🧠:Okay, let's see. I need to find the boat's speed in still water. The problem says the boat traveled a km across a lake and then half that distance along a river. The entire trip took 1 hour. The river's current speed is c km/h. Hmm.First, let me make sure I understand the problem correctly. The boat goes across a lake, which I assume means going from one point to another directly across the lake, so the distance is a km. Then, it goes half of that distance along a river. Since a river has a current, the boat's effective speed will depend on whether it's going upstream or downstream. Wait, the problem doesn't specify which direction the boat is going on the river. Hmm. But the river is flowing into the lake, so if the boat is going along the river from the lake, it would be going downstream, right? Or maybe the boat went upstream? Wait, maybe I need to clarify that.Wait, the problem says "half of that distance along a river flowing into the lake". So the river is flowing into the lake, which means the direction towards the lake is downstream, and away from the lake is upstream. If the boat first went across the lake, then along the river, maybe it went upstream? Because after crossing the lake, going along the river might be going against the current to return? Or maybe it went downstream? Wait, the problem doesn't specify the direction. Hmm. This might be a critical point.Wait, maybe the problem is that the boat went across the lake and then half that distance along the river. So, if the lake is a body of water with a river flowing into it, then after crossing the lake, the boat could enter the river and go either upstream or downstream. But since the problem mentions the entire trip lasted 1 hour, maybe it's a round trip? Wait, no, the problem says "entire trip", which could be one way: across the lake and then along the river. But the wording is "then half of that distance along a river". So first, across the lake a km, then along the river (a/2) km. But direction?Wait, perhaps the boat went across the lake and then up the river? If the river is flowing into the lake, then going up the river would be against the current, so the boat's speed would be reduced. Alternatively, if it went down the river, the speed would be increased. But the problem doesn't specify. Wait, maybe it's a one-way trip: crossing the lake and then going up the river? Or maybe the trip is across the lake and then back along the river? Hmm. Wait, the problem says "entire trip lasted 1 hour". So maybe it's going across the lake and then returning along the river? That would make sense if it's a round trip. Let me check the original problem again."First, a boat traveled a km across a lake and then half of that distance along a river flowing into the lake. The entire trip lasted 1 hour. Find the boat's speed in still water, given that the speed of the river's current is с km/h."Wait, the problem says "first" the boat traveled a km across the lake, and then half that distance along the river. So, it's a one-way trip: first across the lake, then along the river. So after crossing the lake, the boat goes along the river for a/2 km. The river flows into the lake, so if the boat is moving along the river, depending on the direction, the current will affect its speed. If it's moving upstream (against the current), the speed would be (boat speed - current speed). If downstream, (boat speed + current speed). But the problem doesn't specify the direction. Hmm, this is confusing.Wait, maybe the boat is going upstream? Because if the boat went downstream, the problem might not have enough information. Wait, but maybe it's part of the problem to consider both directions? But the problem states "the entire trip lasted 1 hour". So maybe the trip is across the lake and then upstream along the river, and then perhaps back? Wait, no, the problem says "entire trip", which is just the two segments: a km across the lake and then a/2 km along the river. Wait, but if it's a round trip, it would have to return. Hmm. Maybe I need to re-examine the problem.Wait, the problem says: "a boat traveled a km across a lake and then half of that distance along a river flowing into the lake. The entire trip lasted 1 hour." So, it's a one-way trip: first across the lake (a km), then along the river (a/2 km). Then, perhaps the trip is from point A to point B across the lake, then from point B to point C along the river. So, the entire trip is from A to C via B, and that takes 1 hour. So, the direction along the river could be either upstream or downstream. But the problem doesn't specify, so perhaps we need to assume a direction. Wait, maybe the standard assumption is that after crossing the lake, the boat goes upstream along the river? Because rivers flow into lakes, so going from the lake into the river would be upstream. Wait, no, actually, if the river is flowing into the lake, then the direction from the lake into the river would be against the current, i.e., upstream. Wait, no, if the river flows into the lake, then the river's current is towards the lake. So, if the boat is on the lake and then enters the river, moving along the river away from the lake would be going upstream (against the current), and moving towards the lake would be downstream (with the current). So, if the boat travels from the lake into the river, it would be moving upstream. Alternatively, if it's moving from the river into the lake, that's downstream. But the problem states the boat first travels across the lake, then along the river. So starting from some point, crosses the lake (a km), then proceeds along the river (a/2 km). Depending on the path, moving along the river could be upstream or downstream.Wait, this is a critical ambiguity. Since the problem doesn't specify the direction, perhaps we have to assume that the boat goes upstream, as that is the more complex scenario, and often problems like this involve going against the current. Alternatively, maybe the problem expects the time for both legs: across the lake and along the river (upstream or downstream) to sum to 1 hour. But without knowing the direction, we can't set up the equation. Wait, perhaps the problem is intended to have the boat go upstream, so the return trip would be downstream? But the problem doesn't mention a return trip. Wait, maybe the trip is one-way: across the lake and then upstream along the river, with total time 1 hour. Alternatively, maybe the boat goes across the lake, then back along the river, making a total trip, but the distances are different.Wait, let me parse the problem again: "a boat traveled a km across a lake and then half of that distance along a river flowing into the lake. The entire trip lasted 1 hour." So, "entire trip" is the combination of the two legs: a km across the lake and a/2 km along the river. So, the trip is in one direction: first across the lake, then along the river. Since the river flows into the lake, moving along the river away from the lake would be upstream, against the current. So, the boat's speed during the river leg would be (boat speed in still water - current speed). Therefore, the time taken for the lake leg is (distance)/(speed). Since on the lake, the water is still, so the boat's speed is its speed in still water, let's call that v. Then, the time for the lake leg is a / v.Then, for the river leg, the distance is a/2 km, and the speed is (v - c), since it's going upstream. Therefore, the time for the river leg is (a/2) / (v - c). The total time is a / v + (a/2) / (v - c) = 1 hour. So, we can set up the equation: a / v + (a / 2) / (v - c) = 1, and solve for v.Alternatively, if the boat was going downstream on the river, the speed would be (v + c), and the time would be (a/2)/(v + c). But the problem doesn't specify, so maybe we have to assume it's going upstream? Wait, but the problem says "along a river flowing into the lake". If the boat is moving along the river, starting from the lake, it would be moving against the current (upstream). If it was moving towards the lake, it would be with the current (downstream). But the problem just says "along a river". Since the river flows into the lake, the direction along the river could be either upstream or downstream. Hmm.Wait, maybe the problem is expecting the boat to go upstream and downstream, but the problem states only two segments: across the lake and then half that distance along the river. Wait, unless the entire trip is a round trip: going across the lake, then up the river, then back down the river, and back across the lake. But the problem doesn't say that. It specifically mentions "a boat traveled a km across a lake and then half of that distance along a river". So it's a one-way trip with two segments. Therefore, the problem is underdetermined unless we assume the direction along the river.But given that the problem is solvable, and the answer is expected to be in terms of a and c, perhaps the direction is upstream. Let me proceed with that assumption.So, if the boat goes upstream, the time for the river leg is (a/2)/(v - c). Then total time is a/v + (a/2)/(v - c) = 1. Then, solve for v.Alternatively, if the boat went downstream, the time would be (a/2)/(v + c), leading to a different equation. Since the problem says "along a river flowing into the lake", perhaps the direction is into the river, i.e., upstream. So, let's go with upstream.So, the equation is:(a / v) + (a / (2(v - c))) = 1Now, solving for v. Let's write that equation:a / v + a / [2(v - c)] = 1Multiply both sides by 2v(v - c):2a(v - c) + a v = 2v(v - c)Expand:2a v - 2a c + a v = 2v² - 2c vCombine like terms:(2a v + a v) - 2a c = 2v² - 2c v3a v - 2a c = 2v² - 2c vBring all terms to one side:3a v - 2a c - 2v² + 2c v = 0Rearrange terms:-2v² + (3a + 2c)v - 2a c = 0Multiply both sides by -1:2v² - (3a + 2c)v + 2a c = 0Now, this is a quadratic equation in terms of v:2v² - (3a + 2c)v + 2a c = 0Let's solve for v using the quadratic formula. The quadratic is Av² + Bv + C = 0, where:A = 2B = - (3a + 2c)C = 2a cThus,v = [ (3a + 2c) ± sqrt( (3a + 2c)^2 - 16a c ) ] / (4)Let me compute the discriminant:D = (3a + 2c)^2 - 16a cExpand (3a + 2c)^2:= 9a² + 12a c + 4c² - 16a c= 9a² - 4a c + 4c²So,v = [3a + 2c ± sqrt(9a² - 4a c + 4c²)] / 4Hmm, can this be simplified? Let's see.The discriminant 9a² -4ac +4c². Let me check if it's a perfect square.Suppose it factors as (ma + nc)^2 = m²a² + 2mnac + n²c². Comparing to 9a² -4ac +4c².So,m² = 9 => m = 3n² = 4 => n = 2Then, 2mn = 2*3*2 = 12. But in the discriminant, the cross term is -4ac. Doesn't match. So not a perfect square. Therefore, the square root cannot be simplified further. Therefore, the solution is:v = [3a + 2c ± sqrt(9a² -4ac +4c²)] / 4But we need to consider which solution makes physical sense. Since speed must be positive, and greater than the current speed c (because otherwise, the boat couldn't go upstream). Wait, if the boat's speed in still water is v, then to go upstream, v must be greater than c. So v > c. Therefore, we need to check which of the solutions satisfies v > c.Let's compute both roots.First, the positive root:v1 = [3a + 2c + sqrt(9a² -4ac +4c²)] / 4Second, the negative root:v2 = [3a + 2c - sqrt(9a² -4ac +4c²)] / 4We need to see if either of these is greater than c.Alternatively, maybe only one of them is positive. Let's check.Assuming a and c are positive constants.First, compute sqrt(9a² -4ac +4c²). Let's see:9a² -4ac +4c² = 9a² -4ac +4c². Let's write it as 9a² +4c² -4ac. This is equivalent to 9a² -4ac +4c². Let's see if this is always positive. The discriminant of the quadratic in terms of a (if we treat c as a constant):If we consider 9a² -4ac +4c², the discriminant is (-4c)^2 -4*9*4c² = 16c² - 144c² = -128c² < 0. Therefore, the quadratic 9a² -4ac +4c² is always positive for all real a and c. Therefore, sqrt(9a² -4ac +4c²) is a real positive number.Therefore, both roots are real. Now, check if they are positive.First, v1: numerator is 3a + 2c + sqrt(9a² -4ac +4c²). All terms are positive, so v1 is positive.v2: numerator is 3a + 2c - sqrt(9a² -4ac +4c²). Need to check if this is positive.Let me see:Let’s denote S = sqrt(9a² -4ac +4c²). Let's compare 3a + 2c and S.Compute (3a + 2c)^2 = 9a² + 12ac + 4c²Compare to S² = 9a² -4ac +4c²So, (3a + 2c)^2 - S² = (9a² +12ac +4c²) - (9a² -4ac +4c²) = 16acSince a and c are positive, 16ac > 0, so (3a + 2c)^2 > S², so 3a + 2c > S. Therefore, 3a + 2c - S > 0. Therefore, v2 is positive as well.So both roots are positive. Now, we need to determine which one is physically meaningful. Since the boat's speed in still water must be greater than the current speed c to be able to go upstream. Let's check both solutions.Take v1 = [3a + 2c + sqrt(9a² -4ac +4c²)] / 4Compare with c:v1 - c = [3a + 2c + sqrt(9a² -4ac +4c²) - 4c]/4 = [3a - 2c + sqrt(9a² -4ac +4c²)] / 4Similarly for v2:v2 - c = [3a + 2c - sqrt(9a² -4ac +4c²) -4c]/4 = [3a - 2c - sqrt(9a² -4ac +4c²)] /4Now, for v1: numerator is 3a -2c + sqrt(...). Since sqrt(...) is sqrt(9a² -4ac +4c²). Let's see:Let’s denote sqrt(9a² -4ac +4c²) = sqrt((3a)^2 -4ac + (2c)^2). Hmm, not sure. Let's compute if 3a -2c is positive. Let's suppose a and c are positive. If 3a -2c >0, then adding sqrt(...) which is positive, so numerator is positive. If 3a -2c is negative, then sqrt(...) might be greater than |3a -2c|.But let's check for specific values. Let's take a = 2, c =1. Then:sqrt(9*(4) -4*2*1 +4*1) = sqrt(36 -8 +4) = sqrt(32) ≈5.6563a -2c =6 -2=4So numerator v1 -c = (4 +5.656)/4 ≈9.656/4≈2.414>0v1 -c is positive.For v2: numerator is 3a -2c -sqrt(...). For a=2, c=1:3a -2c=6 -2=4sqrt(...)=sqrt(32)=≈5.656So 4 -5.656≈-1.656, divided by 4≈-0.414. So v2 -c ≈-0.414 <0. Therefore, v2 <c in this case, which is impossible because the boat can't go upstream if its speed in still water is less than the current. Therefore, v2 is invalid. Therefore, only v1 is the valid solution.Therefore, the boat's speed in still water is [3a +2c +sqrt(9a² -4ac +4c²)] /4But let's see if we can simplify this expression. Let me factor out the square root:sqrt(9a² -4ac +4c²). Let me write this as sqrt(9a² +4c² -4ac). Perhaps we can write it as sqrt((3a - 2c/3)^2 + ...). Hmm, not sure. Alternatively, maybe factor it as a quadratic in a:9a² -4ac +4c² =9a² -4ac +4c². Hmm. Alternatively, maybe express it as (3a - (2/3)c)^2 + (4c² - (4/9)c²) ?Wait, let me compute:(3a - (2/3)c)^2 =9a² -4ac + (4/9)c²Then, 9a² -4ac +4c² = (3a - (2/3)c)^2 + (4c² - (4/9)c²) = (3a - (2/3)c)^2 + (32/9)c²Not helpful. Alternatively, maybe there's another way.Alternatively, recognize that 9a² -4ac +4c² = (3a)^2 -4ac + (2c)^2. Hmm, doesn't seem like a standard identity. Maybe it's irreducible. Therefore, perhaps the answer is best left as:v = [3a + 2c + sqrt(9a² -4ac +4c²)] /4But let's check dimensions. If a is in km, c in km/h, then inside the sqrt, we have km² - km*(km/h) + km²/h²? Wait, no, wait. Wait, a is a distance in km, c is a speed in km/h. Wait, the equation we set up earlier had units consistent?Wait, in the equation a / v + (a/2)/(v - c) =1, the units for each term are:a (km) / v (km/h) = time (hours)Similarly, (a/2) / (v - c) is also km / (km/h) = hours. So total time is 1 hour. So units are consistent. Therefore, the solution v will have units km/h, as required.Therefore, the expression is correct. However, perhaps the problem expects a simpler answer. Let me check if there's an algebraic mistake.Wait, let's retrace the steps.Original equation:a / v + (a / 2)/(v - c) =1Multiply both sides by 2v(v - c):2a(v - c) +a v = 2v(v - c)Yes:Left side: 2a(v - c) +a v =2a v -2a c +a v=3a v -2a cRight side:2v(v - c)=2v² -2c vBring all terms to left:3a v -2a c -2v² +2c v=0Rearrange:-2v² +3a v +2c v -2a c=0Which is:-2v² +v(3a +2c) -2a c=0Multiply by -1:2v² - (3a +2c)v +2a c=0Quadratic equation, correct.Solutions:v=[3a +2c ±sqrt((3a +2c)^2 -16a c)]/4Wait, discriminant is (3a +2c)^2 -16a c=9a² +12a c +4c² -16a c=9a² -4a c +4c², which is what I had earlier. Correct.Therefore, no algebraic mistake. Therefore, the answer is as above. Therefore, the boat's speed in still water is [3a +2c +sqrt(9a² -4ac +4c²)] /4 km/h.Alternatively, maybe we can factor this expression. Let me see:Let me factor numerator:3a +2c +sqrt(9a² -4ac +4c²)Is there a way to write this as something squared? Hmm. Let me suppose that sqrt(9a² -4ac +4c²) can be expressed as something.Wait, 9a² -4ac +4c². Let me check if this can be expressed as (3a - kc)^2.(3a -kc)^2=9a² -6k a c +k²c². Comparing to 9a² -4ac +4c²:So, -6k = -4 => k=4/6=2/3Then, k²=4/9. But in the original expression, the constant term is 4c², which is 4, but k²=4/9. So, 9a² -4ac +4c²= (3a - (2/3)c)^2 +4c² - (4/9)c²=(3a - (2/3)c)^2 + (32/9)c². Not helpful.Alternatively, maybe as a combination of other terms. Alternatively, perhaps rationalizing or another approach. Alternatively, maybe there's an error in the initial assumption of direction.Wait, if instead the boat went downstream on the river, then the time on the river would be (a/2)/(v +c), leading to the equation:a /v + (a /2)/(v +c) =1Then, following similar steps:Multiply by 2v(v +c):2a(v +c) +a v = 2v(v +c)Expand:2a v +2a c +a v =2v² +2c vCombine:3a v +2a c =2v² +2c vBring all terms to left:3a v +2a c -2v² -2c v=0Rearrange:-2v² + (3a -2c)v +2a c=0Multiply by -1:2v² + (-3a +2c)v -2a c=0Then, quadratic equation:2v² + (-3a +2c)v -2a c=0Solutions:v = [3a -2c ± sqrt(( -3a +2c )² +16a c )]/4Wait, discriminant D=( -3a +2c )² +16a c=9a² -12a c +4c² +16a c=9a² +4a c +4c²Which is (3a + (2/3)c)^2 + ...? Not sure. Wait, 9a² +4a c +4c². Let's check:Is 9a² +4ac +4c² a perfect square? Let's see:Suppose it's (3a + kc)^2=9a² +6k a c +k²c². Comparing:6k=4 =>k=2/3Then k²=4/9, but in the expression we have 4c². So, no. Therefore, not a perfect square. Therefore, the solution would be:v=[3a -2c ±sqrt(9a² +4ac +4c²)]/4Again, two solutions. Then, check which one is positive and greater than zero. However, in this case, if the boat is going downstream, the speed in still water just needs to be positive, but since it's going downstream, even if v <c, it can still proceed, just that the effective speed would be v +c. However, in reality, a boat's speed in still water is always positive, but there's no restriction other than v >0 here. But we have to see which solution makes sense.But in this case, if we take the positive root:v1=[3a -2c + sqrt(9a² +4ac +4c²)] /4And v2=[3a -2c - sqrt(9a² +4ac +4c²)] /4Again, sqrt(9a² +4ac +4c²) is greater than 3a +2c? Let's check:Compute (3a +2c)^2=9a² +12ac +4c². Compare to the sqrt term squared:9a² +4ac +4c². So 9a² +4ac +4c² <9a² +12ac +4c², so sqrt(9a² +4ac +4c²) <3a +2c. Therefore, 3a -2c - sqrt(...) could be positive or negative.Take a=2, c=1 again.Compute sqrt(9*(4)+4*2*1 +4*1)=sqrt(36 +8 +4)=sqrt(48)=≈6.928v1=(6 -2 +6.928)/4=(10.928)/4≈2.732v2=(6 -2 -6.928)/4≈(-2.928)/4≈-0.732So v2 is negative, invalid. So only v1 is valid. Then, v1=2.732 km/h. But in this case, since the boat is going downstream, there's no problem with v being less than c, because downstream speed is v +c. But in this case, v=2.732, c=1, so downstream speed is 3.732, which is fine.But the problem is to find the boat's speed in still water, given that the entire trip took 1 hour. Depending on the direction of the river leg, we get different solutions. Since the problem didn't specify the direction, but the river flows into the lake, the boat's path after the lake could be either upstream or downstream. However, in many such problems, if not specified, it's often assumed that the boat is going upstream, but since both cases give different answers, there might be an issue.Wait, but the problem statement in Russian (assuming the original is similar) might have a standard interpretation. Since the user wrote "the speed of the river's current is с km/h", using "с" which is Cyrillic, maybe the problem is from a Russian source, where typically such problems assume that after crossing the lake, the boat goes upstream along the river. Therefore, the answer would be the first solution.Alternatively, maybe the problem implies a round trip, but the original statement doesn't mention returning. Hmm. Wait, the problem says "the entire trip lasted 1 hour". If "trip" refers to a round trip, then the boat goes across the lake, then upstream, then back downstream, and back across the lake. But the problem states "a boat traveled a km across a lake and then half of that distance along a river". So it's possible that "entire trip" is a round trip: a km across the lake, a/2 km upstream, then a/2 km downstream, and a km back across the lake. Total time 1 hour. But the problem doesn't specify this. It's ambiguous.Alternatively, maybe "entire trip" is one way: a km across the lake and a/2 km along the river, taking 1 hour total. Then the problem is underdetermined unless we know the direction. Since the problem is solvable, the user must have intended a specific direction. Given that in many problems, when a boat goes into a river from a lake, it's assumed to go upstream, I think the first solution is the intended one.Therefore, the answer is [3a +2c +sqrt(9a² -4ac +4c²)] /4Alternatively, perhaps the problem has a different interpretation. Let me consider another approach.Wait, maybe the boat travels across the lake and back along the river. So, the total distance is a km across the lake, and then a/2 km along the river, but as a round trip: a km across, then a/2 km upstream, then a/2 km downstream, then a km back across the lake. But the problem says "entire trip lasted 1 hour". If that's the case, the total time is time upstream + time downstream + 2*(time across the lake). But the problem doesn't mention a return trip. Therefore, I think the original interpretation is correct: one way trip, across the lake and then along the river (upstream), total time 1 hour.Therefore, the speed in still water is [3a +2c +sqrt(9a² -4ac +4c²)] /4But let's check with sample values. Suppose a=2 km, c=1 km/h.Then, the solution is:[3*2 +2*1 +sqrt(9*4 -4*2*1 +4*1)] /4 = [6 +2 +sqrt(36 -8 +4)] /4 = [8 +sqrt(32)]/4 ≈ [8 +5.656]/4≈13.656/4≈3.414 km/hCheck if this satisfies the original equation.Time across lake:2 /3.414≈0.586 hoursTime upstream:1 / (3.414 -1)=1 /2.414≈0.414 hoursTotal time≈0.586 +0.414≈1 hour. Correct.If we take the other direction (downstream), with a=2, c=1:Solution would be [3a -2c +sqrt(9a² +4ac +4c²)] /4= [6 -2 +sqrt(36 +8 +4)] /4= [4 +sqrt(48)]/4≈[4 +6.928]/4≈10.928/4≈2.732 km/hTime across lake:2 /2.732≈0.732 hoursTime downstream:1 / (2.732 +1)=1/3.732≈0.268 hoursTotal≈0.732 +0.268≈1 hour. Correct.So both solutions are valid, but correspond to different directions. Therefore, the problem is ambiguous, but since it's asking for the boat's speed in still water, which is a fixed value, but we get two possible solutions depending on direction. However, in most textbook problems, when a boat goes along a river after a lake without specifying direction, it's usually upstream, as that presents a scenario where the current affects the boat's speed. Therefore, the answer is likely the first solution.Alternatively, the problem might have a typo or require assuming the river portion is downstream. But given that both solutions are mathematically correct, but only the upstream solution enforces v >c, which is a physical necessity, whereas downstream allows v <c, which is physically possible (since downstream speed is v +c, which is positive even if v <c), but the problem asks for the boat's speed in still water, which can be any positive value, but if we consider that the boat has to return to the original point, but the problem doesn't mention a return trip.Given the ambiguity, but the problem being presented as solvable, the answer is likely the upstream case, leading to the solution:v = [3a +2c +sqrt(9a² -4ac +4c²)] /4Alternatively, the problem might have intended the river portion to be downstream, leading to a different solution, but without more information, we can't be sure. However, since in the upstream case, we must have v >c, which gives a unique solution, whereas downstream allows any v >0, but both solutions positive, but only one valid in the upstream case. Given that, and textbook conventions, the answer is the first one.Therefore, the boat's speed in still water is [3a +2c +sqrt(9a² -4ac +4c²)] /4 km/h.

✅:The boat's speed in still water can be determined by analyzing the time spent traveling across the lake and along the river. 1. Lake Segment: The boat travels ( a ) km at speed ( v ) (in still water), taking ( frac{a}{v} ) hours.2. River Segment: It then travels ( frac{a}{2} ) km upstream against a current ( c ) km/h, resulting in an effective speed of ( v - c ). The time for this segment is ( frac{a}{2(v - c)} ).The total time for both segments is 1 hour:[frac{a}{v} + frac{a}{2(v - c)} = 1]Solving the Equation:1. Multiply through by ( 2v(v - c) ):[2a(v - c) + av = 2v(v - c)]2. Simplify and rearrange into a quadratic equation:[2v^2 - (3a + 2c)v + 2ac = 0]3. Apply the quadratic formula:[v = frac{3a + 2c pm sqrt{9a^2 - 4ac + 4c^2}}{4}]Physical Consideration:Only the positive root where ( v > c ) (to navigate upstream) is valid:[v = frac{3a + 2c + sqrt{9a^2 - 4ac + 4c^2}}{4}]Final Answer:The boat's speed in still water is:[boxed{frac{3a + 2c + sqrt{9a^2 - 4ac + 4c^2}}{4}}]

🔑:Given a problem: Initially, a boat traveled a km over a lake and then half of this distance along a river flowing into the lake. The entire journey took 1 hour. Find the boat's speed if the river's current speed is c km/h.1. Define variables: - Let x be the speed of the boat in still water (km/h). - The distance covered on the lake is a km. - The distance covered on the river is frac{a}{2} km.2. Determine the time taken for each segment of the journey: - The time to travel a km on the lake: [ frac{a}{x} text{ hours} ] - The time to travel frac{a}{2} km on the river, considering the current speed c: [ frac{frac{a}{2}}{x - c} = frac{a}{2(x - c)} text{ hours} ]3. Set up the equation for the total travel time which is given as 1 hour: [ frac{a}{x} + frac{a}{2(x - c)} = 1 ]4. Eliminate the fractions by finding a common denominator: [ frac{2a(x - c) + ax}{2x(x - c)} = 1 ]5. Simplify the equation: [ 2a(x - c) + ax = 2x(x - c) ] [ 2ax - 2ac + ax = 2x^2 - 2xc ]6. Combine like terms and reorganize into a quadratic equation: [ 2x^2 - 2xc - 2ax + ax + 2ac = 0 ] [ 2x^2 - (2c + 3a)x + 2ac = 0 ]7. Solve the quadratic equation using the quadratic formula: [ x = frac{-(2c + 3a) pm sqrt{(2c + 3a)^2 - 4(2)(2ac)}}{4} ] [ x = frac{2c + 3a pm sqrt{9a^2 - 4ac + 4c^2}}{4} ]8. Since x > c, select the appropriate root: [ x = frac{3a + 2c + sqrt{9a^2 - 4ac + 4c^2}}{4} ] Conclusion: [ boxed{frac{3a + 2c + sqrt{9a^2 - 4ac + 4c^2}}{4} text{ km/h}} ]

🧠:Okay, so I need to find the area under the curve y = x times g(x), where g(x) is sqrt(x(1 - x)). The problem states that the area is π/16, but I guess I need to verify that. Let me start by understanding what the function looks like and then figure out how to set up the integral.First, let's write down the function clearly. Given g(x) = sqrt(x(1 - x)), so y = x * sqrt(x(1 - x)). That simplifies to y = x * sqrt(x) * sqrt(1 - x) right? Wait, sqrt(x) times x is x^(3/2). So maybe writing y = x^(3/2) * (1 - x)^(1/2). Hmm, that seems a bit complicated, but perhaps manageable.Since the function is defined on the interval [0, 1], because inside the square root, x(1 - x) must be non-negative. So x is between 0 and 1, inclusive. At x = 0, the function is 0, and at x = 1, it's also 0 because of the (1 - x) term. So the graph starts at 0, goes up to some maximum, and then back down to 0. The area under this curve from 0 to 1 is what we need to compute.So the area A should be the integral from 0 to 1 of y dx, which is integral from 0 to 1 of x * sqrt(x(1 - x)) dx. Let me write that out:A = ∫₀¹ x * sqrt(x(1 - x)) dxLet me simplify the integrand. Let's see, x * sqrt(x(1 - x)) can be written as x * x^(1/2) * (1 - x)^(1/2) = x^(3/2) * (1 - x)^(1/2). So the integrand is x^(3/2) * (1 - x)^(1/2). This resembles the form of a Beta function, which is ∫₀¹ t^(a-1)(1 - t)^(b-1) dt = B(a, b) = Γ(a)Γ(b)/Γ(a + b). So if I can express the integral in terms of the Beta function, then perhaps I can use known values of the Gamma function to compute it.Let me recall that the Beta function B(a, b) is equal to 2∫₀^(π/2) (sinθ)^(2a - 1) (cosθ)^(2b - 1) dθ. But maybe that's complicating things. Alternatively, if the integral is from 0 to 1 and has the form t^c (1 - t)^d dt, then it's B(c + 1, d + 1) as long as c and d are greater than -1.In our case, the integrand is x^(3/2) (1 - x)^(1/2). So comparing to t^c (1 - t)^d, we have c = 3/2 and d = 1/2. Therefore, the integral is B(3/2 + 1, 1/2 + 1) = B(5/2, 3/2). Then, using the relationship between Beta and Gamma functions:B(5/2, 3/2) = Γ(5/2)Γ(3/2)/Γ(5/2 + 3/2) = Γ(5/2)Γ(3/2)/Γ(4).Now, I need to compute Γ(5/2), Γ(3/2), and Γ(4). Remember that Γ(n) = (n - 1)! for positive integers, so Γ(4) = 3! = 6. For half-integers, Γ(1/2) = sqrt(π), and Γ(n + 1) = nΓ(n). So let's compute Γ(3/2):Γ(3/2) = (1/2)Γ(1/2) = (1/2)sqrt(π).Similarly, Γ(5/2) = (3/2)Γ(3/2) = (3/2)(1/2)sqrt(π) = (3/4)sqrt(π).Therefore, substituting back into B(5/2, 3/2):Γ(5/2)Γ(3/2)/Γ(4) = (3/4 sqrt(π)) * (1/2 sqrt(π)) / 6Multiplying numerators: (3/4)(1/2) * π = (3/8)πDenominator: 6So (3/8 π) / 6 = (3π)/48 = π/16.So the area is indeed π/16. That matches the given value. Alternatively, if I didn't recall the Beta function, maybe I could use substitution. Let's try that as a check.Let me consider a substitution to evaluate the integral ∫₀¹ x^(3/2) (1 - x)^(1/2) dx. Maybe substitution with trigonometric functions. Let me set x = sin²θ. Then, when x = 0, θ = 0, and when x = 1, θ = π/2. Then dx = 2 sinθ cosθ dθ.Substituting into the integral:∫₀^(π/2) (sin²θ)^(3/2) * (1 - sin²θ)^(1/2) * 2 sinθ cosθ dθSimplify the exponents:(sin²θ)^(3/2) = sin³θ(1 - sin²θ)^(1/2) = cosθTherefore, the integral becomes:∫₀^(π/2) sin³θ * cosθ * 2 sinθ cosθ dθWait, let's verify:Original integrand after substitution: sin³θ * cosθ * 2 sinθ cosθ dθMultiply the terms:2 sin³θ * sinθ * cosθ * cosθ = 2 sin^4θ cos²θ dθSo integral becomes 2 ∫₀^(π/2) sin^4θ cos²θ dθHmm, this seems more complex. Let me see if I can compute this integral using reduction formulas or known integrals.Recall that integrals of the form ∫ sin^nθ cos^mθ dθ can be expressed using the Beta function or Gamma function as well, but maybe we can use a substitution here.Alternatively, use the formula for ∫ sin^nθ cos^mθ dθ from 0 to π/2, which is [Γ((n + 1)/2)Γ((m + 1)/2)] / [2Γ((n + m + 2)/2)]In our case, n = 4, m = 2.So the integral is [Γ((4 + 1)/2)Γ((2 + 1)/2)] / [2Γ((4 + 2 + 2)/2)] = [Γ(5/2)Γ(3/2)] / [2Γ(4)]Which is the same as before, leading to π/16. So regardless of the substitution, we end up with the same result. Therefore, the area is indeed π/16.Alternatively, perhaps integrating by parts. Let me try that.Let me set u = x^(3/2), dv = (1 - x)^(1/2) dxThen du = (3/2) x^(1/2) dx, and v = ?To find v, integrate dv = (1 - x)^(1/2) dx.Let me set w = 1 - x, so dw = -dx. Then ∫(1 - x)^(1/2) dx = -∫w^(1/2) dw = - (2/3) w^(3/2) + C = - (2/3)(1 - x)^(3/2) + CTherefore, v = - (2/3)(1 - x)^(3/2)So integration by parts gives:uv | from 0 to 1 - ∫₀¹ v duCompute uv at 1 and 0:At x = 1: u = 1^(3/2) = 1, v = - (2/3)(0)^(3/2) = 0At x = 0: u = 0^(3/2) = 0, v = - (2/3)(1)^(3/2) = -2/3So uv from 0 to 1 is 0 - (0 * (-2/3)) = 0 - 0 = 0Therefore, the integral becomes - ∫₀¹ v du = - ∫₀¹ [ - (2/3)(1 - x)^(3/2) ] * (3/2) x^(1/2) dxSimplify:= (2/3)(3/2) ∫₀¹ (1 - x)^(3/2) x^(1/2) dxThe constants (2/3)*(3/2) cancel out to 1.So we have ∫₀¹ (1 - x)^(3/2) x^(1/2) dxHmm, but this integral is similar to the original, but with exponents increased and decreased. Let me check:Original integral was ∫ x^(3/2) (1 - x)^(1/2) dxNow, after integrating by parts, we have ∫ x^(1/2) (1 - x)^(3/2) dx. So the exponents on x and (1 - x) have swapped. Maybe integrating by parts again would lead us somewhere?Let me try integrating by parts again. Let me denote this new integral as ∫ x^(1/2) (1 - x)^(3/2) dxSet u = x^(1/2), dv = (1 - x)^(3/2) dxThen du = (1/2) x^(-1/2) dx, and v = ?Again, compute v by integrating dv:∫(1 - x)^(3/2) dx. Let w = 1 - x, dw = -dx. So ∫w^(3/2) (-dw) = - (2/5) w^(5/2) + C = - (2/5)(1 - x)^(5/2) + CTherefore, v = - (2/5)(1 - x)^(5/2)Integration by parts gives:uv | from 0 to 1 - ∫₀¹ v duCompute uv at limits:At x = 1: u = 1^(1/2) = 1, v = - (2/5)(0)^(5/2) = 0At x = 0: u = 0^(1/2) = 0, v = - (2/5)(1)^(5/2) = -2/5Thus, uv from 0 to 1 is 0 - (0 * (-2/5)) = 0Therefore, the integral becomes - ∫₀¹ v du = - ∫₀¹ [ - (2/5)(1 - x)^(5/2) ] * (1/2) x^(-1/2) dxSimplify:= (2/5)(1/2) ∫₀¹ (1 - x)^(5/2) x^(-1/2) dx= (1/5) ∫₀¹ (1 - x)^(5/2) x^(-1/2) dxHmm, this integral is now ∫ x^(-1/2) (1 - x)^(5/2) dx from 0 to 1. Which is another Beta function:B(1/2, 7/2) = Γ(1/2)Γ(7/2)/Γ(4)Again, Γ(1/2) = sqrt(π), Γ(7/2) = (5/2)(3/2)(1/2)sqrt(π) = (15/8)sqrt(π), Γ(4) = 6Therefore, B(1/2, 7/2) = sqrt(π)*(15/8 sqrt(π)) / 6 = (15/8 π) / 6 = (15π)/48 = 5π/16Therefore, the integral is (1/5)*(5π/16) = π/16So after two integrations by parts, we end up with π/16. So that confirms the result again.Alternatively, maybe use substitution t = sqrt(x(1 - x)), but that might complicate things. Let me try substitution x = sin²θ as before. Wait, I already did that and it led to the Beta function. Alternatively, maybe substitution t = x.But perhaps another substitution. Let me think.Alternatively, observe that the integral is ∫₀¹ x^(3/2) (1 - x)^(1/2) dx. Let me write this as ∫₀¹ x^(3/2) (1 - x)^(1/2) dx. Let me make a substitution t = x, so when x goes from 0 to 1, t does the same. Not helpful. Alternatively, substitute u = 1 - x, then when x = 0, u = 1; x = 1, u = 0. Then integral becomes ∫₁⁰ (1 - u)^(3/2) u^(1/2) (-du) = ∫₀¹ (1 - u)^(3/2) u^(1/2) du, which is the same as the original integral. So symmetry here doesn't help.Alternatively, consider expressing the integrand in terms that can be expanded into a series. For example, expand (1 - x)^(1/2) as a binomial series. The binomial expansion for (1 - x)^k is Σ_{n=0}^∞ (k choose n) (-x)^n, where (k choose n) is the generalized binomial coefficient. For k = 1/2, the expansion is valid for |x| < 1, which is okay here since x is in [0,1].So, (1 - x)^(1/2) = Σ_{n=0}^∞ ( (1/2 choose n) ) (-x)^n )Therefore, the integral becomes ∫₀¹ x^(3/2) Σ_{n=0}^∞ ( (1/2 choose n) ) (-1)^n x^n dx = Σ_{n=0}^∞ ( (1/2 choose n) ) (-1)^n ∫₀¹ x^(3/2 + n) dxIntegrate term by term:∫₀¹ x^(3/2 + n) dx = 1/(5/2 + n) = 2/(5 + 2n)Therefore, the integral is Σ_{n=0}^∞ ( (1/2 choose n) ) (-1)^n * 2/(5 + 2n)Hmm, but summing this series might not be straightforward. Let's see if we can compute this sum.Recall that (1/2 choose n) = (1/2)(1/2 - 1)(1/2 - 2)...(1/2 - n + 1)/n! )Which can be written as (-1)^{n-1} ( (2n - 3)!! ) / (2^n n! ) for n ≥ 1, and (1/2 choose 0) = 1But this might get complicated. Let's compute a few terms and see if a pattern emerges.For n=0: term is (1/2 choose 0)(-1)^0 * 2/(5 + 0) = 1 * 1 * 2/5 = 2/5n=1: (1/2 choose 1) = (1/2)(1/2 - 1)/1! = (1/2)(-1/2)/1 = -1/4. Then term is (-1/4)(-1)^1 * 2/(5 + 2) = (-1/4)(-1) * 2/7 = (1/4)(2/7) = 1/14n=2: (1/2 choose 2) = (1/2)(1/2 - 1)(1/2 - 2)/2! = (1/2)(-1/2)(-3/2)/2 = (1/2)(3/4)/2 = 3/16. Then term is (3/16)(-1)^2 * 2/(5 + 4) = (3/16)(1) * 2/9 = 6/(144) = 1/24n=3: (1/2 choose 3) = (1/2)(-1/2)(-3/2)(-5/2)/6 = (1/2)(-1/2)(-3/2)(-5/2)/6 = (-15/16)/6 = -15/96 = -5/32. Term is (-5/32)(-1)^3 * 2/(5 + 6) = (-5/32)(-1)^3 = (-5/32)(-1) * 2/11 = (5/32)(2/11) = 10/(352) = 5/176 ≈ 0.0284So adding these terms:n=0: 2/5 ≈ 0.4n=1: +1/14 ≈ 0.0714, total ≈ 0.4714n=2: +1/24 ≈ 0.0417, total ≈ 0.5131n=3: +5/176 ≈ 0.0284, total ≈ 0.5415Hmm, but we know the actual value is π/16 ≈ 0.19635. Wait, that can't be. There's a contradiction here. Wait, no, wait. Wait, π is approximately 3.1416, so π/16 ≈ 0.19635. But the terms I'm calculating are summing up to something over 0.5. That suggests a mistake in my approach. Where did I go wrong?Wait, perhaps the binomial expansion isn't converging here? Wait, the binomial series for (1 - x)^(1/2) converges for |x| < 1, which is true here since x is in [0,1]. But integrating term by term should work, but maybe I messed up the coefficients.Wait, let me check n=0 term:Original integral after expansion is Σ_{n=0}^infty [ (1/2 choose n) (-1)^n ] * ∫₀¹ x^{3/2 + n} dxWhich is Σ_{n=0}^infty [ (1/2 choose n) (-1)^n ] * [1 / (5/2 + n + 1) ]Wait, wait, wait. Wait, integrating x^{3/2 + n} from 0 to 1 is [ x^{5/2 + n} / (5/2 + n) ) ] from 0 to 1, which is 1/(5/2 + n) = 2/(5 + 2n). So that part is correct.But the coefficients: (1/2 choose n) is the generalized binomial coefficient. Let me verify (1/2 choose n):For n=0: (1/2 choose 0) = 1For n=1: (1/2 choose 1) = (1/2) / 1! = 1/2Wait, no! Wait, the generalized binomial coefficient is defined as:(α choose n) = α(α - 1)(α - 2)...(α - n + 1)/n!So for α = 1/2 and n=1:(1/2 choose 1) = (1/2)/1! = 1/2Similarly, for n=2:(1/2 choose 2) = (1/2)(1/2 - 1)/2! = (1/2)(-1/2)/2 = (-1/4)/2 = -1/8Wait, this contradicts my previous calculation. So I think I made a mistake there.Wait, let's recast the coefficients properly:n=0: 1n=1: (1/2)/1! = 1/2n=2: (1/2)(1/2 - 1)/2! = (1/2)(-1/2)/2 = (-1/4)/2 = -1/8n=3: (1/2)(-1/2)(-3/2)/3! = (1/2)(-1/2)(-3/2)/6 = (3/8)/6 = 1/16Wait, hold on, let's compute them step by step.(1/2 choose 0) = 1(1/2 choose 1) = (1/2)/1! = 1/2(1/2 choose 2) = (1/2)(1/2 - 1)/2! = (1/2)(-1/2)/2 = (-1/4)/2 = -1/8(1/2 choose 3) = (1/2)(-1/2)(-3/2)/3! = (1/2)(-1/2)(-3/2)/6 = (3/8)/6 = 3/48 = 1/16(1/2 choose 4) = (1/2)(-1/2)(-3/2)(-5/2)/4! = (1/2)(-1/2)(-3/2)(-5/2)/24 = (-15/16)/24 = -15/384 = -5/128Therefore, the coefficients alternate in sign starting from n=1: n=1 is +1/2, n=2 is -1/8, n=3 is +1/16, n=4 is -5/128, etc.Therefore, correcting the previous calculations:At n=0: term = (1) * (-1)^0 * 2/(5 + 0) = 2/5 ≈ 0.4n=1: (1/2) * (-1)^1 * 2/(5 + 2*1) = (1/2)(-1) * 2/7 = (-1/2)(2/7) = -1/7 ≈ -0.1429Total after n=1: 0.4 - 0.1429 ≈ 0.2571n=2: (-1/8) * (-1)^2 * 2/(5 + 4) = (-1/8)(1) * 2/9 = (-1/8)(2/9) = -2/72 ≈ -0.0278Total: ≈ 0.2571 - 0.0278 ≈ 0.2293n=3: (1/16) * (-1)^3 * 2/(5 + 6) = (1/16)(-1) * 2/11 = (-2)/(176) ≈ -0.0114Total: ≈ 0.2293 - 0.0114 ≈ 0.2179n=4: (-5/128) * (-1)^4 * 2/(5 + 8) = (-5/128)(1) * 2/13 = (-10)/(1664) ≈ -0.0060Total: ≈ 0.2179 - 0.0060 ≈ 0.2119Continuing:n=4 term ≈ -0.0060n=5: ( (1/2 choose 5) ) = (1/2)(-1/2)(-3/2)(-5/2)(-7/2)/5! = (1/2)(-1/2)(-3/2)(-5/2)(-7/2)/120= (1* -1 * -3 * -5 * -7)/ (2^5 * 120) ) = (105)/(32 * 120) = 105/3840 = 7/256 ≈ 0.02734375But (1/2 choose 5) = 7/256? Wait, let's compute step by step:(1/2 choose 5) = (1/2)(-1/2)(-3/2)(-5/2)(-7/2)/5!Multiply numerators: 1/2 * (-1/2) = -1/4; * (-3/2) = 3/8; * (-5/2) = -15/16; * (-7/2) = 105/32Denominator: 5! = 120So coefficient is (105/32)/120 = 105/(32*120) = 105/3840 = 7/256 ≈ 0.02734375Thus term n=5:(7/256) * (-1)^5 * 2/(5 + 10) = (7/256)(-1) * 2/15 = (-14)/(3840) ≈ -0.00365Total after n=5: ≈ 0.2119 - 0.00365 ≈ 0.20825n=6: Similarly, this term would be positive? Let's check the sign. (1/2 choose 6) would have factors alternating signs. Since previous terms alternated signs starting with n=1: -, +, -, +, etc. Wait, actually, the signs alternate starting with n=0: +, -, +, -, etc. Wait, no, the expansion of (1 - x)^(1/2) is Σ_{n=0}^infty ( (1/2 choose n) ) (-x)^n. So each term is ( (1/2 choose n) ) (-1)^n x^n. So when we multiply by (-1)^n from the expansion, the signs will alternate depending on n.But in our integral, after expanding (1 - x)^(1/2) as Σ_{n=0}^infty (1/2 choose n) (-1)^n x^n, multiplying by x^(3/2) gives Σ_{n=0}^infty (1/2 choose n) (-1)^n x^{n + 3/2}. Then integrating term by term.But in the integral expression, we have:Σ_{n=0}^infty (1/2 choose n) (-1)^n ∫₀¹ x^{n + 3/2} dx = Σ_{n=0}^infty (1/2 choose n) (-1)^n [1/(n + 5/2)]Which is the same as Σ_{n=0}^infty (1/2 choose n) (-1)^n [2/(2n + 5)]So each term is (1/2 choose n) (-1)^n * 2/(2n + 5)Given that, the terms alternate in sign depending on n. But (1/2 choose n) itself has alternating signs as n increases. Let's check:For n=0: (1/2 choose 0) = 1, term: 1 * 1 * 2/5 ≈ +0.4n=1: (1/2 choose 1) = 1/2, term: 1/2 * (-1) * 2/7 ≈ -0.1429n=2: (1/2 choose 2) = -1/8, term: -1/8 * 1 * 2/9 ≈ -0.0278n=3: (1/2 choose 3) = 1/16, term: 1/16 * (-1) * 2/11 ≈ -0.0114n=4: (1/2 choose 4) = -5/128, term: -5/128 * 1 * 2/13 ≈ -0.0060n=5: (1/2 choose 5) = 7/256, term: 7/256 * (-1) * 2/15 ≈ -0.00365n=6: (1/2 choose 6) = -21/1024, term: -21/1024 * 1 * 2/17 ≈ -0.00242So each term alternates sign, but the magnitude is decreasing. Adding these up:After n=0: 0.4n=1: 0.4 - 0.1429 ≈ 0.2571n=2: 0.2571 - 0.0278 ≈ 0.2293n=3: 0.2293 - 0.0114 ≈ 0.2179n=4: 0.2179 - 0.0060 ≈ 0.2119n=5: 0.2119 - 0.00365 ≈ 0.20825n=6: 0.20825 - 0.00242 ≈ 0.20583Continuing this, the series converges towards π/16 ≈ 0.19635. So we can see that the partial sums approach that value, albeit slowly. Therefore, even though summing the series manually is tedious, it does converge to the correct result. This further confirms that the area is indeed π/16.Alternatively, we could use numerical integration to approximate the integral. Let's try that as another verification.Compute the integral ∫₀¹ x^(3/2) (1 - x)^(1/2) dx numerically.First, note that at x=0.5, the integrand is (0.5)^(3/2)*(0.5)^(1/2) = (0.5)^2 = 0.25. So the function peaks somewhere above 0.25.But to approximate the integral, maybe use the trapezoidal rule with a few intervals.But since this is time-consuming, perhaps use Simpson's rule for a better approximation.Let's divide the interval [0,1] into n=4 intervals, so h=(1 - 0)/4=0.25.Compute the function at x=0, 0.25, 0.5, 0.75, 1.At x=0: 0^(3/2)*sqrt(1 - 0)=0x=0.25: (0.25)^(3/2)*sqrt(0.75)= (0.125)*sqrt(0.75) ≈ 0.125*0.8660 ≈ 0.10825x=0.5: 0.5^(3/2)*sqrt(0.5) = (0.35355)*0.7071 ≈ 0.25Wait, 0.5^(3/2) = sqrt(0.5)^3 = (0.7071)^3 ≈ 0.3535, and sqrt(0.5) ≈ 0.7071, so 0.3535 * 0.7071 ≈ 0.25x=0.75: (0.75)^(3/2)*sqrt(0.25) = (0.75)^(3/2)*0.50.75^(3/2) = sqrt(0.75)^3 ≈ (0.8660)^3 ≈ 0.6495; 0.6495 * 0.5 ≈ 0.32475x=1: 0Now applying Simpson's rule, which states that ∫₀¹ f(x)dx ≈ (h/3)[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + f(x4)]With h=0.25, x0=0, x1=0.25, x2=0.5, x3=0.75, x4=1So:≈ (0.25/3)[0 + 4*0.10825 + 2*0.25 + 4*0.32475 + 0]Calculate inside:4*0.10825 = 0.4332*0.25 = 0.54*0.32475 ≈ 1.299Sum: 0 + 0.433 + 0.5 + 1.299 + 0 ≈ 2.232Multiply by 0.25/3 ≈ 2.232 * 0.08333 ≈ 0.186Which is close to π/16 ≈ 0.19635. The error here is because we used only 4 intervals. Using more intervals would give a better approximation, but this already gets us in the ballpark, supporting the analytical result.All these methods—Beta functions, integration by parts, substitution to trigonometric functions, series expansion, and numerical integration—consistently point to the area being π/16. Therefore, the given answer is correct.Final AnswerThe area of the region is boxed{dfrac{pi}{16}}.

✅:Given ( g(x) = sqrt{x(1-x)} ), we need to find the area of the region enclosed by the graph of the function ( y = x g(x) ) and the ( x )-axis over the interval ([0, 1]).The function ( y = x g(x) ) simplifies to ( y = x^{3/2} (1 - x)^{1/2} ). The area under this curve is given by the integral:[A = int_0^1 x^{3/2} (1 - x)^{1/2} , dx]This integral can be recognized as a Beta function ( B(a, b) ), which is defined as:[B(a, b) = int_0^1 t^{a-1} (1 - t)^{b-1} , dt]Comparing the integrand ( x^{3/2} (1 - x)^{1/2} ) with the Beta function form, we identify ( a = frac{5}{2} ) and ( b = frac{3}{2} ). Thus, the integral is:[Bleft(frac{5}{2}, frac{3}{2}right)]The Beta function is related to the Gamma function by:[B(a, b) = frac{Gamma(a) Gamma(b)}{Gamma(a + b)}]Using the properties of the Gamma function, we know:[Gammaleft(frac{5}{2}right) = frac{3}{4} sqrt{pi}, quad Gammaleft(frac{3}{2}right) = frac{1}{2} sqrt{pi}, quad Gamma(4) = 3! = 6]Substituting these values into the Beta function:[Bleft(frac{5}{2}, frac{3}{2}right) = frac{Gammaleft(frac{5}{2}right) Gammaleft(frac{3}{2}right)}{Gamma(4)} = frac{left(frac{3}{4} sqrt{pi}right) left(frac{1}{2} sqrt{pi}right)}{6} = frac{frac{3}{8} pi}{6} = frac{pi}{16}]Thus, the area of the region enclosed by the graph of ( y = x g(x) ) and the ( x )-axis is:[boxed{dfrac{pi}{16}}]

🔑:1. Given the function ( g(x) = sqrt{x (1-x)} ) defined on the interval ([0,1]), we need to find the area of the region enclosed by ( y = x g(x) ) and the x-axis.2. Notice that the function ( g(x) = sqrt{x(1-x)} ) describes the upper half of a semicircle with diameter along the x-axis from (x = 0) to (x = 1).3. The function ( y = x g(x) ) can be rewritten as: [ y = x sqrt{x (1-x)} ]4. We are instructed to calculate the area of the region bounded by ( y = x g(x) ) and the x-axis. To do this, we need to integrate the function over the interval ([0, 1]): [ A = int_{0}^{1} x sqrt{x (1-x)} , dx ]5. To simplify the integration, use the substitution ( u = x(1-x) ), hence ( du = (1-2x) dx ). When ( x = 0 ), ( u = 0 ). When ( x = 1 ), ( u = 0 ). This substitution might complicate the direct integration but notice that: [ 2x = 1 implies x = frac{1}{2} ] So we transform the original integral by splitting it: [ int_{0}^{frac{1}{2}} x sqrt{x(1-x)} , dx + int_{frac{1}{2}}^{1} x sqrt{x(1-x)} , dx ]6. Further examining the geometric properties of the function: [ x g(x) + (1-x) g(1-x) = g(x) quad text{for} quad 0 leq x leq frac{1}{2} ] Because: [ g(1-x) = sqrt{(1-x)x} = sqrt{x(1-x)} = g(x) ]7. This shows the symmetry of the region around ( x = frac{1}{2} ).8. Finally, considering the area formed by the semicircle under transformation: [ text{Area} = 2 times int_{0}^{frac{1}{2}} x sqrt{x ( 1-x)} , dx , text{since it is symmetric} ]9. The semicircle of radius (frac{1}{2}) has an area of: [ A = frac{1}{2} pi left(frac{1}{2}right)^2 = frac{pi}{4} times frac{1}{4} = frac{pi}{16} ]Conclusion: [boxed{frac{pi}{16}}]

🧠:Okay, let's tackle this post office location problem. Hmm, the postman has to deliver mail to n houses on a straight street, and each time he goes from the post office to a house and back. The goal is to place the post office such that the total distance he travels is minimized. For parts a and b, n is 5 and 6, and part c is for any n. Let's start with part a.First, I need to visualize the problem. The street is a straight line with houses positioned at certain points. Let's assume the street is represented by a number line, and each house has a coordinate. The post office needs to be located at a point x such that the sum of the distances from x to each house, multiplied by 2 (since he goes there and back each time), is minimized. Wait, but actually, since he goes to each house once per day, but each trip is a round trip from the post office to the house and back. So for each house, the distance is 2*|x - h_i|, where h_i is the position of the i-th house. The total distance is the sum over all houses of 2*|x - h_i|. Therefore, minimizing the total distance is equivalent to minimizing the sum of |x - h_i|, because the 2 is a constant factor. So the problem reduces to finding the point x that minimizes the sum of absolute distances to all houses. Ah, right! This is a classic problem in optimization. The solution that minimizes the sum of absolute deviations is the median of the data set. So if the post office is located at the median of the house positions, the total distance will be minimized. Let me confirm that. Yes, the median is known to minimize the sum of absolute differences. For example, if you have an odd number of houses, the median is the middle one. For even numbers, any point between the two middle houses would work. Wait, but in this problem, the post office has to be at a specific point, right? So for even n, there might be multiple optimal solutions, but if we need a single point, perhaps the median is still defined as the middle point between the two central houses? Or maybe in the case where n is even, any point between the two middle houses is optimal. Let me think.Suppose the houses are at positions h_1 ≤ h_2 ≤ ... ≤ h_n. For odd n=5, the median is h_3. For even n=6, the median could be any point between h_3 and h_4. However, since the problem asks "where should the post office be located", and it's a straight line (so we can choose any real number between the two points if needed), but maybe the minimal total distance is achieved when the post office is anywhere between those two central points. But if we have to pick a specific point, perhaps the midpoint between h_3 and h_4 for n=6. Wait, but the sum of absolute deviations is minimized over the entire interval between h_3 and h_4. So actually, any x in [h_3, h_4] would give the same minimal sum. But is that true?Wait, no. Wait, let's take a simple example. Suppose n=2, with houses at 0 and 1. The sum of |x - 0| + |x - 1| is minimized when x is between 0 and 1. The minimal sum is 1, which occurs for any x in [0,1]. So in that case, any point between the two houses gives the same total distance. Similarly, for n=4, the sum would be minimized when x is between h_2 and h_3. However, for n=5, the sum is minimized at h_3. So for even n, there's an interval where the sum is minimized, and for odd n, it's a single point.But the problem states that the post office has to be located at a specific point. So in the case of even n, since all points between the two middle houses give the same minimal total distance, perhaps the answer is any point between them. But the question says "Where should the post office be located", so maybe we can specify the entire interval. However, the problem might be expecting the median, which for even n is sometimes considered the average of the two middle numbers. Wait, but in the context of optimization, it's actually any point between the two middle points. Let me verify with an example.Take n=6, with houses at positions 1,2,3,4,5,6. The median would be between 3 and 4. If we place the post office at 3.5, the sum of absolute deviations is |3.5-1| + |3.5-2| + |3.5-3| + |3.5-4| + |3.5-5| + |3.5-6|. Calculating each term: 2.5 + 1.5 + 0.5 + 0.5 + 1.5 + 2.5 = 9. If we instead place it at 3, the sum is |3-1| + |3-2| + |3-3| + |3-4| + |3-5| + |3-6| = 2 +1 +0 +1 +2 +3 = 9. Similarly, at 4: |4-1| + |4-2| + |4-3| + |4-4| + |4-5| + |4-6| = 3 +2 +1 +0 +1 +2 = 9. So indeed, placing the post office anywhere between 3 and 4 gives the same total distance. Therefore, for even n, the optimal post office location is any point between the two middle houses. However, in some contexts, people might refer to the lower median or upper median, but since the problem doesn't specify, maybe we need to state that the post office can be located anywhere between the two middle houses. But in the original problem statement, it says "the post office" implying a single location, so maybe the answer is the interval from the lower median to the upper median. But the problem might expect the median point, which for even n is sometimes considered the average of the two middle numbers. However, in optimization terms, any point in between is acceptable. Let me check.Looking back, the problem says "Where should the post office be located so that the total distance the postman travels is minimized". So mathematically, the set of minimizers is the interval between the two central points when n is even. Therefore, for even n, the optimal post office can be anywhere between the two middle houses. For odd n, it's exactly the middle house.Therefore, for part a) n=5, the post office should be located at the 3rd house (the median). For part b) n=6, it can be anywhere between the 3rd and 4th houses. For part c), arbitrary n, it's the median house when n is odd, and any point between the n/2-th and (n/2 +1)-th houses when n is even.But let me check with specific examples. Let's take n=5, houses at positions 1,2,3,4,5. The median is 3. The total distance is 2*(|3-1| + |3-2| + |3-3| + |3-4| + |3-5|) = 2*(2 +1 +0 +1 +2) = 2*6 = 12. If we place the post office at 2, total distance is 2*(1 +0 +1 +2 +3) = 2*7=14, which is more. Similarly at 4, it's 2*(3 +2 +1 +0 +1)=14. So indeed, 3 is the optimal.For n=6, houses at 1,2,3,4,5,6. Placing post office at 3.5 gives total distance 2*(2.5 +1.5 +0.5 +0.5 +1.5 +2.5) = 2*9=18. If placed at 3, it's 2*(2 +1 +0 +1 +2 +3)=2*9=18. Similarly at 4, same result. So anywhere between 3 and 4 gives the same total distance. Thus, for even n, the optimal is the interval between the two central houses.Therefore, the answers should be:a) For n=5, the median house, which is the 3rd one when ordered.b) For n=6, any point between the 3rd and 4th houses.c) For arbitrary n, if n is odd, the median (the (n+1)/2-th house); if n is even, any point between the n/2-th and (n/2 +1)-th houses.But wait, the problem might expect specific coordinates depending on how the houses are arranged. However, the problem states "the street is a straight line segment" but doesn't specify the positions of the houses. Therefore, the answer should be in terms of the positions of the houses. Wait, the problem is general, not with specific coordinates. So for n=5, the answer is the median of the 5 houses. Since the street is a straight line, we can order the houses from left to right, and the median is the middle one. Similarly for n=6, any point between the two middle houses.But the problem might have houses at specific positions, but it's not given. Wait, the problem states "a street with n houses" but doesn't specify their positions. Therefore, the answer has to be in general terms. So for any n, the optimal location is the median (for odd n) or between the two middle houses (for even n). Therefore, the answer is as above.Wait, but in the problem statement, it says "the street is a straight line segment". Are the houses assumed to be equally spaced? Probably not, because if they were, the answer might be the center, but the problem doesn't specify that. So the houses can be anywhere along the street. Therefore, the solution is purely based on the order statistics: the median minimizes the sum of absolute deviations.Therefore, the answer is as follows:a) For 5 houses, the post office should be located at the position of the third house when the houses are ordered from one end to the other.b) For 6 houses, the post office can be located anywhere between the third and fourth houses when ordered.c) For an arbitrary number of houses n, the optimal location is the median position (the middle house when n is odd) or any point between the two middle houses when n is even.Yes, this aligns with the properties of the median minimizing the sum of absolute deviations. Therefore, the final answers should reflect that.Final Answera) The post office should be located at the median house, which is the third house when sorted. boxed{text{Median of the 5 houses}}b) The post office should be located anywhere between the third and fourth houses when sorted. boxed{text{Between the 3rd and 4th houses}}c) For an arbitrary ( n ), the post office should be located at the median house if ( n ) is odd, or between the two middle houses if ( n ) is even. boxed{text{Median for odd } n, text{ between the two medians for even } n}Note: The exact answers depend on the houses' positions. The solutions provided assume the houses are ordered along the street, and the median (or median interval for even ( n )) is determined based on their positions.

✅:To determine the optimal location for the post office such that the total distance the postman travels is minimized, we need to consider the sum of absolute deviations from the post office to each house. The solution involves finding the median of the house positions, as the median minimizes the sum of absolute deviations. Key Steps:1. Understanding the Problem: - The postman travels from the post office to each house and back, so the total distance for each house is twice the absolute distance from the post office to the house. - Minimizing the total distance is equivalent to minimizing the sum of absolute distances from the post office to each house.2. Median as the Optimal Solution: - The median of a set of points minimizes the sum of absolute deviations. For an odd number of houses, the median is the middle house. For an even number of houses, any point between the two middle houses minimizes the sum.3. Application to Specific Cases: - For ( n = 5 ) (odd), the median is the third house when the houses are ordered. - For ( n = 6 ) (even), the optimal location is anywhere between the third and fourth houses when ordered. - For arbitrary ( n ): - If ( n ) is odd, the median (middle house) is the optimal location. - If ( n ) is even, any point between the two middle houses is optimal. Final Answers:a) For ( n = 5 ), the post office should be located at the median house, which is the third house when sorted. boxed{text{Median of the 5 houses}}b) For ( n = 6 ), the post office should be located anywhere between the third and fourth houses when sorted. boxed{text{Between the 3rd and 4th houses}}c) For an arbitrary ( n ), the post office should be located at the median house if ( n ) is odd, or between the two middle houses if ( n ) is even. boxed{text{Median for odd } n, text{ between the two medians for even } n}

🔑:Let's analyze the given problem step-by-step for each part: # (a) Solve the problem for ( n=5 ):1. Setting up the coordinates: Let's introduce a coordinate axis, on which the houses along Pushkin street are placed with coordinates ( x_1, x_2, x_3, x_4, x_5 ) ordered such that ( x_1 < x_2 < x_3 < x_4 < x_5 ). 2. Define the variable: Let ( t ) denote the coordinate of the post office on this number line.3. Calculate the total distance ( 2S ): Since the postman walks to each house and back to the post office, the total distance ( 2S ) traveled by the postman is given by: [ 2S = 2(left|x_1 - tright| + left|x_2 - tright| + left|x_3 - tright| + left|x_4 - tright| + left|x_5 - tright|) ]4. Objective: We need to find ( t ) such that ( S ) (and thus ( 2S )) is minimized. This means we need to minimize: [ S = left|x_1 - tright| + left|x_2 - tright| + left|x_3 - tright| + left|x_4 - tright| + left|x_5 - tright| ] 5. Analyze the function ( S = S(t) ): The plot of this function ( S(t) ) is a piecewise linear graph with points of change in slope at each ( x_i ) (i.e., ( x_1, x_2, x_3, x_4, x_5 )). The slope decreases as ( t ) moves to the right until it reaches ( x_3 ) and then increases.6. Determine the minimum point: Observing that the function ( S(t) ) decreases until ( x_3 ) and increases afterwards, we see that the minimum value of ( S(t) ) is achieved at ( t = x_3 ), i.e., the median coordinate. Placing the post office at ( t = x_3 ) minimizes the total distance that the postman has to walk.Conclusion for ( n = 5 ):[boxed{x_3}]# (b) Solve the problem for ( n=6 ):1. Similar analysis: Consider the coordinates ( x_1, x_2, x_3, x_4, x_5, x_6 ) ordered such that ( x_1 < x_2 < x_3 < x_4 < x_5 < x_6 ).2. Calculate the total distance ( 2S ): Similarly, [ 2S = 2(left|x_1 - tright| + left|x_2 - tright| + left|x_3 - tright| + left|x_4 - tright| + left|x_5 - tright| + left|x_6 - tright|) ] 3. Minimize ( S ): Minimize: [ S = left|x_1 - tright| + left|x_2 - tright| + left|x_3 - tright| + left|x_4 - tright| + left|x_5 - tright| + left|x_6 - tright| ]4. Determine the minimum point: For ( n = 6 ), the median ( t ) should be any point between ( x_3 ) and ( x_4 ). In other words, ( t ) can be any value such that: [ x_3 le t le x_4 ]Conclusion for ( n = 6 ):[boxed{x_3 leq t leq x_4}]# (c) Generalize the solution for arbitrary ( n ):1. For odd ( n ): If ( n ) is odd, say ( n = 2k + 1 ), the median coordinate is ( x_{k+1} ), meaning [ t = x_{frac{n+1}{2}} ]2. For even ( n ): If ( n ) is even, say ( n = 2k ), ( t ) should be any value between ( x_k ) and ( x_{k+1} ), meaning [ x_{frac{n}{2}} leq t leq x_{frac{n+2}{2}} ]Conclusion for general ( n ):[boxed{begin{cases} t = x_{frac{n+1}{2}} & text{if } n text{ is odd} x_{frac{n}{2}} le t le x_{frac{n+2}{2}} & text{if } n text{ is even}end{cases}}]

🧠:Okay, so I need to prove that in a convex pentagon, if four of the perpendiculars dropped from each vertex to the opposite side intersect at one point, then the fifth perpendicular also passes through that same point. Hmm, interesting problem. Let me try to break this down step by step.First, let me visualize a convex pentagon. A convex pentagon has all its interior angles less than 180 degrees, and all its vertices pointing outwards. Now, each vertex has an opposite side. Wait, in a pentagon, each vertex is connected to two adjacent vertices, so the opposite side would be the one that's not adjacent. Let me confirm: in a pentagon, for any given vertex, there are two sides adjacent to it, and the remaining two sides are non-adjacent. But since a pentagon has five sides, each vertex is opposite to one side? Wait, no. Wait, if it's a regular pentagon, each vertex is opposite to a side, but in a general convex pentagon, is that still the case? Wait, actually, in a pentagon, each vertex is associated with two adjacent sides. The "opposite" side would be the one that is not adjacent. But since there are five sides, each vertex is separated from two sides by one side. Let me count: for vertex A, the adjacent sides are AB and EA. Then the next sides are BC and DE. Wait, no. Wait, a pentagon has vertices labeled A, B, C, D, E, connected in order. So each vertex is connected to the next vertex. So for vertex A, the adjacent sides are AB and EA. The non-adjacent sides would be BC, CD, and DE? Wait, no, that's three sides. Wait, each vertex is part of two sides. The opposite side would be the one that is not adjacent. But in a pentagon, each vertex is two edges away from a side. Wait, maybe in a pentagon, each side has a unique opposite vertex? Let me check.Wait, perhaps the concept of "opposite" in a pentagon isn't as straightforward as in a hexagon, where each side has a clear opposite side. In a pentagon, maybe each vertex has an opposite side. For example, in a regular pentagon, each vertex is opposite to a side. If you draw a line from a vertex perpendicular to the opposite side, that would be an altitude. So maybe in the problem, they are talking about these altitudes from each vertex to their opposite sides. The problem states that if four of these perpendiculars (altitudes) intersect at a single point, then the fifth one must also pass through that point.So the key here is that four of these altitudes concur at a point, and we need to show the fifth does too. This seems like a concurrency problem. In triangle geometry, we know that certain lines (like medians, altitudes, angle bisectors) concur, but in pentagons, such concurrency is not typical unless there's some special condition. So this problem is giving a condition where four are concurrent, and we have to prove the fifth is too.Let me recall if there's a theorem related to this. Hmm, not that I can remember specifically for pentagons, but maybe there's a general principle. Maybe using the converse of the theorem that if all five perpendiculars concur, then they meet at a single point. But here, the problem is that given four concur, show the fifth does too.Alternatively, maybe using coordinate geometry? Assign coordinates to the pentagon's vertices and try to compute equations of the perpendiculars. But that might get complicated with five variables. Alternatively, use vectors?Wait, maybe there's a projective geometry approach? Or using Ceva's theorem? Ceva's theorem is about concurrency of lines in a triangle. But we're dealing with a pentagon. Is there a Ceva-like theorem for pentagons?Alternatively, maybe using the concept of orthocenters. In a triangle, the three altitudes meet at the orthocenter. But in a pentagon, if multiple altitudes concur, maybe there's a similar concept. But I don't know of an orthocenter for a pentagon.Wait, let's think about the properties of perpendiculars. Each perpendicular from a vertex to the opposite side is a line that is orthogonal to that side. If four such lines meet at a point, perhaps we can use some orthogonality conditions to derive that the fifth must also pass through that point.Alternatively, maybe use the concept of cyclic pentagons? But cyclic polygons have vertices on a circle, which relates to perpendicular bisectors, but here we are dealing with perpendiculars from vertices to sides.Wait, perhaps using duality? In projective geometry, points and lines can be dualized. Not sure if that's helpful here.Alternatively, think about the pedal polygon. The pedal polygon of a point with respect to a polygon is formed by the projections of that point onto the sides. If four of the projections lie on the respective sides, does that imply something about the fifth?Wait, the problem states that four perpendiculars from the vertices to the opposite sides meet at a point. Let's call this point O. So for four vertices, the perpendicular from each of them to their respective opposite sides passes through O. We need to show that the fifth perpendicular also passes through O.Perhaps we can use the fact that if a point lies on four such perpendiculars, then by some geometric constraints, it must lie on the fifth.Let me try to model this. Let's denote the pentagon as ABCDE, convex. Let’s suppose that the perpendiculars from A, B, C, D to their opposite sides all pass through point O. We need to show that the perpendicular from E to its opposite side also passes through O.First, let's clarify which side is opposite each vertex. In a pentagon, each vertex is associated with two adjacent sides. The opposite side would be the one that is two edges away. For example, in pentagon ABCDE, the opposite side to vertex A would be the side CD. Wait, is that correct? Let's see: Vertex A is connected to B and E. The sides adjacent to A are AB and EA. The next sides are BC and DE, and the side opposite would be CD? Wait, maybe. Let me check with a regular pentagon. In a regular pentagon, each vertex is opposite to a side. For example, vertex A is opposite side CD. Because in a regular pentagon, if you draw a line from A perpendicular to CD, that would be the altitude. Similarly, from B perpendicular to DE, etc.Assuming that, then in pentagon ABCDE, the opposite sides are:- A opposite CD- B opposite DE- C opposite EA- D opposite AB- E opposite BCIs that the standard opposition? Maybe, but in a general convex pentagon, the "opposite" side might be defined differently. But perhaps the problem defines the opposite side as the one that is two edges away. Let me confirm. For vertex A, moving two edges away from A along the pentagon would lead to side CD. Similarly, from B, two edges away is DE, etc. So if that's the case, then each vertex's opposite side is indeed two edges away.So if the pentagon is labeled in order ABCDE, then:- The opposite side of A is CD- The opposite side of B is DE- The opposite side of C is EA- The opposite side of D is AB- The opposite side of E is BCTherefore, the perpendiculars from each vertex to these sides.Given that four of these perpendiculars concur at O, prove the fifth does as well.So, assuming that the perpendiculars from A, B, C, D to CD, DE, EA, AB respectively all pass through O. Then need to show that the perpendicular from E to BC also passes through O.Alternatively, maybe the problem defines the opposite side differently. But since the user says "the opposite sides", it's important to clarify. However, since it's a convex pentagon, the opposition is likely as I described.Alternatively, maybe each vertex's opposite side is the one that doesn't share the vertex. For example, for vertex A, the sides not containing A are BC, CD, DE, and EA. Wait, EA does contain A. Wait, no: sides are AB, BC, CD, DE, EA. So for vertex A, the sides not containing A are BC, CD, DE. So opposite side might be one of those. But in a pentagon, each vertex is part of two sides, so the other three sides don't contain the vertex. So perhaps "opposite side" is not uniquely defined unless specified. Hmm, this is a problem. Maybe the problem assumes a specific opposition, such that each vertex corresponds to one opposite side. Perhaps in a cyclic order. For example, in a pentagon, if you list the vertices as A, B, C, D, E, then the opposite side to A is the side that is two positions ahead, which would be CD? Wait, not sure.Alternatively, maybe each vertex's opposite side is the one that is non-adjacent and across the pentagon. In a regular pentagon, each vertex is directly opposite a side. So perhaps in a general convex pentagon, the same applies: the side that is "facing" the vertex.Alternatively, perhaps the problem defines the opposite side as follows: For each vertex, the opposite side is the side that is not adjacent and does not share a vertex with the given vertex. Wait, but in a pentagon, each side is adjacent to two vertices. So, for a given vertex, the sides that do not share that vertex are the three sides that are not connected to it. But since each vertex is part of two sides, the remaining three sides don't include it. So the opposite side could be the middle one of those three. For example, for vertex A, the sides not containing A are BC, CD, DE. The middle one would be CD. Similarly, for B, the sides not containing B are CD, DE, EA; the middle one would be DE. For C, sides not containing C are DE, EA, AB; middle is EA. For D, sides not containing D are EA, AB, BC; middle is AB. For E, sides not containing E are AB, BC, CD; middle is BC. So in this case, the opposite sides are CD, DE, EA, AB, BC respectively. So yes, this matches the earlier assignment. So that seems consistent. So in this case, each vertex has an opposite side which is the middle of the three sides not containing the vertex.So, given that, in pentagon ABCDE, the opposite sides are:A: CDB: DEC: EAD: ABE: BCTherefore, the perpendiculars from each vertex to these sides. So, given that four of these perpendiculars concur at O, we need to show the fifth does as well.So, let's assume that the perpendiculars from A to CD, from B to DE, from C to EA, and from D to AB all pass through point O. We need to show that the perpendicular from E to BC also passes through O.How to approach this? Maybe use coordinate geometry. Assign coordinates to the pentagon's vertices and set up equations for the perpendiculars. Then, given that four of them pass through O, derive that the fifth must also pass through O.But coordinate geometry might be messy, but let's try.Let me denote the coordinates as follows:Let’s assign coordinates to the pentagon. Let me label the vertices as A, B, C, D, E in order. Let me assign coordinates such that point O is the origin, (0,0), to simplify calculations. Wait, but we don't know where O is yet. Alternatively, perhaps assign coordinates such that some points are at convenient positions.Alternatively, use vector methods. Let me denote the position vectors of the vertices as vectors a, b, c, d, e. Then, the sides can be represented as lines between these points.The perpendicular from a vertex to its opposite side must satisfy certain conditions. For example, the perpendicular from A to CD: the line CD is defined by points C and D. The equation of line CD can be written in parametric form, and the foot of the perpendicular from A to CD can be found. Then, the condition that the perpendicular passes through O would mean that O lies on that line.But since O is the intersection point of four perpendiculars, maybe we can set up equations based on the fact that O lies on these four perpendiculars and then show that it must lie on the fifth.Alternatively, since we are dealing with perpendiculars, we can use the concept that the vector from O to the vertex is orthogonal to the direction vector of the opposite side.Wait, let's consider that. For the perpendicular from A to CD to pass through O, the vector OA must be orthogonal to the direction vector of side CD.Wait, no. Wait, the line from A perpendicular to CD must pass through O. So, the line OA is the perpendicular from A to CD only if OA is perpendicular to CD.Wait, actually, if the perpendicular from A to CD passes through O, then OA is the perpendicular line from A to CD. Therefore, OA is perpendicular to CD. Similarly, OB is perpendicular to DE, OC is perpendicular to EA, OD is perpendicular to AB, and OE is perpendicular to BC.Therefore, the condition is that each vector from O to the vertex is perpendicular to the direction vector of the opposite side.So, in vector terms, (a - o) · (d - c) = 0, since OA is from O to A, vector a - o, and CD is from C to D, vector d - c. Similarly for the others.But since O is the intersection point, we can set O as the origin for simplicity. Let me assume O is the origin. Then, the vectors from O to A, B, C, D, E are just vectors a, b, c, d, e.Then, the condition that the perpendicular from A to CD passes through O (which is the origin) would mean that the line OA is perpendicular to CD. Therefore, vector a is perpendicular to vector (d - c). So, a · (d - c) = 0.Similarly, the perpendicular from B to DE passes through O, so vector b is perpendicular to (e - d): b · (e - d) = 0.Perpendicular from C to EA passes through O: c · (a - e) = 0.Perpendicular from D to AB passes through O: d · (b - a) = 0.We need to prove that the perpendicular from E to BC also passes through O, which would mean that e · (c - b) = 0.So, given that:1. a · (d - c) = 02. b · (e - d) = 03. c · (a - e) = 04. d · (b - a) = 0We need to show that e · (c - b) = 0.So, we have four equations and need to derive the fifth. Let's see if these equations can be manipulated to get the fifth.Let me write them out more clearly:1. a · d - a · c = 02. b · e - b · d = 03. c · a - c · e = 04. d · b - d · a = 0Need to show that e · c - e · b = 0.Let me see if I can relate these equations.From equation 1: a · d = a · cFrom equation 4: d · b = d · aFrom equation 3: c · a = c · eFrom equation 2: b · e = b · dSo, let me note down:From equation 1: a · d = a · cFrom equation 4: d · b = d · a ⇒ d · b = a · d (since d · a = a · d)But from equation 1, a · d = a · c. Therefore, d · b = a · c.From equation 3: c · a = c · e ⇒ since c · a is the same as a · c, so a · c = c · eSimilarly, from equation 2: b · e = b · dSo, let's see:We have:d · b = a · c (from equation 4 and 1)a · c = c · e (from equation 3)Therefore, d · b = c · eSimilarly, from equation 2: b · e = b · d ⇒ b · e = d · b (since b · d = d · b)Therefore, d · b = b · e = c · e (from above)Therefore, c · e = b · eWhich implies that (c - b) · e = 0Which is exactly the condition we need: e · (c - b) = 0 ⇒ (c - b) · e = 0 ⇒ c · e - b · e = 0 ⇒ c · e = b · eWhich we have just derived from the previous equations.Therefore, e · (c - b) = 0, which means that the vector from E to O (which is e, since O is the origin) is perpendicular to (c - b), the direction vector of side BC. Hence, the perpendicular from E to BC passes through O.Hence, proved.Wait, let me check that again. Let's retrace:From equation 4: d · b = d · aFrom equation 1: a · d = a · c ⇒ d · a = a · cTherefore, d · b = a · cFrom equation 3: c · a = c · e ⇒ a · c = c · eTherefore, d · b = c · eFrom equation 2: b · e = b · d ⇒ b · e = d · bTherefore, d · b = b · e = c · eThus, b · e = c · e ⇒ (c - b) · e = 0 ⇒ e is perpendicular to (c - b)Which is the direction of side BC (from B to C is vector c - b). Therefore, the line OE is perpendicular to BC, which means the perpendicular from E to BC passes through O. Therefore, the fifth perpendicular also passes through O.Therefore, the conclusion follows from the given four conditions. Hence, the fifth perpendicular must pass through the same point.So, the key was to express the orthogonality conditions as dot products and then manipulate the equations to derive the fifth condition. This approach uses vector algebra to systematically relate each condition and arrive at the required result.Final Answerboxed{text{The fifth perpendicular also passes through the common intersection point.}}

✅:Given a convex pentagon where four perpendiculars dropped from the vertices to the opposite sides intersect at one point, we need to prove that the fifth perpendicular also passes through this point.1. Define the pentagon and opposite sides: - Label the pentagon vertices as (A, B, C, D, E). - The opposite sides for each vertex are: - (A) opposite (CD) - (B) opposite (DE) - (C) opposite (EA) - (D) opposite (AB) - (E) opposite (BC)2. Vector conditions for perpendicularity: - Assume the common intersection point (O) is the origin. - The vectors from (O) to the vertices are (mathbf{a}, mathbf{b}, mathbf{c}, mathbf{d}, mathbf{e}). - The perpendicular conditions translate to dot products: 1. (mathbf{a} cdot (mathbf{d} - mathbf{c}) = 0) 2. (mathbf{b} cdot (mathbf{e} - mathbf{d}) = 0) 3. (mathbf{c} cdot (mathbf{a} - mathbf{e}) = 0) 4. (mathbf{d} cdot (mathbf{b} - mathbf{a}) = 0)3. Manipulate the equations: - From equation 1: (mathbf{a} cdot mathbf{d} = mathbf{a} cdot mathbf{c}) - From equation 4: (mathbf{d} cdot mathbf{b} = mathbf{d} cdot mathbf{a}) - From equation 3: (mathbf{c} cdot mathbf{a} = mathbf{c} cdot mathbf{e}) - From equation 2: (mathbf{b} cdot mathbf{e} = mathbf{b} cdot mathbf{d})4. Relate the equations: - (mathbf{d} cdot mathbf{b} = mathbf{a} cdot mathbf{c}) (from equations 1 and 4) - (mathbf{a} cdot mathbf{c} = mathbf{c} cdot mathbf{e}) (from equation 3) - (mathbf{d} cdot mathbf{b} = mathbf{c} cdot mathbf{e}) - (mathbf{b} cdot mathbf{e} = mathbf{d} cdot mathbf{b}) (from equation 2) - Therefore, (mathbf{c} cdot mathbf{e} = mathbf{b} cdot mathbf{e})5. Conclude the fifth condition: - (mathbf{e} cdot (mathbf{c} - mathbf{b}) = 0), which means the perpendicular from (E) to (BC) passes through (O).Thus, the fifth perpendicular also passes through the common intersection point.[boxed{text{The fifth perpendicular also passes through the common intersection point.}}]

🔑:To prove that the fifth perpendicular also passes through the same point, we will approach this geometrically and algebraically.1. Define the Intersection Point: Let O be the point of intersection of the perpendiculars drawn from the vertices A, B, C, and D of the convex pentagon ABCDE.2. Geometrical Insight: Let us assume the perpendiculars from the vertices are dropped to the opposite sides. These perpendiculars intersect at point O. We need to prove that the perpendicular from vertex E also passes through point O.3. Use Vector Notation: Let overrightarrow{OA}, overrightarrow{OB}, overrightarrow{OC}, overrightarrow{OD}, and overrightarrow{OE} be the position vectors of vertices A, B, C, D, and E from the point O respectively.4. Orthogonality Conditions: Using the fact that the perpendiculars are orthogonal, we will have: - OA perp CD implies overrightarrow{OA} cdot (overrightarrow{OC} - overrightarrow{OD}) = 0, - OB perp DE implies overrightarrow{OB} cdot (overrightarrow{OD} - overrightarrow{OE}) = 0, - OC perp EA implies overrightarrow{OC} cdot (overrightarrow{OE} - overrightarrow{OA}) = 0, - OD perp AB implies overrightarrow{OD} cdot (overrightarrow{OA} - overrightarrow{OB}) = 0.5. Translate to Equalities: By orthogonality, the dot products (scalar products) must be zero: [ overrightarrow{OA} cdot overrightarrow{OC} = overrightarrow{OA} cdot overrightarrow{OD}, ] [ overrightarrow{OB} cdot overrightarrow{OD} = overrightarrow{OB} cdot overrightarrow{OE}, ] [ overrightarrow{OC} cdot overrightarrow{OE} = overrightarrow{OC} cdot overrightarrow{OA}, ] [ overrightarrow{OD} cdot overrightarrow{OA} = overrightarrow{OD} cdot overrightarrow{OB}. ]6. Transitivity and Symmetry in Equalities: Using the transitive property of equality and symmetry of dot products: [ overrightarrow{OA} cdot overrightarrow{OC} = overrightarrow{OA} cdot overrightarrow{OD} = overrightarrow{OB} cdot overrightarrow{OD} = overrightarrow{OB} cdot overrightarrow{OE} = overrightarrow{OC} cdot overrightarrow{OE} = overrightarrow{OC} cdot overrightarrow{OA} = overrightarrow{OD} cdot overrightarrow{OA} = overrightarrow{OD} cdot overrightarrow{OB}. ]7. Five Perpendiculars: For the fifth perpendicular from vertex E: - Using our derived equalities: since OE perp AB, we need: [ overrightarrow{OE} cdot (overrightarrow{OA} - overrightarrow{OB}) = 0. ] - Given that overrightarrow{OA} cdot overrightarrow{OB} establishes the position of O, this verifies that O lies on OE.8. Conclusion: From the derived perpendicular conditions and the transitive property of equal dot products, we establish that the point O through which the first four perpendiculars pass must also be a point through which the perpendicular from vertex E must pass.[boxed{text{Thus, the fifth perpendicular also passes through the point } O.}]

🧠:Okay, so I need to prove that when point M moves along the parabola ( y^2 = 2px ), the line ( M_1M_2 ) always passes through a fixed point. Also, I need to find the coordinates of this fixed point. Let me start by understanding the problem step by step.First, let's parse the given information. The parabola is ( y^2 = 2px ). Points A and B are fixed with coordinates ( A(a, b) ) and ( B(-a, 0) ). The conditions are ( ab neq 0 ), which means neither a nor b is zero, so point A isn't on either axis, and point B is on the x-axis but not at the origin. Also, ( b^2 neq 2pa ), which probably means that point A isn't on the parabola. Now, M is a point on the parabola. The lines AM and BM intersect the parabola again at points ( M_1 ) and ( M_2 ), respectively. We need to show that no matter where M is on the parabola (as long as ( M_1 ) and ( M_2 ) exist and are distinct), the line connecting ( M_1 ) and ( M_2 ) passes through a fixed point. Then, find that fixed point's coordinates.Alright, let's start by parametrizing the parabola. For a parabola ( y^2 = 4ax ), a common parametrization is ( (at^2, 2at) ). But here, the equation is ( y^2 = 2px ), so comparing to the standard form ( y^2 = 4ax ), we have 4a = 2p, so a = p/2. Therefore, parametrizing the parabola ( y^2 = 2px ), we can write points on the parabola as ( (frac{p}{2}t^2, pt) ). Alternatively, maybe using a different parameter. Let me confirm.Alternatively, sometimes people use a parameter m such that the slope is m, but maybe it's easier to stick with the standard parametric form. So, if I let t be a parameter, then the coordinates of a general point M on the parabola can be written as ( left( frac{p}{2}t^2, pt right) ). Let me check: if x = (p/2)t², then y² = p²t², which is equal to 2p*(p/2)t² = p²t². Yes, that works. So parametric coordinates for M are ( left( frac{p}{2}t^2, pt right) ).So, let me denote M as ( left( frac{p}{2}t^2, pt right) ), where t is a parameter. Then, lines AM and BM will intersect the parabola again at M₁ and M₂ respectively.I need to find the equations of lines AM and BM, find their other intersection points with the parabola, which are M₁ and M₂, then find the equation of line M₁M₂ and show that this line always passes through a fixed point, regardless of the parameter t (i.e., regardless of where M is on the parabola). Then, find that fixed point's coordinates.So, step by step:1. Find equations for lines AM and BM.2. Find their other intersection points with the parabola, M₁ and M₂.3. Determine the equation of line M₁M₂.4. Show that this line passes through a fixed point, independent of t.5. Find the coordinates of that fixed point.Let me start with step 1: finding equations of lines AM and BM.First, line AM connects point A(a, b) and M( (p/2)t², pt ). Let me find the parametric equations for line AM.Alternatively, since I need to find the other intersection point M₁, maybe I can use parametric equations for the line and substitute into the parabola's equation.Let me recall that if we have a line intersecting a parabola, the points of intersection can be found by solving the system. Since we already know two points on the line (A and M), but we need the other intersection point. Alternatively, since we know one point of intersection (M), and we can parametrize the line in terms of a parameter, then substitute into the parabola equation and solve for the parameter values. The two solutions will correspond to points A and M₁? Wait, no: actually, point A is not necessarily on the parabola. Wait, given that A(a, b) with ( b² neq 2pa ), so point A is not on the parabola. Therefore, line AM passes through A (not on the parabola) and M (on the parabola). Therefore, the line intersects the parabola at M and another point M₁. Similarly, line BM passes through B (which is (-a, 0)) and M, and intersects the parabola again at M₂.Therefore, for each line AM and BM, we need to find the second intersection point with the parabola.To find the second intersection point, perhaps using the parametric form of the line and solving for the points where it intersects the parabola.Alternatively, since we know one point (M) where the line intersects the parabola, we can use the parametric form of the line starting at A, passing through M, and find the other intersection point. Similarly for BM.Alternatively, for line AM: let's parametrize the line. Let me use a parameter s. Let me define the parametric equations for line AM. Let me set s = 0 at point A and s = 1 at point M.But actually, to find the intersections with the parabola, perhaps it's better to use a different parametrization.Alternatively, let's write the equation of line AM. Let me compute the slope first.Slope of AM: ( m_{AM} = frac{pt - b}{frac{p}{2}t^2 - a} )So, the equation of line AM is:( y - b = m_{AM}(x - a) )Similarly, for line BM, slope is ( m_{BM} = frac{pt - 0}{frac{p}{2}t^2 - (-a)} = frac{pt}{frac{p}{2}t^2 + a} )Equation of BM: ( y - 0 = m_{BM}(x + a) )Now, to find the other intersection points M₁ and M₂, we can substitute the line equations into the parabola equation ( y^2 = 2px ) and solve for x and y.This might result in quadratic equations, where one solution is the known point (M), and the other solution is the new point (M₁ or M₂).Alternatively, we can use Vieta's theorem, since for a quadratic equation, the sum of roots is known if we know one root.But let's try with line AM first.Equation of line AM: ( y = m_{AM}(x - a) + b )Substitute into parabola equation ( y^2 = 2px ):( [m_{AM}(x - a) + b]^2 = 2px )Expand this equation:( m_{AM}^2(x - a)^2 + 2m_{AM}b(x - a) + b^2 = 2px )Which is a quadratic equation in x. Since we know that x-coordinate of M is ( frac{p}{2}t^2 ), which is one solution. Let me denote the roots as ( x = frac{p}{2}t^2 ) and ( x = x_{M1} ). Then, by Vieta's formula, the sum of the roots is equal to the coefficient of x divided by the coefficient of x².But let me proceed step by step.First, expand the left-hand side:First, expand ( (x - a)^2 = x^2 - 2ax + a^2 )So,( m_{AM}^2x^2 - 2a m_{AM}^2x + a^2m_{AM}^2 + 2m_{AM}b x - 2a m_{AM}b + b^2 = 2px )Now, collect like terms:Quadratic term: ( m_{AM}^2x^2 )Linear term: ( (-2a m_{AM}^2 + 2m_{AM}b)x )Constant term: ( a^2m_{AM}^2 - 2a m_{AM}b + b^2 )So the equation becomes:( m_{AM}^2x^2 + (-2a m_{AM}^2 + 2m_{AM}b - 2p)x + (a^2m_{AM}^2 - 2a m_{AM}b + b^2) = 0 )Since we know that one of the roots is ( x = frac{p}{2}t^2 ), let's denote this as ( x_M ).Therefore, the sum of the roots (x_M + x_{M1}) is equal to [2a m_{AM}^2 - 2m_{AM}b + 2p]/m_{AM}^2Wait, Vieta's formula says that for quadratic equation ( ax² + bx + c = 0 ), sum of roots is -b/a.Therefore, sum of roots here is:x_M + x_{M1} = [2a m_{AM}^2 - 2m_{AM}b + 2p]/m_{AM}^2So, x_{M1} = [2a m_{AM}^2 - 2m_{AM}b + 2p]/m_{AM}^2 - x_MBut x_M is known: ( x_M = frac{p}{2}t^2 )But this seems complicated. Maybe there's a better way. Alternatively, since we know that the line passes through A(a, b) and M( (p/2)t², pt ), we can parametrize the line in terms of a parameter.Let me parametrize line AM as follows: let parameter s vary from 0 to 1, with s=0 at A and s=1 at M. So, parametric equations:x = a + s( (p/2)t² - a )y = b + s( pt - b )Then, substitute these into the parabola equation y² = 2px.So,[ b + s(pt - b) ]² = 2p [ a + s( (p/2)t² - a ) ]Expand left side:b² + 2s b(pt - b) + s²(pt - b)^2 = 2p a + 2p s( (p/2)t² - a )Now, collect terms in s², s, and constants:Left side:s²(pt - b)^2 + 2s b(pt - b) + b²Right side:2pa + 2p s( (p/2)t² - a )Bring all terms to left side:s²(pt - b)^2 + 2s b(pt - b) + b² - 2pa - 2p s( (p/2)t² - a ) = 0Now, group terms by powers of s:s² term: (pt - b)^2s term: 2b(pt - b) - 2p( (p/2)t² - a )constant term: b² - 2paSo, the quadratic equation in s is:[ (pt - b)^2 ]s² + [ 2b(pt - b) - 2p( (p/2)t² - a ) ]s + (b² - 2pa) = 0We know that s=1 corresponds to point M, so s=1 is one root. Therefore, the product of the roots is (constant term)/(coefficient of s²):s₁ * s₂ = (b² - 2pa)/( (pt - b)^2 )But one root is s=1, so the other root is s = (b² - 2pa)/( (pt - b)^2 )But s corresponds to the parameter along the line from A to M. However, the other intersection point M₁ is when s = s₁, which would give the coordinates:x = a + s₁( (p/2)t² - a )y = b + s₁( pt - b )Therefore, coordinates of M₁ are:x₁ = a + [ (b² - 2pa)/( (pt - b)^2 ) ]( (p/2)t² - a )y₁ = b + [ (b² - 2pa)/( (pt - b)^2 ) ]( pt - b )Similarly for line BM.But this seems quite involved. Maybe there's a smarter way. Let me think.Alternatively, since we have a parametric point M on the parabola, maybe we can use the concept of chord of a parabola. For a parabola, the chord passing through two points can be related to their parameters. But here, since A and B are fixed points not on the parabola, perhaps not directly applicable.Alternatively, perhaps use coordinates in terms of parameters and find M₁ and M₂, then find the equation of M₁M₂, then find the fixed point.Alternatively, perhaps consider inversion or projective geometry concepts, but that might be overcomplicating.Alternatively, consider that for any conic, the polar line of a point with respect to the conic can be used, but not sure.Alternatively, let me consider that since M₁ and M₂ are defined as the second intersections, perhaps there's a relationship between their coordinates. Let me attempt to compute them explicitly.Let me try to compute M₁ first. For line AM.We have point A(a, b) and point M( (p/2)t², pt ). Let me parametrize the line AM as:x = a + k*( (p/2)t² - a )y = b + k*( pt - b )Where k is a parameter. When k=0, we get A(a, b); when k=1, we get M.We can substitute these into the parabola equation y² = 2px:[ b + k(pt - b) ]² = 2p[ a + k( (p/2)t² - a ) ]Expanding:b² + 2k b(pt - b) + k²(pt - b)^2 = 2pa + 2pk( (p/2)t² - a )Rearranging:b² - 2pa + 2k b(pt - b) + k²(pt - b)^2 - 2pk( (p/2)t² - a ) = 0Let me collect like terms:Terms with k²:(pt - b)^2Terms with k:2b(pt - b) - 2p( (p/2)t² - a )Constants:b² - 2paSo the equation is:k²(pt - b)^2 + k[ 2b(pt - b) - 2p( (p/2)t² - a ) ] + (b² - 2pa) = 0We know that k=1 is a solution (since M is on the line and the parabola), so we can factor out (k - 1). Let me perform polynomial division or use the fact that if k=1 is a root, then substituting k=1 into the equation should give 0.Alternatively, using Vieta's formula: the product of the roots is (b² - 2pa)/(pt - b)^2. Since one root is k=1, the other root is k = (b² - 2pa)/(pt - b)^2. Therefore, the parameter value for M₁ is k = (b² - 2pa)/(pt - b)^2.Therefore, coordinates of M₁ are:x₁ = a + [ (b² - 2pa)/(pt - b)^2 ]*( (p/2)t² - a )y₁ = b + [ (b² - 2pa)/(pt - b)^2 ]*( pt - b )Similarly, let's compute coordinates for M₂ by repeating the process for line BM.Point B is (-a, 0), and point M is ( (p/2)t², pt ). The line BM can be parametrized as:x = -a + m*( (p/2)t² + a )y = 0 + m*( pt - 0 ) = m ptSubstitute into parabola equation y² = 2px:(m pt)^2 = 2p[ -a + m( (p/2)t² + a ) ]Simplify:m² p² t² = 2p[ -a + m( (p/2)t² + a ) ]Divide both sides by p:m² p t² = 2[ -a + m( (p/2)t² + a ) ]Expand the right side:-2a + 2m( (p/2)t² + a ) = -2a + m p t² + 2 m aSo:m² p t² = -2a + m p t² + 2 m aBring all terms to left side:m² p t² - m p t² - 2 m a + 2a = 0Factor:m² p t² - m(p t² + 2a) + 2a = 0This is a quadratic in m. Again, we know that m=1 is a root because point M is on line BM and the parabola. So, using Vieta's formula, the product of the roots is (2a)/(p t²). Since one root is m=1, the other root is m = 2a / (p t²). Wait, no. Vieta's formula says that the product of the roots is c/a, where the quadratic is a m² + b m + c = 0. Wait, let's write the quadratic as:p t² m² - (p t² + 2a) m + 2a = 0Therefore, coefficients are:A = p t²B = - (p t² + 2a)C = 2aTherefore, the product of the roots is C/A = 2a / (p t²)Since one root is m=1, the other root is (2a)/(p t²) / 1 = 2a/(p t²). Wait, no. Vieta's formula states that m1 * m2 = C/A. Since one root is m=1, then 1 * m2 = 2a/(p t²), so m2 = 2a/(p t²).Therefore, the parameter value for M₂ is m = 2a/(p t²)Therefore, coordinates of M₂ are:x₂ = -a + [ 2a/(p t²) ]*( (p/2)t² + a )y₂ = [ 2a/(p t²) ]* pt = 2a/(p t²) * pt = 2a/tCompute x₂:x₂ = -a + [ 2a/(p t²) ]*( (p/2)t² + a )First, compute the term inside the brackets:( (p/2)t² + a ) = (p t²)/2 + aMultiply by 2a/(p t²):[2a/(p t²)] * (p t²/2 + a ) = [2a/(p t²)]*(p t²/2) + [2a/(p t²)]*a = a + (2a²)/(p t²)Therefore, x₂ = -a + a + (2a²)/(p t²) = (2a²)/(p t²)So coordinates of M₂ are ( (2a²)/(p t²), 2a/t )Now, we have coordinates for M₁ and M₂.Coordinates of M₁:x₁ = a + [ (b² - 2pa)/(pt - b)^2 ]*( (p/2)t² - a )y₁ = b + [ (b² - 2pa)/(pt - b)^2 ]*( pt - b )Coordinates of M₂:x₂ = (2a²)/(p t² )y₂ = 2a/tHmm, this seems messy. Maybe simplify expressions for M₁ and M₂.Let me try to simplify M₁ first.Let me denote:Let’s compute the term (pt - b)² in the denominator:(pt - b)^2 = p² t² - 2p b t + b²Numerator for the scaling factor in M₁ is (b² - 2pa). So,Scaling factor: (b² - 2pa)/(p² t² - 2p b t + b² )Therefore, x₁ = a + [ (b² - 2pa)/(p² t² - 2p b t + b² ) ]*( (p/2)t² - a )Similarly, y₁ = b + [ (b² - 2pa)/(p² t² - 2p b t + b² ) ]*( pt - b )This seems complicated, but perhaps we can factor numerator and denominator.Wait, the denominator is p² t² - 2p b t + b² = (pt - b)^2Numerator is (b² - 2pa)So, x₁ = a + (b² - 2pa)/(pt - b)^2 * ( (p/2)t² - a )Similarly for y₁.Let me factor out the term (b² - 2pa) from the expression.Let me compute x₁:x₁ = a + (b² - 2pa) * [ ( (p/2)t² - a ) / (pt - b)^2 ]Similarly, y₁ = b + (b² - 2pa) * [ (pt - b ) / (pt - b)^2 ) ] = b + (b² - 2pa)/(pt - b )So y₁ simplifies to:y₁ = b + (b² - 2pa)/(pt - b )Let me compute that:Combine the terms over a common denominator:y₁ = [ b(pt - b) + b² - 2pa ] / (pt - b )= [ b pt - b² + b² - 2pa ] / (pt - b )Simplify numerator:b pt - 2paThus:y₁ = (b pt - 2pa)/(pt - b ) = [ pt b - 2pa ] / (pt - b )Factor numerator:= [ p t b - 2 p a ] / ( p t - b )= p ( bt - 2a ) / ( pt - b )Hmm, interesting. Let me note this: y₁ = p(bt - 2a)/(pt - b )Similarly, let's compute x₁:x₁ = a + (b² - 2pa) * [ ( (p/2)t² - a ) / (pt - b )² ]Let me write this as:x₁ = a + (b² - 2pa)( (p/2)t² - a ) / (pt - b )²Perhaps we can express x₁ and y₁ in terms of t, and similarly for x₂ and y₂, then find the equation of line M₁M₂ and see if it passes through a fixed point.Alternatively, maybe parametrizing the line M₁M₂ and finding its equation, then showing that for all t, there is a common point.Alternatively, since we need to show that all such lines pass through a fixed point, perhaps we can use the concept of the radical axis of two circles, but since it's a parabola and lines, maybe not directly.Alternatively, assume that the fixed point is (h, k). Then, for the line M₁M₂ to pass through (h, k), the determinant of the coordinates of M₁, M₂, and (h, k) should be zero. So, setting up the equation and solving for h and k such that the equation holds for all t.Let me try this approach.Let me denote the coordinates:M₁: ( x₁, y₁ ) = [ a + (b² - 2pa)( (p/2)t² - a ) / (pt - b )² , p(bt - 2a)/(pt - b ) ]M₂: ( x₂, y₂ ) = ( 2a²/(p t² ), 2a/t )We need to find h and k such that the determinant:| x₁ y₁ 1 || x₂ y₂ 1 || h k 1 | = 0This determinant being zero implies that the three points are colinear. So, expanding this determinant gives the equation of line M₁M₂. For this to hold for all t, the coefficients of powers of t in the resulting equation must be zero, leading to equations for h and k.However, computing this determinant with such complicated expressions for x₁ and y₁ might be very involved. Let me see if there's a smarter way.Alternatively, perhaps compute the equation of line M₁M₂ in terms of t, then see if it can be written in a form that shows it passes through a fixed point regardless of t.Let me compute the equation of line M₁M₂.First, let's compute the slope of line M₁M₂:Slope m = (y₂ - y₁)/(x₂ - x₁ )Then, equation of the line is y - y₁ = m(x - x₁ )But this might be complicated. Alternatively, using the two-point form:(y - y₁)/(y₂ - y₁) = (x - x₁)/(x₂ - x₁ )But again, this seems messy. Let me see if I can find parametric expressions or relations.Alternatively, consider expressing the line M₁M₂ parametrically, then attempt to eliminate the parameter t to find the relation between x and y, then observe that the equation is linear and passes through a fixed point.Alternatively, perhaps notice some symmetry or relation between the coordinates.Wait, let me first try to simplify the coordinates of M₁ and M₂.Starting with M₂:M₂ has coordinates ( 2a²/(p t² ), 2a/t )Let me denote s = 1/t. Then, coordinates of M₂ become ( 2a² s² / p, 2a s )Similarly, s is a parameter related to t. Maybe this substitution can help.Similarly, for M₁:y₁ = p(bt - 2a)/(pt - b )Let me write y₁ as:y₁ = p(bt - 2a)/(pt - b ) = [ p bt - 2pa ] / ( pt - b )Similarly, if I let s = 1/t, then t = 1/s, so:y₁ = [ p b (1/s) - 2pa ] / ( p (1/s ) - b ) = [ ( pb/s - 2pa ) / ( p/s - b ) ]Multiply numerator and denominator by s:= ( pb - 2pa s ) / ( p - b s )Similarly, x₁:x₁ = a + (b² - 2pa)( (p/2)(1/s² ) - a ) / ( p(1/s ) - b )²Wait, this seems complicated as well.Alternatively, perhaps instead of substituting s = 1/t, let's consider that M₂ has coordinates ( 2a²/(p t² ), 2a/t ). Let's denote u = 1/t, so M₂ is ( 2a² u² / p, 2a u )Then, perhaps coordinates of M₁ can be expressed in terms of u as well. Let me see.Given that t = 1/u, then:For y₁:y₁ = p(bt - 2a)/(pt - b ) = p( b(1/u) - 2a ) / ( p(1/u ) - b ) = [ ( pb/u - 2pa ) / ( p/u - b ) ] = [ pb - 2pa u ) / ( p - b u ) ]Similarly, x₁:x₁ = a + (b² - 2pa)( (p/2)(1/u² ) - a ) / ( p(1/u ) - b )²= a + (b² - 2pa)( (p/(2u² ) - a ) / ( (p - b u )² / u² ) )= a + (b² - 2pa)( (p - 2a u² )/(2u² ) ) / ( (p - b u )² / u² )= a + (b² - 2pa)(p - 2a u² )/(2 (p - b u )² )Hmm, this seems still complicated.Alternatively, perhaps consider the coordinates of M₁ and M₂ and look for a fixed point (h, k) such that the determinant:| x₁ y₁ 1 || x₂ y₂ 1 || h k 1 | = 0Expanding the determinant:x₁(y₂ - k) - x₂(y₁ - k) + h(y₁ - y₂) = 0But for this to hold for all t, the coefficients of t in this equation must cancel out, leading to equations for h and k.But computing this determinant would require expressing x₁, y₁, x₂, y₂ in terms of t, then expanding, which might be very tedious. Maybe there's a better approach.Wait, maybe consider that since the problem states that the line M₁M₂ passes through a fixed point regardless of M, so maybe this fixed point is related to points A and B. Maybe their midpoint or some harmonic conjugate?Alternatively, think about specific cases. Let me plug in specific values for a, b, p to see if I can find a pattern.Let me choose specific values for a, b, p to simplify calculations. For example, let me set p=2, a=1, b=1. Then, check if the fixed point can be found in this case, and then generalize.Given p=2, a=1, b=1. Then, the parabola is y² = 4x. Points A(1,1) and B(-1,0). Let's pick a point M on the parabola, find M₁ and M₂, then find the line M₁M₂ and see where it passes through.Let me choose a parameter t for point M on the parabola. For parabola y² = 4x, parametric coordinates are (t², 2t). Let me pick t as a parameter. So M is (t², 2t).Compute M₁: intersection of line AM with the parabola again.Line AM connects A(1,1) and M(t², 2t). Let me find the equation of line AM.Slope m = (2t - 1)/(t² - 1)Equation: y - 1 = m(x - 1)Substitute into parabola equation y² = 4x:(y)^2 = 4x=> (m(x - 1) + 1)^2 = 4xExpand:m²(x - 1)^2 + 2m(x - 1) + 1 = 4xm²(x² - 2x + 1) + 2m x - 2m + 1 = 4xm² x² - 2m² x + m² + 2m x - 2m + 1 - 4x = 0Collect like terms:m² x² + (-2m² + 2m - 4)x + (m² - 2m + 1) = 0We know that x = t² is a root. Let me call the other root x₁.By Vieta's formula, sum of roots x + x₁ = [2m² - 2m + 4]/m²Thus, x₁ = [2m² - 2m + 4]/m² - t²But m = (2t - 1)/(t² - 1)This might get complicated. Alternatively, note that in the specific case, with p=2, a=1, b=1, the fixed point might be computable.Alternatively, take a specific t and compute M₁ and M₂, then find the line M₁M₂ and see where it goes.Let me pick t=1. Then, point M is (1, 2).But wait, when t=1, M is (1, 2). But point A is (1,1). So line AM is vertical line x=1, which intersects the parabola at x=1, which is M(1,2) and also (1, -2). But since the parabola is y²=4x, when x=1, y=±2. But point A is (1,1), which is not on the parabola. So line AM from A(1,1) to M(1,2) is vertical line x=1, which intersects the parabola at M(1,2) and M₁(1,-2). Then, line BM connects B(-1,0) to M(1,2). Let's find M₂.Line BM connects (-1,0) to (1,2). The slope is (2-0)/(1 - (-1)) = 2/2 = 1. Equation is y = x +1.Find intersection with parabola y² = 4x:(x + 1)^2 = 4x => x² + 2x +1 = 4x => x² - 2x +1 =0 => (x-1)^2=0. So the only intersection is at x=1, which is M(1,2). So, in this case, the line BM is tangent to the parabola at M, so M₂ coincides with M, but the problem statement requires M₁ and M₂ to exist and M₁ ≠ M₂. Therefore, t=1 is not allowed. So pick a different t.Let me pick t=2. Then, M is (4, 4).Line AM connects A(1,1) to M(4,4). Slope m=(4-1)/(4-1)= 3/3=1. Equation: y -1 = 1*(x -1) → y = x.Intersect with parabola y²=4x: x² =4x → x(x-4)=0. So x=0 (y=0) and x=4 (y=4). Therefore, M is (4,4), so M₁ is (0,0).Line BM connects B(-1,0) to M(4,4). Slope m=(4 -0)/(4 - (-1))=4/5. Equation: y = (4/5)(x +1).Intersect with parabola y²=4x:[(4/5)(x +1)]² =4x → (16/25)(x +1)^2 =4x → 16(x² + 2x +1) =100x →16x² +32x +16 =100x →16x² -68x +16=0Divide by 4:4x² -17x +4=0Using quadratic formula: x = [17 ±√(289 - 64)]/8 = [17 ±√225]/8 = [17 ±15]/8Solutions: (32)/8=4 and (2)/8=0.25. So x=4 (point M) and x=0.25. Then y= (4/5)(0.25 +1)= (4/5)(1.25)= (4/5)(5/4)=1. So M₂ is (0.25, 1).Therefore, M₁ is (0,0), M₂ is (0.25,1). Line M₁M₂ connects (0,0) and (0.25,1). Slope is 4. Equation: y=4x.This line passes through the origin (0,0), but in this case, with t=2, the line M₁M₂ is y=4x, which passes through (0,0). But is (0,0) the fixed point? Let's check another t.Take t=0.5. Then, point M is (0.25, 1).Line AM connects A(1,1) to M(0.25,1). This is horizontal line y=1. Intersection with parabola y²=4x:1²=4x →x=0.25. So only intersection is M(0.25,1). Therefore, line AM is tangent at M, so M₁ does not exist. Therefore, t=0.5 is invalid.Another t, say t=3. Then, M is (9,6).Line AM from A(1,1) to M(9,6). Slope m=(6-1)/(9-1)=5/8. Equation: y -1 = (5/8)(x -1)Substitute into parabola:[(5/8)(x -1) +1]^2 =4xCompute:(5x/8 -5/8 +1)^2 =4x → (5x/8 + 3/8)^2 =4x → (25x² +30x +9)/64 =4x →25x² +30x +9=256x →25x² -226x +9=0Solutions: x=(226 ±√(226² -4*25*9))/50=(226 ±√(51076 -900))/50=(226 ±√50176)/50=(226 ±224)/50Thus, x=(226+224)/50=450/50=9 (point M) and x=(226-224)/50=2/50=0.04. So M₁ is (0.04, y).Compute y from line equation: y= (5/8)(0.04 -1)+1=(5/8)(-0.96)+1= -0.6 +1=0.4. So M₁ is (0.04,0.4). Since parabola is y²=4x, check y²=0.16, 4x=0.16, so x=0.04. Correct.Line BM connects B(-1,0) to M(9,6). Slope m=(6-0)/(9 - (-1))=6/10=3/5. Equation: y= (3/5)(x +1)Intersect with parabola:[(3/5)(x +1)]^2 =4x →9/25(x+1)^2=4x →9(x² +2x +1)=100x →9x² +18x +9=100x →9x² -82x +9=0Solutions: x=(82 ±√(6724 -324))/18=(82 ±√6400)/18=(82 ±80)/18x=(162)/18=9 (point M) and x=(2)/18≈0.1111. So M₂ is (0.1111, y). Compute y= (3/5)(0.1111 +1)= (3/5)(1.1111)=0.6666. So M₂ is (0.1111, 0.6666).Now, line M₁M₂ connects (0.04,0.4) and (0.1111, 0.6666). Let me compute the equation.Slope m=(0.6666 -0.4)/(0.1111 -0.04)=0.2666/0.0711≈3.75. So slope is 15/4=3.75.Equation: y -0.4 = 15/4(x -0.04)Check if this line passes through (0,0):Left side:0 -0.4= -0.4Right side:15/4*(0 -0.04)=15/4*(-0.04)= -0.15Not equal, so doesn't pass through (0,0). Wait, but earlier example with t=2 passed through (0,0). So inconsistency here. Therefore, maybe (0,0) is not the fixed point.Alternatively, maybe in this case, the fixed point is another point. Let me compute the equation of line M₁M₂ for t=3.Points M₁(0.04,0.4) and M₂(0.1111,0.6666)Slope m=(0.6666 -0.4)/(0.1111 -0.04)=0.2666/0.0711≈3.75=15/4Equation: y = 15/4 x + cUsing M₁: 0.4 =15/4 *0.04 + c →0.4=0.15 +c →c=0.25Thus, equation: y=15/4 x + 0.25Now, check if this line passes through a fixed point. Let me see if there's a common point for t=2 and t=3.For t=2, line M₁M₂ was y=4x. For t=3, line is y=15/4 x +0.25. Intersection of these two lines:4x =15/4 x +0.25 →(16/4 x -15/4 x)=0.25 →1/4 x=0.25→x=1, y=4. But point (1,4) is on both lines. However, in the case of t=2, line M₁M₂ passes through (0,0) and (1,4), but in the case of t=3, line passes through (1,4) and (0.1111,0.6666). Wait, but this suggests that (1,4) is the fixed point? But when I checked for t=2, the line was y=4x passing through (0,0) and (1,4). For t=3, the line is y=15/4 x +0.25, which passes through (1,4) because 15/4 *1 +0.25=3.75+0.25=4.So, perhaps (1,4) is the fixed point? But this is specific to the example with a=1, b=1, p=2. Let me check with another t.Take t=4. Point M is (16,8). Compute M₁ and M₂.Line AM connects A(1,1) to M(16,8). Slope m=(8-1)/(16-1)=7/15. Equation: y -1 =7/15(x -1)Intersect with parabola:[(7/15)(x -1) +1]^2=4xExpand:[7x/15 -7/15 +1]^2 =4x →[7x/15 +8/15]^2=4x →(49x² + 112x +64)/225=4x →49x² +112x +64=900x →49x² -788x +64=0Solutions: x=(788 ±√(788² -4*49*64))/98But since this is time-consuming, let me just note that one solution is x=16, the other can be found via Vieta: product is 64/49, so other solution x= (64/49)/16=64/(49*16)=4/49≈0.0816. So M₁ is (4/49, y). Compute y from line equation:y=7/15*(4/49 -1)+1=7/15*(-45/49)+1= -7/15*(45/49)+1= - (315/735)+1= - (3/7)+1=4/7≈0.5714.Line BM connects B(-1,0) to M(16,8). Slope m=(8-0)/(16 -(-1))=8/17. Equation: y=8/17(x +1)Intersect with parabola:(8/17(x +1))^2=4x →64/289(x+1)^2=4x →64(x² +2x +1)=1156x →64x² +128x +64=1156x →64x² -1028x +64=0Divide by 4:16x² -257x +16=0Solutions: x=(257 ±√(257² -4*16*16))/32But again, Vieta: product is 16/16=1, so x=16 and x=1/16. Therefore, M₂ is (1/16, y). Compute y=8/17*(1/16 +1)=8/17*(17/16)=8/16=0.5.Thus, M₁ is (4/49,4/7) and M₂ is (1/16,0.5). Compute the equation of line M₁M₂.Slope m=(0.5 -4/7)/(1/16 -4/49)= ( (7/14 -8/14) ) / ( (49/784 -64/784) )= ( -1/14 ) / ( -15/784 )= (-1/14)*(-784/15)= 784/(14*15)=56/15≈3.7333Equation: y -4/7=56/15(x -4/49)Check if it passes through (1,4):Left side:4 -4/7=24/7≈3.4286Right side:56/15*(1 -4/49)=56/15*(45/49)=56/15*45/49=(56*45)/(15*49)= (8*7*9*5)/(15*7*7)= (8*9*5)/(15*7)= (360)/(105)= 24/7≈3.4286. Yes, it passes through (1,4).Therefore, in this specific case (a=1, b=1, p=2), the fixed point is (1,4).But in the general case, given the problem has parameters a, b, p, I need to find the coordinates of the fixed point in terms of a, b, p.From the specific case, the fixed point was (a, 2b) when p=2, a=1, b=1. Wait, (1,4) is (a, 2b)? No, 2b would be 2*1=2, but here it's 4. Alternatively, in the specific case p=2, the fixed point y-coordinate was 4, which is 2p. Wait, 2p=4, so (a, 2p). But a=1 here, but fixed point was (1,4). So (a, 2p). But in the general case, the fixed point might be (a, 2p). Wait, no. Because in the specific case, p=2, and the fixed point y-coordinate was 4=2p. The x-coordinate was a=1.Wait, in the specific case, fixed point was (a, 2p). But a=1, p=2, so (1,4). That fits.Alternatively, in another example, let me choose different a and p. Suppose a=2, p=3, b=1.Parabola y²=6x. Points A(2,1) and B(-2,0). Choose a point M on the parabola. Let me choose t=1, so coordinates (3/2, 3*1)= (1.5,3). Then line AM connects A(2,1) to M(1.5,3). Slope m=(3-1)/(1.5 -2)=2/(-0.5)=-4. Equation: y -1 = -4(x -2) → y = -4x +9.Intersect with parabola y²=6x:(-4x +9)^2=6x →16x² -72x +81=6x →16x² -78x +81=0Solutions: x=(78 ±√(78² -4*16*81))/32=(78 ±√(6084 -5184))/32=(78 ±√900)/32=(78 ±30)/32x=(108)/32=27/8=3.375 and x=48/32=1.5. So M₁ is (27/8, y). Compute y=-4*(27/8)+9= -27/2 +9= -13.5 +9= -4.5= -9/2. So M₁ is (27/8, -9/2).Line BM connects B(-2,0) to M(1.5,3). Slope m=(3 -0)/(1.5 -(-2))=3/3.5=6/7. Equation: y=6/7(x +2).Intersect with parabola y²=6x:(6/7(x +2))²=6x →36/49(x² +4x +4)=6x →36x² +144x +144=294x →36x² -150x +144=0Divide by 6:6x² -25x +24=0Solutions:x=(25 ±√(625 -576))/12=(25 ±7)/12. So x=32/12=8/3≈2.6667 and x=18/12=1.5. So M₂ is (8/3, y). Compute y=6/7*(8/3 +2)=6/7*(14/3)=6/7*(14/3)=4. So M₂ is (8/3,4).Line M₁M₂ connects (27/8, -9/2) and (8/3,4). Find the equation.Slope m=(4 - (-9/2))/(8/3 -27/8)= (17/2)/(64/24 -81/24)= (17/2)/(-17/24)= (17/2)*(-24/17)= -12.Equation: y -4= -12(x -8/3)Simplify: y= -12x +32 +4= -12x +36.Check if this passes through a fixed point. Let's assume the fixed point is (a, 2p). Here, a=2, p=3, so (2,6). Check if (2,6) is on the line y=-12x +36. Plug in x=2: y= -24 +36=12 ≠6. Not passing through (2,6). So previous guess is incorrect.Alternatively, let me check if there's a fixed point in this example. Solve for h and k such that -12h +36 =k. But this line is y=-12x +36. For it to pass through a fixed point regardless of M, but in this case, with t=1, the line is y=-12x +36. If I take another t, say t=2 for this a=2, p=3, b=1.Point M when t=2: coordinates ( (3/2)(2)^2, 3*2 )= (6,6).Line AM connects A(2,1) to M(6,6). Slope m=(6-1)/(6-2)=5/4. Equation: y -1=5/4(x -2) →y=5/4 x -5/2 +1=5/4 x -3/2.Intersect with parabola y²=6x:(5/4 x -3/2)^2=6x →25/16 x² -15/4 x +9/4=6x →25x² -60x +36=96x →25x² -156x +36=0Solutions: x=(156 ±√(156² -4*25*36))/50=(156 ±√(24336 -3600))/50=(156 ±√20736)/50=(156 ±144)/50.x=(300)/50=6 (point M) and x=12/50=6/25=0.24. So M₁ is (6/25, y). Compute y=5/4*(6/25) -3/2= (30/100) -3/2= 0.3 -1.5= -1.2= -6/5.Line BM connects B(-2,0) to M(6,6). Slope m=(6 -0)/(6 -(-2))=6/8=3/4. Equation: y=3/4(x +2).Intersect with parabola y²=6x:(3/4(x +2))^2=6x →9/16(x² +4x +4)=6x →9x² +36x +36=96x →9x² -60x +36=0→3x² -20x +12=0.Solutions:x=(20 ±√(400 -144))/6=(20 ±16)/6. So x=(36)/6=6 and x=4/6=2/3≈0.6667. So M₂ is (2/3, y). Compute y=3/4*(2/3 +2)=3/4*(8/3)=2. So M₂ is (2/3, 2).Line M₁M₂ connects (6/25, -6/5) and (2/3, 2). Slope m=(2 - (-6/5))/(2/3 -6/25)= (16/5)/(32/75)= (16/5)*(75/32)= (16*75)/(5*32)= (3*75)/32=225/32≈7.03125.Equation: y -2=225/32(x -2/3)Compute when x=2/3: y=2. But check if this line passes through a fixed point. Let's see if there's a common point between this line and the previous line when t=1, which was y=-12x +36.Solve:-12x +36 =225/32 x -225/32*(2/3) +2Multiply both sides by 32 to eliminate denominators:-384x +1152=225x -150 +64-384x +1152=225x -86-609x= -1238x=1238/609≈2.033Then y≈-12*(2.033)+36≈-24.4 +36=11.6. So intersection at≈(2.033,11.6). Not a fixed point. Therefore, in this case with a=2, p=3, b=1, the fixed point is not immediately obvious. Therefore, my previous assumption for the fixed point being (a, 2p) is incorrect.Alternatively, maybe the fixed point is (a, b²/(2p)), but in the first example, with a=1, b=1, p=2, this would be (1,1/4), which is not the case. Earlier fixed point was (1,4). Not matching.Alternatively, perhaps the fixed point is related to point A and B. For instance, in the first example, points A(1,1) and B(-1,0). Fixed point was (1,4). Not sure.Alternatively, let's try to compute the fixed point in the general case.Given the coordinates of M₁ and M₂, we can write parametric equations for line M₁M₂ and then find the fixed point.Recall from earlier steps:For a general parameter t, coordinates of M₁:x₁ = a + (b² - 2pa)( (p/2)t² - a ) / (pt - b )²y₁ = p(bt - 2a ) / (pt - b )Coordinates of M₂:x₂ = 2a²/(p t² )y₂ = 2a/tWe need to show that line M₁M₂ passes through a fixed point (h,k) independent of t.To find such a point (h,k), it must satisfy the equation of line M₁M₂ for all t. Therefore, the equation:(y - y₁)(x₂ - x₁) = (y₂ - y₁)(x - x₁ )must hold for (h,k) for all t.This can be rewritten as:(y - y₁)(x₂ - x₁) - (y₂ - y₁)(x - x₁ ) = 0Substituting (h,k):(k - y₁)(x₂ - x₁) - (y₂ - y₁)(h - x₁ ) = 0This equation must hold for all t. Therefore, we can treat this as an identity in t and solve for h and k.However, expanding this seems very complex given the expressions for x₁, y₁, x₂, y₂. Therefore, perhaps we can find h and k by clever substitution or by assuming certain forms.Alternatively, notice that the fixed point should satisfy the equations of both lines AM and BM for some particular M, but this might not be directly helpful.Alternatively, suppose that the fixed point is the intersection point of the polars of A and B with respect to the parabola. The polar line of a point with respect to a conic is the locus of points whose harmonic conjugates with respect to the intersections of any line through the point lie on the conic. For a parabola, the polar line of a point (x₀,y₀) is given by yy₀ = p(x + x₀). For the general parabola y² = 4ax, the polar of (x₀,y₀) is yy₀ = 2a(x + x₀). In our case, the parabola is y² = 2px, which is similar to y² = 4a x with a = p/2. So, the polar line of a point (x₀,y₀) is yy₀ = 2*(p/2)(x + x₀) = p(x + x₀).Therefore, the polar of point A(a,b) is y*b = p(x + a )Similarly, the polar of point B(-a,0) is y*0 = p(x - a ) →0 = p(x - a )→x = aTherefore, the polars of A and B are the line x = a (polar of B) and the line y*b = p(x + a ) (polar of A). The intersection of these two polars is the point x = a, y*b = p(a + a ) = 2pa ⇒ y = 2pa / b. Therefore, the intersection point of the polars is (a, 2pa/b ).This might be the fixed point.In the first example with a=1, b=1, p=2, the fixed point would be (1, 2*2*1 /1 ) = (1,4), which matches our earlier result. In the second example, a=2, p=3, b=1, the fixed point would be (2, 2*3*2 /1 ) = (2,12). Let me check with the second example.In the second example with a=2, p=3, b=1, the fixed point would be (2, 12). In the case when t=1, line M₁M₂ was y=-12x +36. Does this line pass through (2,12)? Plug in x=2: y=-24 +36=12. Yes. For the case when t=2, line M₁M₂ was y=225/32 x - something. Wait, with t=2, M₂ is (2/3, 2), M₁ is (6/25, -6/5). The equation was y=225/32 x - something. Wait, let me recalculate.Wait, for a=2, p=3, b=1, when t=2, the line M₁M₂ connects (6/25, -6/5) and (2/3, 2). Let's compute the equation:Slope m=(2 - (-6/5))/(2/3 -6/25)= (16/5)/( (50/75 -18/75 ))= (16/5)/(32/75)= (16/5)*(75/32)= (16*75)/(5*32)= (3*75)/32=225/32≈7.03125Equation: y -2 =225/32(x -2/3)Check if (2,12) is on this line:Left side:12 -2=10Right side:225/32*(2 -2/3)=225/32*(4/3)=225/24≈9.375≠10. Not on the line. Therefore, contradiction. Wait, but according to the polar intersection, the fixed point should be (2, 2*3*2 /1 )=(2,12). But in the specific case, when t=1, the line passed through (2,12), but when t=2, it does not. So inconsistency here. Therefore, perhaps my assumption is wrong.Wait, in the second example with a=2, p=3, b=1, the fixed point calculated via polars is (a, 2pa/b )=(2, 12). When t=1, line M₁M₂ was y=-12x +36, which passes through (2,12): y=-24 +36=12. Yes. When t=2, line M₁M₂ is y=225/32 x - 225/32*(2/3) +2=225/32 x - 150/32 + 64/32=225/32 x -86/32. Plug in x=2: y=450/32 -86/32=364/32=11.375≈11.375≠12. So, does not pass through (2,12). Therefore, my assumption is incorrect.But wait, in the first example, it worked for t=2 and t=3, but in the second example, it worked for t=1 but not for t=2. Therefore, the intersection of polars might not be the fixed point in general.Alternatively, perhaps the fixed point is related to the midpoint of A and B or something else.Alternatively, let's try to find the fixed point analytically.Let me consider the line M₁M₂ and try to find (h,k) such that for all t, the point lies on the line.We have coordinates of M₁ and M₂ as functions of t. Let me denote t as a parameter and try to express the line equation in terms of t, then eliminate t to find a relation between h and k.The equation of the line M₁M₂ can be written as:(y - y₁)(x₂ - x₁) = (y₂ - y₁)(x - x₁ )Plugging in the coordinates:(y - y₁)(x₂ - x₁) = (y₂ - y₁)(x - x₁ )Let me substitute x₁, y₁, x₂, y₂:(y - [p(bt - 2a)/(pt - b)])*( [2a²/(p t²)] - [a + (b² - 2pa)( (p/2)t² - a ) / (pt - b )² ]) = ( [2a/t - p(bt - 2a)/(pt - b ) ] )*(x - [a + (b² - 2pa)( (p/2)t² - a ) / (pt - b )² ])This equation is quite complex, but perhaps simplifying step by step.First, let me compute x₂ - x₁:x₂ - x₁ = [2a²/(p t²)] - [a + (b² - 2pa)( (p/2)t² - a ) / (pt - b )² ]= [2a²/(p t²) - a ] - (b² - 2pa)( (p/2)t² - a ) / (pt - b )²Similarly, y₂ - y₁ = [2a/t - p(bt - 2a)/(pt - b ) ]Let me simplify y₂ - y₁:= 2a/t - p(bt - 2a)/(pt - b )= [2a(pt - b ) - p t (bt - 2a ) ] / [ t(pt - b ) ]Expand numerator:2a pt - 2a b - p b t² + 2a p tCombine like terms:(2a pt + 2a pt) - 2a b - p b t²=4a pt - 2a b - p b t²Thus, y₂ - y₁ = [ -p b t² +4a pt -2a b ] / [ t(pt - b ) ]Similarly, compute x₂ - x₁:x₂ - x₁ = 2a²/(p t² ) -a - (b² -2pa)( (p/2)t² -a ) / (pt - b )²Let me write this as:= (2a² - a p t² ) / (p t² ) - (b² -2pa)( (p/2)t² -a ) / (pt - b )²This is quite complicated, but perhaps factor numerator.Alternatively, let me consider specific terms and see if the equation can be simplified.Alternatively, assume that the fixed point (h,k) satisfies the equation for all t, so substituting (h,k) into the line equation:(k - y₁)(x₂ - x₁) = (y₂ - y₁)(h - x₁ )This must hold for all t. Therefore, the left-hand side and right-hand side must be equal as rational functions of t. Therefore, we can equate the coefficients of corresponding powers of t in the numerator and denominator.However, this approach would be very tedious due to the complexity of expressions. Perhaps there is a smarter substitution or a way to express h and k.Alternatively, assume that the fixed point is (h, k) = (a, 2pa/b ). As per the polar intersection.Let me check if this point satisfies the equation for general t.Substitute h = a, k = 2pa/b.Left-hand side: (k - y₁)(x₂ - x₁ ) = (2pa/b - y₁ )(x₂ - x₁ )Right-hand side: (y₂ - y₁ )(h - x₁ ) = (y₂ - y₁ )(a - x₁ )Let me compute both sides.First, compute y₁:y₁ = p(bt - 2a)/(pt - b )k - y₁ = 2pa/b - p(bt -2a)/(pt - b )= (2pa(pt - b ) - p b (bt -2a )) / [ b(pt - b ) ]Expand numerator:2pa pt - 2pa b - p b² t + 2p a b= 2p² a t -2pa b - p b² t + 2pa b= 2p² a t - p b² t= t p (2pa - b² )Thus,k - y₁ = t p (2pa - b² ) / [ b(pt - b ) ]Now, compute x₂ -x₁:x₂ -x₁ = 2a²/(p t² ) - x₁From earlier:x₁ = a + (b² -2pa)( (p/2)t² -a ) / (pt - b )²Thus,x₂ -x₁ = 2a²/(p t² ) -a - (b² -2pa)( (p/2)t² -a ) / (pt - b )²Let me factor out terms:= -a + 2a²/(p t² ) - (b² -2pa)( (p/2)t² -a ) / (pt - b )²This expression is complex, but perhaps substituting the numerator and denominator.Now, the left-hand side is:(k - y₁)(x₂ -x₁ ) = [ t p (2pa - b² ) / ( b(pt - b ) ) ] * [ -a + 2a²/(p t² ) - (b² -2pa)( (p/2)t² -a ) / (pt - b )² ]This seems very complicated.Now, compute the right-hand side:(y₂ - y₁ )(a - x₁ )First, compute a - x₁:a - x₁ = - (x₁ -a ) = - (b² -2pa)( (p/2)t² -a ) / (pt - b )²From earlier, x₁ = a + (b² -2pa)( (p/2)t² -a ) / (pt - b )²Thus,a -x₁ = - (b² -2pa)( (p/2)t² -a ) / (pt - b )²Now, compute (y₂ - y₁ ):From earlier, we found:y₂ - y₁ = [ -p b t² +4a pt -2a b ] / [ t(pt - b ) ]Therefore, the right-hand side is:[ -p b t² +4a pt -2a b ] / [ t(pt - b ) ] * [ - (b² -2pa)( (p/2)t² -a ) / (pt - b )² ]= - [ ( -p b t² +4a pt -2a b ) / [ t(pt - b ) ] ] * [ (b² -2pa)( (p/2)t² -a ) / (pt - b )² ]= [ (p b t² -4a pt +2a b ) / [ t(pt - b ) ] ] * [ (b² -2pa)( (p/2)t² -a ) / (pt - b )² ]= [ (p b t² -4a pt +2a b ) (b² -2pa)( (p/2)t² -a ) ] / [ t(pt - b )³ ]This expression is also extremely complicated.Comparing left-hand side and right-hand side:Left-hand side:[ t p (2pa - b² ) / ( b(pt - b ) ) ] * [ -a + 2a²/(p t² ) - (b² -2pa)( (p/2)t² -a ) / (pt - b )² ]Right-hand side:[ (p b t² -4a pt +2a b ) (b² -2pa)( (p/2)t² -a ) ] / [ t(pt - b )³ ]Given the complexity, it's not obvious whether these two expressions are equal. Therefore, perhaps my assumption that (a, 2pa/b ) is the fixed point is incorrect, or there's a miscalculation.Alternatively, given that in the first example with a=1, b=1, p=2, the fixed point was (1,4), which is (a, 2pa/b ), since 2pa/b = 2*2*1 /1 =4. Similarly, in the second example with a=2, p=3, b=1, the polar intersection gives (2,12), which is 2pa/b= 2*3*2 /1=12. But in that case, the line M₁M₂ for t=2 didn't pass through (2,12). However, perhaps there was a miscalculation.Wait, let me re-examine the second example with a=2, p=3, b=1, t=2.Point M is (6,6). Line AM connects A(2,1) to M(6,6). The line equation is y=5/4 x -3/2. Intersection with parabola y²=6x gives M₁ at (6/25, -6/5). Line BM connects B(-2,0) to M(6,6), equation y=3/4(x+2). Intersection with parabola gives M₂ at (2/3, 2). The line M₁M₂ should pass through (2,12). Let's check.Compute the equation of line M₁M₂:M₁(6/25, -6/5), M₂(2/3, 2)Slope m=(2 - (-6/5))/(2/3 -6/25)= (16/5)/(32/75)= (16/5)*(75/32)=15/2=7.5Equation: y -2=15/2(x -2/3)At x=2:y -2=15/2*(2 -2/3)=15/2*(4/3)=15/2*(4/3)=30/3=10→y=12Therefore, the line passes through (2,12). Earlier mistake was mine in the calculation. Therefore, even in the second example, the fixed point is (a, 2pa/b ). Therefore, the fixed point is indeed (a, 2pa/b ).Thus, returning to the general case, after verifying through examples, the fixed point is (a, 2pa/b ).Therefore, the answer is that the line M₁M₂ always passes through the fixed point with coordinates ( left( a, frac{2pa}{b} right) ).Final AnswerThe fixed point is boxed{left( a, dfrac{2pa}{b} right)}.

✅:Given the parabola ( y^2 = 2px ) and fixed points ( A(a, b) ) and ( B(-a, 0) ) with the conditions ( ab neq 0 ) and ( b^2 neq 2pa ). Let ( M ) be a point on the parabola. Suppose lines ( AM ) and ( BM ) intersect the parabola again at points ( M_1 ) and ( M_2 ), respectively. We need to prove that the line ( M_1M_2 ) always passes through a fixed point and determine its coordinates.1. Parametrization of the Parabola: - A general point ( M ) on the parabola ( y^2 = 2px ) can be parametrized as ( M left( frac{p}{2}t^2, pt right) ) where ( t ) is a parameter.2. Finding Intersection Points ( M_1 ) and ( M_2 ): - Line ( AM ): - The line ( AM ) intersects the parabola again at ( M_1 ). Using Vieta's theorem and solving the quadratic equation derived from the line equation, the coordinates of ( M_1 ) are found to be complex expressions involving ( t ). - Line ( BM ): - The line ( BM ) intersects the parabola again at ( M_2 ). Similarly, using Vieta's theorem, the coordinates of ( M_2 ) are derived.3. Equation of Line ( M_1M_2 ): - The coordinates of ( M_1 ) and ( M_2 ) are substituted into the line equation. Simplification shows that the line equation involves terms that can be factored and simplified to reveal a fixed point.4. Fixed Point Analysis: - By considering the polars of points ( A ) and ( B ) with respect to the parabola, it is found that the intersection of these polars gives the fixed point. The polar line of ( A(a, b) ) is ( yb = p(x + a) ) and the polar line of ( B(-a, 0) ) is ( x = a ). Their intersection is at ( (a, frac{2pa}{b}) ).5. Verification with Examples: - Specific examples with different values of ( a ), ( b ), and ( p ) confirm that the fixed point is indeed ( left( a, frac{2pa}{b} right) ). Final AnswerThe fixed point through which the line ( M_1M_2 ) always passes is (boxed{left( a, dfrac{2pa}{b} right)}).

🔑:1. Let the coordinates of points M, M_1, and M_2 on the parabola y^2 = 2px be (frac{y_0^2}{2p}, y_0), (frac{y_1^2}{2p}, y_1), and (frac{y_2^2}{2p}, y_2) respectively.2. Since A, M, and M_1 are collinear, the slope condition must hold for the points A(a,b), M(frac{y_0^2}{2p}, y_0), and M_1(frac{y_1^2}{2p}, y_1). Therefore, we write the slope equation: [ frac{frac{y_1^2}{2p} - frac{y_0^2}{2p}}{y_1 - y_0} = frac{frac{y_0^2}{2p} - a}{y_0 - b} ]3. Simplifying the left-hand side by combining fractions: [ frac{frac{y_1^2 - y_0^2}{2p}}{y_1 + y_0} = frac{frac{y_0^2}{2p} - a}{y_0 - b} ]4. Rearranging and simplifying the equation: [ frac{y_1 + y_0}{2p} cdot (y_1 - y_0) = frac{y_0^2 - 2pa}{2p} (y_0 - b) ]5. Therefore, we have: [ y_1 y_0 = b(y_1 + y_0) - 2pa quad text{(since } (y_1^2 - y_0^2) = (y_1 + y_0)(y_1 - y_0) text{)} ]6. Solving for y_1: [ y_1 = frac{b y_0 - 2pa}{y_0 - b} ]7. Similarly, considering points B(-a, 0), M(frac{y_0^2}{2p}, y_0), and M_2(frac{y_2^2}{2p}, y_2), since B, M, and M_2 are collinear: [ y_2 = frac{2pa}{y_0} ]8. Let (x, y) be any point on the line M_1M_2. Then the slope condition must hold between M_1(frac{y_1^2}{2p}, y_1) and M_2(frac{y_2^2}{2p}, y_2): [ y_1 y_2 = y (y_1 + y_2) - 2px ]9. Substitute y_1 = frac{b y_0 - 2pa}{y_0 - b} and y_2 = frac{2pa}{y_0} into the equation: [ frac{(b y_0 - 2pa) cdot 2pa}{(y_0 - b) y_0} = y left(frac{b y_0 - 2pa}{y_0 - b} + frac{2pa}{y_0}right) - 2px ]10. Simplify and rearrange to find x: [ frac{2p(b y_0 - 2pa) cdot a}{(y_0 - b)y_0} = yleft(frac{b y_0}{y_0 - b} + frac{2pa}{y_0}right) - 2px ] [ frac{2pa b y_0 - 4p^2 a^2}{y_0 (y_0 - b)} = y left(frac{by_0^2 + 2p a(y_0 - b)}{y_0 (y_0 - b)}right) - 2px ] We need 2py_{0} ^{2}x and 2pb(a-x) and (a-x)(b y_{0}^{2}-2p a) 11. Simplifying further yields: [ y_{0} ^2 (2 p x - by) + 2 p b (a-x) + 2 p a(by - 2 p a) = 0 ]12. Therefore, the line M_1M_2 always passes through the fixed point: [ left(a, frac{2 p a}{b}right) ]Conclusion:[ boxed{left(a, frac{2 p a}{b}right)}]

🧠:Okay, let's try to figure out this problem. So, the question is: for each integer m between 1 and 2000 inclusive, there exists an integer k such that m/2001 < k/n < (m+1)/2002. We need to find the smallest possible n. The options are A. 2003, B. 2008, C. 3106, D. 4003.Hmm, let's start by understanding the problem. For every m from 1 to 2000, there should be some integer k where the fraction k/n is squeezed between m/2001 and (m+1)/2002. Our goal is to find the smallest n that satisfies this condition for all m in that range.First, maybe I should rephrase the inequality to make it clearer. The inequality is m/2001 < k/n < (m+1)/2002. We can rewrite this to find k in terms of m and n. Multiplying all parts by n, we get (m/n)*n < k < [(m+1)/2002]*n. So, k must be an integer strictly between m*n/2001 and (m+1)*n/2002. The key is that for each m, this interval must contain at least one integer k. So, the interval (m*n/2001, (m+1)*n/2002) must contain at least one integer.Therefore, the problem reduces to ensuring that for every m from 1 to 2000, the length of the interval ( (m+1)/2002 - m/2001 ) * n is greater than 1, which would guarantee that there's at least one integer in each interval. Wait, no, that's not exactly correct. The length of the interval times n would give the number of units between the two bounds. But even if the length is greater than 1, if the interval is not aligned properly, it might not contain an integer. However, if the length is greater than 1, then by the pigeonhole principle, there must be at least one integer in the interval. So maybe that's a way to approach this.Let me compute the difference between the upper and lower bounds. Let's compute ( (m+1)/2002 - m/2001 ). Let's compute that:= (m+1)*2001 - m*2002 / (2001*2002)= [ (m*2001 + 2001) - (m*2002) ] / (2001*2002)= [ -m + 2001 ] / (2001*2002)= (2001 - m)/ (2001*2002)So the difference is (2001 - m)/(2001*2002). Therefore, the length of the interval for each m is (2001 - m)/(2001*2002). Then, multiplying by n to get the length in terms of k: n*(2001 - m)/(2001*2002). For this length to be greater than 1, we need n*(2001 - m)/(2001*2002) > 1. Solving for n gives n > (2001*2002)/(2001 - m).But since this has to hold for all m from 1 to 2000, the maximum value of (2001*2002)/(2001 - m) occurs when m is minimized, right? Wait, no, because as m increases, 2001 - m decreases, so the denominator gets smaller, making the entire expression larger. So the maximum of (2001*2002)/(2001 - m) occurs when m is as large as possible. The maximum m is 2000, so plugging in m=2000, we get (2001*2002)/(2001 - 2000) = 2001*2002 /1 = 2001*2002 = 2001*2002 = 2001*2000 + 2001*2 = 4,002,000 + 4,002 = 4,006,002. That's way larger than any of the answer choices. But that can't be right because the answer choices are around 2000-4000.Wait, maybe my approach is incorrect. Because if we require that the length of the interval is greater than 1 for all m, then n would have to be larger than 4,000,000, which is not possible. So perhaps this approach isn't the right way to go. Maybe the length doesn't have to be greater than 1 for all m. Instead, we need that for each m, the interval (m/2001, (m+1)/2002) when scaled by n contains at least one integer. So even if the interval is small, as long as it's positioned such that it "catches" an integer.Alternatively, maybe the problem is equivalent to the requirement that between any two consecutive fractions m/2001 and (m+1)/2002, there must be a fraction k/n. So, in other words, the Farey sequence of order n must lie between these two fractions for each m. Maybe the key is that the mediant of the two fractions m/2001 and (m+1)/2002 must lie between them, but the mediant would be (m + (m+1))/(2001 + 2002) = (2m +1)/4003. But this mediant might not necessarily be the fraction k/n we need. Alternatively, perhaps the condition is similar to the density of fractions k/n in the real numbers, such that they must intersect every interval between m/2001 and (m+1)/2002.Alternatively, maybe we can consider the problem as a system of inequalities. For each m, there must exist k such that:m/2001 < k/n < (m+1)/2002Multiplying all terms by n:m*n/2001 < k < (m+1)*n/2002Since k must be an integer, this implies that the ceiling of (m*n/2001) must be less than (m+1)*n/2002. So:ceil( m*n / 2001 ) < (m+1)*n / 2002This inequality must hold for all m from 1 to 2000. Therefore, for each m:ceil( (m*n)/2001 ) < ( (m+1)*n ) / 2002But how can we translate this into a condition on n?Alternatively, let's rearrange the inequality:( m*n ) / 2001 < k < ( (m+1)*n ) / 2002The smallest k that could satisfy this is k = floor( (m*n)/2001 ) + 1. Similarly, the largest k is floor( ((m+1)*n)/2002 - 1 ). So, for there to exist such a k, we need:floor( (m*n)/2001 ) + 1 < ( (m+1)*n ) / 2002But perhaps more straightforwardly, the existence of k requires that the interval (m*n/2001, (m+1)*n/2002) contains at least one integer. For this to happen, the ceiling of m*n/2001 must be less than (m+1)*n/2002. So:ceil( m*n / 2001 ) < (m+1)*n / 2002Which can be rewritten as:ceil( m*n / 2001 ) <= (m+1)*n / 2002 - ε, for any ε > 0.But since ceil(a) is the smallest integer greater than or equal to a, the inequality ceil(a) < b is equivalent to a < floor(b). Wait, maybe not. Let me think.If ceil(x) < y, then x < ceil(x) < y, which implies x < y. So, ceil( m*n /2001 ) < (m+1)*n /2002 implies that m*n /2001 < (m+1)*n /2002 - δ, for some δ >0. But perhaps this is not the easiest way.Alternatively, if we can ensure that (m+1)*n /2002 - m*n /2001 > 1, then the interval would have length greater than 1, hence containing an integer. Wait, but earlier when we calculated the difference:( (m+1)/2002 - m/2001 ) = (2001 - m)/(2001*2002)Multiply by n to get the length of the interval in terms of k:n*(2001 - m)/(2001*2002) > 1So, solving for n:n > (2001*2002)/(2001 - m)But as m ranges from 1 to 2000, the maximum value of (2001*2002)/(2001 - m) occurs when m is maximum, i.e., m=2000. Then, 2001*2002 / (2001 -2000) = 2001*2002 /1= 2001*2002 = 2001*2000 + 2001*2= 4,002,000 +4,002=4,006,002. So n must be greater than 4 million? But the answer choices are all less than 4003. So this approach is wrong. Because the problem does not require that the interval has length greater than 1, but that there's at least one integer in the interval, even if the interval is smaller than 1. So my initial thought was incorrect. The interval can be less than 1, but as long as it's positioned between two integers, it will contain an integer. Wait, no. If the interval is less than 1 in length, it's possible that it lies entirely between two integers and thus contains no integers. So, for the interval to contain at least one integer, the length of the interval must be at least 1, or it must be positioned such that it crosses an integer.But actually, even if the length is less than 1, it might still contain an integer. For example, if the interval is (1.1, 1.2), it's length 0.1, which is less than 1, but doesn't contain an integer. However, if the interval is (0.9, 1.1), length 0.2, which is less than 1, but it contains integer 1. So, the key is that the interval must overlap with an integer. So even if the length is less than 1, as long as it's positioned such that it includes an integer. Therefore, the necessary and sufficient condition is that the ceiling of the lower bound is less than the upper bound. In other words, for each m, we need that:ceil( m*n / 2001 ) < ( (m+1)*n ) / 2002Because if the smallest integer greater than m*n/2001 is still less than (m+1)*n/2002, then that integer lies within the interval.Therefore, the condition is:ceil( m*n / 2001 ) < ( (m+1)*n ) / 2002Which can be rewritten as:ceil( m*n / 2001 ) <= ( (m+1)*n -1 ) / 2002Because (m+1)*n /2002 is not necessarily an integer, so we subtract 1 to make sure that the ceiling is strictly less than the upper bound.Alternatively, perhaps another approach. Let's consider the fractional part. Let's denote that for each m, there exists an integer k such that:m*n < 2001*k < (m+1)*n*2001/2002Wait, this seems complicated. Alternatively, maybe we can rearrange the original inequality:m/2001 < k/n < (m+1)/2002Cross-multiplying denominators to eliminate fractions:Multiply all terms by 2001*2002*n to get:m*2002*n < k*2001*2002 < (m+1)*2001*nBut maybe that's not helpful. Wait, let's try to cross-multiply the inequalities step by step.Starting with m/2001 < k/n. Multiplying both sides by 2001*n:m*n < 2001*kSimilarly, k/n < (m+1)/2002. Multiplying both sides by n*2002:2002*k < (m+1)*nTherefore, combining both inequalities:m*n < 2001*k < 2002*k < (m+1)*nWait, that seems conflicting. Wait, the two inequalities are:m*n < 2001*k and 2002*k < (m+1)*nTherefore, combining them:m*n < 2001*k < 2002*k < (m+1)*nBut this implies that 2001*k < 2002*k, which is always true because 2001 < 2002 and k is positive. But the problem is whether 2001*k < (m+1)*n. So combining the two inequalities:m*n < 2001*k < (m+1)*nBut 2001*k must be greater than m*n and less than (m+1)*n. Therefore, for each m, there must exist an integer k such that:m*n < 2001*k < (m+1)*nSimilarly, dividing by 2001:(m*n)/2001 < k < ((m+1)*n)/2001But wait, this is different from the previous inequality. Wait, the original cross-multiplied inequalities are:From m/2001 < k/n: m*n < 2001*kFrom k/n < (m+1)/2002: k*2002 < (m+1)*nTherefore, combining:m*n < 2001*k < 2002*k < (m+1)*nWait, so 2001*k < 2002*k is trivial, but 2002*k < (m+1)*n must also hold. Therefore, 2001*k must be greater than m*n, and 2002*k must be less than (m+1)*n. So combining both:m*n < 2001*k < 2002*k < (m+1)*nTherefore, the inequalities can be rewritten as:m*n < 2001*k < (m+1)*nBut since 2001*k < (m+1)*n and 2001*k > m*n, we can write:m*n < 2001*k < (m+1)*nWhich implies that:m/2001 < k/n < (m+1)/2002But perhaps this is circular. Alternatively, perhaps combining the two inequalities:From m*n < 2001*k and 2002*k < (m+1)*n, we can write:m*n < 2001*k < 2002*k < (m+1)*nTherefore, 2001*k < (m+1)*n and 2002*k > m*n. Wait, no, the second inequality is 2002*k < (m+1)*n, which gives k < (m+1)*n /2002. The first inequality gives k > m*n /2001.So the real condition is that k must satisfy:m*n /2001 < k < (m+1)*n /2002And k must be an integer. Therefore, for each m from 1 to 2000, the interval ( m*n /2001 , (m+1)*n /2002 ) must contain at least one integer.Therefore, our problem reduces to finding the smallest n such that for every m in 1 to 2000, the interval ( m*n /2001 , (m+1)*n /2002 ) contains at least one integer.To ensure this, we can use the concept of the floor function. For the interval to contain an integer, there must exist an integer k such that:k - 1 < m*n /2001 < k < (m+1)*n /2002 < k +1But that might not be helpful. Alternatively, the distance between m*n /2001 and (m+1)*n /2002 must be such that they straddle an integer.Alternatively, consider the fractional parts. Let me think.Let’s denote that for each m, there must exist k such that:m*n /2001 < k < (m+1)*n /2002This can be rearranged as:m*n /2001 < k < (m+1)*n /2002Which can be rewritten as:k > m*n /2001andk < (m+1)*n /2002Therefore, subtracting these two inequalities:(m+1)*n /2002 - m*n /2001 > 0Which we already calculated earlier as (2001 - m)/ (2001*2002) *n >0, which is always true since m <2001.But the important part is that the interval between m*n /2001 and (m+1)*n /2002 must contain at least one integer.A theorem related to this is the Dirichlet approximation theorem, which states that for any real number and any positive integer N, there exists integers p and q with 1 ≤ q ≤ N such that |α - p/q| < 1/(qN). But I'm not sure if that's directly applicable here.Alternatively, perhaps considering Beatty sequences or some other form of diophantine approximation.Alternatively, think about the problem in terms of the mediant of two fractions. The mediant of two fractions a/b and c/d is (a+c)/(b+d). The mediant lies between the two fractions if they are positive and a/b < c/d. However, in our case, the fractions are m/2001 and (m+1)/2002. Let's compute the mediant:Mediant = (m + (m+1)) / (2001 + 2002) = (2m +1)/4003We need this mediant to lie between m/2001 and (m+1)/2002. Let's check:Is (2m +1)/4003 > m/2001?Compute (2m +1)/4003 - m/2001 = (2m +1)/4003 - 2m/4002Find common denominator 4003*4002:[(2m +1)*4002 - 2m*4003] / (4003*4002)= [ (2m*4002 + 4002) - 2m*4003 ] / (4003*4002 )= [ 2m*(4002 -4003) + 4002 ] / (4003*4002 )= [ -2m + 4002 ] / (4003*4002 )For m >=1, numerator is -2m +4002. For m <=2001, this is positive when -2m +4002 >0 => m <2001. Since m <2001, then this is positive. Therefore, the mediant is greater than m/2001.Similarly, check if (2m +1)/4003 < (m+1)/2002:Compute (m+1)/2002 - (2m +1)/4003= [(m+1)*4003 - (2m +1)*2002]/(2002*4003)= [ (m*4003 +4003) - (2m*2002 +2002) ] / (2002*4003)= [ m*4003 - 2m*2002 + 4003 -2002 ] / (2002*4003 )= [ m*(4003 -4004) + 2001 ] / denominator= [ -m +2001 ] / denominatorSince m >=1, -m +2001 >=0 when m <=2000. Since our m is from 1 to2000, this is non-negative. Therefore, the mediant is less than or equal to (m+1)/2002. Wait, but when m=2000:- m +2001 =1, which is positive. Therefore, the mediant is less than (m+1)/2002.Therefore, the mediant (2m+1)/4003 lies between m/2001 and (m+1)/2002.Therefore, if we take n=4003, then k=2m+1 would satisfy m/2001 < (2m+1)/4003 < (m+1)/2002. Therefore, n=4003 works. But the problem asks for the smallest possible n. So is 4003 the minimal n, or is there a smaller n?Looking at the options, 4003 is option D. But there are smaller options like 2003, 2008, 3106. So we need to check if a smaller n can satisfy the condition.Wait, but according to the mediant property, if we take n=4003, then for each m, k=2m+1 works. So 4003 is a valid n. But maybe there's a smaller n.Alternatively, perhaps n must satisfy that for each m, the interval (m*n/2001, (m+1)*n/2002) contains an integer. To find the minimal n, perhaps we need that the maximum over m of the minimal gap required to catch an integer is minimized.Alternatively, perhaps using the concept of the greedy algorithm or continued fractions.Alternatively, to ensure that for each m, ceil(m*n/2001) < (m+1)*n/2002.Which can be written as:ceil(m*n/2001) < (m+1)*n/2002Which implies:m*n/2001 +1 <= (m+1)*n/2002Because ceil(x) <= x +1. Therefore, if we can ensure that m*n/2001 +1 <= (m+1)*n/2002, then ceil(m*n/2001) <= (m+1)*n/2002.But let's write this inequality:(m*n)/2001 + 1 <= (m+1)*n /2002Multiply both sides by 2001*2002 to eliminate denominators:m*n*2002 + 2001*2002 <= (m+1)*n*2001Rearranging terms:m*n*2002 - (m+1)*n*2001 <= -2001*2002Factor n:n*(m*2002 - (m+1)*2001) <= -2001*2002Compute the coefficient:m*2002 - (m+1)*2001 = m*(2002 -2001) -2001 = m -2001Therefore, the inequality becomes:n*(m -2001) <= -2001*2002Since m ranges from 1 to2000, m -2001 is negative. Therefore, dividing both sides by (m -2001), which is negative, reverses the inequality:n >= (-2001*2002)/(m -2001) = (2001*2002)/(2001 - m)Thus, for each m, n >= (2001*2002)/(2001 -m)Therefore, the minimal n must be at least the maximum of (2001*2002)/(2001 -m) over m=1,2,...,2000.Since m ranges from1 to2000, 2001 -m ranges from2000 down to1. Therefore, (2001*2002)/(2001 -m) is maximized when 2001 -m is minimized, i.e., when m is maximized. So when m=2000, then 2001 -m=1, so (2001*2002)/1=2001*2002=4,006,002. Which is way too big, but the answer options are much smaller, so clearly this approach is leading us astray.Wait, but earlier we derived that n must be greater than or equal to (2001*2002)/(2001 -m) for each m, so the minimal n must be the maximum over m of these values. But the maximum is 2001*2002=4,006,002, which is not among the options. Therefore, this suggests that our initial approach is wrong. Perhaps the error comes from assuming that ceil(m*n/2001) <= (m+1)*n/2002 implies m*n/2001 +1 <= (m+1)*n/2002, but in reality, ceil(m*n/2001) could be much less than (m*n/2001) +1. For example, if m*n/2001 is already close to an integer, ceil(m*n/2001) would be that integer, which could be much less than (m*n/2001) +1. Therefore, the inequality we derived is a sufficient condition but not necessary. Therefore, even if n is smaller than 4 million, it might still satisfy the original condition.Therefore, we need another approach.Let’s consider that for each m, there must exist k such that:m/2001 < k/n < (m+1)/2002Rearranging for k:m*n/2001 < k < (m+1)*n/2002So k must be an integer strictly between m*n/2001 and (m+1)*n/2002.Therefore, for each m, the interval ( m*n/2001 , (m+1)*n/2002 ) must contain at least one integer.To ensure this, we can use a theorem from number theory: if the interval (a, b) has length b - a > 1, then it contains at least one integer. If the length is <=1, it might not contain an integer. However, even if the length is less than 1, depending on the positions of a and b, the interval might still contain an integer.But in our case, the length of each interval is:[(m+1)/2002 - m/2001] *n = [ (2001 -m)/(2001*2002) ] *nThis length must be such that for each m, this interval contains an integer. The minimal n that ensures that for all m, this interval contains at least one integer.Alternatively, instead of considering the length, think about the fractional parts. Let’s denote that for each m, there must exist k such that:k/n > m/2001 and k/n < (m+1)/2002Which can be rewritten as:k > m*n/2001 and k < (m+1)*n/2002Therefore, the interval for k is ( m*n/2001 , (m+1)*n/2002 )To have at least one integer in this interval, we need that the ceiling of m*n/2001 is less than (m+1)*n/2002. So:ceil( m*n/2001 ) < (m+1)*n/2002This must hold for all m from1 to2000.Now, let's denote m*n = q*2001 + r, where q is the quotient and r is the remainder when m*n is divided by2001, i.e., 0 <= r <2001.Then, m*n/2001 = q + r/2001.Therefore, ceil(m*n/2001) = q +1 if r >0, else q.Therefore, the condition becomes:If r >0, then q +1 < (m+1)*n /2002If r =0, then q < (m+1)*n /2002But r = m*n mod2001, so unless m*n is a multiple of2001, which for m and n being integers, unless n is a multiple of2001/gcd(m,2001). Since2001=3*23*29, which is square-free.But this might be complicated. Let's try to work with the inequality.ceil(m*n/2001) < (m+1)*n/2002Let’s denote:ceil(m*n/2001) <= (m+1)*n/2002 - ε for some ε >0.But since ceil(m*n/2001) is an integer, the inequality can be written as:ceil(m*n/2001) <= floor( (m+1)*n/2002 )But floor( (m+1)*n/2002 ) >= ceil(m*n/2001 )Thus, for each m:floor( (m+1)*n/2002 ) >= ceil(m*n/2001 )Which implies that:(m+1)*n/2002 -1 >= ceil(m*n/2001 )But ceil(m*n/2001 ) <= m*n/2001 +1Therefore:(m+1)*n/2002 -1 >= m*n/2001 +1Rearranging:(m+1)*n/2002 - m*n/2001 >=2Multiply both sides by2001*2002 to eliminate denominators:(m+1)*n*2001 -m*n*2002 >=2*2001*2002Factor n:n[ (m+1)*2001 -m*2002 ] >=2*2001*2002Compute the coefficient inside the brackets:(m+1)*2001 -m*2002= m*2001 +2001 -m*2002=2001 -m*(2002 -2001)=2001 -mTherefore, the inequality becomes:n*(2001 -m) >=2*2001*2002Thus:n >= (2*2001*2002)/(2001 -m)Again, this must hold for all m from1 to2000. Therefore, n must be at least the maximum of (2*2001*2002)/(2001 -m) over m=1,2,...,2000.Similarly to before, the maximum occurs when m is maximized, i.e., m=2000:n >= (2*2001*2002)/1=2*2001*2002=8,012,004Which is even larger than before, which is impossible since the answer choices are much smaller. Therefore, this approach is also leading us to an incorrect conclusion.Thus, it's clear that this method of deriving inequalities is not yielding a useful bound. Maybe my initial approach is flawed.Let’s try considering specific values of n given in the options and see if they work.Option A: n=2003We need to check for each m from1 to2000, whether there exists k such that m/2001 <k/2003<(m+1)/2002.Alternatively, multiply all terms by 2001*2002*2003 to eliminate denominators:m*2002*2003 <k*2001*2003 <(m+1)*2001*2003Wait, this might not be helpful. Alternatively, cross-multiplying the inequalities:For m/2001 <k/2003:m*2003 <2001*kFor k/2003 < (m+1)/2002:k*2002 < (m+1)*2003So combining:m*2003 <2001*k <2002*k < (m+1)*2003But since 2001 <2002, 2001*k <2002*k, so this doesn't create a problem. The key is whether there exists k such that:m*2003 <2001*k < (m+1)*2003Dividing by2001:(m*2003)/2001 <k < ( (m+1)*2003 )/2001Similarly, k must be an integer in this interval. Let's compute the length of the interval:( (m+1)*2003/2001 - m*2003/2001 )=2003/2001Which is approximately 1.00099. So the length is just over 1. Therefore, for each m, the interval has length slightly more than1, so by the pigeonhole principle, there must be at least one integer in each interval. Therefore, n=2003 might work? But why is 2003 an option, but the correct answer is different. Wait, but 2003 is option A.Wait, but if the length is approximately1.00099, which is just over1, then yes, the interval will contain at least one integer. However, this reasoning might not hold for all m. For example, if the lower bound is an integer, then the interval would be from integer to integer +1.00099, which would contain the next integer. But if the lower bound is just below an integer, then the interval could be from just below integer to just below integer +1.00099, which might contain integer. Hmm.Wait, let's take a concrete example. Let's take m=1.For m=1, n=2003:Lower bound:1*2003/2001≈1.0009985Upper bound:2*2003/2002≈2.000998So the interval is approximately(1.0009985, 2.000998). So the integer k=2 is in this interval. Wait, but k must be greater than1.0009985 and less than2.000998. So k=2 is inside. Similarly, for m=2:Lower bound:2*2003/2001≈2.001997Upper bound:3*2003/2002≈3.001496So interval (2.001997,3.001496). So k=3 is inside. Similarly, this seems to work.But wait, what about m=2000:Lower bound:2000*2003/2001≈2000*1.000998≈2000+2000*0.000998≈2000+1.996≈2001.996Upper bound:2001*2003/2002≈(2001/2002)*2003≈ (approx 0.9995)*2003≈2001.4985Wait, wait, this can't be. Wait, m=2000:Lower bound:2000*2003/2001≈(2000/2001)*2003≈(0.9995)*2003≈2001.4985Upper bound:2001*2003/2002≈(2001/2002)*2003≈0.9995*2003≈2001.4985Wait, no, this can't be. Wait, let's compute upper bound for m=2000:Upper bound=(m+1)*2003/2002=2001*2003/2002= (2002 -1)*2003/2002=2003 -2003/2002≈2003 -1.0005≈2001.9995Wait, so for m=2000:Lower bound=2000*2003/2001≈(2000*2003)/2001= (2000/2001)*2003≈0.9995*2003≈2001.4985Upper bound=2001*2003/2002≈(2001/2002)*2003≈0.9995*2003≈2001.4985Wait, that can't be. Both lower and upper bounds are approximately2001.4985? That doesn't make sense. Wait, no:Wait, m=2000:Lower bound=2000*2003/2001= (2000/2001)*2003≈(0.9995)*2003≈2001.4985Upper bound=(2000+1)*2003/2002=2001*2003/2002= (2001/2002)*2003≈0.9995*2003≈2001.4985Wait, so the interval is (≈2001.4985,≈2001.4985), which is a single point? That can't be right. There must be a miscalculation here.Wait, let's compute it more accurately.Compute lower bound for m=2000, n=2003:Lower bound=2000*2003/2001= (2000*2003)/2001= (2000/2001)*2003= (2000/2001)*2003=2003 - (2003/2001)=2003 -1.0009985...≈2001.9990015Wait, that's different from what I thought. Wait, 2000*2003=4,006,000Divide by2001:4,006,000 /2001≈2001*(2001 -1)/2001=2001 -1/2001≈2000.9995Wait, wait, no. 2000*2003=2000*(2000+3)=2000² +6000=4,000,000 +6,000=4,006,000Divide by2001:4,006,000 ÷2001≈2001*2=4002, which is 4002, but 2001*2001=4,004,001, which is less than4,006,000. So 4,006,000 -4,004,001=1,999. Therefore,4,006,000=2001*2001 +1,999. Therefore,4,006,000/2001=2001 +1,999/2001≈2001 +0.999≈2001.999Similarly, upper bound:(m+1)*n /2002=2001*2003/20022001*2003=4,006,003Divide by2002:4,006,003/2002≈2002*2=4004, but 2002*2002=4,008,004, which is more than4,006,003. Wait, 2002*2001=2002*2000 +2002=4,004,000 +2002=4,006,002. So4,006,003 -4,006,002=1. Therefore,4,006,003=2002*2001 +1. Therefore,4,006,003/2002=2001 +1/2002≈2001.0005Therefore, for m=2000, the interval is approximately (2001.999,2001.0005). Wait, that can't be possible. Wait, no, the upper bound is2001.0005 and lower bound is2001.999? That would mean the interval is empty. But that's impossible. There must be a mistake in calculations.Wait, no, I think there's a confusion in m. When m=2000, the upper bound is (2000+1)/2002=2001/2002≈0.9995. Multiply by n=2003:2001/2002*2003≈2003*(1 -1/2002)=2003 -2003/2002≈2003 -1.0005≈2001.9995Similarly, lower bound is m/2001 *n=2000/2001*2003≈2000*(1 +2/2001)≈2000 +4000/2001≈2000 +1.998≈2001.998So the interval is approximately (2001.998,2001.9995), which is a very small interval of length≈0.0015. So between2001.998 and2001.9995. Therefore, the integer k=2002 lies outside this interval. But there are no integers between2001.998 and2001.9995. So the interval does not contain any integer. Therefore, for m=2000 and n=2003, there is no integer k satisfying the inequality. Therefore, n=2003 does not work. So option A is invalid.So this shows that n=2003 fails for m=2000. Therefore, our initial assumption that the length being slightly over1 would ensure an integer in the interval is incorrect, because in reality, the interval could be shifted such that even with a length slightly over1, it might not contain an integer if it's positioned between two integers. For example, an interval from 2001.998 to2001.9995 has length≈0.0015, which is less than1, so it doesn't contain any integer. Therefore, n=2003 is invalid.Similarly, we can check n=2008. Maybe similar issues occur.Alternatively, let's consider n=4003. For m=2000:Lower bound=2000*4003/2001≈2000*2.001≈4002Upper bound=2001*4003/2002≈2001*2≈4002Wait, exact computation:Lower bound=2000*4003/2001= (2000/2001)*4003=4003*(1 -1/2001)≈4003 -2≈4001Upper bound=2001*4003/2002= (2001/2002)*4003≈4003*(1 -1/2002)≈4003 -2≈4001Wait, exact calculation:2000*4003=8,006,000Divide by2001:8,006,000 ÷2001≈4001 (since2001*4001=2001*4000 +2001=8,004,000 +2001=8,006,001). So8,006,000/2001=4001 -1/2001≈4000.9995Upper bound=2001*4003/2002= (2001*4003)/2002= (2001/2002)*4003≈4003*(1 -1/2002)≈4003 -4003/2002≈4003 -2.001≈4000.999So the interval is approximately(4000.9995,4000.999), which is actually empty. Wait, that can't be. Wait, the lower bound is4000.9995 and upper bound is4000.999? That's impossible, since lower bound should be less than upper bound.Wait, no, actually:Lower bound is m/2001 *n=2000/2001*4003Upper bound is (m+1)/2002 *n=2001/2002*4003Wait, compute exact values:2000/2001*4003=4003*(1 -1/2001)=4003 -4003/2001≈4003 -2.001≈4000.999Similarly, 2001/2002*4003=4003*(1 -1/2002)=4003 -4003/2002≈4003 -2.001≈4000.999Wait, so both bounds are approximately4000.999, so the interval is actually from≈4000.999 to≈4000.999, which is a point. Therefore, there is no integer k in this interval. Therefore, even n=4003 doesn't work? But earlier, using the mediant, we found that k=2m+1 would work. Wait, confusion arises.Wait, no, wait, when n=4003, the fraction k/n is (2m+1)/4003. For m=2000, k=4001. Then, check if 2000/2001 <4001/4003 <2001/2002.Compute 2000/2001≈0.9995, 4001/4003≈0.9995, and2001/2002≈0.9995.Let me compute exact values:4001/4003=1 -2/4003≈0.9995006242000/2001≈0.9995002492001/2002≈0.999500499Therefore, 2000/2001≈0.999500249 <4001/4003≈0.999500624 <2001/2002≈0.999500499?Wait, no, 0.999500624 is greater than0.999500499. Therefore,4001/4003 >2001/2002. So this does not satisfy the inequality. Therefore, the mediant doesn't lie between the two fractions for m=2000. Therefore, my earlier assumption was wrong.Therefore, n=4003 doesn't work. Then why is it an option?Wait, maybe the problem is that for m=2000, the mediant is (2*2000 +1)/4003=4001/4003≈0.999500624. But (m+1)/2002=2001/2002≈0.999500499. So the mediant is greater than (m+1)/2002, which violates the upper bound. Therefore, the mediant does not lie between m/2001 and (m+1)/2002 for m=2000. Therefore, our earlier reasoning was incorrect.Therefore, maybe the mediant approach is invalid for m=2000.This suggests that n=4003 is invalid. But option D is4003. Hmm, confusion arises.Wait, let's re-examine the problem statement:"For each integer m satisfying 0 < m <2001, there exists an integer k such that m/2001 <k/n < (m+1)/2002 holds, where n is a positive integer. What is the smallest possible value of n?"So for m=2000, we need k/n to be between2000/2001 and2001/2002.Compute the difference between2001/2002 and2000/2001:2001/2002 -2000/2001=(2001² -2000*2002)/(2001*2002)= (2001² - (2001 -1)(2001 +1))/ (2001*2002)= (2001² - (2001² -1))/ (2001*2002)=1/(2001*2002)Therefore, the interval between2000/2001 and2001/2002 has length1/(2001*2002). So to have a fraction k/n in that interval, n must be at least2001*2002 +1, by the Farey sequence property. The Farey sequence of order n includes all fractions with denominator <=n, and between any two consecutive terms a/b and c/d in the Farey sequence, we have bc -ad=1. Therefore, to have a fraction between two fractions with denominators2001 and2002, n would need to be at least2001 +2002=4003. But according to Farey's theorem, if two fractions a/b and c/d are consecutive in Farey sequence of order n, then their difference is at least1/(n(n+1)), but I might be misremembering.Alternatively, the mediant of two fractions a/b and c/d is (a+c)/(b+d). If the mediant is in reduced form, then it lies between the two fractions. So for m=2000, the mediant is (2000+2001)/(2001+2002)=4001/4003≈0.999500624, which is greater than2001/2002≈0.999500499. Therefore, the mediant does not lie between them. Hence, the mediant approach doesn't work here.Therefore, perhaps the minimal n is such that1/n <1/(2001*2002). Therefore, n>2001*2002=4,006,002. But that's not possible given the options.This suggests that there's a misunderstanding in the problem approach.Alternatively, let's think differently. For each m, we need the interval (m/2001, (m+1)/2002) to contain a fraction k/n. The minimal n would be the least common multiple or something related to the denominators2001 and2002.But2001 and2002 are coprime? 2002=2*7*11*13, and2001=3*23*29. They share no common factors. Therefore, lcm(2001,2002)=2001*2002=4,006,002. Which is again too large.Alternatively, maybe the problem is related to the concept of a "reduced" interval. The interval between m/2001 and (m+1)/2002 is a very small interval, and to have a fraction k/n within that interval requires that n is large enough such that1/n is smaller than the interval's length. But the length of the interval is1/(2001*2002), so to have1/n <1/(2001*2002), which impliesn>2001*2002, which is again4 million.But the answer choices are much smaller, so there must be a different approach.Wait, perhaps the intervals overlap in such a way that a single n can cover multiple m's? But the problem states "for each integer m", so each m must have its own k.Alternatively, perhaps consider the continued fraction expansion of2001/2002 and find a convergent that gives the minimal n.Alternatively, consider the inequality:For each m, there exists k such that m/2001 <k/n <(m+1)/2002This can be rewritten as:m <2001*k/n <2001*(m+1)/2002Wait, not sure.Alternatively, consider the inequality in terms of k:k >m*n/2001k <(m+1)*n/2002Therefore, the difference between these two bounds is:(m+1)*n/2002 -m*n/2001 =n*( (m+1)*2001 -m*2002 )/(2001*2002 )=n*(2001 -m)/(2001*2002 )To ensure that this difference is positive, which it is since m <2001.But we need this interval to contain at least one integer. So the length of the interval is n*(2001 -m)/(2001*2002 ). To have at least one integer in this interval, the length must be at least1, or the interval must be positioned such that it contains an integer even with length less than1.If we require the length to be at least1, then:n*(2001 -m)/(2001*2002 ) >=1=>n >= (2001*2002)/(2001 -m )As before, the maximum required n is4,006,002, which is too big.But perhaps if we don't require the length to be at least1, but instead use the fact that the intervals for different m's might overlap in such a way that a clever choice of n allows coverage.Alternatively, consider that the problem is equivalent to the following: the set {k/n |k∈ℕ} intersects every interval (m/2001, (m+1)/2002 ) for m=1,...,2000.This is similar to the problem of the additive sequence modulo1 covering certain intervals.Alternatively, think of it as a covering problem: the fractions k/n must cover all intervals (m/2001, (m+1)/2002).The minimal n is the smallest number such that the spacing between consecutive fractions k/n is smaller than the minimal length of the intervals (m/2001, (m+1)/2002). The minimal interval length is when m is maximum, m=2000, length=1/(2001*2002 ). The spacing between consecutive fractions k/n is1/n. So we need1/n <1/(2001*2002 ), which impliesn>2001*2002=4,006,002, which is not among the options. Therefore, again conflicting.Therefore, it's clear that the problem requires a different approach.Let’s consider the following: For the inequality m/2001 <k/n <(m+1)/2002 to hold for some k, we can rearrange the inequalities:From m/2001 <k/n:m < (2001/n)*kFrom k/n <(m+1)/2002:k < (n/2002)*(m+1)Therefore, combining both:m < (2001/n)*k < (2001/n)*(n/2002)*(m+1 )= (2001/2002)*(m+1 )So we have:m < (2001/2002)*(m+1 )Multiply both sides by2002:2002*m <2001*(m+1 )2002m <2001m +20012002m -2001m <2001m <2001Which is true since m <2001. Therefore, the inequalities are consistent.But this doesn't give us information about n.Alternatively, we can consider the problem as requiring that for each m, there exists k such that:k ∈ ( m*n/2001 , (m+1)*n/2002 )Which is equivalent to:There exists integer k where:m*n/2001 <k <(m+1)*n/2002This is equivalent to:The set of integers k must intersect each such interval for m=1 to2000.Another approach: For a fixed n, the intervals (m*n/2001, (m+1)*n/2002 ) for m=1 to2000 must each contain at least one integer.To find the smallest n such that all these intervals contain at least one integer.Perhaps we can model this as a lattice problem, where we need the line n = ... to intersect certain regions.Alternatively, perhaps think of the problem in terms of avoiding bad n's. For a given n, if there exists an m where the interval (m*n/2001, (m+1)*n/2002 ) contains no integer, then n is invalid. So we need the smallest n such that for all m, the interval contains an integer.This is similar to the problem of n being such that the Beatty sequence or some complementary sequence covers all m.Alternatively, we can think of the problem as a system of inequalities. For each m, there must exist k such that:k > m*n/2001andk < (m+1)*n/2002Which implies that:There exists k where:m*n/2001 <k < (m+1)*n/2002The number of integers in the interval is floor((m+1)*n/2002 ) - ceil(m*n/2001 ) +1. We need this to be at least1 for all m.But how to translate this into a condition on n?Perhaps for each m, floor((m+1)*n/2002 ) - ceil(m*n/2001 ) >=0Which would mean that there's at least one integer in the interval.Therefore, the condition is:floor((m+1)*n/2002 ) >= ceil(m*n/2001 )This must hold for all m=1,2,...,2000.To find the smallest n for which this holds.This seems like a problem that can be approached using the theory of diophantine approximations or the Chinese Remainder Theorem, but I'm not sure.Alternatively, perhaps we can model this as a linear programming problem. For each m, the difference between (m+1)*n/2002 and m*n/2001 must be greater than the difference between their ceilings and floors.But this is vague.Alternatively, consider the worst-case m, which would be where the interval (m*n/2001, (m+1)*n/2002 ) is least likely to contain an integer. The worst case would be when the interval is between two integers, i.e., when the fractional parts of m*n/2001 and (m+1)*n/2002 are such that the interval doesn't contain an integer.To prevent this, we need that for all m, the fractional part of m*n/2001 is not too close to1, and the fractional part of (m+1)*n/2002 is not too close to0, in such a way that the interval doesn't wrap around an integer.Alternatively, the key is to ensure that for each m, the interval overlaps with an integer. Therefore, the fractional part of m*n/2001 plus the length of the interval must exceed1. Wait, the length of the interval is ((m+1)/2002 -m/2001 )*n=(2001 -m )n/(2001*2002 ). The fractional part of m*n/2001 plus this length must be >=1.But I'm not sure.Alternatively, let's denote that for each m, there exists k such that:k > m*n/2001andk < (m+1)*n/2002Let’s denote x_m = m*n/2001 and y_m = (m+1)*n/2002. We need there to exist an integer k in (x_m, y_m ) for all m.The distance between x_m and y_m is y_m -x_m =n*(2001 -m )/(2001*2002 ). To ensure that this interval contains an integer, we need that the ceiling ofx_m is less than y_m, i.e., ceil(x_m ) <y_m.Therefore, for each m, ceil(m*n/2001 ) < (m+1)*n/2002.This is the key inequality.Let’s write this as:ceil(m*n/2001 ) < (m+1)*n/2002.Let’s denote q_m = floor(m*n/2001 ), then ceil(m*n/2001 )=q_m +1.Thus, the inequality becomes:q_m +1 < (m+1)*n/2002Multiply both sides by2002:2002*q_m +2002 < (m+1)*nBut q_m = floor(m*n/2001 ) >= m*n/2001 -1Therefore:2002*q_m +2002 >=2002*(m*n/2001 -1 ) +2002= (2002/2001)*m*n -2002 +2002= (2002/2001)*m*nThus:(2002/2001)*m*n < (m+1)*nDivide both sides byn (since n>0):(2002/2001)*m <m+1Multiply both sides by2001:2002*m <2001*(m+1 )2002m <2001m +2001m <2001Which is true since m <2001. Therefore, the inequality holds, but this doesn't give us a bound on n.Alternatively, this shows that the condition ceil(m*n/2001 ) < (m+1)*n/2002 is equivalent to m <2001, which is always true. Therefore, this approach isn't helpful.Alternatively, perhaps the problem requires that the inequality (m+1)*n/2002 -m*n/2001 >1 - (fractional part of m*n/2001 )But this seems too vague.Given that the answer choices are given, and after testing n=2003 and finding it invalid for m=2000, perhaps we need to find the minimal n such that for all m, ceil(m*n/2001 ) < (m+1)*n/2002.To find such n, perhaps we can set up the inequality:(m+1)*n/2002 -m*n/2001 >=1Which would imply that the interval has length >=1, ensuring an integer inside. However, we saw that this requires n >=4,006,002, which is not among the options. Therefore, this approach is invalid.Alternatively, perhaps the minimal n is the minimal n for which the inequality (m+1)*n/2002 -m*n/2001 >1 - epsilon, where epsilon is the minimal fractional part required. But this is unclear.Given that this problem is multiple choice, perhaps the correct answer is4003 (option D) because it is the next year after2001 and2002, but our earlier calculation showed that n=4003 doesn't work for m=2000. However, this might be an error in calculation.Wait, let's recheck for n=4003 and m=2000:Lower bound:2000*4003/2001= (2000*4003)/2001= (2000/2001)*4003=4003*(1 -1/2001)≈4003 -2.001≈4000.999Upper bound:2001*4003/2002= (2001*4003)/2002= (2001/2002)*4003≈4003*(1 -1/2002)≈4003 -2.001≈4000.999Wait, both bounds are approximately4000.999, so there's no integer k in between. Therefore, n=4003 doesn't work. This suggests that none of the options work, but the problem must have a correct answer among the choices. Therefore, there must be a flaw in our reasoning.Alternatively, perhaps our initial approach is incorrect, and there is a different way to model the problem.Let's consider that for each m, there must exist k such that:m/2001 <k/n < (m+1)/2002This is equivalent to:k must be in the interval ( m*n/2001, (m+1)*n/2002 )For this interval to contain at least one integer, the length of the interval must be >=1 - the fractional part of m*n/2001. Wait, not sure.Alternatively, the interval must contain an integer, so the fractional parts must overlap.Suppose we denote the fractional part of m*n/2001 as {m*n/2001}=m*n/2001 -floor(m*n/2001 ). Then, the interval (m*n/2001, (m+1)*n/2002 ) can be written as (q + f, q + f + Δ ), where q=floor(m*n/2001 ) and f={m*n/2001 }, and Δ=( (m+1)*n/2002 -m*n/2001 )For this interval to contain an integer, the next integer q+1 must lie within the interval. Therefore, q +1 must be less than (m+1)*n/2002. So:q +1 < (m+1)*n/2002But q=floor(m*n/2001 ), so q <=m*n/2001 <q +1Therefore, the condition becomes:m*n/2001 <q +1 < (m+1)*n/2002Which implies:q +1 < (m+1)*n/2002But q +1 >m*n/2001So, combining:m*n/2001 <q +1 < (m+1)*n/2002Which implies:m*n/2001 < (m+1)*n/2002Subtract m*n/2001 from both sides:0 < (m+1)*n/2002 -m*n/2001Which we already know is true.But how to ensure that q +1 < (m+1)*n/2002.Given that q=floor(m*n/2001 ), then q +1=ceil(m*n/2001 )So the condition is:ceil(m*n/2001 ) < (m+1)*n/2002Which we can rewrite as:ceil(m*n/2001 ) < (m+1)*n/2002This is the same condition as before.To find the minimal n such that for all m=1,2,...,2000:ceil(m*n/2001 ) < (m+1)*n/2002This seems to be the key inequality.Let’s try to find n such that this inequality holds for all m.Rearranging the inequality:ceil(m*n/2001 ) < (m+1)*n/2002This is equivalent to:m*n/2001 < (m+1)*n/2002 -1Wait, no. Because ceil(a) < b implies a < b and ceil(a) is the smallest integer greater than a. So ceil(a) < b => a < ceil(a) < b => a < b -1. Because if ceil(a) < b, then a < ceil(a) <= b -1. Therefore, ceil(a) < b => a < b -1.Therefore, the inequality ceil(m*n/2001 ) < (m+1)*n/2002 is equivalent to:m*n/2001 < (m+1)*n/2002 -1Therefore:m*n/2001 +1 < (m+1)*n/2002Multiply both sides by2001*2002:m*n*2002 +2001*2002 < (m+1)*n*2001Rearranged:m*n*2002 - (m+1)*n*2001 < -2001*2002Factor n:n*(m*2002 - (m+1)*2001 ) < -2001*2002As before, compute the coefficient:m*2002 - (m+1)*2001=2002m -2001m -2001= m -2001Therefore:n*(m -2001 ) < -2001*2002Since m -2001 <0 for m <2001, we can divide both sides by (m -2001 ), reversing the inequality:n > (2001*2002 )/(2001 -m )So for each m, n > (2001*2002 )/(2001 -m )The maximal value of the right-hand side occurs when m is as large as possible. The largest m is2000, so:n >2001*2002 / (2001 -2000 )=2001*2002 /1=2001*2002=4,006,002Therefore, n must be greater than4,006,002. But this contradicts the answer choices. Therefore, our approach must be incorrect.This suggests that the condition ceil(m*n/2001 ) < (m+1)*n/2002 is too strict, and that even if n is smaller than4,006,002, there might be values of m for which the interval contains an integer due to the alignment of the fractions.Given that this approach leads to a required n of over4 million, which is not among the answer choices, we must have made a wrong assumption.Let’s think differently. Suppose n is such that the denominators2001 and2002 divide n in some way. For instance, suppose n is a multiple of2001 and2002, but since they are coprime, n would need to be a multiple of2001*2002=4,006,002, which is again too large.Alternatively, perhaps consider the problem as a system of congruences. For each m, we need:k ≡ something mod n, but this seems vague.Alternatively, think of the problem in terms of the spacing between consecutive terms of the form k/n. The spacing is1/n. The interval (m/2001, (m+1)/2002 ) has length1/(2001*2002 ). To ensure that this interval contains a multiple of1/n, we need that1/n <1/(2001*2002 ), which again impliesn >2001*2002.But since the answer choices are smaller, this suggests that the problem requires a different interpretation.Perhaps the question allows for the same k to be used for multiple m's, but the problem states "for each integer m", so each m must have its own k.Wait, but the problem says "there exists an integer k". So for each m, there exists some k (dependent on m) such that the inequality holds.Therefore, the k can vary with m.This means that for each m, the interval (m/2001, (m+1)/2002 ) must contain at least one fraction k/n. So we need the set {k/n |k∈ℤ} to intersect every interval (m/2001, (m+1)/2002 ) for m=1,...,2000.This is equivalent to saying that the grid of fractions k/n must have a point in each of these intervals.The minimal such n is known in mathematics as the minimal number such that the Farey sequence of order n covers all these intervals.Unfortunately, there isn't a straightforward formula for this, but based on some research, this problem is related to the so-called "covering problem" in number theory, where you want a set of fractions to cover a set of intervals.Given that this is a multiple-choice question, and the options are2003,2008,3106,4003.Given that our initial approach with n=4003 didn't work for m=2000, but perhaps there was an error in the calculation.Wait, let's recompute n=4003 for m=2000:Lower bound: m*n/2001 =2000*4003/2001= (2000/2001)*4003=4003 -4003/2001≈4003 -2.001≈4000.999Upper bound: (m+1)*n/2002=2001*4003/2002= (2001/2002)*4003=4003 -4003/2002≈4003 -2.001≈4000.999Wait, so both lower and upper bounds are approximately4000.999, so there is no integer k between them. Therefore, n=4003 doesn't work. Therefore, none of the answer choices work? But the problem must have a solution.This suggests that there's a mistake in our understanding.Wait, perhaps the original inequality is m/2001 < k/n < (m+1)/2002, but when m=2000, the upper bound is2001/2002≈0.9995, and n=4003 would have k=4003*0.9995≈4001. Therefore, k=4001 and check4001/4003≈0.9995, which is indeed between2000/2001≈0.9995 and2001/2002≈0.9995. Wait, but numerically:Compute2000/2001≈0.999500499...Compute4001/4003≈0.999500624...Compute2001/2002≈0.999500499...Thus,4001/4003≈0.999500624 is greater than2001/2002≈0.999500499, so the inequality4001/4003 <2001/2002 is false. Therefore, k=4001 does not satisfy the upper bound for m=2000 when n=4003.Therefore, n=4003 is invalid. But option D is4003. This suggests that there's a mistake in the problem statement or the answer choices.Alternatively, perhaps the problem statement allows m=0, but it says0 <m <2001, so m=1 to2000.Alternatively, perhaps the question had a typo and the upper bound is(m+1)/2001 instead of2002. But the user provided the question as-is.Alternatively, maybe the answer is D.4003, despite the calculation showing otherwise, due to a different reasoning.Another approach: For n=4003, the fractions k/n can be written as (2m+1)/4003 for m=1 to2000. Let's check if this lies between m/2001 and (m+1)/2002.Compute (2m+1)/4003 - m/2001:= (2m+1)*2001 -m*4003 / (4003*2001 )= (4002m +2001 -4003m ) / denominator= (-m +2001 ) / denominator= (2001 -m ) / (4003*2001 )Which is positive since m <2001. Therefore, (2m+1)/4003 >m/2001.Now, compute (m+1)/2002 - (2m+1)/4003:= (m+1)*4003 - (2m+1)*2002 / (2002*4003 )= (4003m +4003 -4004m -2002 ) / denominator= (-m +2001 ) / denominator= (2001 -m ) / (2002*4003 )Also positive since m <2001. Therefore, (2m+1)/4003 < (m+1)/2002.Therefore, for n=4003, choosing k=2m+1 gives a fraction between m/2001 and (m+1)/2002 for all m=1 to2000. Therefore, n=4003 works.But earlier computation for m=2000:k=2*2000 +1=4001. So check4001/4003≈0.999500624.Compute m/2001=2000/2001≈0.999500499.Compute (m+1)/2002=2001/2002≈0.999500499.But4001/4003≈0.999500624 >0.999500499. Therefore, the upper bound is violated. However, according to the algebra above, (2m+1)/4003 < (m+1)/2002.But numerically, for m=2000:Left side:4001/4003≈0.999500624Right side:2001/2002≈0.999500499Thus,4001/4003 >2001/2002, which contradicts the algebraic result.Wait, this suggests a mistake in the algebraic manipulation.Let me recompute:(m+1)/2002 - (2m+1)/4003= [(m+1)*4003 - (2m+1)*2002 ] / (2002*4003 )= [4003m +4003 -4004m -2002 ] / denominator= [ -m +2001 ] / denominatorFor m=2000, this is ( -2000 +2001 ) / denominator=1/(2002*4003 )>0Therefore, (m+1)/2002 - (2m+1)/4003=1/(2002*4003 )>0. Therefore, (2m+1)/4003 < (m+1)/2002.But numerically, for m=2000:Left side=4001/4003≈0.999500624Right side=2001/2002≈0.999500499But0.999500624 >0.999500499, so the inequality is not holding. This contradicts the algebraic result. What's the issue?Wait, the calculation:(m+1)*4003 - (2m+1)*2002= (2000+1)*4003 - (2*2000 +1)*2002=2001*4003 -4001*2002Compute this exactly:2001*4003=2001*(4000+3)=2001*4000 +2001*3=8,004,000 +6,003=8,010,0034001*2002=4001*(2000+2)=4001*2000 +4001*2=8,002,000 +8,002=8,010,002Therefore,2001*4003 -4001*2002=8,010,003 -8,010,002=1Therefore, (m+1)/2002 - (2m+1)/4003=1/(2002*4003 )Therefore, (2m+1)/4003= (m+1)/2002 -1/(2002*4003 )Therefore, (2m+1)/4003 is less than (m+1)/2002 by a very small amount.But numerically, for m=2000:(2m+1)/4003=4001/4003≈0.999500624(m+1)/2002=2001/2002≈0.999500499But0.999500624 is greater than0.999500499, which contradicts the algebra. However, the algebraic result shows that:(2m+1)/4003= (m+1)/2002 -1/(2002*4003 )Therefore, (2m+1)/4003=0.999500499 -1/(2002*4003 )But wait, no:(m+1)/2002=2001/2002≈0.999500499Subtract1/(2002*4003 )≈1/(8,016,006 )≈0.0000001247Therefore, (2m+1)/4003≈0.999500499 -0.0000001247≈0.9995003743But4001/4003≈0.999500624. This contradicts. Therefore, there's a miscalculation.Wait, no, according to the exact computation:2001*4003 -4001*2002=1Therefore, (m+1)/2002 - (2m+1)/4003=1/(2002*4003 )Therefore, (2m+1)/4003= (m+1)/2002 -1/(2002*4003 )Therefore, numerically:(m+1)/2002=2001/2002≈0.999500499Subtract1/(2002*4003 )≈0.0000001247Gives≈0.9995003743But4001/4003= (4003 -2)/4003=1 -2/4003≈1 -0.00049975=0.99950025Wait, this is conflicting. Compute4001/4003:4001/4003=1 -2/4003≈1 -0.0004997502=0.9995002498Wait, that's approximately0.99950025, which is less than0.9995003743. Therefore, there's a discrepancy.Wait, according to the exact fractions:4001/4003=4001/4003≈0.9995002498(m+1)/2002 -1/(2002*4003 )=2001/2002 -1/(2002*4003 )= (2001*4003 -1 )/(2002*4003 )= (8,010,003 -1 )/(8,016,006 )=8,010,002/8,016,006≈0.9995002498Which matches4001/4003≈0.9995002498. Therefore, the correct value of(2m+1)/4003 is≈0.9995002498, which is less than(m+1)/2002≈0.999500499. Therefore, the previous numerical calculation was incorrect due to approximation errors. Therefore,4001/4003≈0.99950025 is indeed less than2001/2002≈0.9995005.Therefore, for m=2000, k=4001/4003≈0.99950025 lies between2000/2001≈0.999500499 and2001/2002≈0.999500499. Wait, no,2000/2001≈0.999500499 is approximately0.9995005, and4001/4003≈0.99950025 is less than0.9995005. Therefore,4001/4003 <2000/2001, which contradicts.Wait, compute2000/2001≈0.9995004994001/4003≈0.9995002498Therefore,4001/4003 <2000/2001, which means that k=4001/4003 is less than m/2001=2000/2001≈0.9995005, which violates the lower bound.But according to the algebra, (2m+1)/4003= (m+1)/2002 -1/(2002*4003 )For m=2000:(m+1)/2002=2001/2002≈0.999500499Subtract1/(2002*4003 )≈0.0000001247Gives≈0.9995003743But4001/4003≈0.9995002498, which is different.Wait, there's a mistake in the algebraic manipulation.Let's redo the algebra:We claimed that for k=2m+1, (2m+1)/4003 lies between m/2001 and (m+1)/2002.But the algebra shows:(m+1)/2002 - (2m+1)/4003=1/(2002*4003 )Therefore, (2m+1)/4003= (m+1)/2002 -1/(2002*4003 )Similarly, compute (2m+1)/4003 -m/2001:= [ (m+1)/2002 -1/(2002*4003 ) ] -m/2001= (m+1)/2002 -m/2001 -1/(2002*4003 )= [ (m+1)*2001 -m*2002 ] / (2001*2002 ) -1/(2002*4003 )= [2001m +2001 -2002m ] / (2001*2002 ) -1/(2002*4003 )= ( -m +2001 ) / (2001*2002 ) -1/(2002*4003 )= (2001 -m )/(2001*2002 ) -1/(2002*4003 )For this to be positive, we need:(2001 -m )/(2001*2002 ) >1/(2002*4003 )Multiply both sides by2002*4003*2001:(2001 -m )*4003 >2001Which is true for m <2001 -2001/4003≈2001 -0.5≈2000.5Since m is an integer up to2000, this holds for all m=1,2,...,2000.Therefore, (2m+1)/4003 -m/2001>0 => (2m+1)/4003 >m/2001.Therefore, for all m=1,2,...,2000, (2m+1)/4003 lies between m/2001 and (m+1)/2002.Therefore, despite the previous numerical approximation error, the exact values show that k=2m+1 satisfies the inequality for n=4003.The confusion arose due to the very small differences between the fractions, which made manual approximations misleading. Therefore, n=4003 is indeed a valid solution, and the correct answer is D.4003.

✅:The problem requires finding the smallest positive integer ( n ) such that for every integer ( m ) with ( 0 < m < 2001 ), there exists an integer ( k ) satisfying:[frac{m}{2001} < frac{k}{n} < frac{m+1}{2002}.]Key Steps:1. Interval Analysis: For each ( m ), the interval (left( frac{m}{2001}, frac{m+1}{2002} right)) must contain at least one fraction (frac{k}{n}). This means the interval's length must be sufficient to ensure an integer ( k ) exists within it.2. Mediant Fraction Approach: The mediant of two fractions (frac{a}{b}) and (frac{c}{d}) is (frac{a+c}{b+d}). For ( frac{m}{2001} ) and ( frac{m+1}{2002} ), the mediant is ( frac{2m+1}{4003} ). This mediant lies strictly between the two fractions: [ frac{m}{2001} < frac{2m+1}{4003} < frac{m+1}{2002}. ]3. Verification: For ( n = 4003 ), choosing ( k = 2m + 1 ) ensures the fraction ( frac{2m+1}{4003} ) satisfies the inequality for all ( m ). Algebraic verification confirms that: [ frac{2m+1}{4003} - frac{m}{2001} > 0 quad text{and} quad frac{m+1}{2002} - frac{2m+1}{4003} > 0, ] proving the fraction lies within the interval.4. Conclusion: Smaller values of ( n ) (like 2003, 2008, 3106) fail to cover all intervals, especially for ( m ) near 2000. The mediant approach with ( n = 4003 ) ensures coverage for all ( m ).Answer: (boxed{D})

🔑: Problem:Given that for every integer ( m ) which satisfies ( 0 < m < 2001 ), there exists an integer ( k ) such that [frac{m}{2001} < frac{k}{n} < frac{m+1}{2002}]where ( n ) is a positive integer, the smallest possible value of ( n ) is ( boxed{4003} ).1. Let's first understand the given inequality: - Given: ( frac{m}{2001} < frac{k}{n} < frac{m+1}{2002} ) - We need to find the minimum possible value of ( n ).2. Consider the inequality ( frac{m}{2001} < frac{k}{n} < frac{m+1}{2002} ).3. To ensure ( n ) is minimized, let us consider specific values of ( m ).4. We start examining with the end cases: let ( m = 2000 ). - Then the given inequality becomes: [ frac{2000}{2001} < frac{k}{n} < frac{2001}{2002} ]5. Now, we'll derive some insights from properties of fractions.6. For any ( x > 0 ), the function ( frac{x}{x+1} ) is strictly increasing. So: [ frac{2000}{2001} < frac{2001}{2002} ]7. It’s clear that the middle term ( frac{k}{n} ) must fit between those two fractions.8. To solve this, start by checking the boundary constraints: - As ( n to 2002 ), ( frac{n-2}{n} ) approaches ( frac{2000}{2002} ), which is slightly less than ( frac{2000}{2001} ) - Therefore, for ( frac{2000}{2001} < frac{n-2}{n} ): [ frac{2000}{2001} < frac{n-2}{n} implies 2000n < 2001(n-2) implies 2000n < 2001n - 4002 ] [ 0 < 2001 - 4002 implies 4002 < n implies n > 4002 ] - The smallest ( n ) greater than 4002 is 4003.9. Thus, combining the conditions from both the checks, we see that ( n = 4003 ) is the smallest integer that fulfills the inequality for all ( 0 < m < 2001 ). Conclusion:The smallest possible value for ( n ) is (boxed{4003}).