So, perhaps that suggests that to satisfy the condition, there must be a common club.But I need to think more formally. Maybe I can use some principles from set theory or combinatorics.Let me think about the problem in terms of sets. Each student is represented by a set of two clubs. The condition is that for any three students, the intersection of their sets is non-empty. So, in other words, for any three sets (each of size two), their intersection is non-empty.I need to show that under these conditions, there is a club that is present in all sets.Hmm, so maybe I can use the pigeonhole principle or something like that.Alternatively, maybe I can think about it as a graph problem. Each club is a vertex, and each student is an edge connecting two vertices. Then, the condition is that any three edges share a common vertex. So, in graph terms, any three edges must have a common vertex. Then, the problem reduces to showing that there is a vertex that is incident to all edges.Wait, that might be a useful perspective. So, if I model the clubs as vertices and the students as edges, then the condition is that any three edges share a common vertex. So, in graph theory terms, this is a 3-uniform hypergraph where every three edges share a common vertex. But I'm not sure if that's the exact terminology.Alternatively, in graph terms, if every three edges share a common vertex, then the graph has a common vertex for any three edges. So, does that imply that all edges are incident to a single vertex?Wait, that seems similar to what we need to prove. If every three edges share a common vertex, then all edges must be incident to that vertex. So, in other words, there is a universal vertex that is connected to all edges.But is that necessarily true? Let me think.Suppose we have a graph where not all edges are incident to a single vertex. Then, can we have three edges that don't share a common vertex?Yes, for example, in a triangle graph, each edge is connected to two vertices, but no single vertex is shared by all three edges. So, in that case, the three edges don't share a common vertex, which violates the condition.Therefore, if in our graph, every three edges share a common vertex, then the graph must be such that all edges are incident to a single vertex. Hence, there is a universal vertex connected to all edges.Translating back to the original problem, that means there is a club (vertex) that every student (edge) is a member of.Therefore, this seems to be a valid approach. By modeling the problem as a graph where clubs are vertices and students are edges, the condition that any three students share a common club translates to any three edges sharing a common vertex. This implies that all edges must be incident to a single vertex, meaning there is a club that every student attends.But let me try to formalize this argument to make sure it's rigorous.Assume, for contradiction, that there is no club that every student attends. Then, the graph (as defined above) does not have a universal vertex. Therefore, there exist at least two edges (students) that are not incident to the same vertex (club). Let's say edge A is connected to vertices C1 and C2, and edge B is connected to vertices C1 and C3, where C2 ≠ C3.Now, consider a third edge (student) C. If edge C is connected to C1 and some other club, say C4, then the three edges A, B, and C all share vertex C1. But if edge C is connected to, say, C2 and C3, then edges A, B, and C do not share a common vertex, which contradicts the given condition.Wait, but in the problem statement, it's given that any three students share at least one common club. So, in terms of the graph, any three edges must share a common vertex. Therefore, edge C cannot be connected to C2 and C3 because then edges A, B, and C would not share a common vertex. Therefore, edge C must be connected to C1 and some other club.But if every edge must be connected to C1, then C1 is the universal vertex, meaning every student is a member of club C1. This contradicts our initial assumption that there is no such universal club.Therefore, our assumption must be wrong, and there must exist a club that every student attends.Hmm, that seems to make sense. So, by assuming the opposite and reaching a contradiction, we've proven that there must be a club that every student is a part of.Let me see if I can think of another way to approach this problem, maybe using set theory directly without the graph model.Each student is in two clubs, so we can represent each student as a pair of clubs. The condition is that for any three students, their sets of clubs have a non-empty intersection. That is, there is at least one club common to all three.We need to show that there exists a club that is common to all students.Suppose, for contradiction, that no such club exists. Then, for every club, there is at least one student not in that club.Since each student is in two clubs, if no club is universal, then for every club, there is at least one student not in it. Let's try to construct three students such that they don't share a common club, which would contradict the given condition.Let me pick a club, say Club 1. Since it's not universal, there exists a student not in Club 1. Let's call this student A. Student A is in two other clubs, say Club 2 and Club 3.Similarly, pick another club, say Club 2. Since it's not universal, there exists a student not in Club 2. Let's call this student B. Student B is in two other clubs, say Club 1 and Club 4.Now, consider student A (in Club 2 and 3) and student B (in Club 1 and 4). Now, let's pick a third student, C. Student C is in two clubs. If student C is in Club 1, then students A, B, and C share Club 1. If student C is in Club 2, then students A, B, and C share Club 2. If student C is in Club 3, then students A, B, and C share Club 3. If student C is in Club 4, then students A, B, and C share Club 4. Wait, but student C can only be in two clubs. So, if student C is in, say, Club 5 and Club 6, then students A, B, and C don't share a common club, which would violate the condition.But in our setup, student C must be in two clubs, but if we've already used up Club 1, 2, 3, 4, then student C could be in Club 5 and 6, which are new clubs. But then, students A, B, and C don't share a common club, which contradicts the given condition.Therefore, our assumption that no club is universal leads to a contradiction, meaning there must be a universal club that every student is a part of.Hmm, that seems to reinforce the earlier conclusion. So, whether I model it as a graph or approach it directly through set theory, the conclusion is the same: there must be a club that every student attends.Let me try to think of another angle. Maybe using the principle of inclusion-exclusion or something like that.Each student is in two clubs, so the total number of club memberships is twice the number of students. Let's denote the number of students as n, so the total number of club memberships is 2n.Now, if there are k clubs, then the sum of the sizes of all clubs is 2n. The problem states that any three students share at least one common club. So, for any three students, there is at least one club that contains all three.Wait, but actually, the problem says any group of three children shares at least one common club. So, it's not necessarily that all three are in the same club, but that there is at least one club that all three are in. Wait, no, actually, if they share a common club, that means all three are in that club. So, yes, for any three students, there exists a club that contains all three.Wait, that's a stronger condition than I initially thought. So, not only do any two students share a club, but any three students share a club. That's a much stronger condition.So, in that case, maybe the conclusion is even more straightforward. Because if any three students must be in a common club, then certainly, if you take all students, they must all be in that common club.Wait, but that might not necessarily be the case. Let me think.Suppose there are multiple clubs, each of which contains a large number of students, but not all. However, any three students must be in at least one of these clubs. So, is it possible to have multiple clubs, each missing some students, but arranged in such a way that any three students are covered by at least one club?Hmm, that seems tricky. Let me try to think of an example.Suppose we have two clubs, Club A and Club B. Suppose Club A has all students except one, say student X, and Club B has all students except another one, say student Y. Now, if I take any three students, unless all three are X, Y, and someone else, they will be in at least one club. But wait, if I take students X, Y, and Z, then Z is in both clubs, but X is only in Club B, and Y is only in Club A. So, there is no club that contains all three, which violates the condition.Therefore, such an arrangement doesn't satisfy the condition. So, maybe having two clubs isn't sufficient unless one of them is universal.Alternatively, suppose we have three clubs, each missing a different student. But then, similar to the two-club case, if I take three students each missing a different club, they won't have a common club, violating the condition.Therefore, it seems that to satisfy the condition that any three students share a common club, there must be a club that contains all students. Because otherwise, you can find three students, each missing a different club, and thus not sharing a common club.Wait, but in the problem, each student is in exactly two clubs. So, if there are multiple clubs, each student is in two of them. So, if a club is not universal, then there are students not in it. But if you have multiple non-universal clubs, you can potentially have students missing different clubs, leading to the possibility of three students not sharing a common club.Therefore, to prevent that, there must be a club that includes all students, ensuring that any three students will at least be in that club.So, this seems to be another way to see why a universal club must exist.Let me try to formalize this argument.Assume that no club is universal. Then, for each club, there exists at least one student not in that club. Since each student is in two clubs, the number of clubs must be at least three. Let's say we have clubs C1, C2, C3, etc.Now, consider students A, B, and C. If A is not in C1, B is not in C2, and C is not in C3, then the three students A, B, and C do not share a common club, which contradicts the given condition.Therefore, our assumption that no club is universal must be false. Hence, there must be a club that every student attends.Hmm, that seems to make sense. So, by assuming that no club is universal, we can find three students who don't share a common club, which violates the problem's condition. Therefore, there must be a universal club.I think I've covered multiple angles now: graph theory, set theory, and direct contradiction. All of them seem to lead to the same conclusion that there must be a club that every student attends.Just to make sure, let me think about a specific example with more students.Suppose there are five students: A, B, C, D, E. Each attends two clubs. Let's say there are three clubs: X, Y, Z.If every three students must share a common club, then let's see how the clubs can be arranged.If club X has all five students, then obviously, any three students will share club X. So, that satisfies the condition.But suppose club X has four students, and club Y has four students, and club Z has four students. But each student is in two clubs. So, for example, A is in X and Y, B is in X and Z, C is in Y and Z, D is in X and Y, E is in X and Z.Now, let's check if any three students share a common club.Take A, B, C: A is in X and Y; B is in X and Z; C is in Y and Z. They don't all share a common club.Wait, that violates the condition. So, this arrangement doesn't work.Alternatively, if I make sure that every three students share a common club, perhaps by having one club that everyone is in.So, if club X has all five students, and clubs Y and Z have some overlapping, but since each student is in two clubs, they can be in X and Y or X and Z.In this case, any three students will share club X, satisfying the condition.Therefore, this seems to reinforce the idea that having a universal club is necessary to satisfy the condition.Another thought: if there are multiple clubs, each missing some students, but arranged in such a way that any three students are covered by at least one club, is that possible?But as I saw earlier, if each club is missing at least one student, then you can find three students each missing a different club, leading to no common club among them. Therefore, it's not possible to have multiple non-universal clubs without violating the condition.So, in conclusion, the only way to satisfy the condition that any three students share a common club is if there is a club that every student attends.I think I've thought through this problem from multiple perspectives: graph theory, set theory, specific examples, and direct contradiction. All of them lead to the same conclusion. Therefore, I'm confident that the statement is true: there must be at least one club that every child attends.