answer：Alright, so I have this problem here: "If 15 machines can finish a job in a certain number of days, then 5 more machines would be needed to finish the job in one-fourth less time. How many days does it take for the 15 machines to finish the job?" Okay, let's try to break this down step by step.First, I need to understand what's being asked. We have 15 machines that can finish a job in a certain number of days, which we'll call D days. Then, if we add 5 more machines, making it 20 machines in total, the job can be finished in one-fourth less time than D. So, the new time would be D minus one-fourth of D, which is (3/4)D.Now, I remember that when dealing with machines and time, it's often about work rates. The more machines you have, the less time it takes to finish the job, assuming all machines work at the same rate. So, the number of machines and the time taken are inversely proportional. That means if you double the number of machines, the time should be halved, and so on.Let me write down what I know:- 15 machines take D days to finish the job.- 20 machines take (3/4)D days to finish the same job.Since the amount of work done is the same in both cases, I can set up an equation based on the work done. Work is usually calculated as the number of machines multiplied by the time they take. So, for 15 machines, the work done is 15 * D. For 20 machines, it's 20 * (3/4)D.Since the work done is the same, I can set these equal to each other:15 * D = 20 * (3/4)DHmm, let's see. If I simplify the right side first:20 * (3/4)D = (20 * 3/4) * D = 15 * DSo, 15 * D = 15 * DWait, that's interesting. Both sides are equal, which means this equation is always true, regardless of the value of D. That suggests that the relationship holds for any D, which doesn't help me find the specific value of D.But the problem is asking for the number of days it takes for the 15 machines to finish the job, so I must be missing something. Maybe I need to think about this differently.Perhaps I should consider the rate at which each machine works. Let's say each machine has a work rate of R jobs per day. Then, 15 machines would have a combined rate of 15R, and 20 machines would have a combined rate of 20R.The time taken is inversely proportional to the rate, so:For 15 machines: Time = 1 / (15R) = DFor 20 machines: Time = 1 / (20R) = (3/4)DSo, I can set up the equation:1 / (20R) = (3/4) * (1 / (15R))Let's simplify this:1 / (20R) = (3/4) / (15R)Multiply both sides by 20R to eliminate the denominator:1 = (3/4) * (20R) / (15R)Simplify the right side:(3/4) * (20R / 15R) = (3/4) * (4/3) = 1So, again, I end up with 1 = 1, which is always true but doesn't help me find D.Hmm, maybe I need to approach this by considering the total work done. Let's denote the total work as W.For 15 machines taking D days:W = 15 * DFor 20 machines taking (3/4)D days:W = 20 * (3/4)D = 15DSo, again, both expressions for W are equal, which just confirms the relationship but doesn't give me the value of D.Wait a minute, maybe the problem is designed in such a way that D cancels out, meaning that the number of days isn't uniquely determined by the information given. That could be possible, but the problem is asking for a specific number of days, so perhaps I'm missing a step.Let me try to think differently. Maybe I should express D in terms of the number of machines.If 15 machines take D days, then 20 machines take (3/4)D days.Since the number of machines and time are inversely proportional, the product of machines and time should be constant.So, 15 * D = 20 * (3/4)DSimplify the right side:20 * (3/4)D = 15DSo, 15D = 15DAgain, this doesn't help me find D.Perhaps the problem is missing some information, or maybe I'm interpreting it incorrectly. Let me read it again:"If 15 machines can finish a job in a certain number of days, then 5 more machines would be needed to finish the job in one-fourth less time. How many days does it take for the 15 machines to finish the job?"Wait, it says "5 more machines would be needed to finish the job in one-fourth less time." So, does that mean that adding 5 machines reduces the time by one-fourth, or that the time is one-fourth less than the original time?I think it's the latter: the time with 20 machines is one-fourth less than the time with 15 machines, so it's (3/4)D.But as I saw earlier, this leads to an identity, not a solvable equation.Maybe I need to consider that the time is reduced by one-fourth, meaning the new time is D - (1/4)D = (3/4)D, which is what I did before.But still, it doesn't give me a specific value for D.Perhaps the problem assumes that the work done is the same, so the total work W = 15D = 20*(3/4)D = 15D, which again doesn't help.Wait, maybe I need to think about the rate per machine.Let’s denote the work rate of one machine as R (jobs per day). Then:15 machines: 15R * D = 1 job20 machines: 20R * (3/4)D = 1 jobSo, both equal to 1 job.From the first equation:15R * D = 1 => R = 1 / (15D)From the second equation:20R * (3/4)D = 1Substitute R from the first equation:20 * (1 / (15D)) * (3/4)D = 1Simplify:(20 * 3/4) / 15 = 1(15) / 15 = 11 = 1Again, it's an identity, not helpful for finding D.This suggests that the problem as stated doesn't provide enough information to determine D uniquely. It only establishes a relationship between the number of machines and the time, but without a specific value or additional information, D remains arbitrary.However, the problem is asking for a specific number of days, so perhaps there's an assumption I'm missing. Maybe the job is considered as 1 unit of work, and we need to find D such that the rates align.But as shown, the rates align for any D, so it's impossible to determine D from the given information alone.Wait, maybe I misinterpreted the problem. Let me read it again carefully:"If 15 machines can finish a job in a certain number of days, then 5 more machines would be needed to finish the job in one-fourth less time. How many days does it take for the 15 machines to finish the job?"So, it's saying that if you have 15 machines, they take D days. If you add 5 more machines (total 20), they take one-fourth less time, which is (3/4)D.But as I've shown, this relationship holds for any D, so D cannot be determined from the given information.Perhaps the problem expects me to recognize that the number of days is the same regardless of the number of machines, but that doesn't make sense because more machines should reduce the time.Alternatively, maybe the problem is designed to show that D cannot be determined from the given information, but that seems unlikely since it's asking for a specific number.Wait, perhaps I need to set up the equation differently. Let's try expressing the work done in terms of machine-days.Work = Number of machines * Number of daysSo, for 15 machines: Work = 15DFor 20 machines: Work = 20*(3/4)D = 15DSo, 15D = 15D, which is always true.This again shows that D cannot be determined from the given information.Therefore, the problem as stated doesn't provide enough data to find the exact number of days D.But since the problem is asking for D, maybe there's a standard assumption or perhaps a typo in the problem statement. Alternatively, maybe I'm supposed to assume that the time with 20 machines is one-fourth of the original time, not one-fourth less.Let me try that interpretation.If the time with 20 machines is one-fourth of D, then:15D = 20*(1/4)D = 5DSo, 15D = 5D => 15 = 5, which is not possible.So, that interpretation doesn't work.Alternatively, maybe one-fourth less time means the time is reduced by one-fourth, so the new time is D - (1/4)D = (3/4)D, which is what I did before.But as shown, that doesn't help find D.Perhaps the problem is expecting me to recognize that the number of machines and time are inversely proportional, so:15 machines * D days = 20 machines * (3/4)D daysWhich simplifies to:15D = 15DAgain, no help.Wait, maybe I need to consider that the time is inversely proportional to the number of machines, so:D2 = D1 * (N1 / N2)Where D2 is the new time, N1 is the original number of machines, and N2 is the new number.So, D2 = D * (15 / 20) = (3/4)DWhich is consistent with the problem statement.But again, this doesn't help find D.I think the conclusion is that the problem doesn't provide enough information to determine D uniquely. It only establishes a proportional relationship between the number of machines and the time taken, but without a specific value or additional information, D remains arbitrary.Therefore, the answer is that the number of days cannot be determined from the given information.

answer：Okay, so I have this system of equations to solve:[left{begin{matrix}sin x + cos x = frac{sqrt{n}}{2} tan frac{x}{2} = frac{sqrt{n} - 2}{3}end{matrix}right.]And I need to find all solutions ( x ) in the interval (left[0, frac{pi}{4}right)). Hmm, let's see. I think I should start by analyzing each equation separately and then see how they can be connected.First, looking at the first equation: (sin x + cos x = frac{sqrt{n}}{2}). I remember that expressions like (sin x + cos x) can be rewritten using a trigonometric identity. Maybe I can square both sides to simplify it. Let me try that.Squaring both sides:[(sin x + cos x)^2 = left(frac{sqrt{n}}{2}right)^2]Expanding the left side:[sin^2 x + 2 sin x cos x + cos^2 x = frac{n}{4}]I know that (sin^2 x + cos^2 x = 1), so substituting that in:[1 + 2 sin x cos x = frac{n}{4}]Hmm, (2 sin x cos x) is equal to (sin 2x), right? So, substituting that:[1 + sin 2x = frac{n}{4}]Which simplifies to:[sin 2x = frac{n}{4} - 1]Okay, so that's one equation involving (sin 2x). Now, let's look at the second equation: (tan frac{x}{2} = frac{sqrt{n} - 2}{3}).I remember that (tan frac{x}{2}) can be expressed in terms of (sin x) and (cos x). The identity is:[tan frac{x}{2} = frac{sin x}{1 + cos x}]Alternatively, another identity is:[tan frac{x}{2} = sqrt{frac{1 - cos x}{1 + cos x}}]Maybe that's useful. Let me use that one because it's in terms of (cos x), which might help me connect it to the first equation.So, setting that equal to the given expression:[sqrt{frac{1 - cos x}{1 + cos x}} = frac{sqrt{n} - 2}{3}]Let me square both sides to eliminate the square root:[frac{1 - cos x}{1 + cos x} = left(frac{sqrt{n} - 2}{3}right)^2]Simplify the right side:[frac{1 - cos x}{1 + cos x} = frac{(sqrt{n} - 2)^2}{9}]Let me denote ( k = frac{(sqrt{n} - 2)^2}{9} ) to make it simpler. Then:[frac{1 - cos x}{1 + cos x} = k]Cross-multiplying:[1 - cos x = k (1 + cos x)]Expanding the right side:[1 - cos x = k + k cos x]Bring all terms involving (cos x) to one side:[1 - k = cos x (1 + k)]Solving for (cos x):[cos x = frac{1 - k}{1 + k}]Substituting back ( k = frac{(sqrt{n} - 2)^2}{9} ):[cos x = frac{1 - frac{(sqrt{n} - 2)^2}{9}}{1 + frac{(sqrt{n} - 2)^2}{9}}]Simplify numerator and denominator:[cos x = frac{9 - (sqrt{n} - 2)^2}{9 + (sqrt{n} - 2)^2}]Let me compute ((sqrt{n} - 2)^2):[(sqrt{n} - 2)^2 = n - 4 sqrt{n} + 4]So substituting back:[cos x = frac{9 - (n - 4 sqrt{n} + 4)}{9 + (n - 4 sqrt{n} + 4)} = frac{9 - n + 4 sqrt{n} - 4}{9 + n - 4 sqrt{n} + 4}]Simplify numerator and denominator:Numerator: (5 - n + 4 sqrt{n})Denominator: (13 + n - 4 sqrt{n})So,[cos x = frac{5 - n + 4 sqrt{n}}{13 + n - 4 sqrt{n}}]Hmm, that's a bit complicated. Maybe I can find a relationship between this and the first equation.From the first equation, I had:[sin 2x = frac{n}{4} - 1]But (sin 2x = 2 sin x cos x). So, if I can express (sin x) in terms of (cos x), I can substitute.Since (sin^2 x + cos^2 x = 1), then:[sin x = sqrt{1 - cos^2 x}]So,[sin 2x = 2 sqrt{1 - cos^2 x} cdot cos x]But from earlier, (sin 2x = frac{n}{4} - 1). So,[2 sqrt{1 - cos^2 x} cdot cos x = frac{n}{4} - 1]But I already have an expression for (cos x) in terms of (n). Let me substitute that in.Let me denote ( C = cos x = frac{5 - n + 4 sqrt{n}}{13 + n - 4 sqrt{n}} )Then,[2 sqrt{1 - C^2} cdot C = frac{n}{4} - 1]This seems quite involved. Maybe there's a better way. Alternatively, perhaps I can express (sin x + cos x) in terms of (tan frac{x}{2}).Wait, another approach: Let me denote ( t = tan frac{x}{2} ). Then, using the identities:[sin x = frac{2t}{1 + t^2}, quad cos x = frac{1 - t^2}{1 + t^2}]So, substituting into the first equation:[frac{2t}{1 + t^2} + frac{1 - t^2}{1 + t^2} = frac{sqrt{n}}{2}]Combine the terms:[frac{2t + 1 - t^2}{1 + t^2} = frac{sqrt{n}}{2}]Multiply both sides by (1 + t^2):[2t + 1 - t^2 = frac{sqrt{n}}{2} (1 + t^2)]Multiply both sides by 2 to eliminate the fraction:[4t + 2 - 2t^2 = sqrt{n} (1 + t^2)]Bring all terms to one side:[4t + 2 - 2t^2 - sqrt{n} - sqrt{n} t^2 = 0]Combine like terms:[(-2 - sqrt{n}) t^2 + 4t + (2 - sqrt{n}) = 0]So, this is a quadratic equation in terms of ( t ):[(-2 - sqrt{n}) t^2 + 4t + (2 - sqrt{n}) = 0]Let me write it as:[(2 + sqrt{n}) t^2 - 4t + (sqrt{n} - 2) = 0]Multiplying both sides by -1 to make the coefficient of ( t^2 ) positive:[(2 + sqrt{n}) t^2 - 4t + (sqrt{n} - 2) = 0]Now, from the second equation, we have:[t = tan frac{x}{2} = frac{sqrt{n} - 2}{3}]So, substituting ( t = frac{sqrt{n} - 2}{3} ) into the quadratic equation:[(2 + sqrt{n}) left( frac{sqrt{n} - 2}{3} right)^2 - 4 left( frac{sqrt{n} - 2}{3} right) + (sqrt{n} - 2) = 0]Let me compute each term step by step.First, compute ( left( frac{sqrt{n} - 2}{3} right)^2 ):[left( frac{sqrt{n} - 2}{3} right)^2 = frac{(sqrt{n} - 2)^2}{9} = frac{n - 4 sqrt{n} + 4}{9}]So, the first term:[(2 + sqrt{n}) cdot frac{n - 4 sqrt{n} + 4}{9} = frac{(2 + sqrt{n})(n - 4 sqrt{n} + 4)}{9}]Let me expand the numerator:[(2 + sqrt{n})(n - 4 sqrt{n} + 4) = 2n - 8 sqrt{n} + 8 + n sqrt{n} - 4n + 4 sqrt{n}]Wait, let me do it step by step:Multiply 2 by each term:- (2 cdot n = 2n)- (2 cdot (-4 sqrt{n}) = -8 sqrt{n})- (2 cdot 4 = 8)Multiply (sqrt{n}) by each term:- (sqrt{n} cdot n = n^{3/2})- (sqrt{n} cdot (-4 sqrt{n}) = -4n)- (sqrt{n} cdot 4 = 4 sqrt{n})So, combining all terms:[2n - 8 sqrt{n} + 8 + n^{3/2} - 4n + 4 sqrt{n}]Combine like terms:- (2n - 4n = -2n)- (-8 sqrt{n} + 4 sqrt{n} = -4 sqrt{n})- The constants: 8- The term with (n^{3/2}): (n^{3/2})So, the numerator becomes:[n^{3/2} - 2n - 4 sqrt{n} + 8]Therefore, the first term is:[frac{n^{3/2} - 2n - 4 sqrt{n} + 8}{9}]Now, the second term:[-4 left( frac{sqrt{n} - 2}{3} right) = -frac{4 sqrt{n} - 8}{3} = frac{-4 sqrt{n} + 8}{3}]The third term is:[sqrt{n} - 2]So, putting all together:[frac{n^{3/2} - 2n - 4 sqrt{n} + 8}{9} + frac{-4 sqrt{n} + 8}{3} + (sqrt{n} - 2) = 0]To combine these, let me get a common denominator of 9:First term remains as is:[frac{n^{3/2} - 2n - 4 sqrt{n} + 8}{9}]Second term:[frac{-4 sqrt{n} + 8}{3} = frac{(-4 sqrt{n} + 8) cdot 3}{9} = frac{-12 sqrt{n} + 24}{9}]Third term:[sqrt{n} - 2 = frac{9 sqrt{n} - 18}{9}]So, combining all three terms:[frac{n^{3/2} - 2n - 4 sqrt{n} + 8 - 12 sqrt{n} + 24 + 9 sqrt{n} - 18}{9} = 0]Simplify the numerator:Combine like terms:- (n^{3/2}): (n^{3/2})- (n) terms: (-2n)- (sqrt{n}) terms: (-4 sqrt{n} - 12 sqrt{n} + 9 sqrt{n} = (-4 -12 +9) sqrt{n} = (-7) sqrt{n})- Constants: (8 + 24 - 18 = 14)So, numerator becomes:[n^{3/2} - 2n - 7 sqrt{n} + 14]Thus, the equation is:[frac{n^{3/2} - 2n - 7 sqrt{n} + 14}{9} = 0]Multiply both sides by 9:[n^{3/2} - 2n - 7 sqrt{n} + 14 = 0]Hmm, this is a complicated equation. Maybe I can substitute ( m = sqrt{n} ), so that ( n = m^2 ). Let's try that.Substituting:[(m^2)^{3/2} - 2m^2 - 7m + 14 = 0]Simplify ( (m^2)^{3/2} = m^3 ):[m^3 - 2m^2 - 7m + 14 = 0]So, now we have a cubic equation in ( m ):[m^3 - 2m^2 - 7m + 14 = 0]Let me try to factor this. Maybe rational roots? The possible rational roots are factors of 14 over factors of 1: ±1, ±2, ±7, ±14.Let me test m=2:[2^3 - 2*2^2 -7*2 +14 = 8 - 8 -14 +14 = 0Yes! So, m=2 is a root.Therefore, we can factor out (m - 2):Using polynomial division or synthetic division.Divide ( m^3 - 2m^2 -7m +14 ) by (m - 2):Using synthetic division:2 | 1 -2 -7 14 2 0 -14 1 0 -7 0So, the cubic factors as:[(m - 2)(m^2 - 7) = 0]Therefore, the roots are:- ( m = 2 )- ( m = sqrt{7} )- ( m = -sqrt{7} )But since ( m = sqrt{n} ) and ( n ) is a natural number, ( m ) must be positive. So, possible solutions are ( m = 2 ) and ( m = sqrt{7} ).Thus, ( sqrt{n} = 2 ) or ( sqrt{n} = sqrt{7} ).Therefore, ( n = 4 ) or ( n = 7 ).Now, let's check each case.**Case 1: n = 4**From the second equation:[tan frac{x}{2} = frac{sqrt{4} - 2}{3} = frac{2 - 2}{3} = 0]So, ( tan frac{x}{2} = 0 ). Since ( x ) is in ([0, frac{pi}{4})), ( frac{x}{2} ) is in ([0, frac{pi}{8})). The only solution is ( frac{x}{2} = 0 ), so ( x = 0 ).Check the first equation:[sin 0 + cos 0 = 0 + 1 = 1]And ( frac{sqrt{4}}{2} = frac{2}{2} = 1 ). So, it holds. So, ( x = 0 ) is a solution when ( n = 4 ).**Case 2: n = 7**From the second equation:[tan frac{x}{2} = frac{sqrt{7} - 2}{3}]Let me compute ( frac{sqrt{7} - 2}{3} ). Since ( sqrt{7} approx 2.6458 ), so ( sqrt{7} - 2 approx 0.6458 ), so ( frac{0.6458}{3} approx 0.2153 ).So, ( tan frac{x}{2} approx 0.2153 ). Let me compute ( frac{x}{2} ):[frac{x}{2} = arctan(0.2153) approx 0.211 text{ radians}]Therefore, ( x approx 0.422 ) radians. Since ( frac{pi}{4} approx 0.7854 ), this is within the interval ([0, frac{pi}{4})).Now, check the first equation:[sin x + cos x = frac{sqrt{7}}{2} approx frac{2.6458}{2} approx 1.3229]Compute ( sin x + cos x ) with ( x approx 0.422 ):[sin(0.422) approx 0.410, quad cos(0.422) approx 0.912][sin x + cos x approx 0.410 + 0.912 = 1.322]Which is approximately equal to ( frac{sqrt{7}}{2} approx 1.3229 ). So, it holds.Therefore, ( x approx 0.422 ) radians is a solution when ( n = 7 ).But let me express ( x ) more precisely. Since ( tan frac{x}{2} = frac{sqrt{7} - 2}{3} ), then:[frac{x}{2} = arctanleft( frac{sqrt{7} - 2}{3} right)][x = 2 arctanleft( frac{sqrt{7} - 2}{3} right)]Alternatively, using the identity ( sin 2x = frac{n - 4}{8} ) from earlier:For ( n = 7 ):[sin 2x = frac{7 - 4}{8} = frac{3}{8}][2x = arcsinleft( frac{3}{8} right)][x = frac{1}{2} arcsinleft( frac{3}{8} right)]Which is another way to express the solution.So, summarizing, we have two solutions:1. ( x = 0 ) when ( n = 4 )2. ( x = frac{1}{2} arcsinleft( frac{3}{8} right) ) when ( n = 7 )I should check if there are any other possible values of ( n ). Since ( n ) is a natural number, and from the cubic equation, the only solutions are ( n = 4 ) and ( n = 7 ). So, these are the only possible natural numbers that satisfy the system.Therefore, the solutions are:- For ( n = 4 ), ( x = 0 )- For ( n = 7 ), ( x = frac{1}{2} arcsinleft( frac{3}{8} right) )I think that's all. Let me just verify once more.For ( n = 4 ):- ( sin 0 + cos 0 = 1 ), which equals ( frac{sqrt{4}}{2} = 1 )- ( tan frac{0}{2} = 0 ), which equals ( frac{sqrt{4} - 2}{3} = 0 )Perfect.For ( n = 7 ):- ( sin x + cos x approx 1.3229 ), which equals ( frac{sqrt{7}}{2} approx 1.3229 )- ( tan frac{x}{2} approx 0.2153 ), which equals ( frac{sqrt{7} - 2}{3} approx 0.2153 )So both equations are satisfied.Therefore, these are indeed the solutions.**Final Answer**The solutions are ( boxed{0} ) when ( n = 4 ) and ( boxed{frac{1}{2} arcsinleft( frac{3}{8} right)} ) when ( n = 7 ).

answer：First, identify the number of blocks Martha initially has, which is 4.Next, determine the number of additional blocks Martha finds, which is 80.Finally, add the initial number of blocks to the additional blocks to find the total number of blocks Martha ends with: 4 + 80 = 84.

answer：First, I need to determine the number of technicians in the workshop. I'll start by defining the variables: let ( T ) represent the number of technicians and ( R ) represent the number of the remaining workers.From the problem, I know that the total number of workers is 42, so I can write the equation:[T + R = 42]Next, I'll use the information about the average salaries to set up another equation. The total salary of all workers is the sum of the total salaries of the technicians and the remaining workers. This gives me:[18000T + 6000R = 8000 times 42][18000T + 6000R = 336000]To simplify this equation, I'll divide the entire equation by 6000:[3T + R = 56]Now I have a system of two equations:1. ( T + R = 42 )2. ( 3T + R = 56 )I'll subtract the first equation from the second to eliminate ( R ):[(3T + R) - (T + R) = 56 - 42][2T = 14][T = 7]Therefore, there are 7 technicians in the workshop.

answer：Alright, so I have this problem about triangle centers and homothety. Let me try to wrap my head around it step by step. First, the problem mentions triangle ABC with its circumcenter O, centroid G, and orthocenter H. Then, it defines A', B', and C' as the midpoints of the sides BC, CA, and AB respectively. Also, D, E, F are the feet of the altitudes from A, B, and C. Part 1 asks to show that triangle A'B'C' is the image of triangle ABC under a homothety centered at G. Hmm, homothety is a dilation (scaling) transformation centered at a point. So, I need to find a scaling factor and show that each point A, B, C maps to A', B', C' respectively under this transformation centered at G.I remember that the centroid G divides each median in the ratio 2:1. So, if I consider the midpoint A' of BC, the centroid G is located 2/3 of the way from A to A'. That suggests that the scaling factor might be 1/2 because moving from G to A' is half the distance from G to A. Let me test this idea.If I take point A, the centroid G is (A + B + C)/3. The midpoint A' is (B + C)/2. So, vector GA' would be A' - G = (B + C)/2 - (A + B + C)/3. Let me compute that:GA' = (3(B + C) - 2(A + B + C))/6 = (3B + 3C - 2A - 2B - 2C)/6 = (B + C - 2A)/6.Similarly, vector GA is A - G = A - (A + B + C)/3 = (3A - A - B - C)/3 = (2A - B - C)/3.Hmm, so GA' is (B + C - 2A)/6 and GA is (2A - B - C)/3. Notice that GA' is exactly -1/2 of GA. So, GA' = -1/2 GA. That means that A' is the image of A under a homothety centered at G with a ratio of -1/2. Similarly, this should hold for B' and C' as well. So, triangle A'B'C' is indeed the image of ABC under a homothety centered at G with a ratio of -1/2. That makes sense because midpoints are halfway along the sides, and the centroid is the balance point.Okay, so part 1 seems manageable. Now, moving on to part 2: deduce that O, G, and H are collinear. Hmm, I remember something called the Euler line, which states that the centroid, circumcenter, and orthocenter are colinear. So, maybe this homothety can help us see why.Since A'B'C' is the image of ABC under a homothety centered at G, perhaps the orthocenter of ABC maps to the circumcenter of A'B'C' under this homothety. Let me think.The homothety centered at G with ratio -1/2 maps ABC to A'B'C'. So, it should map the orthocenter H of ABC to the orthocenter of A'B'C'. But wait, what is the orthocenter of A'B'C'? Since A'B'C' is the medial triangle, its orthocenter is actually the centroid of ABC, which is G. Wait, no, that doesn't seem right.Wait, no, the orthocenter of the medial triangle is the nine-point center of the original triangle, which is the midpoint of OH. Hmm, maybe I'm getting confused here.Alternatively, since A'B'C' is similar to ABC scaled down by 1/2, maybe their circumcenters are related. The circumradius of A'B'C' should be half that of ABC. But where is the circumcenter of A'B'C'? It should be the image of O under the homothety.Since homothety maps O to the circumcenter of A'B'C', and the homothety is centered at G with ratio -1/2, then the circumcenter of A'B'C' is the midpoint between O and H? Wait, no, let's see.Wait, if the homothety maps ABC to A'B'C', then it maps the orthocenter H of ABC to the orthocenter of A'B'C'. But the orthocenter of A'B'C' is actually the nine-point center of ABC, which is the midpoint of OH. So, if H maps to the nine-point center, which is the midpoint of OH, then the homothety centered at G with ratio -1/2 sends H to the midpoint of OH. But the homothety also sends O to the circumcenter of A'B'C', which is the nine-point center. So, the nine-point center is the image of both O and H under this homothety? That can't be unless O, G, and H are colinear. Because if you have a homothety centered at G mapping O to the nine-point center and H to the nine-point center, then O, G, H must lie on a straight line.Wait, maybe I'm overcomplicating. Let me recall that in homothety, if a point P is mapped to P', then the line PP' passes through the center of homothety. So, since O is mapped to the nine-point center (let's call it N), and H is also mapped to N, then both O and H lie on the line through G and N. So, O, G, and H are colinear.Yes, that makes sense. So, the Euler line is established through this homothety.Now, part 3: using the same homothety, show that the points A', B', C', D, E, and F are concyclic. So, these six points lie on a circle.I remember that the nine-point circle passes through the midpoints of the sides (A', B', C'), the feet of the altitudes (D, E, F), and the midpoints of the segments from each vertex to the orthocenter. So, the nine-point circle is exactly the circle that passes through these points.But how does homothety come into play here? Since A'B'C' is the image of ABC under homothety centered at G with ratio -1/2, and the nine-point circle is related to the circumcircle scaled down by 1/2.Wait, the nine-point circle has a radius half of the circumradius. So, if the homothety centered at G with ratio -1/2 maps the circumcircle of ABC to the nine-point circle of ABC. Therefore, the image of the circumcircle under this homothety is the nine-point circle, which passes through A', B', C', D, E, F.Therefore, all these points lie on the image of the circumcircle under the homothety, which is the nine-point circle. Hence, they are concyclic.Alternatively, since A'B'C' is the medial triangle, its circumcircle is the nine-point circle of ABC, which passes through D, E, F as well. So, all six points lie on this circle.I think that's the gist of it. So, summarizing:1. A'B'C' is the image of ABC under homothety at G with ratio -1/2.2. This homothety maps O to the nine-point center and H to the nine-point center, hence O, G, H are colinear.3. The image of the circumcircle under this homothety is the nine-point circle, which passes through A', B', C', D, E, F, so they are concyclic.**Final Answer**1. boxed{A'B'C'} is the image of (ABC) under a homothety centered at (G).2. boxed{O}, boxed{G}, and boxed{H} are collinear.3. boxed{A'}, boxed{B'}, boxed{C'}, boxed{D}, boxed{E}, and boxed{F} are concyclic.**Final Answer**1. boxed{A'B'C'} is the image of (ABC) under a homothety centered at (G).2. boxed{O}, boxed{G}, and boxed{H} are collinear.3. boxed{A'}, boxed{B'}, boxed{C'}, boxed{D}, boxed{E}, and boxed{F} are concyclic.

answer：First, substitute ( q = 7 ) into the expressions for the number of quarters Alice and Bob have.Alice has ( 3q + 4 ) quarters:[3(7) + 4 = 21 + 4 = 25 text{ quarters}]Bob has ( 2q - 3 ) quarters:[2(7) - 3 = 14 - 3 = 11 text{ quarters}]Next, find the difference in the number of quarters:[25 - 11 = 14 text{ quarters}]Since 1 quarter is equivalent to 5 nickels, convert the difference to nickels:[14 times 5 = 70 text{ nickels}]