answer：Okay, so I'm trying to solve this problem about macronutrient balance for athletes. It's part a and part b, and I need to figure out both. Let me start with part a.The equation given is ( C^2 - 4C + 3 = P ), where ( C ) is carbohydrates in grams and ( P ) is protein in grams. For part a, I need to solve for ( C ) when ( P = 2 ) grams. So, I'll substitute 2 into the equation for ( P ).That gives me:[ C^2 - 4C + 3 = 2 ]Hmm, okay, so I can subtract 2 from both sides to set the equation to zero:[ C^2 - 4C + 3 - 2 = 0 ]Simplifying that:[ C^2 - 4C + 1 = 0 ]Now, this is a quadratic equation in the form ( ax^2 + bx + c = 0 ). I can use the quadratic formula to solve for ( C ). The quadratic formula is:[ C = frac{-b pm sqrt{b^2 - 4ac}}{2a} ]In this case, ( a = 1 ), ( b = -4 ), and ( c = 1 ). Plugging these into the formula:[ C = frac{-(-4) pm sqrt{(-4)^2 - 4(1)(1)}}{2(1)} ]Simplify the numerator:[ C = frac{4 pm sqrt{16 - 4}}{2} ][ C = frac{4 pm sqrt{12}}{2} ]Wait, ( sqrt{12} ) can be simplified. ( sqrt{12} = 2sqrt{3} ), so:[ C = frac{4 pm 2sqrt{3}}{2} ]I can factor out a 2 in the numerator:[ C = frac{2(2 pm sqrt{3})}{2} ]The 2s cancel out:[ C = 2 pm sqrt{3} ]So, the solutions are:[ C = 2 + sqrt{3} ] and [ C = 2 - sqrt{3} ]Let me approximate ( sqrt{3} ) to check if these make sense. ( sqrt{3} ) is approximately 1.732. So:- ( 2 + 1.732 = 3.732 ) grams- ( 2 - 1.732 = 0.268 ) gramsHmm, 0.268 grams of carbohydrates seems really low for an athlete. Maybe that's a minimum? Or perhaps it's just a mathematical solution, and in reality, athletes would need more carbs. But since the problem doesn't specify any constraints on ( C ), both solutions are valid mathematically.Moving on to part b. I need to determine the range of carbohydrate intake ( C ) that ensures protein intake ( P ) remains between 1 and 3 grams, inclusive. So, ( 1 leq P leq 3 ).Given the equation ( P = C^2 - 4C + 3 ), I can set up the inequality:[ 1 leq C^2 - 4C + 3 leq 3 ]I can break this into two separate inequalities:1. ( C^2 - 4C + 3 geq 1 )2. ( C^2 - 4C + 3 leq 3 )Let me solve each one separately.Starting with the first inequality:[ C^2 - 4C + 3 geq 1 ]Subtract 1 from both sides:[ C^2 - 4C + 2 geq 0 ]This is another quadratic inequality. Let me find the roots of the equation ( C^2 - 4C + 2 = 0 ) using the quadratic formula again.Here, ( a = 1 ), ( b = -4 ), ( c = 2 ):[ C = frac{-(-4) pm sqrt{(-4)^2 - 4(1)(2)}}{2(1)} ][ C = frac{4 pm sqrt{16 - 8}}{2} ][ C = frac{4 pm sqrt{8}}{2} ]Simplify ( sqrt{8} = 2sqrt{2} ):[ C = frac{4 pm 2sqrt{2}}{2} ]Factor out a 2:[ C = frac{2(2 pm sqrt{2})}{2} ]Cancel the 2s:[ C = 2 pm sqrt{2} ]So, the roots are ( 2 + sqrt{2} ) and ( 2 - sqrt{2} ). Approximately, ( sqrt{2} ) is about 1.414, so:- ( 2 + 1.414 = 3.414 )- ( 2 - 1.414 = 0.586 )Since the coefficient of ( C^2 ) is positive, the parabola opens upwards. Therefore, the inequality ( C^2 - 4C + 2 geq 0 ) holds when ( C leq 2 - sqrt{2} ) or ( C geq 2 + sqrt{2} ).Now, moving to the second inequality:[ C^2 - 4C + 3 leq 3 ]Subtract 3 from both sides:[ C^2 - 4C leq 0 ]Factor the left side:[ C(C - 4) leq 0 ]This is a quadratic inequality. The roots are ( C = 0 ) and ( C = 4 ). The parabola opens upwards, so the inequality ( C(C - 4) leq 0 ) holds between the roots, i.e., when ( 0 leq C leq 4 ).Now, combining both inequalities from the first and second parts:From the first inequality, ( C leq 2 - sqrt{2} ) or ( C geq 2 + sqrt{2} ).From the second inequality, ( 0 leq C leq 4 ).So, the solution is the intersection of these two results.Let me visualize this. The first inequality gives two intervals: ( (-infty, 2 - sqrt{2}] ) and ( [2 + sqrt{2}, infty) ). The second inequality restricts ( C ) to ( [0, 4] ).Therefore, the intersection is:- ( [0, 2 - sqrt{2}] ) and ( [2 + sqrt{2}, 4] )So, the range of ( C ) that satisfies both inequalities is ( 0 leq C leq 2 - sqrt{2} ) or ( 2 + sqrt{2} leq C leq 4 ).But wait, in the context of the problem, ( C ) represents carbohydrate intake in grams. It's unlikely that an athlete would have zero carbohydrates, but since the problem doesn't specify a lower bound, I guess mathematically 0 is allowed. However, in reality, athletes need some carbs, but perhaps the model allows for 0.But let me check if ( C = 0 ) is acceptable. Plugging into the original equation:[ P = 0^2 - 4(0) + 3 = 3 ]So, ( P = 3 ), which is within the desired range. So, ( C = 0 ) is acceptable mathematically.Similarly, ( C = 4 ):[ P = 16 - 16 + 3 = 3 ]Which is also within the range.So, the range is from 0 up to ( 2 - sqrt{2} ) and from ( 2 + sqrt{2} ) up to 4.But wait, let me make sure about the endpoints. For the first inequality, ( C leq 2 - sqrt{2} ) or ( C geq 2 + sqrt{2} ). But the second inequality restricts ( C ) to be between 0 and 4. So, the combined solution is ( 0 leq C leq 2 - sqrt{2} ) and ( 2 + sqrt{2} leq C leq 4 ).But let me check if ( C = 2 - sqrt{2} ) is approximately 0.586, which is greater than 0, so that's fine. And ( 2 + sqrt{2} ) is approximately 3.414, which is less than 4, so that's also fine.Therefore, the range of ( C ) is two intervals: from 0 to approximately 0.586 grams and from approximately 3.414 grams to 4 grams.But let me express this exactly using square roots. So, the exact values are ( 2 - sqrt{2} ) and ( 2 + sqrt{2} ).Therefore, the range is:[ 0 leq C leq 2 - sqrt{2} quad text{or} quad 2 + sqrt{2} leq C leq 4 ]But wait, the problem says "the range of carbohydrate intake that ensures the protein intake remains between 1 and 3 grams, inclusive." So, I need to make sure that for all ( C ) in these intervals, ( P ) is between 1 and 3.Let me test a value in each interval to confirm.First interval: Let's pick ( C = 1 ). Then:[ P = 1 - 4 + 3 = 0 ]Wait, that's 0, which is below 1. Hmm, that's a problem. Did I make a mistake?Wait, no, because in the first interval, ( C ) is between 0 and ( 2 - sqrt{2} ) (approximately 0.586). So, if I pick ( C = 0.5 ), which is within 0 and 0.586:[ P = 0.25 - 2 + 3 = 1.25 ]Which is within 1 and 3. Okay, that works.If I pick ( C = 0 ):[ P = 0 - 0 + 3 = 3 ]Which is exactly 3, so that's fine.If I pick ( C = 2 - sqrt{2} ) (approximately 0.586):[ P = (2 - sqrt{2})^2 - 4(2 - sqrt{2}) + 3 ]Let me compute that:First, expand ( (2 - sqrt{2})^2 ):[ 4 - 4sqrt{2} + 2 = 6 - 4sqrt{2} ]Then, subtract ( 4(2 - sqrt{2}) ):[ 6 - 4sqrt{2} - 8 + 4sqrt{2} = -2 ]Then add 3:[ -2 + 3 = 1 ]So, at ( C = 2 - sqrt{2} ), ( P = 1 ), which is the lower bound.Similarly, at ( C = 2 + sqrt{2} ):[ P = (2 + sqrt{2})^2 - 4(2 + sqrt{2}) + 3 ]Expand ( (2 + sqrt{2})^2 ):[ 4 + 4sqrt{2} + 2 = 6 + 4sqrt{2} ]Subtract ( 4(2 + sqrt{2}) ):[ 6 + 4sqrt{2} - 8 - 4sqrt{2} = -2 ]Add 3:[ -2 + 3 = 1 ]Wait, that's also 1. But earlier, I thought ( C = 4 ) gives ( P = 3 ). Let me check ( C = 4 ):[ P = 16 - 16 + 3 = 3 ]Yes, that's correct.Wait, so at ( C = 2 + sqrt{2} ), ( P = 1 ), and as ( C ) increases beyond that up to 4, ( P ) increases from 1 to 3.Wait, but when I tested ( C = 3 ), which is between ( 2 + sqrt{2} ) (~3.414) and 4, let's see:[ P = 9 - 12 + 3 = 0 ]Wait, that can't be right. Wait, ( C = 3 ) is less than ( 2 + sqrt{2} ) (~3.414), so it's actually in the middle region where ( P ) is less than 1. But according to our earlier solution, the valid ranges are ( 0 leq C leq 2 - sqrt{2} ) and ( 2 + sqrt{2} leq C leq 4 ). So, ( C = 3 ) is not in the valid range, which is why ( P = 0 ) there, which is below 1.Wait, but earlier when I solved the inequalities, I thought that for ( C ) between ( 2 + sqrt{2} ) and 4, ( P ) would be between 1 and 3. But when I plug in ( C = 3.5 ), let's see:[ P = (3.5)^2 - 4(3.5) + 3 = 12.25 - 14 + 3 = 1.25 ]Which is within 1 and 3. Okay, that works.Wait, but when I plug in ( C = 3 ), which is less than ( 2 + sqrt{2} ) (~3.414), I get ( P = 0 ), which is outside the desired range. So, that's why our solution is correct: only ( C ) values in ( [0, 2 - sqrt{2}] ) and ( [2 + sqrt{2}, 4] ) give ( P ) between 1 and 3.So, to express the range as an inequality, it's two separate intervals. But the problem says "the range of carbohydrate intake," so I think it's acceptable to write it as two inequalities combined with 'or'.So, the final answer for part b is:[ 0 leq C leq 2 - sqrt{2} quad text{or} quad 2 + sqrt{2} leq C leq 4 ]But let me write it using inequalities without the 'or' for clarity. So, combining both intervals, it's:[ C in [0, 2 - sqrt{2}] cup [2 + sqrt{2}, 4] ]But since the question asks to express it as an inequality, I think writing both parts with 'or' is appropriate.Wait, but in the problem statement, it's part b, and the answer should be in boxed notation. So, I need to present it as an inequality.Alternatively, since it's two separate intervals, I can write it as:[ 0 leq C leq 2 - sqrt{2} quad text{or} quad 2 + sqrt{2} leq C leq 4 ]But perhaps the problem expects a single inequality, but since it's two separate ranges, I think it's okay to present both.Wait, let me check if I made any mistakes in solving the inequalities.First inequality: ( C^2 - 4C + 3 geq 1 ) led to ( C^2 - 4C + 2 geq 0 ), which has roots at ( 2 pm sqrt{2} ). Since the parabola opens upwards, the solution is ( C leq 2 - sqrt{2} ) or ( C geq 2 + sqrt{2} ).Second inequality: ( C^2 - 4C + 3 leq 3 ) led to ( C^2 - 4C leq 0 ), which factors to ( C(C - 4) leq 0 ), so solution is ( 0 leq C leq 4 ).Intersection of these is ( 0 leq C leq 2 - sqrt{2} ) and ( 2 + sqrt{2} leq C leq 4 ). So, that's correct.Therefore, the range is as above.So, summarizing:a) When ( P = 2 ), ( C = 2 pm sqrt{3} ), approximately 0.268g and 3.732g.b) The range of ( C ) is from 0 to ( 2 - sqrt{2} ) and from ( 2 + sqrt{2} ) to 4.I think that's it.

answer：Okay, so I have this problem about a hyperbola, and I need to find the radius of the incircle of a triangle formed by a point P on the hyperbola and the two foci. Let me try to break this down step by step.First, let me recall what a hyperbola is. The equation given is ( frac{x^{2}}{4} - frac{y^{2}}{12} = 1 ). I know that the standard form of a hyperbola is ( frac{(x - h)^2}{a^2} - frac{(y - k)^2}{b^2} = 1 ) for a hyperbola that opens left and right. Here, the center is at (h, k), which in this case is (0, 0) because there are no shifts in the equation. So, the center is at the origin.Next, I need to find the foci of this hyperbola. For hyperbolas, the distance from the center to each focus is given by ( c ), where ( c^2 = a^2 + b^2 ). In this equation, ( a^2 = 4 ) and ( b^2 = 12 ). So, ( c^2 = 4 + 12 = 16 ), which means ( c = 4 ). Therefore, the foci are located at ( -4, 0 ) and ( 4, 0 ). Let me denote them as ( F_1 = (-4, 0) ) and ( F_2 = (4, 0) ).Now, the problem mentions a point P on the hyperbola. Let me denote the coordinates of P as (x, y). Since P lies on the hyperbola, it must satisfy the equation ( frac{x^2}{4} - frac{y^2}{12} = 1 ).The triangle in question is ( triangle PF_1F_2 ). I need to find the centroid (M) and the incenter (I) of this triangle. The condition given is that the line MI is perpendicular to the x-axis. Since MI is perpendicular to the x-axis, that means MI is a vertical line. So, the x-coordinates of M and I must be the same.Let me recall what the centroid and incenter are. The centroid M of a triangle is the intersection point of the medians, and its coordinates are the average of the coordinates of the three vertices. So, for triangle ( PF_1F_2 ), the centroid M will have coordinates:( M_x = frac{x + (-4) + 4}{3} = frac{x}{3} )( M_y = frac{y + 0 + 0}{3} = frac{y}{3} )So, M is at ( left( frac{x}{3}, frac{y}{3} right) ).Now, the incenter I is the point where the angle bisectors of the triangle meet, and it is also the center of the incircle. The coordinates of the incenter can be found using the formula:( I_x = frac{a x_A + b x_B + c x_C}{a + b + c} )( I_y = frac{a y_A + b y_B + c y_C}{a + b + c} )where a, b, c are the lengths of the sides opposite to vertices A, B, C respectively.In triangle ( PF_1F_2 ), let me denote the vertices as follows:- A = P = (x, y)- B = F1 = (-4, 0)- C = F2 = (4, 0)So, the sides opposite to these vertices are:- a = length of BC- b = length of AC- c = length of ABLet me compute the lengths of these sides.First, the length of BC. Since B is (-4, 0) and C is (4, 0), the distance between them is:( a = |BC| = sqrt{(4 - (-4))^2 + (0 - 0)^2} = sqrt{8^2} = 8 )Next, the length of AC. A is (x, y) and C is (4, 0). So,( b = |AC| = sqrt{(x - 4)^2 + (y - 0)^2} = sqrt{(x - 4)^2 + y^2} )Similarly, the length of AB. A is (x, y) and B is (-4, 0). So,( c = |AB| = sqrt{(x + 4)^2 + (y - 0)^2} = sqrt{(x + 4)^2 + y^2} )So, the incenter coordinates are:( I_x = frac{a x_A + b x_B + c x_C}{a + b + c} = frac{8 cdot x + sqrt{(x - 4)^2 + y^2} cdot (-4) + sqrt{(x + 4)^2 + y^2} cdot 4}{8 + sqrt{(x - 4)^2 + y^2} + sqrt{(x + 4)^2 + y^2}} )( I_y = frac{a y_A + b y_B + c y_C}{a + b + c} = frac{8 cdot y + sqrt{(x - 4)^2 + y^2} cdot 0 + sqrt{(x + 4)^2 + y^2} cdot 0}{8 + sqrt{(x - 4)^2 + y^2} + sqrt{(x + 4)^2 + y^2}} = frac{8y}{8 + sqrt{(x - 4)^2 + y^2} + sqrt{(x + 4)^2 + y^2}} )So, the incenter I has coordinates:( I = left( frac{8x - 4sqrt{(x - 4)^2 + y^2} + 4sqrt{(x + 4)^2 + y^2}}{8 + sqrt{(x - 4)^2 + y^2} + sqrt{(x + 4)^2 + y^2}}, frac{8y}{8 + sqrt{(x - 4)^2 + y^2} + sqrt{(x + 4)^2 + y^2}} right) )Now, the centroid M is at ( left( frac{x}{3}, frac{y}{3} right) ), and the incenter I is at the coordinates above. The condition given is that MI is perpendicular to the x-axis, which means that the line connecting M and I is vertical. Therefore, the x-coordinates of M and I must be equal. So,( frac{x}{3} = frac{8x - 4sqrt{(x - 4)^2 + y^2} + 4sqrt{(x + 4)^2 + y^2}}{8 + sqrt{(x - 4)^2 + y^2} + sqrt{(x + 4)^2 + y^2}} )Let me denote ( sqrt{(x - 4)^2 + y^2} = d ) and ( sqrt{(x + 4)^2 + y^2} = e ) to simplify the equation.So, substituting, the equation becomes:( frac{x}{3} = frac{8x - 4d + 4e}{8 + d + e} )Let me cross-multiply to eliminate the denominator:( x(8 + d + e) = 3(8x - 4d + 4e) )Expanding both sides:Left side: ( 8x + x d + x e )Right side: ( 24x - 12d + 12e )Bring all terms to the left side:( 8x + x d + x e - 24x + 12d - 12e = 0 )Simplify:( (8x - 24x) + (x d + 12d) + (x e - 12e) = 0 )( (-16x) + d(x + 12) + e(x - 12) = 0 )So,( -16x + d(x + 12) + e(x - 12) = 0 )But remember that ( d = sqrt{(x - 4)^2 + y^2} ) and ( e = sqrt{(x + 4)^2 + y^2} ). Let me substitute back:( -16x + sqrt{(x - 4)^2 + y^2}(x + 12) + sqrt{(x + 4)^2 + y^2}(x - 12) = 0 )This seems complicated, but maybe we can find a way to simplify it. Let me think about the properties of the hyperbola.Given that P is on the hyperbola ( frac{x^2}{4} - frac{y^2}{12} = 1 ), which is a hyperbola centered at the origin, opening to the left and right. The foci are at (-4, 0) and (4, 0), as we found earlier.One important property of hyperbolas is that the difference of distances from any point on the hyperbola to the two foci is constant and equal to ( 2a ). In this case, ( a^2 = 4 ), so ( a = 2 ). Therefore, ( |PF_2 - PF_1| = 2a = 4 ).So, ( |e - d| = 4 ), where ( e = PF_2 ) and ( d = PF_1 ). Since the hyperbola opens to the right and left, for points on the right branch, ( PF_2 - PF_1 = 4 ), and for points on the left branch, ( PF_1 - PF_2 = 4 ). Let me assume that P is on the right branch, so ( e - d = 4 ). If that's not the case, maybe we can adjust later.So, ( e = d + 4 ). Let me substitute this into our equation:( -16x + d(x + 12) + (d + 4)(x - 12) = 0 )Expanding:( -16x + d x + 12 d + d x - 12 d + 4x - 48 = 0 )Simplify term by term:-16x + d x + 12 d + d x - 12 d + 4x - 48Combine like terms:For x terms: (-16x + d x + d x + 4x) = (-16x + 2d x + 4x) = (-12x + 2d x)For d terms: (12 d - 12 d) = 0Constants: -48So, the equation becomes:( -12x + 2d x - 48 = 0 )Factor out 2x:( 2x(-6 + d) - 48 = 0 )Wait, maybe another approach. Let me write it as:( (-12x + 4x) + 2d x - 48 = 0 )Wait, that might not be the best way. Let me go back.Wait, after expanding, I have:-16x + d x + 12 d + d x - 12 d + 4x - 48Combine x terms:-16x + d x + d x + 4x = (-16x + 4x) + (d x + d x) = (-12x) + 2d xCombine d terms:12 d - 12 d = 0Constants: -48So, overall:( (-12x + 2d x) - 48 = 0 )Factor x:( x(-12 + 2d) - 48 = 0 )So,( x(2d - 12) = 48 )Divide both sides by 2:( x(d - 6) = 24 )So,( x = frac{24}{d - 6} )But d is ( sqrt{(x - 4)^2 + y^2} ). So, substituting back:( x = frac{24}{sqrt{(x - 4)^2 + y^2} - 6} )This seems a bit messy, but maybe we can square both sides to eliminate the square root.Let me denote ( sqrt{(x - 4)^2 + y^2} = d ), so:( x = frac{24}{d - 6} )Multiply both sides by (d - 6):( x(d - 6) = 24 )Which is the same as before.But since ( d = sqrt{(x - 4)^2 + y^2} ), let me express y^2 from the hyperbola equation.From the hyperbola equation:( frac{x^2}{4} - frac{y^2}{12} = 1 )Multiply both sides by 12:( 3x^2 - y^2 = 12 )So,( y^2 = 3x^2 - 12 )Therefore, ( d = sqrt{(x - 4)^2 + 3x^2 - 12} )Simplify inside the square root:( (x - 4)^2 + 3x^2 - 12 = x^2 - 8x + 16 + 3x^2 - 12 = 4x^2 - 8x + 4 )So,( d = sqrt{4x^2 - 8x + 4} )Factor out 4:( d = sqrt{4(x^2 - 2x + 1)} = sqrt{4(x - 1)^2} = 2|x - 1| )Since P is on the hyperbola, and we assumed it's on the right branch, x should be greater than or equal to 2 (since a = 2). So, x >= 2, so x - 1 >= 1, which is positive. Therefore, |x - 1| = x - 1, so:( d = 2(x - 1) )So, d = 2x - 2Now, going back to the equation:( x = frac{24}{d - 6} )Substitute d = 2x - 2:( x = frac{24}{(2x - 2) - 6} = frac{24}{2x - 8} = frac{24}{2(x - 4)} = frac{12}{x - 4} )So,( x = frac{12}{x - 4} )Multiply both sides by (x - 4):( x(x - 4) = 12 )Expand:( x^2 - 4x = 12 )Bring all terms to one side:( x^2 - 4x - 12 = 0 )Solve this quadratic equation:Using quadratic formula:( x = frac{4 pm sqrt{16 + 48}}{2} = frac{4 pm sqrt{64}}{2} = frac{4 pm 8}{2} )So,( x = frac{4 + 8}{2} = 6 ) or ( x = frac{4 - 8}{2} = -2 )But since we assumed P is on the right branch, x >= 2, so x = 6 is acceptable, and x = -2 is on the left branch, which we initially didn't consider. Let me check if x = -2 can be a solution.If x = -2, then from the hyperbola equation:( frac{(-2)^2}{4} - frac{y^2}{12} = 1 )Simplify:( frac{4}{4} - frac{y^2}{12} = 1 implies 1 - frac{y^2}{12} = 1 implies - frac{y^2}{12} = 0 implies y = 0 )So, P would be (-2, 0). But then, the triangle ( PF_1F_2 ) would have vertices at (-2, 0), (-4, 0), and (4, 0). This is a degenerate triangle because all points are on the x-axis. So, the inradius would be zero, which doesn't make sense in this context. Therefore, we discard x = -2.So, x = 6 is the valid solution. Let me find y.From the hyperbola equation:( y^2 = 3x^2 - 12 = 3(6)^2 - 12 = 3*36 - 12 = 108 - 12 = 96 )So, y = ±√96 = ±4√6Therefore, point P is (6, 4√6) or (6, -4√6). Since the hyperbola is symmetric with respect to the x-axis, both points will give the same result for the inradius. Let me take P = (6, 4√6) for simplicity.Now, let me compute the inradius of triangle ( PF_1F_2 ). The inradius r is given by the formula:( r = frac{A}{s} )where A is the area of the triangle, and s is the semi-perimeter.First, let me find the lengths of the sides of the triangle.We already have:- a = |F1F2| = 8- b = |PF2| = e = d + 4 = 2x - 2 + 4 = 2x + 2. Wait, hold on. Earlier, we found that d = 2x - 2, and since e = d + 4, then e = 2x - 2 + 4 = 2x + 2.But let me compute them directly.Point P is (6, 4√6), F1 is (-4, 0), F2 is (4, 0).Compute |PF1|:( |PF1| = sqrt{(6 - (-4))^2 + (4√6 - 0)^2} = sqrt{(10)^2 + (4√6)^2} = sqrt{100 + 16*6} = sqrt{100 + 96} = sqrt{196} = 14 )Compute |PF2|:( |PF2| = sqrt{(6 - 4)^2 + (4√6 - 0)^2} = sqrt{(2)^2 + (4√6)^2} = sqrt{4 + 96} = sqrt{100} = 10 )Compute |F1F2|:We already know it's 8.So, sides are:- a = |F1F2| = 8- b = |PF2| = 10- c = |PF1| = 14Wait, hold on. In the inradius formula, the sides are usually denoted as a, b, c opposite to angles A, B, C. But in our case, the triangle is ( PF_1F_2 ), so sides opposite to P, F1, F2 are |F1F2|, |PF2|, |PF1| respectively.But for the inradius, it doesn't matter how we denote them; the formula is r = A / s, where s is the semi-perimeter.So, let me compute the semi-perimeter s:( s = frac{a + b + c}{2} = frac{8 + 10 + 14}{2} = frac{32}{2} = 16 )Now, compute the area A of triangle ( PF_1F_2 ). Since the triangle has vertices at (6, 4√6), (-4, 0), and (4, 0), I can use the shoelace formula or compute the area as the base times height.Alternatively, since two of the points are on the x-axis, the base is |F1F2| = 8, and the height is the y-coordinate of P, which is 4√6.So, area A = (base * height)/2 = (8 * 4√6)/2 = (32√6)/2 = 16√6Therefore, the inradius r = A / s = 16√6 / 16 = √6Wait, so the inradius is √6.But let me double-check my calculations to make sure.First, sides:- |PF1|: distance from (6,4√6) to (-4,0):√[(6 - (-4))² + (4√6 - 0)²] = √[10² + (4√6)²] = √[100 + 96] = √196 = 14. Correct.- |PF2|: distance from (6,4√6) to (4,0):√[(6 - 4)² + (4√6 - 0)²] = √[2² + (4√6)²] = √[4 + 96] = √100 = 10. Correct.- |F1F2|: distance between (-4,0) and (4,0): 8. Correct.Semi-perimeter s = (14 + 10 + 8)/2 = 32/2 = 16. Correct.Area A: Using base 8 and height 4√6, area is (8 * 4√6)/2 = 16√6. Correct.Therefore, r = A / s = 16√6 / 16 = √6. So, the inradius is √6.Wait, but let me also confirm this using another method, just to be thorough.Alternatively, I can compute the area using Heron's formula.Heron's formula states that A = √[s(s - a)(s - b)(s - c)]So, plugging in:A = √[16(16 - 14)(16 - 10)(16 - 8)] = √[16 * 2 * 6 * 8] = √[16 * 2 * 6 * 8]Compute inside the square root:16 * 2 = 3232 * 6 = 192192 * 8 = 1536So, A = √1536Simplify √1536:1536 = 256 * 6, since 256 * 6 = 1536√1536 = √(256 * 6) = √256 * √6 = 16√6Which matches our previous calculation. So, A = 16√6, s = 16, so r = √6.Therefore, the radius of the incircle is √6.I think this makes sense. Let me just recap the steps to ensure I didn't skip anything important.1. Found the hyperbola's foci at (-4, 0) and (4, 0).2. Expressed the centroid M in terms of P's coordinates.3. Expressed the incenter I using the formula, which led to a complicated equation.4. Used the property of hyperbola that |PF2 - PF1| = 4, assuming P is on the right branch.5. Substituted and simplified to find x = 6, y = ±4√6.6. Calculated the sides of triangle PF1F2 as 8, 10, 14.7. Found semi-perimeter s = 16 and area A = 16√6.8. Computed inradius r = A / s = √6.Everything seems consistent. I don't see any mistakes in the calculations, so I'm confident the answer is √6.**Final Answer**The radius of the incircle is boxed{sqrt{6}}.

answer：First, I'll define the original number of CDs as ( x ). Since the number of vinyl records is twice the number of CDs, the original number of records is ( 2x ).Next, I'll calculate the new totals after the additions. The new number of records is 50% more than the original, so it becomes ( 2x + 0.5 times 2x = 3x ). Similarly, the new number of CDs is 75% more than the original, making it ( x + 0.75x = 1.75x ).Adding these together, the total number of items after the additions is ( 3x + 1.75x = 4.75x ).Given that the total collection after additions is 175 items, I'll set up the equation ( 4.75x = 175 ) and solve for ( x ). Dividing both sides by 4.75 gives ( x = 36.84 ). Since the number of CDs must be a whole number, I'll round this to 37 CDs. Consequently, the original number of records is ( 2 times 37 = 74 ).

answer：Okay, so I have this arithmetic sequence problem here. Let me try to figure it out step by step. The problem says that in the arithmetic sequence {a_n}, a_3 is equal to 4, and the sum of a_4 and a_9 is 22. I need to find the sum of the first 11 terms, S_11. Hmm, okay.First, let me recall what an arithmetic sequence is. It's a sequence where each term after the first is obtained by adding a constant difference. So, if I denote the first term as a_1 and the common difference as d, then the nth term can be written as:a_n = a_1 + (n - 1)dRight, so each term is just the first term plus (n-1) times the common difference. Got that.Now, the problem gives me two specific pieces of information: a_3 = 4 and a_4 + a_9 = 22. I need to use these to find S_11. Let me write down what I know.First, let's express a_3 using the formula for the nth term:a_3 = a_1 + (3 - 1)d = a_1 + 2dAnd we know that a_3 is 4, so:a_1 + 2d = 4 ...(1)Next, let's express a_4 and a_9 similarly.a_4 = a_1 + (4 - 1)d = a_1 + 3da_9 = a_1 + (9 - 1)d = a_1 + 8dThe problem states that a_4 + a_9 = 22, so:(a_1 + 3d) + (a_1 + 8d) = 22Let me simplify that:a_1 + 3d + a_1 + 8d = 22Combine like terms:2a_1 + 11d = 22 ...(2)So now I have two equations:1) a_1 + 2d = 42) 2a_1 + 11d = 22I need to solve this system of equations to find a_1 and d. Once I have those, I can find S_11.Let me solve equation (1) for a_1:a_1 = 4 - 2dNow, substitute this expression for a_1 into equation (2):2*(4 - 2d) + 11d = 22Let me compute that step by step.First, multiply out the 2:2*4 = 8, and 2*(-2d) = -4dSo, 8 - 4d + 11d = 22Combine like terms (-4d + 11d):8 + 7d = 22Now, subtract 8 from both sides:7d = 22 - 87d = 14Divide both sides by 7:d = 14 / 7d = 2Okay, so the common difference d is 2. Now, let's find a_1 using equation (1):a_1 = 4 - 2d = 4 - 2*2 = 4 - 4 = 0So, the first term a_1 is 0. Interesting, so the sequence starts at 0 and each term increases by 2.Let me write out the first few terms to check:a_1 = 0a_2 = a_1 + d = 0 + 2 = 2a_3 = a_2 + d = 2 + 2 = 4 (which matches the given a_3 = 4)a_4 = a_3 + d = 4 + 2 = 6a_9 would be a_1 + 8d = 0 + 16 = 16So, a_4 + a_9 = 6 + 16 = 22, which also matches the given condition. Good, so my values for a_1 and d seem correct.Now, I need to find the sum of the first 11 terms, S_11.I remember that the formula for the sum of the first n terms of an arithmetic sequence is:S_n = n/2 * (2a_1 + (n - 1)d)Alternatively, it can also be written as:S_n = n/2 * (a_1 + a_n)Since I know a_1 and d, I can use either formula. Let me use the first one because I already have a_1 and d.So, plugging in n = 11, a_1 = 0, and d = 2:S_11 = 11/2 * (2*0 + (11 - 1)*2)Simplify inside the parentheses first:2*0 = 0(11 - 1)*2 = 10*2 = 20So, inside the parentheses, it's 0 + 20 = 20Now, multiply by 11/2:S_11 = (11/2)*20Let me compute that:11 divided by 2 is 5.5, and 5.5 multiplied by 20 is 110.Alternatively, 20 divided by 2 is 10, and 11*10 is 110.Either way, S_11 is 110.Wait, just to double-check, maybe I can use the other formula for the sum:S_n = n/2 * (a_1 + a_n)I need a_11 for this. Let me compute a_11:a_11 = a_1 + (11 - 1)d = 0 + 10*2 = 20So, a_11 is 20.Then, S_11 = 11/2 * (0 + 20) = 11/2 * 20 = (11*20)/2 = 220/2 = 110Same result. So, that confirms it.Alternatively, I can think about the sequence. Since a_1 is 0 and d is 2, the sequence is 0, 2, 4, 6, ..., up to a_11 which is 20.So, the terms are 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.Let me add them up:0 + 2 = 22 + 4 = 66 + 6 = 1212 + 8 = 2020 + 10 = 3030 + 12 = 4242 + 14 = 5656 + 16 = 7272 + 18 = 9090 + 20 = 110Yep, that adds up to 110 as well.So, all methods give me S_11 = 110. That must be the correct answer.**Final Answer**The sum of the first 11 terms is boxed{110}.

answer：Alright, so I'm a college student studying international business, and I have this problem about exchange rates for two different countries I'm planning to study in. Let me try to work through this step by step.First, for the first country, they use currency A, and the exchange rate is given by the polynomial equation ( f(x) = x^3 - 4x + 3 ). I need to solve ( f(x) = 0 ) to find the critical points where the value of currency A in my home currency changes direction. These points will help me determine the budget limits for my stay there.Okay, so solving ( x^3 - 4x + 3 = 0 ). Hmm, this is a cubic equation. I remember that to solve cubic equations, sometimes you can factor them by finding rational roots. The Rational Root Theorem says that any possible rational root, p/q, is a factor of the constant term over a factor of the leading coefficient. Here, the constant term is 3, and the leading coefficient is 1, so possible roots are ±1, ±3.Let me test these. Starting with x=1: ( 1 - 4 + 3 = 0 ). Oh, that works! So (x - 1) is a factor. Now, I can perform polynomial division or use synthetic division to factor this cubic.Using synthetic division with root 1:1 | 1 0 -4 3 1 1 -3 1 1 -3 0So, the cubic factors into (x - 1)(x² + x - 3). Now, I need to solve ( x² + x - 3 = 0 ). Using the quadratic formula: ( x = frac{-b pm sqrt{b² - 4ac}}{2a} ). Here, a=1, b=1, c=-3.Calculating discriminant: ( 1 + 12 = 13 ). So, roots are ( frac{-1 pm sqrt{13}}{2} ). So, the solutions are x=1, x=( -1 + sqrt(13) )/2, and x=( -1 - sqrt(13) )/2.Let me approximate these roots to understand them better. sqrt(13) is approximately 3.6055. So, ( -1 + 3.6055 ) / 2 ≈ 2.6055 / 2 ≈ 1.30275. And ( -1 - 3.6055 ) / 2 ≈ -4.6055 / 2 ≈ -2.30275.So, the roots are approximately x ≈ -2.303, x=1, and x≈1.303. These are the critical points where the exchange rate function f(x) crosses zero. Since exchange rates can't be negative, I think we can ignore the negative root because you can't have negative currency. So, the relevant critical points are x=1 and x≈1.303.Wait, but in the context of exchange rates, x is the amount of currency A, so it must be positive. So, the critical points where the function changes direction are at x=1 and x≈1.303. These points are where the function f(x) crosses the x-axis, meaning the exchange rate changes from positive to negative or vice versa. But since exchange rates are positive, maybe these points indicate where the rate is zero, which would be the limits of the budget.So, if f(x) represents the equivalent amount in home currency, when f(x)=0, that would mean the value is zero, which might not be practical. Maybe I need to consider where the function is increasing or decreasing, so the critical points where the derivative is zero.Wait, the problem says "critical points where the value of currency A in your home currency changes direction." So, maybe they mean where the function f(x) has local maxima or minima, which are found by taking the derivative and setting it to zero.Oh, right! Critical points in calculus are where the derivative is zero or undefined. So, maybe I was supposed to take the derivative of f(x) to find where the function changes direction, i.e., from increasing to decreasing or vice versa.Let me double-check. The original problem says: "Solve the polynomial equation f(x) = 0 to find the critical points where the value of currency A in your home currency changes direction." Hmm, it says solve f(x)=0, but critical points are typically where f'(x)=0.Wait, maybe the problem is using "critical points" in a different sense, like where the function crosses zero, which are the roots. But in calculus, critical points are where the derivative is zero or undefined, which are the local maxima or minima.I think the problem might be mixing terminology. If it's asking for critical points in the calculus sense, then I need to find f'(x) and set it to zero. Let me try that.f(x) = x³ - 4x + 3f'(x) = 3x² - 4Set f'(x) = 0: 3x² - 4 = 0 => x² = 4/3 => x = ±√(4/3) ≈ ±1.1547So, critical points at x≈1.1547 and x≈-1.1547. Again, negative x doesn't make sense here, so the critical point is at x≈1.1547.This is where the function changes direction, from increasing to decreasing or vice versa. So, at x≈1.1547, the function has a local maximum or minimum.Wait, let me check the second derivative to see if it's a max or min.f''(x) = 6xAt x≈1.1547, f''(x) is positive, so it's a local minimum.So, the function f(x) has a local minimum at x≈1.1547. So, the value of currency A in home currency changes direction here from decreasing to increasing.But how does this help determine the budget limits? Maybe the roots of f(x)=0 indicate where the exchange rate is zero, which would be the limits beyond which you can't exchange anymore? Or perhaps the critical points indicate the optimal exchange amounts.Wait, maybe I need to graph f(x) to understand better. f(x) is a cubic, so it goes from negative infinity to positive infinity. It crosses the x-axis at x≈-2.303, x=1, and x≈1.303. The critical points are at x≈-1.1547 (local maximum) and x≈1.1547 (local minimum).So, for positive x, the function f(x) starts at x=0, f(0)=3. Then, it decreases until x≈1.1547, reaching a local minimum, then increases beyond that.So, the critical point at x≈1.1547 is where the function stops decreasing and starts increasing. So, if I'm exchanging currency A, the value in home currency decreases until x≈1.1547, then increases. So, maybe the budget limit is around that point?But the problem says to solve f(x)=0. So, the roots are at x=1, x≈1.303, and x≈-2.303. Since x must be positive, the relevant roots are x=1 and x≈1.303. So, at x=1, f(x)=0, and at x≈1.303, f(x)=0 again.Wait, but f(1)=1 -4 +3=0, and f(1.303)= approximately (1.303)^3 -4*(1.303) +3. Let me compute that:1.303^3 ≈ 2.208, 4*1.303≈5.212, so 2.208 -5.212 +3≈0. So, yes, that's correct.So, the function f(x) is zero at x=1 and x≈1.303. So, between x=1 and x≈1.303, f(x) is negative, and outside of that, it's positive.But exchange rates can't be negative, so maybe the function f(x) represents the equivalent home currency, so when f(x)=0, that's the point where the value is zero, which would be a limit.But in reality, exchange rates don't go negative, so perhaps the function f(x) is only valid for certain ranges. So, if f(x) is positive, that's the exchange rate, and when it's negative, it's not applicable.So, the critical points where the function changes direction are at x=1 and x≈1.303. So, for x <1, f(x) is positive, decreasing until x≈1.1547, then increasing beyond x≈1.1547, but since f(x) is zero at x=1 and x≈1.303, the function is positive before x=1, negative between x=1 and x≈1.303, and positive again after x≈1.303.But since exchange rates can't be negative, the practical range is x <1 and x >1.303. So, the critical points at x=1 and x≈1.303 are the limits where the exchange rate is zero, beyond which it becomes positive again.So, for budget limits, I think the relevant points are x=1 and x≈1.303. So, if I exchange less than 1 unit of currency A, the equivalent home currency is positive, but between 1 and 1.303, it's negative, which is not possible, and beyond 1.303, it's positive again.But that seems a bit odd. Maybe I need to interpret this differently. Perhaps the function f(x) is only valid for certain ranges, and the critical points are where the exchange rate changes from increasing to decreasing or vice versa.Wait, the problem says "critical points where the value of currency A in your home currency changes direction." So, maybe it's referring to where the function f(x) changes from increasing to decreasing or vice versa, which are the critical points found by f'(x)=0, which is at x≈1.1547.So, the function f(x) is increasing before x≈1.1547 and decreasing after that? Wait, no, the derivative f'(x)=3x² -4. So, for x < sqrt(4/3)≈1.1547, f'(x) is negative, so f(x) is decreasing. For x > sqrt(4/3), f'(x) is positive, so f(x) is increasing.So, the function decreases until x≈1.1547, then increases after that. So, the critical point at x≈1.1547 is a local minimum.So, in terms of exchange rate, the equivalent home currency decreases as you exchange more currency A until x≈1.1547, then starts increasing again. So, the minimum point is at x≈1.1547, which is the point where the exchange rate is the lowest.So, for budget purposes, if I exchange around x≈1.1547, I get the least value in home currency, which would be the minimum. So, maybe I should avoid exchanging around that point to maximize my budget.But the problem says to solve f(x)=0 to find critical points. So, maybe the critical points are the roots, which are x=1 and x≈1.303, where the function crosses zero. So, these are the points where the exchange rate is zero, meaning beyond these points, the exchange rate becomes positive again.But in reality, exchange rates don't go negative, so maybe the function f(x) is only valid between x=1 and x≈1.303, where it's negative, which doesn't make sense. So, perhaps the function is only valid for x <1 and x >1.303, where it's positive.So, the critical points where the function changes direction (from positive to negative and back to positive) are at x=1 and x≈1.303. So, these are the points where the exchange rate is zero, and beyond these points, the exchange rate becomes positive again.Therefore, for budget limits, I should consider exchanging currency A either below x=1 or above x≈1.303. But since x=1 is a root, and beyond that, the function becomes negative until x≈1.303, where it becomes positive again.But this is getting a bit confusing. Maybe I need to consider that the exchange rate is given by f(x), so f(x) must be positive, so x must be less than 1 or greater than approximately 1.303.So, the critical points where the exchange rate changes direction (from positive to negative and back to positive) are at x=1 and x≈1.303. So, these are the limits where the exchange rate is zero, beyond which it becomes positive again.Therefore, for budget purposes, I should consider exchanging currency A either below 1 unit or above approximately 1.303 units to get a positive exchange rate.But I'm not entirely sure if this is the correct interpretation. Maybe the critical points are the roots, and they indicate the points where the exchange rate is zero, so beyond these points, the exchange rate is positive again.In any case, the problem asks to solve f(x)=0, so the solutions are x=1, x=( -1 + sqrt(13) )/2 ≈1.303, and x=( -1 - sqrt(13) )/2≈-2.303. Since negative currency doesn't make sense, the critical points are x=1 and x≈1.303.So, for part 1, the critical points are x=1 and x≈1.303.Now, moving on to part 2. The exchange rate for currency B is given by the quadratic equation ( g(y) = 2y² - 3y + 1 ). I need to solve the inequality ( g(y) < 5 ) to find the range of currency B that I can exchange to stay within my budget for the second country.So, solving ( 2y² - 3y + 1 < 5 ). Let's rewrite this inequality:2y² - 3y + 1 - 5 < 0 => 2y² - 3y - 4 < 0.So, now we have the quadratic inequality ( 2y² - 3y - 4 < 0 ). To solve this, I need to find the roots of the quadratic equation ( 2y² - 3y - 4 = 0 ) and then determine the intervals where the quadratic is negative.Using the quadratic formula: y = [3 ± sqrt(9 + 32)] / 4 = [3 ± sqrt(41)] / 4.Calculating sqrt(41) ≈6.4031.So, the roots are approximately:y = (3 + 6.4031)/4 ≈9.4031/4≈2.3508andy = (3 - 6.4031)/4≈(-3.4031)/4≈-0.8508.So, the quadratic crosses the y-axis at y≈-0.8508 and y≈2.3508.Since the coefficient of y² is positive (2), the parabola opens upwards. Therefore, the quadratic is negative between its roots.So, the solution to ( 2y² - 3y - 4 < 0 ) is y between approximately -0.8508 and 2.3508.But since y represents the amount of currency B, it can't be negative. So, the valid range is y >0 and y <2.3508.Therefore, the range of y is 0 < y < approximately 2.3508.So, to stay within the budget, I can exchange between 0 and approximately 2.3508 units of currency B.But let me express the exact roots instead of approximations. The roots are y = [3 ± sqrt(41)] / 4.So, the exact solution is y ∈ ( (3 - sqrt(41))/4 , (3 + sqrt(41))/4 ). Since (3 - sqrt(41))/4 is negative, the valid range is y ∈ (0, (3 + sqrt(41))/4 ).Calculating (3 + sqrt(41))/4 exactly is approximately 2.3508, as before.So, the range of y is from 0 to (3 + sqrt(41))/4.Therefore, to maximize my budget, I should exchange between 0 and approximately 2.3508 units of currency B.Wait, but the problem says "to stay within your budget for the second country." So, if g(y) is the equivalent home currency, then g(y) <5 means that the home currency equivalent is less than 5 units. So, I want to exchange y units of currency B such that the equivalent home currency is less than 5.Therefore, the range of y is between the two roots, but since y can't be negative, it's from 0 to (3 + sqrt(41))/4.So, summarizing:1. For currency A, the critical points where the exchange rate changes direction are at x=1 and x=( -1 + sqrt(13) )/2 ≈1.303.2. For currency B, the range of y that keeps the exchange within budget is 0 < y < (3 + sqrt(41))/4 ≈2.3508.I think that's it. Let me just double-check my calculations.For part 1, solving f(x)=0:x³ -4x +3=0Factored as (x-1)(x² +x -3)=0Solutions: x=1, x=(-1 ± sqrt(13))/2Approximate roots: x≈1, x≈1.303, x≈-2.303Ignoring negative, so x=1 and x≈1.303.For part 2, solving 2y² -3y +1 <52y² -3y -4 <0Roots at y=(3 ± sqrt(41))/4 ≈-0.8508 and 2.3508Since y>0, the range is 0<y<2.3508.Yes, that seems correct.

answer：Okay, so I have this problem where I need to find the distance from a point in polar coordinates to a line described by another polar equation. The point is given as (2, π/6) and the line is given by ρ sin(θ - π/6) = 1. Hmm, let me think about how to approach this.First, I remember that in polar coordinates, the distance from a point to a line can be found using some formula, but I'm not exactly sure what it is. Maybe I should convert both the point and the line into Cartesian coordinates because I'm more familiar with distance formulas in Cartesian systems.Alright, let's start by converting the point (2, π/6) to Cartesian coordinates. The conversion formulas are x = ρ cos θ and y = ρ sin θ. So, plugging in the values:x = 2 cos(π/6)y = 2 sin(π/6)I know that cos(π/6) is √3/2 and sin(π/6) is 1/2. So,x = 2*(√3/2) = √3y = 2*(1/2) = 1So the Cartesian coordinates of the point are (√3, 1). Got that down.Now, let's tackle the line equation ρ sin(θ - π/6) = 1. I need to convert this into Cartesian form. I remember that in polar coordinates, equations can sometimes be converted using the relationships x = ρ cos θ, y = ρ sin θ, and ρ² = x² + y².Looking at the equation ρ sin(θ - π/6) = 1, I think I can use the sine subtraction formula. The formula is sin(A - B) = sin A cos B - cos A sin B. So, applying that here:sin(θ - π/6) = sin θ cos(π/6) - cos θ sin(π/6)Plugging this back into the equation:ρ [sin θ cos(π/6) - cos θ sin(π/6)] = 1Now, let's substitute sin θ and cos θ with y/ρ and x/ρ respectively because sin θ = y/ρ and cos θ = x/ρ.So,ρ [ (y/ρ) cos(π/6) - (x/ρ) sin(π/6) ] = 1Simplify this:ρ [ (y cos(π/6) - x sin(π/6)) / ρ ] = 1The ρ in the numerator and denominator cancels out:y cos(π/6) - x sin(π/6) = 1So, the Cartesian equation of the line is y cos(π/6) - x sin(π/6) = 1.Now, let's compute the numerical values of cos(π/6) and sin(π/6). As before, cos(π/6) is √3/2 and sin(π/6) is 1/2. Plugging these in:y*(√3/2) - x*(1/2) = 1Let me write this as:(-1/2)x + (√3/2)y = 1To make it look more familiar, I can multiply both sides by 2 to eliminate the denominators:-1x + √3 y = 2So, the equation becomes:- x + √3 y = 2Alternatively, I can write it as:√3 y - x = 2Now, the standard form of a line is Ax + By + C = 0. So, rearranging the equation:- x + √3 y - 2 = 0Or, multiplying both sides by -1 to make the x coefficient positive:x - √3 y + 2 = 0Wait, but actually, the standard form is usually written with A positive, so let's stick with:- x + √3 y - 2 = 0But to make it consistent, maybe I should write it as:x - √3 y + 2 = 0Wait, no, because if I multiply both sides by -1, the equation becomes x - √3 y + 2 = 0, but that would change the signs of all terms. Let me double-check.Original equation after multiplying by 2: -x + √3 y = 2So, moving all terms to the left: -x + √3 y - 2 = 0Alternatively, x - √3 y + 2 = 0 is not the same because the signs are different. So, perhaps it's better to keep it as -x + √3 y - 2 = 0.But actually, for the distance formula, the sign doesn't matter because it's squared or absolute valued. So, maybe I can just write it as:√3 y - x - 2 = 0Yes, that's another way. So, the line is √3 y - x - 2 = 0.Now, I need to find the distance from the point (√3, 1) to this line. The distance from a point (x₀, y₀) to the line Ax + By + C = 0 is given by:Distance = |A x₀ + B y₀ + C| / sqrt(A² + B²)So, let's identify A, B, and C from the line equation. The line is √3 y - x - 2 = 0, which can be rewritten as -x + √3 y - 2 = 0.So, A = -1, B = √3, C = -2.Alternatively, if I write it as √3 y - x - 2 = 0, then A = -1, B = √3, C = -2.Either way, the coefficients are A = -1, B = √3, C = -2.Now, plug in the point (√3, 1):Distance = |A x₀ + B y₀ + C| / sqrt(A² + B²)= |(-1)(√3) + (√3)(1) + (-2)| / sqrt((-1)² + (√3)²)= | -√3 + √3 - 2 | / sqrt(1 + 3)= | (-√3 + √3) - 2 | / sqrt(4)= |0 - 2| / 2= | -2 | / 2= 2 / 2= 1Wait, so the distance is 1? That seems straightforward. Let me verify my steps.First, converting the point: correct, (√3, 1).Converting the line equation: started with ρ sin(θ - π/6) = 1, used sine subtraction formula, substituted sin θ and cos θ with y/ρ and x/ρ, simplified correctly to y cos(π/6) - x sin(π/6) = 1, which became (-1/2)x + (√3/2)y = 1, multiplied by 2 to get -x + √3 y = 2, then rearranged to standard form.Then, using the distance formula: correctly identified A, B, C as -1, √3, -2. Plugged in (√3, 1):(-1)(√3) + (√3)(1) + (-2) = -√3 + √3 - 2 = -2, absolute value is 2, divided by sqrt(1 + 3) = 2, so 2/2 = 1.Yes, seems correct.Alternatively, I remember that in polar coordinates, the distance from a point (ρ, θ) to the line ρ sin(θ - α) = d is |ρ sin(θ - α) - d|. Wait, is that a formula?Let me think. The general equation of a line in polar coordinates is ρ sin(θ - α) = p, where p is the distance from the origin to the line, and α is the angle between the positive x-axis and the line connecting the origin to the closest point on the line.So, if we have a point (ρ, θ) and a line ρ sin(θ - α) = p, then the distance from the point to the line is |ρ sin(θ - α) - p|.Wait, is that right? Let me check.Yes, actually, I think that's a formula. So, in this case, the line is ρ sin(θ - π/6) = 1, so p = 1 and α = π/6.The point is (2, π/6). So, plugging into the formula:Distance = |2 sin(π/6 - π/6) - 1| = |2 sin(0) - 1| = |0 - 1| = 1.So, same result. That's reassuring.Therefore, whether I convert to Cartesian coordinates and use the distance formula or use the polar coordinate formula directly, I get the distance as 1.So, I think the answer is 1.**Final Answer**The distance is boxed{1}.