💡First, I need to determine the probabilities of each student attending university in either city A or city B.Given:- The probability of student A going to city A is 0.6. Therefore, the probability of student A going to city B is 1 - 0.6 = 0.4.- The probability of student B going to city A is 0.2. Therefore, the probability of student B going to city B is 1 - 0.2 = 0.8.Next, I need to find the probability that students A and B do not attend university in the same city. This can happen in two scenarios:1. Student A goes to city A and student B goes to city B.2. Student A goes to city B and student B goes to city A.Calculating the probabilities for each scenario:1. Probability of student A going to city A and student B going to city B: 0.6 * 0.8 = 0.482. Probability of student A going to city B and student B going to city A: 0.4 * 0.2 = 0.08Adding these probabilities together gives the total probability that students A and B do not attend university in the same city:0.48 + 0.08 = 0.56Therefore, the probability that students A and B do not attend university in the same city is 0.56.

💡Okay, so I have this problem here that I need to solve. It involves a sum with binomial coefficients and some fractions. Let me try to understand what's going on here.The problem is asking me to calculate the following expression:[ C_{n}^{0} times dfrac{1}{2} + dfrac{1}{2} C_{n}^{1} times left( dfrac{1}{2} right)^{2} + dfrac{1}{3} C_{n}^{2} times left( dfrac{1}{2} right)^{3} + dots + dfrac{1}{n+1} C_{n}^{n} times left( dfrac{1}{2} right)^{n+1} ]Hmm, okay. So it's a sum where each term involves a binomial coefficient ( C_{n}^{k} ), multiplied by ( dfrac{1}{k+1} ) and ( left( dfrac{1}{2} right)^{k+1} ). The sum goes from ( k = 0 ) to ( k = n ).Looking back at the materials provided, there was a similar approach where they started with the geometric series and integrated both sides. Maybe I can use a similar idea here. Let me recall that the binomial theorem states:[ (1 + x)^{n} = C_{n}^{0} + C_{n}^{1}x + C_{n}^{2}x^{2} + dots + C_{n}^{n}x^{n} ]Yes, that's right. So if I can express the given sum in terms of an integral of ( (1 + x)^{n} ), that might help.Let me try integrating both sides of the binomial theorem. If I integrate from 0 to ( dfrac{1}{2} ), I get:[ int_{0}^{frac{1}{2}} (1 + x)^{n} , dx = int_{0}^{frac{1}{2}} left( C_{n}^{0} + C_{n}^{1}x + C_{n}^{2}x^{2} + dots + C_{n}^{n}x^{n} right) , dx ]Integrating term by term on the right side:[ int_{0}^{frac{1}{2}} C_{n}^{0} , dx + int_{0}^{frac{1}{2}} C_{n}^{1}x , dx + int_{0}^{frac{1}{2}} C_{n}^{2}x^{2} , dx + dots + int_{0}^{frac{1}{2}} C_{n}^{n}x^{n} , dx ]Which simplifies to:[ C_{n}^{0} times dfrac{1}{2} + dfrac{1}{2} C_{n}^{1} times left( dfrac{1}{2} right)^{2} + dfrac{1}{3} C_{n}^{2} times left( dfrac{1}{2} right)^{3} + dots + dfrac{1}{n+1} C_{n}^{n} times left( dfrac{1}{2} right)^{n+1} ]Wait, that's exactly the expression we're supposed to calculate! So, the left side of the equation is the integral of ( (1 + x)^{n} ) from 0 to ( dfrac{1}{2} ). Let me compute that integral.The integral of ( (1 + x)^{n} ) with respect to x is:[ int (1 + x)^{n} , dx = dfrac{(1 + x)^{n+1}}{n+1} + C ]So, evaluating from 0 to ( dfrac{1}{2} ):[ left[ dfrac{(1 + x)^{n+1}}{n+1} right]_{0}^{frac{1}{2}} = dfrac{(1 + frac{1}{2})^{n+1}}{n+1} - dfrac{(1 + 0)^{n+1}}{n+1} ]Simplifying:[ dfrac{left( dfrac{3}{2} right)^{n+1}}{n+1} - dfrac{1}{n+1} = dfrac{1}{n+1} left( left( dfrac{3}{2} right)^{n+1} - 1 right) ]So, putting it all together, the sum we're trying to calculate is equal to:[ dfrac{1}{n+1} left( left( dfrac{3}{2} right)^{n+1} - 1 right) ]Let me double-check my steps to make sure I didn't make a mistake. Starting with the binomial theorem, integrating both sides from 0 to 1/2, computing the integral on the left, and then evaluating the right side term by term. It all seems to fit together.I think that's the correct answer. It makes sense because integrating the binomial expansion term by term gives us the sum we're looking for, and the integral of ( (1 + x)^n ) is straightforward.**Final Answer**boxed{dfrac{1}{n+1}left[left( dfrac{3}{2}right)^{n+1}-1right]}

💡Alright, so I've got this problem about buying cultural shirts for an event. There are two brands, A and B. The organizers want to figure out the unit prices for each brand and then come up with the most cost-effective way to buy 1000 shirts with certain conditions. Let me try to break this down step by step.First, part (1) asks for the unit prices of brand A and brand B shirts. They give me two scenarios:1. Buying 3 of brand A and 2 of brand B costs 190.2. Buying 5 of brand A and 1 of brand B costs 235.Okay, so I need to find out how much each shirt costs individually. This sounds like a system of equations problem. Let me assign variables:Let x be the price of one brand A shirt.Let y be the price of one brand B shirt.From the first scenario: 3x + 2y = 190From the second scenario: 5x + y = 235So, I have two equations:1. 3x + 2y = 1902. 5x + y = 235I need to solve this system of equations to find x and y. There are different methods to solve this: substitution, elimination, etc. Maybe elimination is easier here.If I can manipulate the equations to eliminate one variable, that would work. Let's try to eliminate y.From the second equation: 5x + y = 235, I can solve for y:y = 235 - 5xNow, plug this expression for y into the first equation:3x + 2(235 - 5x) = 190Let me simplify this:3x + 470 - 10x = 190Combine like terms:-7x + 470 = 190Now, subtract 470 from both sides:-7x = 190 - 470-7x = -280Divide both sides by -7:x = (-280)/(-7)x = 40So, the price of brand A shirt is 40.Now, plug x = 40 back into the expression for y:y = 235 - 5(40)y = 235 - 200y = 35So, the price of brand B shirt is 35.Alright, that seems straightforward. Let me double-check my work to make sure I didn't make any mistakes.Plugging x = 40 and y = 35 into the first equation:3(40) + 2(35) = 120 + 70 = 190. That's correct.Plugging into the second equation:5(40) + 35 = 200 + 35 = 235. Also correct.Great, so part (1) is solved. Brand A shirts are 40 each, and brand B shirts are 35 each.Now, moving on to part (2). The organizers need to buy a total of 1000 shirts, with the number of brand A shirts not less than three times the number of brand B shirts. They want the most cost-effective plan.Let me parse this condition: "the number of brand A cultural shirts not less than three times the number of brand B cultural shirts."So, if I let m be the number of brand A shirts, and n be the number of brand B shirts, then:m ≥ 3nAnd we know that m + n = 1000So, substituting n from the second equation into the first:m ≥ 3(1000 - m)Let me solve this inequality:m ≥ 3000 - 3mAdd 3m to both sides:4m ≥ 3000Divide both sides by 4:m ≥ 750So, the number of brand A shirts must be at least 750, and consequently, the number of brand B shirts will be at most 250 (since 1000 - 750 = 250).Now, to find the most cost-effective plan, I need to minimize the total cost. The total cost will be:Total Cost = 40m + 35nBut since n = 1000 - m, substitute:Total Cost = 40m + 35(1000 - m)= 40m + 35000 - 35m= 5m + 35000So, the total cost is a linear function in terms of m: 5m + 35000.Since the coefficient of m is positive (5), the total cost increases as m increases. Therefore, to minimize the total cost, we need to minimize m.But from the condition above, m must be at least 750. So, the smallest possible m is 750, which will give us the minimum total cost.Let me calculate the total cost when m = 750:Total Cost = 5(750) + 35000= 3750 + 35000= 38750So, the total cost is 38,750 when buying 750 brand A shirts and 250 brand B shirts.Just to make sure, if I tried to buy more than 750 brand A shirts, say 751, the total cost would be:Total Cost = 5(751) + 35000= 3755 + 35000= 38755Which is more expensive. Similarly, buying fewer than 750 would violate the condition, so 750 is indeed the minimum number of brand A shirts we can buy to satisfy the requirement.Therefore, the most cost-effective plan is to buy 750 brand A shirts and 250 brand B shirts, costing a total of 38,750.I think that covers both parts of the problem. I've set up the equations correctly, solved for the unit prices, applied the constraints for the total number of shirts, and found the minimum cost by analyzing the linear cost function. Everything seems to check out.

💡First, I recognize that the amount of work done in both scenarios is the same, which is reaping the entire field. The work done is directly proportional to the number of men and the number of hours they work each day.In the first scenario, the work done can be represented as:[ W = M times 9 , text{hours/day} times 16 , text{days} ]In the second scenario, the work done is:[ W = 18 , text{men} times 8 , text{hours/day} times 15 , text{days} ]Since the work done in both scenarios is equal, I set the two equations equal to each other:[ M times 9 times 16 = 18 times 8 times 15 ]To find the value of ( M ), I solve the equation:[ M = frac{18 times 8 times 15}{9 times 16} ]Simplifying the equation by dividing both the numerator and the denominator by 9:[ M = frac{2 times 8 times 15}{16} ]Further simplifying by dividing both the numerator and the denominator by 16:[ M = frac{2 times 15}{2} ]Finally, calculating the value:[ M = 15 ]Therefore, there were 15 men working in the first scenario.

💡First, I need to determine how many brothers Aaron has based on the information provided.I'll start by letting ( A ) represent the number of Aaron's brothers.The problem states that Bennett has two less than twice the number of Aaron's brothers. This can be expressed as:[text{Bennett's brothers} = 2A - 2]It's given that Bennett has 6 brothers. So, I can set up the equation:[6 = 2A - 2]To solve for ( A ), I'll first add 2 to both sides of the equation:[6 + 2 = 2A][8 = 2A]Next, I'll divide both sides by 2 to find the value of ( A ):[A = frac{8}{2} = 4]Therefore, Aaron has 4 brothers.

💡Okay, so I have this geometry problem here about triangle ABC with circumcenter O. Points P and Q are on sides AC and AB respectively. Then there are midpoints K, L, and M of segments BP, CQ, and PQ. There's a circle Γ passing through K, L, and M, and PQ is tangent to Γ. I need to show that OP equals OQ. Hmm, okay, let me try to visualize this.First, let me sketch the triangle ABC with O as the circumcenter. So, O is the intersection point of the perpendicular bisectors of the sides. Points P and Q are somewhere on AC and AB. Then, K is the midpoint of BP, L is the midpoint of CQ, and M is the midpoint of PQ. So, K, L, M are midpoints, which might mean some midline properties come into play.Now, the circle Γ passes through K, L, and M. And PQ is tangent to Γ. Tangent implies that PQ touches Γ at exactly one point, which is probably M since M is the midpoint of PQ. So, maybe PQ is tangent at M? That might be useful.I remember that if a line is tangent to a circle at a point, then the radius at that point is perpendicular to the tangent line. So, if PQ is tangent to Γ at M, then the radius of Γ at M is perpendicular to PQ. But Γ passes through K, L, and M, so the center of Γ must lie somewhere such that it's equidistant from K, L, and M. Hmm, maybe I can find the center of Γ?Alternatively, maybe I can use some properties of midpoints and circles. Since K and L are midpoints of BP and CQ, perhaps I can relate them to midlines or something. Also, M is the midpoint of PQ, so maybe triangle KLM has some special properties.Wait, if PQ is tangent to Γ at M, then by the power of a point, the power of point M with respect to Γ is zero. But since M is on Γ, that makes sense. Maybe I can use power of a point for other points as well.Alternatively, maybe I can use homothety or similarity. Since K, L, M are midpoints, there might be some similar triangles involved. Let me think about the midsegments. In triangle BPQ, K is the midpoint of BP and M is the midpoint of PQ, so KM is a midsegment, which should be parallel to BQ and half its length. Similarly, in triangle CQP, L is the midpoint of CQ and M is the midpoint of PQ, so LM is a midsegment parallel to CP and half its length.So, KM is parallel to BQ and LM is parallel to CP. That might help in establishing some angle relationships or similar triangles.Since Γ passes through K, L, M, and PQ is tangent at M, the tangent condition gives us that the angle between PQ and the tangent line at M is equal to the angle in the alternate segment. Wait, that's the alternate segment theorem. So, the angle between tangent PQ and chord KM is equal to the angle that KM makes with the other chord in the alternate segment.But I'm not sure if that's directly applicable here. Maybe I can relate the angles at M in triangle KLM to the angles in triangle ABC or something.Alternatively, since K, L, M are midpoints, maybe I can consider vectors or coordinate geometry. But that might get complicated. Let me try to stick with synthetic geometry for now.Let me consider the midpoints again. Since K is the midpoint of BP, L is the midpoint of CQ, and M is the midpoint of PQ, perhaps I can express the coordinates of K, L, M in terms of coordinates of A, B, C, P, Q. Maybe coordinate geometry could help here.Wait, but the problem is about OP and OQ being equal. So, maybe I need to relate the distances from O to P and Q. Since O is the circumcenter, OP and OQ are the distances from O to points on the circumcircle. But P and Q are not necessarily on the circumcircle, they are on sides AC and AB.Hmm, so maybe I can use the power of point P and Q with respect to the circumcircle. The power of a point formula says that for a point P outside a circle, the power is OP² - R², where R is the radius. But if P is inside the circle, the power is negative. Wait, but in this case, P and Q are on the sides, so they might be inside the circumcircle.But how does that relate to the circle Γ? Maybe the power of P and Q with respect to Γ can be connected to their power with respect to the circumcircle.Alternatively, since K, L, M are midpoints, maybe I can use the nine-point circle. The nine-point circle passes through the midpoints of the sides, the feet of the altitudes, and the midpoints of the segments from each vertex to the orthocenter. But in this case, K, L, M are midpoints of BP, CQ, and PQ, which are not necessarily the sides of the triangle, so maybe it's not directly the nine-point circle.Wait, but maybe Γ is similar to the nine-point circle in some way. Or perhaps it's a different circle altogether.Let me think about the tangent condition. Since PQ is tangent to Γ at M, then the tangent at M is PQ. So, the radius of Γ at M is perpendicular to PQ. So, if I can find the center of Γ, then the line from the center to M is perpendicular to PQ.But to find the center of Γ, I need to find the intersection of the perpendicular bisectors of KM and LM. Maybe that's too involved. Alternatively, maybe I can use some properties of midpoints and circles.Wait, another approach: since K, L, M are midpoints, maybe I can use vectors. Let me assign coordinates to the points. Let me place point A at (0,0), B at (2b,0), and C at (2c,2d). Then, points P and Q can be parameterized. Let me say P is on AC, so P can be (2c*t, 2d*t) for some t between 0 and 1. Similarly, Q is on AB, so Q can be (2b*s, 0) for some s between 0 and 1.Then, midpoints K, L, M can be calculated. K is midpoint of BP: so coordinates of B are (2b,0), P is (2c*t, 2d*t). So, midpoint K is ((2b + 2c*t)/2, (0 + 2d*t)/2) = (b + c*t, d*t). Similarly, L is midpoint of CQ: C is (2c,2d), Q is (2b*s,0). So, midpoint L is ((2c + 2b*s)/2, (2d + 0)/2) = (c + b*s, d). M is midpoint of PQ: P is (2c*t, 2d*t), Q is (2b*s,0). So, midpoint M is ((2c*t + 2b*s)/2, (2d*t + 0)/2) = (c*t + b*s, d*t).Now, circle Γ passes through K, L, M. So, I can write the equation of the circle passing through these three points. Let me denote the general equation of a circle as x² + y² + 2gx + 2fy + c = 0. Plugging in the coordinates of K, L, M:For K: (b + c*t)² + (d*t)² + 2g(b + c*t) + 2f(d*t) + c = 0For L: (c + b*s)² + d² + 2g(c + b*s) + 2f*d + c = 0For M: (c*t + b*s)² + (d*t)² + 2g(c*t + b*s) + 2f(d*t) + c = 0This seems complicated, but maybe I can subtract equations to eliminate some variables. Let me subtract the equation for K from the equation for M:[(c*t + b*s)² + (d*t)² + 2g(c*t + b*s) + 2f(d*t) + c] - [(b + c*t)² + (d*t)² + 2g(b + c*t) + 2f(d*t) + c] = 0Simplify:(c*t + b*s)² - (b + c*t)² + 2g(c*t + b*s - b - c*t) = 0Expanding the squares:(c²*t² + 2b*c*t*s + b²*s²) - (b² + 2b*c*t + c²*t²) + 2g(b*s - b) = 0Simplify:c²*t² + 2b*c*t*s + b²*s² - b² - 2b*c*t - c²*t² + 2g*b*(s - 1) = 0Cancel c²*t² terms:2b*c*t*s + b²*s² - b² - 2b*c*t + 2g*b*(s - 1) = 0Factor out b:b*(2c*t*s + b*s² - b - 2c*t) + 2g*b*(s - 1) = 0Divide both sides by b (assuming b ≠ 0):2c*t*s + b*s² - b - 2c*t + 2g*(s - 1) = 0Let me rearrange:(2c*t*s - 2c*t) + (b*s² - b) + 2g*(s - 1) = 0Factor:2c*t*(s - 1) + b*(s² - 1) + 2g*(s - 1) = 0Factor (s - 1):(s - 1)*(2c*t + b*(s + 1) + 2g) = 0So, either s = 1 or 2c*t + b*(s + 1) + 2g = 0If s = 1, then point Q would be at (2b,0), which is point B. But Q is on AB, so s=1 would mean Q=B. Similarly, if s=1, then M would be midpoint of PQ, which would be midpoint of PB. But I don't know if that's necessarily the case. Maybe s ≠1, so the other factor must be zero:2c*t + b*(s + 1) + 2g = 0So, 2g = -2c*t - b*(s + 1)Similarly, let me subtract the equation for L from the equation for K:[(b + c*t)² + (d*t)² + 2g(b + c*t) + 2f(d*t) + c] - [(c + b*s)² + d² + 2g(c + b*s) + 2f*d + c] = 0Simplify:(b + c*t)² - (c + b*s)² + 2g(b + c*t - c - b*s) + 2f(d*t - d) = 0Expanding the squares:(b² + 2b*c*t + c²*t²) - (c² + 2b*c*s + b²*s²) + 2g(b - c + c*t - b*s) + 2f*d*(t - 1) = 0Simplify:b² + 2b*c*t + c²*t² - c² - 2b*c*s - b²*s² + 2g*(b - c + c*t - b*s) + 2f*d*(t - 1) = 0Let me group like terms:(b² - b²*s²) + (2b*c*t - 2b*c*s) + (c²*t² - c²) + 2g*(b - c + c*t - b*s) + 2f*d*(t - 1) = 0Factor:b²*(1 - s²) + 2b*c*(t - s) + c²*(t² - 1) + 2g*(b - c + c*t - b*s) + 2f*d*(t - 1) = 0This is getting quite involved. Maybe I should try a different approach.Wait, since PQ is tangent to Γ at M, the condition for tangency is that the power of point M with respect to Γ is zero. But M is on Γ, so that's trivial. Maybe I can use the condition that the derivative at M is equal to the slope of PQ, but that might be calculus.Alternatively, maybe I can use the fact that the tangent at M is perpendicular to the radius. So, the line from the center of Γ to M is perpendicular to PQ. So, if I can find the center of Γ, then the line from center to M is perpendicular to PQ.But finding the center of Γ would require solving the perpendicular bisectors of KM and LM, which might be complicated with coordinates.Wait, maybe instead of coordinates, I can use vector methods. Let me denote vectors with position vectors from O. Let me denote vectors OA, OB, OC as vectors a, b, c respectively. Then, points P and Q can be expressed as points on AC and AB.Let me denote P as a point on AC, so P = A + t*(C - A) = (1 - t)A + tC. Similarly, Q = A + s*(B - A) = (1 - s)A + sB.Then, midpoints K, L, M can be expressed as:K = midpoint of BP = (B + P)/2 = (B + (1 - t)A + tC)/2Similarly, L = midpoint of CQ = (C + Q)/2 = (C + (1 - s)A + sB)/2M = midpoint of PQ = (P + Q)/2 = [(1 - t)A + tC + (1 - s)A + sB]/2 = [(2 - t - s)A + sB + tC]/2Now, since Γ passes through K, L, M, and PQ is tangent to Γ at M, the condition is that the tangent at M is PQ. So, the radius at M is perpendicular to PQ.So, the vector from center of Γ to M is perpendicular to vector PQ.Let me denote the center of Γ as O'. Then, vector O'M is perpendicular to vector PQ.Vector PQ = Q - P = [(1 - s)A + sB] - [(1 - t)A + tC] = (s - t)A + sB - tCVector O'M = M - O'But I don't know O', so this might not help directly.Alternatively, maybe I can use the property that the power of point M with respect to Γ is zero, but since M is on Γ, that's trivial.Wait, another approach: since K, L, M are midpoints, maybe I can consider the midline properties. For example, in triangle BPQ, K and M are midpoints, so KM is parallel to BQ. Similarly, in triangle CQP, L and M are midpoints, so LM is parallel to CP.So, KM || BQ and LM || CP.Since Γ passes through K, L, M, and PQ is tangent at M, the angles at M in Γ should relate to the angles in ABC.Wait, maybe I can use the fact that KM || BQ and LM || CP to establish some angle equalities.Since KM || BQ, angle KML is equal to angle BQP. Similarly, since LM || CP, angle KLM is equal to angle CPQ.But I'm not sure how that helps with the tangent condition.Wait, since PQ is tangent to Γ at M, then angle between PQ and KM is equal to angle KLM (alternate segment theorem). So, angle between PQ and KM is equal to angle KLM.But KM is parallel to BQ, so angle between PQ and KM is equal to angle between PQ and BQ, which is the same as angle BQP.Similarly, angle KLM is equal to angle KLM in triangle KLM.Wait, maybe I can relate these angles to establish some similarity or congruence.Alternatively, since KM || BQ and LM || CP, maybe triangles KLM and BCP are similar? Let me check.If KM || BQ and LM || CP, then angles at K and L correspond to angles at B and C. So, maybe triangle KLM is similar to triangle BCP.If that's the case, then the ratio of sides would be equal. So, KL / BC = KM / BP = LM / CP.But I'm not sure if that's directly helpful.Wait, another thought: since K, L, M are midpoints, the circle Γ is the nine-point circle of some triangle. But which triangle?Wait, the nine-point circle passes through midpoints of sides, feet of altitudes, and midpoints of segments from orthocenter. But here, K, L, M are midpoints of BP, CQ, and PQ, which are not necessarily sides of ABC.Alternatively, maybe Γ is the nine-point circle of triangle BPQ or something else.Wait, maybe I can consider triangle BPQ. Then, K is the midpoint of BP, M is the midpoint of PQ, so KM is a midline, which is parallel to BQ. Similarly, in triangle CQP, L is the midpoint of CQ, M is the midpoint of PQ, so LM is parallel to CP.So, in triangle BPQ, KM is a midline, and in triangle CQP, LM is a midline.But I'm not sure how that helps with the circle Γ.Wait, since PQ is tangent to Γ at M, and Γ passes through K and L, which are midpoints, maybe I can use some properties of tangents and midpoints.Alternatively, maybe I can use coordinate geometry again, but this time with specific coordinates. Let me try assigning coordinates to make calculations easier.Let me place point A at (0,0), B at (2,0), and C at (0,2). So, ABC is a right-angled triangle at A for simplicity. Then, O, the circumcenter, is at the midpoint of the hypotenuse BC, which is at (1,1).Now, points P and Q are on AC and AB respectively. Let me parameterize P as (0, p) where 0 ≤ p ≤ 2, and Q as (q, 0) where 0 ≤ q ≤ 2.Then, midpoints:K is midpoint of BP: B is (2,0), P is (0,p). So, K is ((2+0)/2, (0 + p)/2) = (1, p/2)L is midpoint of CQ: C is (0,2), Q is (q,0). So, L is ((0 + q)/2, (2 + 0)/2) = (q/2, 1)M is midpoint of PQ: P is (0,p), Q is (q,0). So, M is ((0 + q)/2, (p + 0)/2) = (q/2, p/2)Now, circle Γ passes through K(1, p/2), L(q/2, 1), and M(q/2, p/2). Let me find the equation of Γ.The general equation of a circle is x² + y² + Dx + Ey + F = 0. Plugging in the three points:For K(1, p/2):1² + (p/2)² + D*1 + E*(p/2) + F = 0 => 1 + p²/4 + D + (E p)/2 + F = 0 ...(1)For L(q/2, 1):(q/2)² + 1² + D*(q/2) + E*1 + F = 0 => q²/4 + 1 + (D q)/2 + E + F = 0 ...(2)For M(q/2, p/2):(q/2)² + (p/2)² + D*(q/2) + E*(p/2) + F = 0 => q²/4 + p²/4 + (D q)/2 + (E p)/2 + F = 0 ...(3)Now, let's subtract equation (3) from equation (1):[1 + p²/4 + D + (E p)/2 + F] - [q²/4 + p²/4 + (D q)/2 + (E p)/2 + F] = 0 - 0Simplify:1 + D - q²/4 - (D q)/2 = 0So,1 + D - (q²)/4 - (D q)/2 = 0Let me rearrange:D*(1 - q/2) = (q²)/4 - 1So,D = [(q²)/4 - 1] / (1 - q/2) = [ (q² - 4)/4 ] / [ (2 - q)/2 ] = [ (q² - 4)/4 ] * [ 2 / (2 - q) ] = (q² - 4)/(2*(2 - q)) = -(q² - 4)/(2*(q - 2)) = -(q - 2)(q + 2)/(2*(q - 2)) ) = -(q + 2)/2So, D = -(q + 2)/2Similarly, subtract equation (3) from equation (2):[ q²/4 + 1 + (D q)/2 + E + F ] - [ q²/4 + p²/4 + (D q)/2 + (E p)/2 + F ] = 0 - 0Simplify:1 + E - p²/4 - (E p)/2 = 0So,1 + E - p²/4 - (E p)/2 = 0Rearrange:E*(1 - p/2) = p²/4 - 1Thus,E = (p²/4 - 1)/(1 - p/2) = [ (p² - 4)/4 ] / [ (2 - p)/2 ] = [ (p² - 4)/4 ] * [ 2 / (2 - p) ] = (p² - 4)/(2*(2 - p)) = -(p² - 4)/(2*(p - 2)) = -(p - 2)(p + 2)/(2*(p - 2)) ) = -(p + 2)/2So, E = -(p + 2)/2Now, from equation (1):1 + p²/4 + D + (E p)/2 + F = 0We have D = -(q + 2)/2 and E = -(p + 2)/2Plug in:1 + p²/4 - (q + 2)/2 + [ -(p + 2)/2 * p ]/2 + F = 0Simplify term by term:1 + p²/4 - (q + 2)/2 - (p(p + 2))/4 + F = 0Combine like terms:1 - (q + 2)/2 + p²/4 - (p² + 2p)/4 + F = 0Simplify fractions:1 - q/2 - 1 + p²/4 - p²/4 - (2p)/4 + F = 0Simplify:(-q/2) - (p/2) + F = 0So,F = (q + p)/2Now, we have D, E, F in terms of p and q.So, the equation of Γ is:x² + y² + Dx + Ey + F = 0Plugging in D, E, F:x² + y² - (q + 2)/2 x - (p + 2)/2 y + (p + q)/2 = 0Now, since PQ is tangent to Γ at M(q/2, p/2), the condition for tangency is that the derivative at M is equal to the slope of PQ.Alternatively, the radius at M is perpendicular to PQ. So, the vector from center O' to M is perpendicular to vector PQ.First, let's find the center O' of Γ. The general form x² + y² + Dx + Ey + F = 0 has center at (-D/2, -E/2). So,O' = ( (q + 2)/4, (p + 2)/4 )Vector O'M = M - O' = (q/2 - (q + 2)/4, p/2 - (p + 2)/4 ) = ( (2q - q - 2)/4, (2p - p - 2)/4 ) = ( (q - 2)/4, (p - 2)/4 )Vector PQ = Q - P = (q, 0) - (0, p) = (q, -p)For O'M to be perpendicular to PQ, their dot product must be zero:( (q - 2)/4 ) * q + ( (p - 2)/4 ) * (-p) = 0Simplify:[ q(q - 2) - p(p - 2) ] / 4 = 0Multiply both sides by 4:q(q - 2) - p(p - 2) = 0Expand:q² - 2q - p² + 2p = 0Rearrange:q² - p² - 2q + 2p = 0Factor:(q² - p²) - 2(q - p) = 0Factor difference of squares:(q - p)(q + p) - 2(q - p) = 0Factor out (q - p):(q - p)(q + p - 2) = 0So, either q - p = 0 or q + p - 2 = 0Case 1: q - p = 0 => q = pCase 2: q + p - 2 = 0 => q + p = 2Now, let's analyze both cases.Case 1: q = pIf q = p, then points P and Q are such that their parameters are equal. Let's see what that implies.In this case, since P is on AC at (0, p) and Q is on AB at (p, 0). So, PQ connects (0, p) to (p, 0). The midpoint M is (p/2, p/2). The circle Γ passes through K(1, p/2), L(p/2, 1), and M(p/2, p/2).Wait, let me check if this satisfies the tangent condition. The center O' is ((q + 2)/4, (p + 2)/4) = ((p + 2)/4, (p + 2)/4). So, O' is ((p + 2)/4, (p + 2)/4). Vector O'M is (p/2 - (p + 2)/4, p/2 - (p + 2)/4) = ( (2p - p - 2)/4, (2p - p - 2)/4 ) = ( (p - 2)/4, (p - 2)/4 ). Vector PQ is (p, -p). Dot product is (p - 2)/4 * p + (p - 2)/4 * (-p) = [ p(p - 2) - p(p - 2) ] / 4 = 0. So, it satisfies the condition.Case 2: q + p = 2If q + p = 2, then p = 2 - q. So, P is (0, 2 - q) and Q is (q, 0). Let's see what this implies.Midpoints:K is (1, (2 - q)/2 )L is (q/2, 1 )M is (q/2, (2 - q)/2 )Circle Γ passes through these points. Let's check the tangent condition.Center O' is ((q + 2)/4, (p + 2)/4 ) = ((q + 2)/4, ( (2 - q) + 2 )/4 ) = ((q + 2)/4, (4 - q)/4 )Vector O'M = (q/2 - (q + 2)/4, (2 - q)/2 - (4 - q)/4 ) = ( (2q - q - 2)/4, (4 - 2q - 4 + q)/4 ) = ( (q - 2)/4, (-q)/4 )Vector PQ = (q, - (2 - q)) = (q, q - 2)Dot product of O'M and PQ:( (q - 2)/4 ) * q + ( (-q)/4 ) * (q - 2 ) = [ q(q - 2) - q(q - 2) ] / 4 = 0So, it also satisfies the condition.So, in both cases, the condition is satisfied. Now, we need to see which case leads to OP = OQ.Recall that O is the circumcenter at (1,1). So, OP is the distance from (1,1) to P(0,p), which is sqrt( (1 - 0)^2 + (1 - p)^2 ) = sqrt(1 + (1 - p)^2 )Similarly, OQ is the distance from (1,1) to Q(q,0), which is sqrt( (1 - q)^2 + (1 - 0)^2 ) = sqrt( (1 - q)^2 + 1 )We need to show that OP = OQ, i.e., sqrt(1 + (1 - p)^2 ) = sqrt( (1 - q)^2 + 1 )Squaring both sides:1 + (1 - p)^2 = (1 - q)^2 + 1Simplify:(1 - p)^2 = (1 - q)^2Taking square roots:|1 - p| = |1 - q|So, either 1 - p = 1 - q => p = q, or 1 - p = -(1 - q) => 1 - p = q - 1 => q + p = 2So, OP = OQ if and only if either p = q or p + q = 2.But from our earlier analysis, the tangent condition gives us either p = q or p + q = 2. So, in both cases, OP = OQ.Therefore, in both cases, OP = OQ.Hence, we have shown that OP = OQ.