answer：Okay, so I have this problem here about a system of linear equations, and I need to find the value of k. Let me try to figure this out step by step. First, the system of equations is:[begin{cases}x + y = 5k x - y = 9kend{cases}]And it's given that the solution to this system is also a solution to the equation ( x - 2y = 22 ). So, I need to find the value of k that makes this true.Hmm, okay. So, I think the first step is to solve the system of equations for x and y in terms of k. Then, substitute those expressions into the third equation ( x - 2y = 22 ) and solve for k. That sounds like a plan.Let me write down the two equations again:1. ( x + y = 5k )2. ( x - y = 9k )I notice that both equations have x and y, and the coefficients are similar. Maybe I can add or subtract the equations to eliminate one of the variables. Let me try adding them together.Adding equation 1 and equation 2:( (x + y) + (x - y) = 5k + 9k )Simplify the left side:( x + y + x - y = 2x )And the right side:( 5k + 9k = 14k )So, ( 2x = 14k ). Dividing both sides by 2:( x = 7k )Okay, so x is 7k. Now, let's find y. I can substitute x back into one of the original equations. Let's use equation 1: ( x + y = 5k ).Substituting x = 7k:( 7k + y = 5k )Subtract 7k from both sides:( y = 5k - 7k )( y = -2k )Alright, so now I have expressions for both x and y in terms of k:( x = 7k )( y = -2k )Now, the problem says that this solution also satisfies the equation ( x - 2y = 22 ). So, let's substitute x and y into that equation.Substituting:( 7k - 2(-2k) = 22 )Simplify the left side:First, multiply out the -2 and -2k:( 7k + 4k = 22 )Combine like terms:( 11k = 22 )Now, solve for k:Divide both sides by 11:( k = 22 / 11 )( k = 2 )So, the value of k is 2. Let me just double-check my steps to make sure I didn't make a mistake.1. Solved the system by adding the two equations, got x = 7k.2. Substituted x into the first equation to find y = -2k.3. Plugged x and y into the third equation: 7k - 2*(-2k) = 22.4. Simplified to 7k + 4k = 22, which is 11k = 22, so k = 2.Looks good. I think that's the correct answer.**Final Answer**The value of ( k ) is boxed{2}.

answer：Okay, so I have this sequence problem here. It says that the sequence {a_n} satisfies a_1 = 2, and each term is defined by the recurrence relation a_n = 1 - 1/a_{n-1}. I need to find the value of a_{2017}. Hmm, that seems a bit tricky, but let me try to figure it out step by step.First, let me write down what I know. The first term is given as a_1 = 2. Then, for each subsequent term, it's defined as 1 minus the reciprocal of the previous term. So, a_2 would be 1 - 1/a_1, which is 1 - 1/2. Let me compute that: 1 - 1/2 is 1/2. So, a_2 is 1/2.Now, moving on to a_3. That would be 1 - 1/a_2. Since a_2 is 1/2, 1/a_2 is 2. So, a_3 is 1 - 2, which is -1. Okay, so a_3 is -1.Next, a_4 is 1 - 1/a_3. Since a_3 is -1, 1/a_3 is -1. So, a_4 is 1 - (-1) which is 1 + 1 = 2. Hmm, interesting. So, a_4 is 2.Wait a second, that's the same as a_1. So, a_4 = a_1. Does that mean the sequence is periodic? Let me check the next few terms to see if it repeats.a_5 would be 1 - 1/a_4. Since a_4 is 2, 1/a_4 is 1/2. So, a_5 is 1 - 1/2 = 1/2, which is the same as a_2.Then, a_6 is 1 - 1/a_5. Since a_5 is 1/2, 1/a_5 is 2. So, a_6 is 1 - 2 = -1, which is the same as a_3.a_7 would be 1 - 1/a_6. Since a_6 is -1, 1/a_6 is -1. So, a_7 is 1 - (-1) = 2, which is the same as a_4 and a_1.So, it seems like the sequence is repeating every 3 terms: 2, 1/2, -1, 2, 1/2, -1, and so on. So, the period here is 3. That means the sequence cycles every 3 terms.Therefore, if the sequence repeats every 3 terms, then to find a_{2017}, I can figure out where 2017 falls in the cycle. In other words, I can compute 2017 modulo 3 to see which term it corresponds to.Let me compute 2017 divided by 3. Let's see, 3 goes into 2017 how many times? Well, 3 times 672 is 2016, right? So, 3*672 = 2016. Then, 2017 is 2016 + 1, so 2017 modulo 3 is 1. That means 2017 is 1 more than a multiple of 3.So, in terms of the cycle, the first term is a_1 = 2, which corresponds to n ≡ 1 mod 3. The second term is a_2 = 1/2, which corresponds to n ≡ 2 mod 3. The third term is a_3 = -1, which corresponds to n ≡ 0 mod 3. So, since 2017 ≡ 1 mod 3, it should correspond to a_1, which is 2.Wait, let me double-check that. So, n=1: 2, n=2: 1/2, n=3: -1, n=4: 2, n=5: 1/2, n=6: -1, etc. So, every time n is 1 more than a multiple of 3, it's 2. So, n=1,4,7,...,2017. So, yes, a_{2017} should be 2.But just to make sure I didn't make a mistake in my reasoning, let me test it with a smaller number. For example, n=4: 2017 is like n=4, which is 1 more than 3, so 2. Similarly, n=7: 1 more than 6, which is 2. So, that seems consistent.Alternatively, maybe I can express the sequence in terms of modular arithmetic. Since the period is 3, the value of a_n depends on n mod 3. So, if n ≡ 1 mod 3, a_n = 2; if n ≡ 2 mod 3, a_n = 1/2; and if n ≡ 0 mod 3, a_n = -1.So, 2017 divided by 3: 3*672=2016, so 2017=3*672 +1, so 2017 ≡1 mod3. Therefore, a_{2017}=2.Wait, just to make sure, let me compute a few more terms manually to confirm the cycle.a_1=2a_2=1 -1/2=1/2a_3=1 -1/(1/2)=1 -2=-1a_4=1 -1/(-1)=1 +1=2a_5=1 -1/2=1/2a_6=1 -1/(1/2)=1 -2=-1a_7=1 -1/(-1)=2Yes, so it's definitely cycling every 3 terms: 2, 1/2, -1, 2, 1/2, -1,...Therefore, the cycle length is 3, and since 2017 mod3=1, a_{2017}=a_1=2.I think that makes sense. So, the answer should be 2.**Final Answer**The value of (a_{2017}) is boxed{2}.

answer：Okay, so I have this problem where I need to find the possible range of values for `a` such that `a`, `a+1`, and `a+2` are the sides of an obtuse triangle. Hmm, let me think about how to approach this.First, I remember that for any triangle, the sum of any two sides must be greater than the third side. This is called the triangle inequality theorem. So, before even considering whether it's obtuse, I need to make sure that these three lengths can form a triangle.Let me write down the triangle inequalities:1. `a + (a + 1) > a + 2`2. `a + (a + 2) > a + 1`3. `(a + 1) + (a + 2) > a`Let me simplify each of these:1. `a + a + 1 > a + 2` simplifies to `2a + 1 > a + 2`, subtract `a` from both sides: `a + 1 > 2`, so `a > 1`.2. `a + a + 2 > a + 1` simplifies to `2a + 2 > a + 1`, subtract `a` from both sides: `a + 2 > 1`, so `a > -1`.3. `a + 1 + a + 2 > a` simplifies to `2a + 3 > a`, subtract `a` from both sides: `a + 3 > 0`, so `a > -3`.Now, looking at these inequalities, the most restrictive one is `a > 1`. So, for the sides to form a triangle, `a` must be greater than 1.But the problem specifies that it's an obtuse triangle. I need to recall the condition for a triangle to be obtuse. I remember that in an obtuse triangle, the square of the longest side is greater than the sum of the squares of the other two sides.So, first, I need to figure out which side is the longest. Since `a`, `a+1`, and `a+2` are consecutive numbers, the longest side is `a + 2`.Therefore, the condition for the triangle to be obtuse is:`(a + 2)^2 > a^2 + (a + 1)^2`Let me expand and simplify this inequality.First, expand `(a + 2)^2`:`(a + 2)^2 = a^2 + 4a + 4`Now, expand `(a + 1)^2`:`(a + 1)^2 = a^2 + 2a + 1`So, plugging back into the inequality:`a^2 + 4a + 4 > a^2 + (a^2 + 2a + 1)`Simplify the right side:`a^2 + 4a + 4 > 2a^2 + 2a + 1`Now, let's bring all terms to one side:`a^2 + 4a + 4 - 2a^2 - 2a - 1 > 0`Combine like terms:`-a^2 + 2a + 3 > 0`Multiply both sides by -1 to make the quadratic coefficient positive, remembering to reverse the inequality sign:`a^2 - 2a - 3 < 0`Now, let's factor this quadratic:Looking for two numbers that multiply to -3 and add to -2. Hmm, -3 and +1.So, `(a - 3)(a + 1) < 0`To solve this inequality, we can find the critical points where the expression equals zero: `a = 3` and `a = -1`.These points divide the number line into three intervals:1. `a < -1`2. `-1 < a < 3`3. `a > 3`We can test each interval:1. For `a < -1`, say `a = -2`: `(-2 - 3)(-2 + 1) = (-5)(-1) = 5 > 0`. So, the inequality is not satisfied here.2. For `-1 < a < 3`, say `a = 0`: `(0 - 3)(0 + 1) = (-3)(1) = -3 < 0`. So, the inequality is satisfied here.3. For `a > 3`, say `a = 4`: `(4 - 3)(4 + 1) = (1)(5) = 5 > 0`. So, the inequality is not satisfied here.Therefore, the inequality `a^2 - 2a - 3 < 0` holds when `-1 < a < 3`.But remember earlier, from the triangle inequality, we had `a > 1`. So, combining these two results, the range of `a` must satisfy both conditions.So, `a` must be greater than 1 and less than 3. Therefore, the possible range of values for `a` is `1 < a < 3`.Wait, let me double-check. If `a` is between 1 and 3, does that ensure that the triangle is obtuse?Let me pick a value in this range, say `a = 2`. Then the sides are 2, 3, 4.Check if it's obtuse:Calculate `4^2 = 16` and `2^2 + 3^2 = 4 + 9 = 13`. Since 16 > 13, it is obtuse.Another test: `a = 1.5`. Sides are 1.5, 2.5, 3.5.Check: `3.5^2 = 12.25` and `1.5^2 + 2.5^2 = 2.25 + 6.25 = 8.5`. 12.25 > 8.5, so it's obtuse.What about `a` approaching 3, say `a = 2.9`. Sides: 2.9, 3.9, 4.9.Check: `4.9^2 = 24.01` and `2.9^2 + 3.9^2 = 8.41 + 15.21 = 23.62`. 24.01 > 23.62, so it's still obtuse.Wait, but when `a` approaches 3 from below, the inequality is still satisfied. But when `a = 3`, let's see:Sides: 3, 4, 5. Wait, that's a right triangle, not obtuse. So, at `a = 3`, it's a right triangle, which is not obtuse. So, our upper limit is indeed less than 3.Similarly, when `a` approaches 1 from above, say `a = 1.1`. Sides: 1.1, 2.1, 3.1.Check: `3.1^2 = 9.61` and `1.1^2 + 2.1^2 = 1.21 + 4.41 = 5.62`. 9.61 > 5.62, so it's obtuse.But at `a = 1`, the sides would be 1, 2, 3. Wait, can 1, 2, 3 form a triangle? Let's check the triangle inequality:1 + 2 = 3, which is not greater than 3. So, it's degenerate, not a valid triangle. Hence, `a` must be greater than 1.Therefore, combining both the triangle inequality and the obtuse condition, the range is `1 < a < 3`.Wait, but let me think again. The quadratic inequality gave us `-1 < a < 3`, but the triangle inequality restricted it to `a > 1`. So, the overlap is indeed `1 < a < 3`.Is there any other condition I might have missed? Let me think.In an obtuse triangle, only one angle is obtuse, and the other two are acute. So, the condition is solely based on the sides. Since we've already considered the longest side, which is `a + 2`, and set up the inequality accordingly, I think we're good.Just to be thorough, let me test `a = 0.5` which is less than 1. The sides would be 0.5, 1.5, 2.5.Check triangle inequality: 0.5 + 1.5 = 2, which is equal to 2.5, so it's not a valid triangle. So, indeed, `a` must be greater than 1.Another test: `a = 4`, which is greater than 3. Sides: 4, 5, 6.Check if it's obtuse: 6^2 = 36, 4^2 + 5^2 = 16 + 25 = 41. 36 < 41, so it's acute. Hence, when `a` is greater than 3, the triangle becomes acute, which is why our upper limit is 3.Therefore, I'm confident that the possible range of `a` is between 1 and 3, not including the endpoints.**Final Answer**The possible range of values for `a` is boxed{(1, 3)}.

answer：Okay, so I have this problem about two boats, A and B, departing from points A and B, which are 150 km apart. They're traveling towards each other on a river. Boat A is going downstream from point A for 90 km and meets boat B, which is going upstream from point B. Both boats have a speed of 30 km/h in still water. I need to find the speed of the river in km/h.Hmm, let me try to visualize this. There's a river flowing from point A to point B, which are 150 km apart. Boat A starts at A and goes downstream, while Boat B starts at B and goes upstream. They meet after Boat A has traveled 90 km. Both boats have a speed of 30 km/h in still water, so their actual speeds will be affected by the river's current.I think I need to figure out the time it takes for each boat to meet each other and set up an equation based on that. Since they start at the same time and meet at the same point, their travel times should be equal.Let me denote the speed of the river as 'v' km/h. So, for Boat A, which is going downstream, its effective speed will be its still water speed plus the river's speed. That would be 30 + v km/h. For Boat B, which is going upstream, its effective speed will be its still water speed minus the river's speed, so 30 - v km/h.Boat A travels 90 km downstream. So, the time taken by Boat A to meet Boat B is the distance divided by its effective speed, which is 90 / (30 + v).Boat B starts at point B, which is 150 km away from point A. But since they meet after Boat A has gone 90 km, Boat B must have traveled 150 - 90 = 60 km upstream. So, the distance Boat B covers is 60 km. Therefore, the time taken by Boat B is 60 / (30 - v).Since both boats meet at the same time, these two times should be equal. So, I can set up the equation:90 / (30 + v) = 60 / (30 - v)Now, I need to solve for 'v'. Let me write that equation again:90 / (30 + v) = 60 / (30 - v)To solve for 'v', I can cross-multiply:90 * (30 - v) = 60 * (30 + v)Let me compute both sides:Left side: 90 * 30 - 90v = 2700 - 90vRight side: 60 * 30 + 60v = 1800 + 60vSo, the equation becomes:2700 - 90v = 1800 + 60vNow, let's bring like terms together. I'll add 90v to both sides:2700 = 1800 + 150vThen, subtract 1800 from both sides:900 = 150vNow, divide both sides by 150:v = 900 / 150Simplify that:v = 6So, the speed of the river is 6 km/h.Wait, let me double-check my calculations to make sure I didn't make a mistake.Starting with the equation:90 / (30 + v) = 60 / (30 - v)Cross-multiplying:90*(30 - v) = 60*(30 + v)Calculating:90*30 = 2700, 90*(-v) = -90v60*30 = 1800, 60*v = 60vSo, 2700 - 90v = 1800 + 60vAdding 90v to both sides:2700 = 1800 + 150vSubtracting 1800:900 = 150vDividing:v = 6Yes, that seems correct. So, the river's speed is 6 km/h.Just to make sure, let me plug it back into the original times.Boat A's speed downstream: 30 + 6 = 36 km/hTime for Boat A: 90 / 36 = 2.5 hoursBoat B's speed upstream: 30 - 6 = 24 km/hTime for Boat B: 60 / 24 = 2.5 hoursYes, both times are equal, so that checks out.Therefore, the speed of the river is 6 km/h.**Final Answer**The speed of the river is boxed{6} km/h.

answer：To find the greatest common divisor (GCD) and the least common multiple (LCM) of 336 and 1260, I'll start by determining their prime factorizations.First, I'll factorize 336. Dividing by 2 repeatedly, I get 336 = 2^4 × 3 × 7.Next, I'll factorize 1260. Dividing by 2, then 3, and continuing with 5 and 7, I obtain 1260 = 2^2 × 3^2 × 5 × 7.For the GCD, I'll identify the common prime factors and take the lowest exponents. The common primes are 2, 3, and 7. The lowest exponents are 2^2, 3^1, and 7^1. Multiplying these together gives the GCD: 4 × 3 × 7 = 84.To find the LCM, I'll take all the prime factors from both numbers, using the highest exponents for each prime. The primes involved are 2, 3, 5, and 7. The highest exponents are 2^4, 3^2, 5^1, and 7^1. Multiplying these together gives the LCM: 16 × 9 × 5 × 7 = 5040.

answer：Alright, so I have this problem about an ellipse, and I need to find the maximum area of the inscribed circle of triangle F1PQ. Hmm, okay, let me try to break this down step by step.First, I know the equation of the ellipse is given as (frac{x^2}{4} + frac{y^2}{3} = 1). So, comparing this to the standard form of an ellipse, which is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), I can see that (a^2 = 4) and (b^2 = 3). Therefore, (a = 2) and (b = sqrt{3}).Now, for an ellipse, the distance from the center to each focus is given by (c), where (c^2 = a^2 - b^2). Plugging in the values, (c^2 = 4 - 3 = 1), so (c = 1). That means the foci are located at ((pm 1, 0)). So, the left focus (F_1) is at ((-1, 0)), and the right focus (F_2) is at ((1, 0)).The problem says that a line (l) passes through the right focus (F_2) and intersects the ellipse at points (P) and (Q). So, line (l) goes through ((1, 0)) and cuts the ellipse at two points (P) and (Q). Then, we need to consider triangle (F_1PQ) and find the maximum area of its inscribed circle.Wait, inscribed circle... that's the incircle. The area of the incircle depends on the inradius, right? The formula for the area of the incircle is (pi r^2), where (r) is the inradius. So, to maximize the area, we need to maximize the inradius (r).But how do we find the inradius of triangle (F_1PQ)? I remember that the inradius (r) of a triangle is related to its area (A) and semi-perimeter (s) by the formula (r = frac{A}{s}). So, if I can find expressions for the area and semi-perimeter of triangle (F_1PQ), I can find (r), and then maximize it.But before that, maybe I should parametrize the line (l) passing through (F_2). Let me think about how to represent line (l). Since it passes through ((1, 0)), I can write its equation in the form (y = m(x - 1)), where (m) is the slope. Alternatively, if the line is vertical, it would be (x = 1), but in that case, it would only intersect the ellipse at one point, so it's not a secant. So, I think the line must have a slope (m), and we can consider all possible slopes except vertical.So, let's assume the line is (y = m(x - 1)). Now, I need to find the points (P) and (Q) where this line intersects the ellipse.To find the intersection points, substitute (y = m(x - 1)) into the ellipse equation:[frac{x^2}{4} + frac{[m(x - 1)]^2}{3} = 1]Let me expand this:[frac{x^2}{4} + frac{m^2(x^2 - 2x + 1)}{3} = 1]Multiply both sides by 12 to eliminate denominators:[3x^2 + 4m^2(x^2 - 2x + 1) = 12]Expanding further:[3x^2 + 4m^2x^2 - 8m^2x + 4m^2 = 12]Combine like terms:[(3 + 4m^2)x^2 - 8m^2x + (4m^2 - 12) = 0]So, this is a quadratic equation in (x). Let me denote this as:[Ax^2 + Bx + C = 0]Where:- (A = 3 + 4m^2)- (B = -8m^2)- (C = 4m^2 - 12)Now, the solutions to this equation will give me the x-coordinates of points (P) and (Q). Let's denote the roots as (x_1) and (x_2). Then, using quadratic formula:[x = frac{-B pm sqrt{B^2 - 4AC}}{2A}]Plugging in the values:[x = frac{8m^2 pm sqrt{(64m^4) - 4(3 + 4m^2)(4m^2 - 12)}}{2(3 + 4m^2)}]Let me compute the discriminant (D):[D = 64m^4 - 4(3 + 4m^2)(4m^2 - 12)]First, compute (4(3 + 4m^2)(4m^2 - 12)):Let me compute ((3 + 4m^2)(4m^2 - 12)):Multiply term by term:(3 * 4m^2 = 12m^2)(3 * (-12) = -36)(4m^2 * 4m^2 = 16m^4)(4m^2 * (-12) = -48m^2)So, adding all together:(12m^2 - 36 + 16m^4 - 48m^2 = 16m^4 - 36m^2 - 36)Multiply by 4:(4 * (16m^4 - 36m^2 - 36) = 64m^4 - 144m^2 - 144)So, discriminant (D = 64m^4 - (64m^4 - 144m^2 - 144))Wait, no, discriminant is (64m^4 - 4AC), which is (64m^4 - [64m^4 - 144m^2 - 144]). So,(D = 64m^4 - 64m^4 + 144m^2 + 144 = 144m^2 + 144 = 144(m^2 + 1))So, discriminant simplifies to (144(m^2 + 1)). That's nice.Therefore, the roots are:[x = frac{8m^2 pm sqrt{144(m^2 + 1)}}{2(3 + 4m^2)} = frac{8m^2 pm 12sqrt{m^2 + 1}}{2(3 + 4m^2)}]Simplify numerator and denominator:Divide numerator and denominator by 2:[x = frac{4m^2 pm 6sqrt{m^2 + 1}}{3 + 4m^2}]So, the x-coordinates of (P) and (Q) are:[x = frac{4m^2 + 6sqrt{m^2 + 1}}{3 + 4m^2} quad text{and} quad x = frac{4m^2 - 6sqrt{m^2 + 1}}{3 + 4m^2}]Hmm, that looks a bit complicated. Maybe I can denote ( sqrt{m^2 + 1} = t ), but not sure if that helps. Alternatively, perhaps I can find parametric expressions for (P) and (Q).Alternatively, maybe instead of using slope (m), I can parametrize the line in terms of an angle or something else. But perhaps it's better to proceed as is.Once I have the x-coordinates, I can find the corresponding y-coordinates using (y = m(x - 1)).So, let me denote (x_1 = frac{4m^2 + 6sqrt{m^2 + 1}}{3 + 4m^2}) and (x_2 = frac{4m^2 - 6sqrt{m^2 + 1}}{3 + 4m^2}).Then, (y_1 = m(x_1 - 1)) and (y_2 = m(x_2 - 1)).So, points (P) and (Q) are ((x_1, y_1)) and ((x_2, y_2)).Now, I need to find the area and semi-perimeter of triangle (F_1PQ). Let's denote (F_1 = (-1, 0)), (P = (x_1, y_1)), and (Q = (x_2, y_2)).First, let's find the lengths of the sides of triangle (F_1PQ). The sides are (F_1P), (F_1Q), and (PQ).Compute (F_1P):[F_1P = sqrt{(x_1 + 1)^2 + (y_1 - 0)^2} = sqrt{(x_1 + 1)^2 + y_1^2}]Similarly, (F_1Q = sqrt{(x_2 + 1)^2 + y_2^2})And (PQ = sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2})But this seems really complicated with all these square roots. Maybe there's a smarter way.Wait, maybe instead of computing everything in terms of (m), I can parametrize the line differently. Maybe using parametric equations for the ellipse.Alternatively, perhaps I can use properties of ellipses and triangles to find a relationship.Wait, another thought: the area of triangle (F_1PQ) can be found using the determinant formula:[A = frac{1}{2} |(x_1(y_2 - 0) + x_2(0 - y_1) + (-1)(y_1 - y_2))|]Simplify:[A = frac{1}{2} |x_1 y_2 - x_2 y_1 - y_1 + y_2|]But since (y_1 = m(x_1 - 1)) and (y_2 = m(x_2 - 1)), substitute these into the expression:[A = frac{1}{2} |x_1 [m(x_2 - 1)] - x_2 [m(x_1 - 1)] - [m(x_1 - 1)] + [m(x_2 - 1)]|]Let me expand this:First term: (x_1 m(x_2 - 1))Second term: (-x_2 m(x_1 - 1))Third term: (-m(x_1 - 1))Fourth term: (m(x_2 - 1))So, combining all terms:[A = frac{1}{2} | m x_1 x_2 - m x_1 - m x_1 x_2 + m x_2 - m x_1 + m + m x_2 - m |]Wait, let me compute each term step by step.First term: (x_1 m(x_2 - 1) = m x_1 x_2 - m x_1)Second term: (-x_2 m(x_1 - 1) = -m x_1 x_2 + m x_2)Third term: (-m(x_1 - 1) = -m x_1 + m)Fourth term: (m(x_2 - 1) = m x_2 - m)Now, adding all four terms together:1. (m x_1 x_2 - m x_1)2. (-m x_1 x_2 + m x_2)3. (-m x_1 + m)4. (m x_2 - m)Now, combine like terms:- (m x_1 x_2 - m x_1 x_2 = 0)- (-m x_1 + m x_2 - m x_1 + m x_2 = (-2m x_1) + (2m x_2))- (m - m = 0)So, overall:[A = frac{1}{2} | -2m x_1 + 2m x_2 | = frac{1}{2} | 2m(x_2 - x_1) | = |m(x_2 - x_1)|]So, the area simplifies to ( |m(x_2 - x_1)| ). Interesting, that's a much simpler expression.Now, let's compute (x_2 - x_1). From earlier, we have:(x_1 = frac{4m^2 + 6sqrt{m^2 + 1}}{3 + 4m^2})(x_2 = frac{4m^2 - 6sqrt{m^2 + 1}}{3 + 4m^2})So, (x_2 - x_1 = frac{4m^2 - 6sqrt{m^2 + 1} - 4m^2 - 6sqrt{m^2 + 1}}{3 + 4m^2} = frac{-12sqrt{m^2 + 1}}{3 + 4m^2})Therefore, (x_2 - x_1 = frac{-12sqrt{m^2 + 1}}{3 + 4m^2}). So, the absolute value is (frac{12sqrt{m^2 + 1}}{3 + 4m^2}).Thus, the area (A) is:[A = |m| cdot frac{12sqrt{m^2 + 1}}{3 + 4m^2}]Simplify:[A = frac{12 |m| sqrt{m^2 + 1}}{3 + 4m^2}]Since area is positive, we can drop the absolute value:[A = frac{12 m sqrt{m^2 + 1}}{3 + 4m^2} quad text{for } m geq 0]Because if (m) is negative, the area would be the same as for (|m|), so we can consider (m geq 0) without loss of generality.Okay, so that's the area. Now, I need the semi-perimeter (s) of triangle (F_1PQ). The semi-perimeter is (frac{F_1P + F_1Q + PQ}{2}).But computing (F_1P), (F_1Q), and (PQ) seems complicated. Maybe there's a smarter way.Wait, another thought: for an ellipse, the sum of distances from any point on the ellipse to the two foci is constant and equal to (2a). So, for points (P) and (Q), we have:(F_1P + F_2P = 2a = 4)Similarly, (F_1Q + F_2Q = 4)But in our case, line (l) passes through (F_2), so (F_2) is one of the points on line (l), but (P) and (Q) are the other intersection points. So, (F_2) is on line (l), but not necessarily on the ellipse? Wait, no, (F_2) is a focus, which is inside the ellipse, not on it. So, line (l) passes through (F_2) and intersects the ellipse at (P) and (Q). So, (F_2) is inside the ellipse, so the line passes through (F_2) and cuts the ellipse at (P) and (Q). So, (F_2) is not one of the points (P) or (Q), unless the line is tangent, but in this case, it's a secant.So, maybe I can use the property that for any point on the ellipse, the sum of distances to the foci is constant. But since (P) and (Q) are on the ellipse, we have:(F_1P + F_2P = 4)(F_1Q + F_2Q = 4)But (F_2) is on line (l), which passes through (P) and (Q). So, perhaps I can express (F_2P) and (F_2Q) in terms of the distances along the line.Wait, maybe I can parametrize the line (l) in terms of a parameter (t), such that (F_2) is at (t = 0), and (P) and (Q) are at some (t = t_1) and (t = t_2). Then, the distances (F_2P) and (F_2Q) would be (|t_1|) and (|t_2|), but I'm not sure if that helps directly.Alternatively, perhaps I can use coordinates to find the distances.Wait, let me think differently. The semi-perimeter (s) is (frac{F_1P + F_1Q + PQ}{2}). If I can express (F_1P + F_1Q) and (PQ) in terms of (m), then I can find (s).But (F_1P + F_1Q) can be related to the properties of the ellipse. Wait, for each point (P) and (Q), (F_1P + F_2P = 4) and (F_1Q + F_2Q = 4). So, (F_1P = 4 - F_2P) and (F_1Q = 4 - F_2Q). So, (F_1P + F_1Q = 8 - (F_2P + F_2Q)).But (F_2P + F_2Q) is the sum of distances from (F_2) to (P) and (Q). Since (P) and (Q) lie on line (l) passing through (F_2), the distance (F_2P + F_2Q) is just the length of segment (PQ), because (F_2) lies between (P) and (Q). Wait, is that true?Wait, no. If (F_2) is on line (l), which intersects the ellipse at (P) and (Q), then (F_2) is between (P) and (Q), so (PQ = PF_2 + F_2Q). Therefore, (F_2P + F_2Q = PQ). So, (F_1P + F_1Q = 8 - PQ).Therefore, the semi-perimeter (s) is:[s = frac{F_1P + F_1Q + PQ}{2} = frac{(8 - PQ) + PQ}{2} = frac{8}{2} = 4]Wait, that's interesting! So, the semi-perimeter (s) is always 4, regardless of the line (l). That simplifies things a lot.So, since (s = 4), and the area (A = frac{12 m sqrt{m^2 + 1}}{3 + 4m^2}), then the inradius (r = frac{A}{s} = frac{A}{4}).Therefore, (r = frac{12 m sqrt{m^2 + 1}}{4(3 + 4m^2)} = frac{3 m sqrt{m^2 + 1}}{3 + 4m^2}).So, to maximize the area of the incircle, which is (pi r^2), we need to maximize (r). Since (pi) is a constant, maximizing (r^2) will maximize the area.Therefore, let's define (f(m) = left( frac{3 m sqrt{m^2 + 1}}{3 + 4m^2} right)^2 = frac{9 m^2 (m^2 + 1)}{(3 + 4m^2)^2}).We need to find the maximum of (f(m)) for (m geq 0).So, let's compute (f(m)):[f(m) = frac{9 m^2 (m^2 + 1)}{(3 + 4m^2)^2}]Let me simplify this expression. Let me denote (u = m^2), so (u geq 0). Then,[f(u) = frac{9 u (u + 1)}{(3 + 4u)^2}]Now, we can consider (f(u)) as a function of (u), and find its maximum for (u geq 0).To find the maximum, take the derivative of (f(u)) with respect to (u) and set it equal to zero.First, compute (f'(u)):Using the quotient rule:If (f(u) = frac{N(u)}{D(u)}), then (f'(u) = frac{N'(u) D(u) - N(u) D'(u)}{[D(u)]^2}).So, let me compute (N(u) = 9u(u + 1) = 9u^2 + 9u), so (N'(u) = 18u + 9).(D(u) = (3 + 4u)^2), so (D'(u) = 2(3 + 4u)(4) = 8(3 + 4u)).Therefore,[f'(u) = frac{(18u + 9)(3 + 4u)^2 - (9u^2 + 9u)(8)(3 + 4u)}{(3 + 4u)^4}]Factor out common terms:First, note that both terms in the numerator have a factor of ((3 + 4u)). Let's factor that out:[f'(u) = frac{(3 + 4u)[(18u + 9)(3 + 4u) - 8(9u^2 + 9u)]}{(3 + 4u)^4} = frac{(18u + 9)(3 + 4u) - 8(9u^2 + 9u)}{(3 + 4u)^3}]Now, compute the numerator:First term: ((18u + 9)(3 + 4u))Let me expand this:(18u * 3 = 54u)(18u * 4u = 72u^2)(9 * 3 = 27)(9 * 4u = 36u)So, adding together:(54u + 72u^2 + 27 + 36u = 72u^2 + 90u + 27)Second term: (8(9u^2 + 9u) = 72u^2 + 72u)Now, subtract the second term from the first term:(72u^2 + 90u + 27 - 72u^2 - 72u = (72u^2 - 72u^2) + (90u - 72u) + 27 = 18u + 27)Therefore, numerator is (18u + 27), so:[f'(u) = frac{18u + 27}{(3 + 4u)^3}]Set (f'(u) = 0):[18u + 27 = 0 implies u = -frac{27}{18} = -frac{3}{2}]But (u = m^2 geq 0), so this critical point is at (u = -frac{3}{2}), which is not in our domain. Therefore, the function (f(u)) has no critical points for (u geq 0). So, the maximum must occur at the boundary.But as (u to infty), let's see the behavior of (f(u)):[f(u) = frac{9u^2 + 9u}{(3 + 4u)^2} approx frac{9u^2}{16u^2} = frac{9}{16}]So, as (u) becomes large, (f(u)) approaches (frac{9}{16}).At (u = 0):[f(0) = frac{0}{9} = 0]So, the function starts at 0 when (u = 0), increases, and approaches (frac{9}{16}) as (u) increases. Therefore, the maximum value of (f(u)) is (frac{9}{16}), achieved as (u to infty).But wait, that can't be right because as (u) increases, (m) increases, and the line becomes steeper. But in reality, the maximum might be achieved somewhere in between.Wait, hold on, perhaps I made a mistake in computing the derivative.Let me double-check the derivative computation.We had:(f(u) = frac{9u(u + 1)}{(3 + 4u)^2})Then,(N(u) = 9u^2 + 9u), so (N'(u) = 18u + 9)(D(u) = (3 + 4u)^2), so (D'(u) = 8(3 + 4u))Then,(f'(u) = frac{(18u + 9)(3 + 4u)^2 - (9u^2 + 9u)(8)(3 + 4u)}{(3 + 4u)^4})Factor out ((3 + 4u)):[f'(u) = frac{(3 + 4u)[(18u + 9)(3 + 4u) - 8(9u^2 + 9u)]}{(3 + 4u)^4} = frac{(18u + 9)(3 + 4u) - 8(9u^2 + 9u)}{(3 + 4u)^3}]Then, expanding:First term: ((18u + 9)(3 + 4u) = 54u + 72u^2 + 27 + 36u = 72u^2 + 90u + 27)Second term: (8(9u^2 + 9u) = 72u^2 + 72u)Subtracting: (72u^2 + 90u + 27 - 72u^2 - 72u = 18u + 27)So, numerator is (18u + 27), which is positive for all (u geq 0). Therefore, (f'(u) > 0) for all (u geq 0). So, the function (f(u)) is strictly increasing on (u geq 0).Therefore, as (u) increases, (f(u)) increases, approaching (frac{9}{16}) as (u to infty). So, the maximum value of (f(u)) is (frac{9}{16}), but it's a limit as (u) approaches infinity.But in reality, (u = m^2) can't be infinity because the line can't have an infinite slope. So, perhaps the maximum is achieved as the line becomes more and more vertical, but in that case, the points (P) and (Q) would approach the top and bottom of the ellipse.Wait, let me think about the ellipse. The ellipse is (frac{x^2}{4} + frac{y^2}{3} = 1), so its major axis is along the x-axis with semi-major axis 2, and minor axis along y-axis with semi-minor axis (sqrt{3}).So, the ellipse extends from ((-2, 0)) to ((2, 0)) along the x-axis, and from ((0, -sqrt{3})) to ((0, sqrt{3})) along the y-axis.So, as the slope (m) approaches infinity, the line (l) becomes vertical, but since the ellipse only extends up to (x = 2), the vertical line (x = 1) (which is the right focus) intersects the ellipse at (y = pm sqrt{3(1 - (1)^2/4)} = pm sqrt{3(3/4)} = pm frac{3}{2}). So, points (P) and (Q) would be ((1, 3/2)) and ((1, -3/2)).Wait, but in that case, the line is vertical, so the slope (m) is undefined. But in our parametrization, we considered (m) finite. So, perhaps as (m) approaches infinity, the points (P) and (Q) approach ((1, pm frac{3}{2})).Let me compute (f(u)) as (u to infty):[f(u) = frac{9u^2 + 9u}{(3 + 4u)^2} approx frac{9u^2}{16u^2} = frac{9}{16}]So, the limit is (frac{9}{16}). Therefore, the maximum value of (f(u)) is (frac{9}{16}), but it's not achieved for any finite (u), only approached as (u) tends to infinity.But in reality, the maximum inradius occurs when the line is vertical, i.e., when (m) approaches infinity, but in that case, the points (P) and (Q) are ((1, pm frac{3}{2})). So, let's compute the inradius in that case.Wait, but if (m) approaches infinity, the area (A) approaches (frac{9}{16}), but actually, (f(u)) approaches (frac{9}{16}), but (r^2 = f(u)), so (r) approaches (frac{3}{4}). Therefore, the maximum inradius is (frac{3}{4}), and the maximum area of the incircle is (pi (frac{3}{4})^2 = frac{9}{16}pi).But wait, let me verify this by actually computing the inradius when the line is vertical.So, if the line is vertical, (x = 1), then points (P) and (Q) are ((1, frac{3}{2})) and ((1, -frac{3}{2})). So, triangle (F_1PQ) has vertices at ((-1, 0)), ((1, frac{3}{2})), and ((1, -frac{3}{2})).Let me compute the sides of this triangle.First, compute (F_1P):Distance between ((-1, 0)) and ((1, frac{3}{2})):[F_1P = sqrt{(1 - (-1))^2 + left(frac{3}{2} - 0right)^2} = sqrt{(2)^2 + left(frac{3}{2}right)^2} = sqrt{4 + frac{9}{4}} = sqrt{frac{25}{4}} = frac{5}{2}]Similarly, (F_1Q = frac{5}{2}).Compute (PQ):Distance between ((1, frac{3}{2})) and ((1, -frac{3}{2})):[PQ = sqrt{(1 - 1)^2 + left(-frac{3}{2} - frac{3}{2}right)^2} = sqrt{0 + (-3)^2} = sqrt{9} = 3]So, sides are (F_1P = frac{5}{2}), (F_1Q = frac{5}{2}), and (PQ = 3).Compute semi-perimeter (s):[s = frac{frac{5}{2} + frac{5}{2} + 3}{2} = frac{5 + 3}{2} = frac{8}{2} = 4]Which matches our earlier conclusion that (s = 4).Compute area (A):Using coordinates, the area can be found via the shoelace formula.Coordinates:(F_1 = (-1, 0)), (P = (1, frac{3}{2})), (Q = (1, -frac{3}{2}))Using shoelace formula:Arrange the points:((-1, 0)), ((1, frac{3}{2})), ((1, -frac{3}{2})), ((-1, 0))Compute the sum:[(-1 cdot frac{3}{2} + 1 cdot (-frac{3}{2}) + 1 cdot 0) - (0 cdot 1 + frac{3}{2} cdot 1 + (-frac{3}{2}) cdot (-1))]Wait, let me recall the shoelace formula:[A = frac{1}{2} |x_1y_2 + x_2y_3 + x_3y_1 - x_2y_1 - x_3y_2 - x_1y_3|]Plugging in:(x_1 = -1), (y_1 = 0)(x_2 = 1), (y_2 = frac{3}{2})(x_3 = 1), (y_3 = -frac{3}{2})So,[A = frac{1}{2} |(-1 cdot frac{3}{2} + 1 cdot (-frac{3}{2}) + 1 cdot 0) - (1 cdot 0 + 1 cdot frac{3}{2} + (-1) cdot (-frac{3}{2}))|]Compute term by term:First part:(-1 cdot frac{3}{2} = -frac{3}{2})(1 cdot (-frac{3}{2}) = -frac{3}{2})(1 cdot 0 = 0)Sum: (-frac{3}{2} - frac{3}{2} + 0 = -3)Second part:(1 cdot 0 = 0)(1 cdot frac{3}{2} = frac{3}{2})(-1 cdot (-frac{3}{2}) = frac{3}{2})Sum: (0 + frac{3}{2} + frac{3}{2} = 3)Therefore,[A = frac{1}{2} | -3 - 3 | = frac{1}{2} | -6 | = frac{1}{2} times 6 = 3]So, the area (A = 3).Then, the inradius (r = frac{A}{s} = frac{3}{4}).Therefore, the area of the incircle is (pi r^2 = pi left(frac{3}{4}right)^2 = frac{9}{16}pi).So, this confirms that when the line is vertical, the inradius is (frac{3}{4}), and the area is (frac{9}{16}pi).But earlier, we saw that as (m) increases, (f(u)) approaches (frac{9}{16}), so the maximum area is indeed (frac{9}{16}pi).Wait, but is this the maximum? Because when the line is vertical, we get an inradius of (frac{3}{4}), but could there be another line where the inradius is larger?Wait, from our earlier analysis, since (f(u)) is strictly increasing and approaches (frac{9}{16}) as (u to infty), the maximum is achieved in the limit as the line becomes vertical. So, the maximum area is (frac{9}{16}pi).But let me check another case to be sure. For example, take (m = 0), which is the horizontal line through (F_2).So, if (m = 0), the line is (y = 0), which is the x-axis. It intersects the ellipse at points where (y = 0), so (x^2/4 = 1), so (x = pm 2). But since the line passes through (F_2 = (1, 0)), the points (P) and (Q) are ((2, 0)) and ((0, 0)). Wait, but ((0, 0)) is not on the ellipse, because plugging into the ellipse equation: (0 + 0 = 0 neq 1). Wait, that can't be.Wait, no, if the line is the x-axis, it intersects the ellipse at (x = pm 2), but since the line passes through (F_2 = (1, 0)), which is inside the ellipse, so the two intersection points are ((2, 0)) and ((-2, 0)). Wait, but ((-2, 0)) is on the ellipse, yes.Wait, but if the line is the x-axis, it passes through (F_2 = (1, 0)) and intersects the ellipse at ((2, 0)) and ((-2, 0)). So, points (P) and (Q) are ((2, 0)) and ((-2, 0)). But then triangle (F_1PQ) would have vertices at ((-1, 0)), ((2, 0)), and ((-2, 0)). But all three points are colinear on the x-axis, so the area is zero. Therefore, the inradius is zero.So, in this case, the area is zero, which is consistent with our earlier result that (f(0) = 0).Therefore, the area of the incircle increases as the slope (m) increases, approaching (frac{9}{16}pi) as the line becomes vertical.Hence, the maximum area is (frac{9}{16}pi).But just to be thorough, let me check another slope, say (m = 1), and compute the inradius.So, if (m = 1), then (u = 1).Compute (f(u)):[f(1) = frac{9 * 1 * (1 + 1)}{(3 + 4 * 1)^2} = frac{18}{49} approx 0.367]So, (r = sqrt{f(1)} = sqrt{frac{18}{49}} = frac{3sqrt{2}}{7} approx 0.606). Then, the area is (pi (frac{3sqrt{2}}{7})^2 = frac{18}{49}pi approx 0.367pi), which is less than (frac{9}{16}pi approx 1.767pi). Wait, no, wait, (frac{9}{16}pi) is approximately 1.767, but that can't be because the area of the incircle can't be larger than the area of the triangle.Wait, hold on, I think I made a mistake here. The area of the incircle is (pi r^2), but (r) is the inradius, which is related to the area of the triangle. However, the area of the incircle is separate from the area of the triangle.Wait, but in our case, the area of the incircle is (pi r^2), where (r) is the inradius. So, if (r = frac{3}{4}), then the area is (frac{9}{16}pi), which is approximately 1.767. But the area of the triangle when the line is vertical is 3, so the incircle area is about 1.767, which is less than the triangle area, which makes sense.But when (m = 1), the inradius is approximately 0.606, so the incircle area is approximately 1.147, which is less than 1.767. So, indeed, as (m) increases, the incircle area increases, approaching (frac{9}{16}pi).Therefore, the maximum area is (frac{9}{16}pi).Thus, after all this computation and verification, I conclude that the maximum area of the inscribed circle of triangle (F_1PQ) is (frac{9}{16}pi).**Final Answer**The maximum area of the inscribed circle is boxed{dfrac{9}{16}pi}.