answer：Okay, so I have this problem where I need to prove that for an equilateral triangle ABC, if P is any point on its incircle, then the sum of the squares of the distances from P to each of the vertices A, B, and C is a constant. That is, PA² + PB² + PC² is constant. Hmm, interesting. I remember that in geometry, especially with regular figures like equilateral triangles, there are often symmetries and properties that can be exploited to prove such results. Let me try to break this down step by step.First, let me recall some properties of an equilateral triangle. All sides are equal, all angles are 60 degrees, and the centroid, circumcenter, incenter, and orthocenter all coincide at the same point. That point is the center of the incircle and the circumcircle. So, in this case, the center of the incircle is also the centroid of the triangle. Let me denote this center as O.Since ABC is equilateral, its inradius (radius of the incircle) can be calculated if I know the side length. Let me denote the side length as 'a'. The formula for the inradius 'r' of an equilateral triangle is r = (a√3)/6. So, the incircle has radius r, and point P lies somewhere on this circle.Now, I need to find PA² + PB² + PC² for any point P on the incircle. Since P is on the incircle, its distance from the center O is equal to the inradius r. So, OP = r. Maybe I can express PA², PB², and PC² in terms of OP and the distances from O to A, B, and C.Wait, but in an equilateral triangle, OA = OB = OC, right? Because O is the centroid and the circumradius is equal for all vertices. The circumradius R of an equilateral triangle is given by R = (a√3)/3. So, OA = OB = OC = R.So, OA = R = 2r, since r = (a√3)/6 and R = (a√3)/3. So, R is twice the inradius. That's a useful relationship.Now, let me think about how to express PA², PB², and PC². I remember that in coordinate geometry, the distance squared between two points can be expressed using the coordinates of those points. Maybe I can set up a coordinate system to simplify this problem.Let me place the equilateral triangle ABC in a coordinate system such that the centroid O is at the origin (0, 0). Since ABC is equilateral, I can position the vertices at convenient coordinates. Let me recall that in such a case, the coordinates can be set as follows:- Let’s place vertex A at (0, h), vertex B at (-a/2, 0), and vertex C at (a/2, 0), where h is the height of the triangle. For an equilateral triangle, the height h is (√3/2)a.So, substituting, vertex A is at (0, (√3/2)a), vertex B is at (-a/2, 0), and vertex C is at (a/2, 0). The centroid O is at the average of the coordinates of A, B, and C, which is indeed (0, 0) in this setup.Now, point P is on the incircle with center O and radius r = (a√3)/6. So, the coordinates of P can be parameterized using an angle θ. Let me write P as (r cos θ, r sin θ). That is, P is ( (a√3/6) cos θ, (a√3/6) sin θ ).Now, I can write expressions for PA², PB², and PC² using the distance formula.First, PA² is the squared distance between P and A. So,PA² = (x_P - x_A)² + (y_P - y_A)²= [ (a√3/6 cos θ - 0) ]² + [ (a√3/6 sin θ - (√3/2 a) ) ]²= (a² * 3 / 36 cos² θ) + (a² * 3 / 36 sin² θ - 2 * a√3/6 * √3/2 a sin θ + (√3/2 a)² )Wait, that seems a bit messy. Let me compute each term step by step.Compute PA²:x_P = (a√3/6) cos θx_A = 0So, (x_P - x_A)² = (a√3/6 cos θ)² = (a² * 3 / 36) cos² θ = (a² / 12) cos² θSimilarly, y_P = (a√3/6) sin θy_A = (√3/2) aSo, (y_P - y_A)² = (a√3/6 sin θ - √3/2 a)²= [ (√3 a / 6 sin θ - √3 a / 2 ) ]²Factor out √3 a / 6:= [ √3 a / 6 (sin θ - 3) ]²= (3 a² / 36) (sin θ - 3)²= (a² / 12) (sin θ - 3)²So, PA² = (a² / 12) cos² θ + (a² / 12)(sin θ - 3)²= (a² / 12)[ cos² θ + (sin θ - 3)² ]Let me expand (sin θ - 3)²:= sin² θ - 6 sin θ + 9So, PA² = (a² / 12)[ cos² θ + sin² θ - 6 sin θ + 9 ]But cos² θ + sin² θ = 1, so:= (a² / 12)[ 1 - 6 sin θ + 9 ]= (a² / 12)(10 - 6 sin θ )= (a² / 12)(10 - 6 sin θ )= (5a² / 6) - (a² / 2) sin θOkay, so PA² = (5a² / 6) - (a² / 2) sin θNow, let me compute PB² and PC² similarly.Compute PB²:Point B is at (-a/2, 0)Point P is at (a√3/6 cos θ, a√3/6 sin θ )So, (x_P - x_B)² = [ a√3/6 cos θ - (-a/2) ]²= [ a√3/6 cos θ + a/2 ]²= [ a (√3/6 cos θ + 1/2 ) ]²= a² ( (√3/6 cos θ + 1/2 )² )Similarly, (y_P - y_B)² = [ a√3/6 sin θ - 0 ]²= (a√3/6 sin θ )²= (3 a² / 36) sin² θ= (a² / 12) sin² θSo, PB² = a² ( (√3/6 cos θ + 1/2 )² ) + (a² / 12) sin² θLet me expand (√3/6 cos θ + 1/2 )²:= ( (√3/6 cos θ )² + 2*(√3/6 cos θ)*(1/2) + (1/2)² )= ( (3/36 cos² θ ) + (√3/6 cos θ ) + 1/4 )= ( (1/12 cos² θ ) + (√3/6 cos θ ) + 1/4 )So, PB² = a² [ (1/12 cos² θ ) + (√3/6 cos θ ) + 1/4 ] + (a² / 12) sin² θ= a² [ (1/12 cos² θ + 1/12 sin² θ ) + (√3/6 cos θ ) + 1/4 ]= a² [ (1/12)(cos² θ + sin² θ ) + (√3/6 cos θ ) + 1/4 ]Again, cos² θ + sin² θ = 1:= a² [ 1/12 + √3/6 cos θ + 1/4 ]= a² [ (1/12 + 3/12 ) + √3/6 cos θ ]= a² [ 4/12 + √3/6 cos θ ]= a² [ 1/3 + √3/6 cos θ ]= (a² / 3) + (a² √3 / 6 ) cos θSimilarly, compute PC².Point C is at (a/2, 0)Point P is at (a√3/6 cos θ, a√3/6 sin θ )So, (x_P - x_C)² = [ a√3/6 cos θ - a/2 ]²= [ a (√3/6 cos θ - 1/2 ) ]²= a² ( (√3/6 cos θ - 1/2 )² )Similarly, (y_P - y_C)² = [ a√3/6 sin θ - 0 ]²= (a√3/6 sin θ )²= (3 a² / 36 ) sin² θ= (a² / 12 ) sin² θSo, PC² = a² ( (√3/6 cos θ - 1/2 )² ) + (a² / 12 ) sin² θExpanding (√3/6 cos θ - 1/2 )²:= ( (√3/6 cos θ )² - 2*(√3/6 cos θ)*(1/2) + (1/2 )² )= ( 3/36 cos² θ - √3/6 cos θ + 1/4 )= ( 1/12 cos² θ - √3/6 cos θ + 1/4 )So, PC² = a² [ 1/12 cos² θ - √3/6 cos θ + 1/4 ] + (a² / 12 ) sin² θ= a² [ (1/12 cos² θ + 1/12 sin² θ ) - √3/6 cos θ + 1/4 ]= a² [ 1/12 (cos² θ + sin² θ ) - √3/6 cos θ + 1/4 ]= a² [ 1/12 - √3/6 cos θ + 1/4 ]= a² [ (1/12 + 3/12 ) - √3/6 cos θ ]= a² [ 4/12 - √3/6 cos θ ]= a² [ 1/3 - √3/6 cos θ ]= (a² / 3 ) - (a² √3 / 6 ) cos θOkay, so now I have expressions for PA², PB², and PC²:PA² = (5a² / 6 ) - (a² / 2 ) sin θPB² = (a² / 3 ) + (a² √3 / 6 ) cos θPC² = (a² / 3 ) - (a² √3 / 6 ) cos θNow, let me sum them up:PA² + PB² + PC² = [ (5a² / 6 ) - (a² / 2 ) sin θ ] + [ (a² / 3 ) + (a² √3 / 6 ) cos θ ] + [ (a² / 3 ) - (a² √3 / 6 ) cos θ ]Let me compute term by term:First, the constants:5a² / 6 + a² / 3 + a² / 3Convert a² / 3 to 2a² / 6, so:5a² / 6 + 2a² / 6 + 2a² / 6 = (5 + 2 + 2 )a² / 6 = 9a² / 6 = 3a² / 2Now, the sine term:- (a² / 2 ) sin θThe cosine terms:(a² √3 / 6 ) cos θ - (a² √3 / 6 ) cos θ = 0So, the cosine terms cancel each other out.Therefore, PA² + PB² + PC² = 3a² / 2 - (a² / 2 ) sin θWait, but this is supposed to be a constant, right? But here, I have a term with sin θ, which depends on θ, the angle parameterizing point P on the incircle. That suggests that PA² + PB² + PC² is not constant, which contradicts the problem statement. Hmm, that can't be right. Did I make a mistake in my calculations?Let me double-check my computations.Starting with PA²:PA² = (a² / 12 ) cos² θ + (a² / 12 )( sin² θ - 6 sin θ + 9 )Wait, hold on, earlier I expanded (sin θ - 3 )² as sin² θ - 6 sin θ + 9, but actually, (sin θ - 3 )² = sin² θ - 6 sin θ + 9. That seems correct.Then, PA² = (a² / 12 )( cos² θ + sin² θ - 6 sin θ + 9 ) = (a² / 12 )(1 - 6 sin θ + 9 ) = (a² / 12 )(10 - 6 sin θ ) = (5a² / 6 ) - (a² / 2 ) sin θ. That seems correct.Now, PB²:PB² = (a² / 3 ) + (a² √3 / 6 ) cos θPC² = (a² / 3 ) - (a² √3 / 6 ) cos θSo, when adding PB² + PC², the cosine terms cancel:PB² + PC² = (a² / 3 + a² / 3 ) + (a² √3 / 6 cos θ - a² √3 / 6 cos θ ) = (2a² / 3 ) + 0 = 2a² / 3Then, PA² + PB² + PC² = PA² + (PB² + PC² ) = [ (5a² / 6 ) - (a² / 2 ) sin θ ] + (2a² / 3 )Convert 2a² / 3 to 4a² / 6:= (5a² / 6 + 4a² / 6 ) - (a² / 2 ) sin θ= (9a² / 6 ) - (a² / 2 ) sin θ= (3a² / 2 ) - (a² / 2 ) sin θHmm, so the sum is 3a² / 2 - (a² / 2 ) sin θ. But this still has a sin θ term, which depends on θ, meaning the sum varies as P moves around the incircle. But the problem states that it's a constant. So, clearly, I have made a mistake somewhere.Wait, perhaps my coordinate system is not the best choice. Maybe I should have chosen a different coordinate system or used a different approach.Let me think again. Maybe instead of placing the centroid at the origin, I should use vector coordinates or exploit symmetry differently.Alternatively, perhaps I can use complex numbers. Let me try that.Let me represent the points as complex numbers. Let me place the centroid O at the origin. Then, the vertices A, B, and C can be represented as complex numbers with magnitude equal to the circumradius R = (a√3)/3.Let me denote the complex numbers as follows:Let’s set A = R e^{i 90°}, B = R e^{-i 30°}, and C = R e^{i 210°}, to form an equilateral triangle. Wait, actually, for simplicity, let me align the triangle such that vertex A is at (0, R), vertex B at (-R/2, -R√3/6 ), and vertex C at (R/2, -R√3/6 ). Hmm, maybe that complicates things.Alternatively, perhaps using vectors. Let me denote vectors OA, OB, OC, and OP.Since O is the centroid, OA, OB, and OC are vectors from the centroid to the vertices. In an equilateral triangle, these vectors have equal magnitude R = (a√3)/3, and the angles between them are 120 degrees.Point P is on the incircle, so vector OP has magnitude r = (a√3)/6.I need to compute PA² + PB² + PC², where PA is the distance from P to A, etc.In vector terms, PA² = |P - A|² = |OP - OA|² = |OP|² + |OA|² - 2 OP · OASimilarly for PB² and PC².So, PA² + PB² + PC² = (|OP|² + |OA|² - 2 OP · OA ) + (|OP|² + |OB|² - 2 OP · OB ) + (|OP|² + |OC|² - 2 OP · OC )Simplify:= 3|OP|² + (|OA|² + |OB|² + |OC|² ) - 2 OP · (OA + OB + OC )But in an equilateral triangle, OA + OB + OC = 0, because the centroid is the origin, so the sum of the position vectors of the vertices is zero.Therefore, the last term is zero.So, PA² + PB² + PC² = 3|OP|² + (|OA|² + |OB|² + |OC|² )Since OA, OB, and OC all have the same magnitude R, this simplifies to:= 3r² + 3R²Because |OP| = r, and |OA| = |OB| = |OC| = R.So, PA² + PB² + PC² = 3r² + 3R²But earlier, we found that R = 2r, since R = (a√3)/3 and r = (a√3)/6.Therefore, R = 2r, so R² = 4r².Substituting back:PA² + PB² + PC² = 3r² + 3*(4r² ) = 3r² + 12r² = 15r²So, PA² + PB² + PC² = 15r², which is a constant, independent of the position of P on the incircle.Wait, that's different from what I got earlier when using coordinate geometry. So, where did I go wrong in the coordinate approach?Looking back, in the coordinate approach, I ended up with PA² + PB² + PC² = 3a² / 2 - (a² / 2 ) sin θ, which depends on θ. But using vectors, I get a constant value of 15r². There must be a mistake in the coordinate approach.Let me check the coordinate approach again.In the coordinate approach, I set up the triangle with vertices at (0, (√3/2)a), (-a/2, 0), and (a/2, 0). The centroid is at (0, 0). The inradius r = (a√3)/6, so point P is at (r cos θ, r sin θ ).Calculating PA², PB², PC², and summing them up gave me 3a² / 2 - (a² / 2 ) sin θ, which is not constant.But according to the vector approach, it should be 15r², which is constant.So, let me compute 15r² in terms of a:r = (a√3)/6, so r² = (3a²)/36 = a² / 12.Thus, 15r² = 15*(a² / 12 ) = (15/12 )a² = (5/4 )a².Wait, 15r² = (5/4 )a².But in the coordinate approach, I got 3a² / 2 - (a² / 2 ) sin θ. Let me compute 3a² / 2:3a² / 2 = 1.5a², and 5/4 a² = 1.25a². These are different.So, clearly, there is a discrepancy. Therefore, my coordinate approach must have an error.Wait, let me compute 15r² again:r = (a√3)/6, so r² = (3a²)/36 = a² / 12.15r² = 15*(a² / 12 ) = (15/12 )a² = (5/4 )a².But in the coordinate approach, I have PA² + PB² + PC² = 3a² / 2 - (a² / 2 ) sin θ.But 3a² / 2 is 1.5a², which is larger than 5/4 a².So, something is wrong.Wait, perhaps my coordinate setup is incorrect. Let me verify the positions of the vertices and the centroid.In an equilateral triangle with side length a, the height h = (√3/2 )a.If I place vertex A at (0, h ), and vertices B and C at (-a/2, 0 ) and (a/2, 0 ), respectively, then the centroid O is indeed at (0, h/3 ).Wait a minute! Earlier, I placed the centroid at (0, 0 ), but in reality, the centroid is at (0, h/3 ). So, my coordinate system was incorrect. I mistakenly placed the centroid at (0, 0 ), but in reality, it's at (0, h/3 ). That was a mistake.So, I need to correct this. Let me redefine the coordinate system properly.Let me place vertex A at (0, h ), vertex B at (-a/2, 0 ), vertex C at (a/2, 0 ), and centroid O at (0, h/3 ). So, O is not at (0, 0 ), but at (0, h/3 ).Given that, the inradius r is the distance from O to any side. In an equilateral triangle, the inradius is h/3, since the centroid divides the height into a ratio of 2:1.So, h = (√3 / 2 )a, so r = h / 3 = (√3 / 6 )a, which matches the earlier formula.So, the incircle is centered at O = (0, h/3 ) = (0, (√3 / 6 )a ), with radius r = (√3 / 6 )a.Therefore, point P on the incircle can be parameterized as:P = ( r cos θ, (√3 / 6 )a + r sin θ ) = ( (√3 / 6 )a cos θ, (√3 / 6 )a + (√3 / 6 )a sin θ )So, P = ( (√3 a / 6 ) cos θ, (√3 a / 6 )(1 + sin θ ) )Now, let me recompute PA², PB², and PC² with this corrected coordinate system.First, PA²:Point A is at (0, (√3 / 2 )a )Point P is at ( (√3 a / 6 ) cos θ, (√3 a / 6 )(1 + sin θ ) )So, PA² = (x_P - x_A )² + (y_P - y_A )²= [ (√3 a / 6 cos θ - 0 ) ]² + [ (√3 a / 6 (1 + sin θ ) - √3 a / 2 ) ]²Simplify each term:x-component squared: ( √3 a / 6 cos θ )² = (3 a² / 36 ) cos² θ = a² / 12 cos² θy-component squared: [ √3 a / 6 (1 + sin θ ) - √3 a / 2 ]²= [ √3 a / 6 (1 + sin θ - 3 ) ]²= [ √3 a / 6 (sin θ - 2 ) ]²= ( 3 a² / 36 )( sin θ - 2 )²= ( a² / 12 )( sin² θ - 4 sin θ + 4 )So, PA² = a² / 12 cos² θ + a² / 12 ( sin² θ - 4 sin θ + 4 )= a² / 12 ( cos² θ + sin² θ - 4 sin θ + 4 )= a² / 12 ( 1 - 4 sin θ + 4 )= a² / 12 ( 5 - 4 sin θ )= (5a² / 12 ) - (a² / 3 ) sin θOkay, PA² = (5a² / 12 ) - (a² / 3 ) sin θNow, compute PB²:Point B is at (-a/2, 0 )Point P is at ( (√3 a / 6 ) cos θ, (√3 a / 6 )(1 + sin θ ) )So, PB² = (x_P - x_B )² + (y_P - y_B )²= [ (√3 a / 6 cos θ + a/2 ) ]² + [ (√3 a / 6 (1 + sin θ ) - 0 ) ]²Compute each term:x-component squared: [ √3 a / 6 cos θ + a/2 ]²= [ a ( √3 / 6 cos θ + 1/2 ) ]²= a² ( ( √3 / 6 cos θ + 1/2 )² )Expand the square:= a² ( (3 / 36 cos² θ ) + ( √3 / 6 * 1/2 * 2 cos θ ) + 1/4 )Wait, let me compute it step by step.( √3 / 6 cos θ + 1/2 )² = ( √3 / 6 cos θ )² + 2*(√3 / 6 cos θ )*(1/2 ) + (1/2 )²= ( 3 / 36 cos² θ ) + ( √3 / 6 cos θ ) + 1/4= ( 1 / 12 cos² θ ) + ( √3 / 6 cos θ ) + 1/4So, x-component squared = a² [ 1 / 12 cos² θ + √3 / 6 cos θ + 1/4 ]y-component squared: [ √3 a / 6 (1 + sin θ ) ]²= ( 3 a² / 36 )(1 + sin θ )²= ( a² / 12 )(1 + 2 sin θ + sin² θ )So, PB² = a² [ 1 / 12 cos² θ + √3 / 6 cos θ + 1/4 ] + a² / 12 (1 + 2 sin θ + sin² θ )Combine terms:= a² [ 1 / 12 cos² θ + √3 / 6 cos θ + 1/4 + 1 / 12 + (2 sin θ ) / 12 + sin² θ / 12 ]Simplify:1 / 12 cos² θ + sin² θ / 12 = (cos² θ + sin² θ ) / 12 = 1 / 12√3 / 6 cos θ remains.1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3(2 sin θ ) / 12 = sin θ / 6So, PB² = a² [ 1 / 12 + √3 / 6 cos θ + 1/3 + sin θ / 6 ]Combine constants:1 / 12 + 1 / 3 = 1 / 12 + 4 / 12 = 5 / 12So, PB² = a² [ 5 / 12 + √3 / 6 cos θ + sin θ / 6 ]Similarly, compute PC².Point C is at (a/2, 0 )Point P is at ( (√3 a / 6 ) cos θ, (√3 a / 6 )(1 + sin θ ) )So, PC² = (x_P - x_C )² + (y_P - y_C )²= [ (√3 a / 6 cos θ - a/2 ) ]² + [ (√3 a / 6 (1 + sin θ ) - 0 ) ]²Compute each term:x-component squared: [ √3 a / 6 cos θ - a/2 ]²= [ a ( √3 / 6 cos θ - 1/2 ) ]²= a² ( ( √3 / 6 cos θ - 1/2 )² )Expand the square:= ( √3 / 6 cos θ )² - 2*(√3 / 6 cos θ )*(1/2 ) + (1/2 )²= ( 3 / 36 cos² θ ) - ( √3 / 6 cos θ ) + 1/4= ( 1 / 12 cos² θ ) - ( √3 / 6 cos θ ) + 1/4So, x-component squared = a² [ 1 / 12 cos² θ - √3 / 6 cos θ + 1/4 ]y-component squared is the same as in PB²:= a² / 12 (1 + 2 sin θ + sin² θ )So, PC² = a² [ 1 / 12 cos² θ - √3 / 6 cos θ + 1/4 ] + a² / 12 (1 + 2 sin θ + sin² θ )Combine terms:= a² [ 1 / 12 cos² θ - √3 / 6 cos θ + 1/4 + 1 / 12 + (2 sin θ ) / 12 + sin² θ / 12 ]Simplify:1 / 12 cos² θ + sin² θ / 12 = 1 / 12- √3 / 6 cos θ remains.1/4 + 1/12 = 1/3(2 sin θ ) / 12 = sin θ / 6So, PC² = a² [ 1 / 12 - √3 / 6 cos θ + 1/3 + sin θ / 6 ]Combine constants:1 / 12 + 1 / 3 = 5 / 12So, PC² = a² [ 5 / 12 - √3 / 6 cos θ + sin θ / 6 ]Now, let me sum PA² + PB² + PC²:PA² = (5a² / 12 ) - (a² / 3 ) sin θPB² = a² [ 5 / 12 + √3 / 6 cos θ + sin θ / 6 ]PC² = a² [ 5 / 12 - √3 / 6 cos θ + sin θ / 6 ]Adding them up:PA² + PB² + PC² = [ (5a² / 12 ) - (a² / 3 ) sin θ ] + [ 5a² / 12 + (a² √3 / 6 ) cos θ + (a² / 6 ) sin θ ] + [ 5a² / 12 - (a² √3 / 6 ) cos θ + (a² / 6 ) sin θ ]Let me compute term by term:First, the constants:5a² / 12 + 5a² / 12 + 5a² / 12 = 15a² / 12 = 5a² / 4Now, the sine terms:- (a² / 3 ) sin θ + (a² / 6 ) sin θ + (a² / 6 ) sin θ= [ - (a² / 3 ) + (a² / 6 ) + (a² / 6 ) ] sin θ= [ - (2a² / 6 ) + (a² / 6 ) + (a² / 6 ) ] sin θ= [ (-2a² + a² + a² ) / 6 ] sin θ= 0 sin θSo, the sine terms cancel out.The cosine terms:(a² √3 / 6 ) cos θ - (a² √3 / 6 ) cos θ = 0So, the cosine terms cancel out.Therefore, PA² + PB² + PC² = 5a² / 4Which is a constant, as expected.Now, let me express this in terms of r, the inradius.We know that r = (√3 / 6 )a, so a = (6 / √3 )r = 2√3 rTherefore, a² = (2√3 r )² = 12 r²So, 5a² / 4 = 5*(12 r² ) / 4 = (60 r² ) / 4 = 15 r²Thus, PA² + PB² + PC² = 15 r², which is a constant, independent of θ.Therefore, the sum PA² + PB² + PC² is indeed a constant value for any point P on the incircle of the equilateral triangle ABC.I think this resolves the discrepancy. Initially, I made a mistake in placing the centroid at the origin, which led to incorrect calculations. After correcting the coordinate system to place the centroid at (0, h/3 ), the calculations aligned with the vector approach, confirming that the sum is constant.

answer：Okay, so I have this problem where a line passes through the point P₀(-4, 0) and has parametric equations:x = -4 + (√3 / 2) ty = (1/2) tThis line intersects a circle defined by x² + y² = 7 at points A and B. I need to find the length of the chord AB.Hmm, let me think. I remember that to find the points where a line intersects a circle, I can substitute the parametric equations of the line into the equation of the circle. That should give me a quadratic equation in terms of the parameter t. Then, solving for t will give me the parameter values corresponding to points A and B. Once I have those, I can find the distance between A and B, which is the length of the chord.Alright, let's start by substituting the parametric equations into the circle equation.The parametric equations are:x = -4 + (√3 / 2) ty = (1/2) tSo, substituting x and y into x² + y² = 7:[ -4 + (√3 / 2) t ]² + [ (1/2) t ]² = 7Let me expand this step by step.First, expand [ -4 + (√3 / 2) t ]²:= (-4)² + 2*(-4)*(√3 / 2) t + (√3 / 2 t)²= 16 - 4√3 t + (3/4) t²Next, expand [ (1/2) t ]²:= (1/4) t²Now, add these two results together:16 - 4√3 t + (3/4) t² + (1/4) t² = 7Combine like terms:(3/4 + 1/4) t² - 4√3 t + 16 = 7That simplifies to:(4/4) t² - 4√3 t + 16 = 7Which is:t² - 4√3 t + 16 = 7Subtract 7 from both sides:t² - 4√3 t + 9 = 0So, the quadratic equation in t is:t² - 4√3 t + 9 = 0Now, I need to solve this quadratic equation for t. The solutions will give me the parameter values t₁ and t₂ corresponding to points A and B.I can use the quadratic formula:t = [4√3 ± √( (4√3)² - 4*1*9 ) ] / 2Let me compute the discriminant first:(4√3)² - 4*1*9 = 16*3 - 36 = 48 - 36 = 12So, the solutions are:t = [4√3 ± √12 ] / 2Simplify √12:√12 = 2√3So, plug that back in:t = [4√3 ± 2√3 ] / 2Factor out 2√3 in the numerator:t = [2√3 (2 ± 1) ] / 2Simplify by dividing numerator and denominator by 2:t = √3 (2 ± 1)So, the two solutions are:t₁ = √3 (2 + 1) = 3√3t₂ = √3 (2 - 1) = √3Therefore, the parameter values are t₁ = 3√3 and t₂ = √3.Now, to find the coordinates of points A and B, I can plug these t values back into the parametric equations.First, for t = 3√3:x = -4 + (√3 / 2)(3√3) = -4 + ( (√3)(3√3) ) / 2Simplify:(√3)(3√3) = 3*(√3)^2 = 3*3 = 9So, x = -4 + 9/2 = (-8/2 + 9/2) = 1/2Similarly, y = (1/2)(3√3) = (3√3)/2So, point A is (1/2, (3√3)/2)Now, for t = √3:x = -4 + (√3 / 2)(√3) = -4 + ( (√3)(√3) ) / 2Simplify:(√3)(√3) = 3So, x = -4 + 3/2 = (-8/2 + 3/2) = (-5/2)Similarly, y = (1/2)(√3) = √3 / 2So, point B is (-5/2, √3 / 2)Now, I have points A(1/2, (3√3)/2) and B(-5/2, √3 / 2). I need to find the distance between these two points.The distance formula is:|AB| = √[ (x₂ - x₁)² + (y₂ - y₁)² ]Let me compute the differences first.x₂ - x₁ = (-5/2) - (1/2) = (-6/2) = -3y₂ - y₁ = (√3 / 2) - (3√3 / 2) = (-2√3 / 2) = -√3Now, square these differences:(-3)² = 9(-√3)² = 3Add them together:9 + 3 = 12Take the square root:√12 = 2√3So, the length of chord AB is 2√3.Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, when I substituted the parametric equations into the circle equation, I got t² - 4√3 t + 9 = 0. Solving this quadratic gave me t = 3√3 and t = √3, which seems correct.Then, plugging t = 3√3 into x and y:x = -4 + (√3 / 2)(3√3) = -4 + (3*3)/2 = -4 + 9/2 = 1/2y = (1/2)(3√3) = 3√3 / 2Similarly, for t = √3:x = -4 + (√3 / 2)(√3) = -4 + 3/2 = -5/2y = (1/2)(√3) = √3 / 2So, points A(1/2, 3√3/2) and B(-5/2, √3/2) are correct.Calculating the distance:Δx = (-5/2 - 1/2) = -6/2 = -3Δy = (√3/2 - 3√3/2) = (-2√3)/2 = -√3Then, distance squared is (-3)^2 + (-√3)^2 = 9 + 3 = 12Square root of 12 is 2√3.Yes, that seems correct.Alternatively, I remember that for a quadratic equation at² + bt + c = 0, the difference of roots is √(b² - 4ac)/a. But in this case, since the quadratic is t² - 4√3 t + 9 = 0, the difference of roots is √[(4√3)^2 - 4*1*9] = √[48 - 36] = √12 = 2√3. So, the length of the chord is 2√3.Wait, that's another way to think about it. Since the parameter t corresponds to the points along the line, the difference in t values gives the length of the chord scaled by the direction vector's magnitude.But in this case, the direction vector of the line is ( √3 / 2, 1/2 ). The magnitude of this vector is √[ (√3 / 2)^2 + (1/2)^2 ] = √[ 3/4 + 1/4 ] = √1 = 1. So, each unit of t corresponds to 1 unit of distance along the line. Therefore, the difference in t values directly gives the distance between A and B.So, since t₁ - t₂ = 3√3 - √3 = 2√3, the length of AB is 2√3.Yes, that's consistent with what I found earlier.So, both methods give me the same result, which is reassuring.Therefore, I'm confident that the length of chord AB is 2√3.**Final Answer**The length of the chord (|AB|) is (boxed{2sqrt{3}}).

answer：Okay, so I've got this problem here about renting a car. Let me try to understand what it's asking. It says that a car rents for a certain amount per day plus 18 cents per mile. I'm on a budget of 75 dollars and can drive 250 miles. I need to find out what the daily rental cost of the car is.Alright, so first, I need to figure out how much the rental costs per day. There's a fixed cost per day, and then there's a variable cost based on how many miles I drive. The variable cost is 18 cents per mile, and I'm driving 250 miles. So, I think I need to calculate the total cost for the miles driven and then see how much of my budget is left for the daily rental cost.Let me write this down step by step.First, the cost per mile is 18 cents. To make it easier, I should convert that to dollars because my budget is in dollars. So, 18 cents is 0.18.Next, I need to find out how much it costs to drive 250 miles. That would be the cost per mile multiplied by the number of miles. So, that's 250 miles times 0.18 per mile.Let me do that multiplication: 250 * 0.18. Hmm, 250 times 0.1 is 25, and 250 times 0.08 is 20. So, adding those together, 25 + 20 = 45. So, the total cost for driving 250 miles is 45.Now, my total budget is 75. If 45 of that is going to the mileage cost, then the remaining amount must be the daily rental cost. So, I subtract the mileage cost from the total budget to find the daily rental cost.That would be 75 minus 45, which equals 30.Wait, so does that mean the daily rental cost is 30? Let me double-check my calculations to make sure I didn't make a mistake.First, converting 18 cents to dollars: 18 cents is indeed 0.18.Then, calculating the mileage cost: 250 miles * 0.18 per mile. Let me do this multiplication again to be sure. 250 * 0.18. Breaking it down, 250 * 0.1 is 25, and 250 * 0.08 is 20. Adding those together gives 45. So, that part checks out.Subtracting the mileage cost from the total budget: 75 - 45 = 30. That seems correct.So, it looks like the daily rental cost of the car is 30.I think I've got it. The daily rental cost is 30.

answer：First, calculate the current average of Josanna's test scores by summing them up and dividing by the number of tests. Then, determine the desired average by adding 4 points to the current average. Set up an equation where the new average, including the next test score, equals the desired average. Solve for the required score on the next test to achieve this new average.

answer：Alright, so I have this triangle ABC, and I need to find angle C and the length of side c. Let me see what information I have. First, the problem states that in triangle ABC, the sides opposite angles A, B, and C are a, b, and c respectively. It also gives me two equations: a + b equals the square root of 3 times c, which is written as a + b = √3 c. The second equation is 2 sin²C = 3 sin A sin B. Okay, so I need to find angle C first. Hmm, let me think about the relationships in a triangle. I remember the Law of Sines, which says that a / sin A = b / sin B = c / sin C = 2R, where R is the radius of the circumscribed circle. Maybe that can help me relate the sides and angles.Also, the Law of Cosines might come in handy here. It states that c² = a² + b² - 2ab cos C. I wonder if I can use that somehow with the given equations.Let me write down what I know:1. a + b = √3 c2. 2 sin²C = 3 sin A sin BI need to find angle C. Maybe I can express sin A and sin B in terms of sin C using the Law of Sines. Let me try that.From the Law of Sines, we have:a = 2R sin A b = 2R sin B c = 2R sin CSo, substituting these into the first equation:2R sin A + 2R sin B = √3 * 2R sin CDivide both sides by 2R:sin A + sin B = √3 sin CHmm, that's a useful equation. Now, let's look at the second equation given: 2 sin²C = 3 sin A sin B.I have two equations involving sin A, sin B, and sin C. Maybe I can solve for sin A and sin B in terms of sin C.Let me denote sin A as x and sin B as y for simplicity. Then, the equations become:1. x + y = √3 sin C 2. 2 sin²C = 3xyI need to find angle C, so I need to relate these equations somehow. Maybe I can express y from the first equation and substitute into the second.From equation 1: y = √3 sin C - xSubstitute into equation 2:2 sin²C = 3x(√3 sin C - x) 2 sin²C = 3√3 sin C x - 3x²Hmm, this is a quadratic in terms of x. Let me rearrange it:3x² - 3√3 sin C x + 2 sin²C = 0This is a quadratic equation: ax² + bx + c = 0, where:a = 3 b = -3√3 sin C c = 2 sin²CTo solve for x, I can use the quadratic formula:x = [3√3 sin C ± √( (3√3 sin C)² - 4*3*2 sin²C )] / (2*3)Let me compute the discriminant:D = (3√3 sin C)² - 4*3*2 sin²C D = 27 sin²C - 24 sin²C D = 3 sin²CSo, the square root of D is √(3 sin²C) = sin C √3Therefore, x = [3√3 sin C ± sin C √3] / 6Factor out sin C √3:x = sin C √3 [3 ± 1] / 6So, two possibilities:1. x = sin C √3 (3 + 1)/6 = sin C √3 * 4 /6 = (2/3) sin C √3 2. x = sin C √3 (3 - 1)/6 = sin C √3 * 2 /6 = (1/3) sin C √3So, x = (2√3 / 3) sin C or x = (√3 / 3) sin CBut x is sin A, so sin A = (2√3 / 3) sin C or sin A = (√3 / 3) sin CSimilarly, since y = √3 sin C - x, let's find y for both cases.Case 1: sin A = (2√3 / 3) sin C Then, y = √3 sin C - (2√3 / 3) sin C = (√3 - 2√3 / 3) sin C = (√3 / 3) sin C So, sin B = (√3 / 3) sin CCase 2: sin A = (√3 / 3) sin C Then, y = √3 sin C - (√3 / 3) sin C = (2√3 / 3) sin C So, sin B = (2√3 / 3) sin CSo, in both cases, sin A and sin B are either (2√3 / 3) sin C and (√3 / 3) sin C or vice versa.But in a triangle, the sum of angles is π, so A + B + C = π. Therefore, A + B = π - C.Also, from the Law of Sines, a / sin A = b / sin B = c / sin C.Let me denote k = c / sin C, so a = k sin A, b = k sin B.Given that a + b = √3 c, substituting:k sin A + k sin B = √3 k sin C sin A + sin B = √3 sin CWhich is consistent with what we had earlier.Now, let's consider the two cases.Case 1: sin A = (2√3 / 3) sin C, sin B = (√3 / 3) sin CThen, A + B = π - CSo, sin(A + B) = sin(π - C) = sin CBut sin(A + B) = sin A cos B + cos A sin BLet me compute sin A cos B + cos A sin B.First, sin A = (2√3 / 3) sin C, sin B = (√3 / 3) sin CLet me denote sin C as s for simplicity.So, sin A = (2√3 / 3) s, sin B = (√3 / 3) sWe can find cos A and cos B using the identity cos²θ + sin²θ = 1.So, cos A = √(1 - sin²A) = √(1 - (4*3 / 9) s²) = √(1 - (4/3) s²)Similarly, cos B = √(1 - sin²B) = √(1 - (3 / 9) s²) = √(1 - (1/3) s²)Therefore, sin(A + B) = sin A cos B + cos A sin B = (2√3 / 3 s) * √(1 - (1/3) s²) + √(1 - (4/3) s²) * (√3 / 3 s)This seems complicated. Maybe there's a better approach.Alternatively, since we have expressions for sin A and sin B in terms of sin C, and we know that A + B = π - C, maybe we can use the sine addition formula.Wait, sin(A + B) = sin(π - C) = sin CBut sin(A + B) = sin A cos B + cos A sin BSo, sin C = sin A cos B + cos A sin BWe have sin A and sin B in terms of sin C, so let's plug those in:sin C = (2√3 / 3 sin C) cos B + cos A (√3 / 3 sin C)Divide both sides by sin C (assuming sin C ≠ 0, which it isn't in a triangle):1 = (2√3 / 3) cos B + (√3 / 3) cos ASo, 1 = (√3 / 3)(2 cos B + cos A)Hmm, not sure if this helps directly. Maybe I can express cos A and cos B in terms of sin C.From earlier, cos A = √(1 - (4/3) sin²C), cos B = √(1 - (1/3) sin²C)So, plug these into the equation:1 = (√3 / 3)[2 √(1 - (1/3) sin²C) + √(1 - (4/3) sin²C)]This looks messy. Maybe instead of going this route, I can use the fact that in a triangle, the sides are proportional to the sines of their opposite angles.Given that a + b = √3 c, and from the Law of Sines, a = k sin A, b = k sin B, c = k sin C.So, a + b = k (sin A + sin B) = √3 k sin CWhich simplifies to sin A + sin B = √3 sin C, which we already have.Alternatively, maybe using the Law of Cosines. Let's recall that c² = a² + b² - 2ab cos CWe also have from the second equation: 2 sin²C = 3 sin A sin BFrom the Law of Sines, sin A = a / (2R), sin B = b / (2R), sin C = c / (2R)So, substituting into 2 sin²C = 3 sin A sin B:2 (c² / (4R²)) = 3 (a / (2R)) (b / (2R)) 2 c² / (4R²) = 3 ab / (4R²) Multiply both sides by 4R²:2 c² = 3 abSo, 2 c² = 3 abThat's a useful relation.Also, from a + b = √3 c, we can square both sides:(a + b)² = 3 c² a² + 2ab + b² = 3 c²But from the Law of Cosines, a² + b² = c² + 2ab cos CSo, substitute into the equation:(c² + 2ab cos C) + 2ab = 3 c² c² + 2ab cos C + 2ab = 3 c² 2ab cos C + 2ab = 2 c² Factor out 2ab:2ab (cos C + 1) = 2 c² Divide both sides by 2:ab (cos C + 1) = c²But from earlier, we have 2 c² = 3 ab, so c² = (3/2) abSubstitute into the equation:ab (cos C + 1) = (3/2) ab Divide both sides by ab (assuming ab ≠ 0, which it isn't in a triangle):cos C + 1 = 3/2 So, cos C = 3/2 - 1 = 1/2Therefore, cos C = 1/2, which means angle C is π/3 or 60 degrees.Okay, so that answers the first part: angle C is 60 degrees or π/3 radians.Now, moving on to part 2: If the area of triangle ABC is √3, find c.The area of a triangle can be given by (1/2)ab sin C.Given that the area is √3, so:(1/2) ab sin C = √3We know that sin C = sin(π/3) = √3 / 2So, plug that in:(1/2) ab (√3 / 2) = √3 Simplify:(√3 / 4) ab = √3 Multiply both sides by 4:√3 ab = 4√3 Divide both sides by √3:ab = 4So, ab = 4Earlier, we had from 2 c² = 3 ab, so:2 c² = 3 * 4 = 12 So, c² = 6 Therefore, c = √6So, the length of side c is √6.Let me just recap to make sure I didn't make any mistakes.1. We started with the given equations: a + b = √3 c and 2 sin²C = 3 sin A sin B.2. Using the Law of Sines, we related sin A and sin B to sin C, leading to expressions for sin A and sin B in terms of sin C.3. Then, using the Law of Cosines and the given equations, we derived that cos C = 1/2, leading to angle C = π/3.4. For the second part, using the area formula, we found ab = 4, and then using the relation 2 c² = 3 ab, we found c = √6.Everything seems to check out. I think that's the correct solution.

answer：Alright, so I've been given this problem about the sum of reciprocals of prime numbers. The sum is:1 + frac{1}{2} + frac{1}{3} + frac{1}{5} + frac{1}{7} + frac{1}{11} + frac{1}{13} + frac{1}{17} + frac{1}{19} + ldots + frac{1}{p}And I need to prove that this sum can be made greater than any given number ( N ) by choosing a sufficiently large prime ( p ). Hmm, okay. So, essentially, I need to show that as we include more and more primes in the denominator, the sum grows without bound—it diverges.First, I recall that the harmonic series, which is the sum of reciprocals of all natural numbers, diverges. That is:sum_{k=1}^{infty} frac{1}{k} = 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + ldotsThis series grows without bound, even though the terms get smaller and smaller. So, if the harmonic series diverges, maybe the sum over primes, which is a subset of the harmonic series, also diverges? But wait, primes are less dense than natural numbers. So, does that mean the sum over primes converges? Or does it still diverge?I think it still diverges, but I need to find a way to prove it. Maybe I can relate the sum over primes to the harmonic series somehow. Let me think about how primes are distributed. The prime number theorem tells us that the number of primes less than a number ( x ) is approximately ( frac{x}{ln x} ). So, primes become less frequent as numbers get larger, but there are still infinitely many of them.If I can show that the sum of reciprocals of primes grows without bound, that would solve the problem. One approach might be to use the fact that the harmonic series diverges and compare the sum over primes to it. But since primes are less dense, I need a different strategy.I remember something about using the product of primes or maybe using logarithms. Wait, there's a famous result that the sum of reciprocals of primes diverges, which is related to the infinitude of primes. Maybe I can use that.Let me try to recall the proof. I think it involves considering the product over primes of ( 1 - frac{1}{p} ) and showing that this product converges to zero, implying that the sum of reciprocals diverges. But I'm not entirely sure how that works.Alternatively, I remember that the sum of reciprocals of primes can be compared to the harmonic series using some inequalities. Maybe I can use the fact that every natural number can be expressed as a product of primes, and use that to relate the two sums.Wait, here's an idea. If I take the harmonic series and factor it into primes, I can express it as a product over primes. Specifically, the harmonic series can be written as:sum_{n=1}^{infty} frac{1}{n} = prod_{p text{ prime}} left(1 + frac{1}{p} + frac{1}{p^2} + frac{1}{p^3} + ldots right)This is because every natural number can be uniquely factored into primes, so the harmonic series is the product of geometric series over each prime. Now, if I take the logarithm of both sides, I get:ln left( sum_{n=1}^{infty} frac{1}{n} right) = sum_{p text{ prime}} ln left(1 + frac{1}{p} + frac{1}{p^2} + frac{1}{p^3} + ldots right)But the harmonic series diverges, so the left side is infinite. Therefore, the sum on the right must also be infinite. Now, let's look at each term in the sum on the right:ln left(1 + frac{1}{p} + frac{1}{p^2} + frac{1}{p^3} + ldots right) = ln left( frac{1}{1 - frac{1}{p}} right) = -ln left(1 - frac{1}{p}right)So, the sum becomes:sum_{p text{ prime}} -ln left(1 - frac{1}{p}right)Which is the same as:sum_{p text{ prime}} ln left( frac{1}{1 - frac{1}{p}} right)Since the original sum diverges, this sum must also diverge. Therefore, the sum of ( ln left( frac{1}{1 - frac{1}{p}} right) ) over primes diverges. But ( ln left( frac{1}{1 - frac{1}{p}} right) ) is approximately ( frac{1}{p} ) for large primes ( p ), because ( ln left( frac{1}{1 - x} right) approx x ) when ( x ) is small.Therefore, if the sum of ( ln left( frac{1}{1 - frac{1}{p}} right) ) diverges, and each term is approximately ( frac{1}{p} ), then the sum of ( frac{1}{p} ) must also diverge. Hence, the sum of reciprocals of primes diverges.Wait, but I'm not sure if that's rigorous enough. Maybe I need a more direct comparison. Let me think about the partial sums.Suppose I consider the partial sum of reciprocals of primes up to some prime ( p ). I need to show that this partial sum can be made larger than any given ( N ) by choosing a sufficiently large ( p ).I know that the harmonic series diverges, so for any ( N ), there exists some ( M ) such that the harmonic series up to ( M ) exceeds ( N ). But since primes are less dense, the sum over primes up to ( M ) will be less than the harmonic series up to ( M ). So, maybe I need a different approach.Alternatively, I can use the fact that the sum of reciprocals of primes is comparable to the logarithm of the product of primes. Let me recall that:sum_{p leq x} frac{1}{p} sim ln ln xAs ( x ) tends to infinity, the sum of reciprocals of primes up to ( x ) behaves like the natural logarithm of the natural logarithm of ( x ). Since ( ln ln x ) grows without bound as ( x ) increases, this implies that the sum of reciprocals of primes diverges.Therefore, for any given ( N ), no matter how large, there exists a prime ( p ) such that the sum up to ( p ) exceeds ( N ).Wait, but I'm not sure if I can just cite that asymptotic result without proving it. Maybe I need to provide a more elementary proof.Let me try another approach. Consider grouping the primes into blocks where each block contains primes in a certain range. For example, consider primes between ( 2^{k} ) and ( 2^{k+1} ). The number of primes in this interval is approximately ( frac{2^{k}}{k ln 2} ) by the prime number theorem.Then, the sum of reciprocals of primes in this interval is at least ( frac{2^{k}}{k ln 2} times frac{1}{2^{k+1}} ) since each prime is at least ( 2^{k} ). Simplifying, this is approximately ( frac{1}{2 k ln 2} ).So, the contribution of each block to the sum is roughly ( frac{1}{2 k ln 2} ). Summing over ( k ), we get:sum_{k=1}^{infty} frac{1}{2 k ln 2} = frac{1}{2 ln 2} sum_{k=1}^{infty} frac{1}{k}But the harmonic series ( sum_{k=1}^{infty} frac{1}{k} ) diverges, so the sum of reciprocals of primes must also diverge.Therefore, for any given ( N ), we can choose a sufficiently large ( p ) such that the sum up to ( p ) exceeds ( N ).Hmm, I think this makes sense. By grouping the primes and showing that each group contributes a significant amount to the sum, and since there are infinitely many such groups, the total sum must diverge.Alternatively, I could use the integral test or comparison test. Since the sum of reciprocals of primes is larger than the sum of reciprocals of numbers of the form ( 2^k ), which diverges, but that might not be accurate because primes are less dense.Wait, no, actually, the sum of reciprocals of primes is smaller than the harmonic series, but we've already established that it still diverges.I think the key idea is that even though primes are less dense, their reciprocals still sum to infinity because the density isn't too sparse. The prime number theorem tells us that primes thin out, but not enough to make the sum converge.So, putting it all together, the sum of reciprocals of primes diverges, meaning that for any ( N ), there exists a prime ( p ) such that the sum up to ( p ) exceeds ( N ).I think that's a solid argument. I might need to write it out more formally, but the intuition is there.