💡Okay, so I have this problem about two points on a Cartesian plane, A and B, with coordinates (a, b) and (c, d) respectively. Their midpoint is M(m, n). Then, both points are translated—A is moved 6 units up and 4 units to the right, while B is moved 4 units down and 5 units to the left. I need to find the distance between the original midpoint M and the new midpoint M' after these translations.Alright, let me start by recalling what a midpoint is. The midpoint M of two points A and B is given by the average of their coordinates. So, M(m, n) = ((a + c)/2, (b + d)/2). That makes sense.Now, when we translate a point, we're essentially sliding it around the plane without rotating or resizing it. So, translating A 6 units up and 4 units to the right means we're adding 4 to its x-coordinate and 6 to its y-coordinate. Similarly, translating B 4 units down and 5 units to the left means subtracting 5 from its x-coordinate and 4 from its y-coordinate.Let me write that down:- Original point A: (a, b)- Translated A: (a + 4, b + 6)- Original point B: (c, d)- Translated B: (c - 5, d - 4)Okay, so now I need to find the new midpoint M' after these translations. The midpoint formula still applies, so M' will be the average of the translated coordinates of A and B.Calculating M':M' = [( (a + 4) + (c - 5) ) / 2, ( (b + 6) + (d - 4) ) / 2]Let me simplify the x-coordinate first:(a + 4 + c - 5) / 2 = (a + c - 1) / 2Similarly, the y-coordinate:(b + 6 + d - 4) / 2 = (b + d + 2) / 2So, M' = ( (a + c - 1)/2, (b + d + 2)/2 )But wait, the original midpoint M was ( (a + c)/2, (b + d)/2 ). So, let me express M' in terms of M.M = (m, n) = ( (a + c)/2, (b + d)/2 )So, M' = ( m - 0.5, n + 1 )That's interesting. So, the new midpoint is shifted by (-0.5, 1) from the original midpoint.Now, to find the distance between M and M', I can use the distance formula between two points.Distance = sqrt[ (x2 - x1)^2 + (y2 - y1)^2 ]Substituting the coordinates:Distance = sqrt[ (m - 0.5 - m)^2 + (n + 1 - n)^2 ]Simplifying inside the square roots:= sqrt[ (-0.5)^2 + (1)^2 ]Calculating the squares:= sqrt[ 0.25 + 1 ]= sqrt[ 1.25 ]Hmm, 1.25 can be written as 5/4, so sqrt(5/4) is sqrt(5)/2.So, the distance between M and M' is sqrt(5)/2 units.Let me double-check my steps to make sure I didn't make a mistake.1. Calculated translated coordinates correctly: A moved right 4, up 6; B moved left 5, down 4. So, A becomes (a+4, b+6), B becomes (c-5, d-4). That seems right.2. Midpoint formula applied correctly: ( (a+4 + c-5)/2, (b+6 + d-4)/2 ). Simplified correctly to ( (a + c -1)/2, (b + d + 2)/2 ). Then expressed in terms of M: (m - 0.5, n + 1). That makes sense because (a + c)/2 is m, so subtracting 1/2 gives m - 0.5, and (b + d)/2 is n, so adding 1 gives n + 1.3. Distance formula applied correctly: sqrt[ (-0.5)^2 + (1)^2 ] = sqrt[0.25 + 1] = sqrt[1.25] = sqrt(5)/2. That seems correct.I think all the steps are solid. So, the distance between the original midpoint M and the new midpoint M' is sqrt(5)/2 units.**Final Answer**The distance between the original midpoint ( M ) and the new midpoint ( M' ) is boxed{dfrac{sqrt{5}}{2}}.

💡Alright, so I have this geometry problem to tackle. It says that for two circles, the contact points of their common external tangents, the contact points of their common internal tangents, and the intersection points of the external tangents with the internal tangents all lie on three different circles. Moreover, these three circles are concentric, meaning they share the same center, and I need to determine the geometric location of this common center.Hmm, okay. Let me start by visualizing two circles. Let's say they're not overlapping, so they're separate from each other. They can be of different sizes, so one might be bigger than the other. I need to think about their external and internal tangents.First, external tangents. These are lines that touch both circles but don't cross the line segment joining their centers. The points where these tangents touch the circles are the contact points. Similarly, internal tangents are the ones that cross between the two circles, touching each one on the side facing the other circle.So, for external tangents, I can imagine two lines that just graze both circles without crossing the line connecting the centers. For internal tangents, the lines cross between the circles, touching each one on the side closer to the other circle.Now, the problem mentions that the contact points of these external and internal tangents lie on circles. Also, the intersection points of external and internal tangents lie on another circle. And all three of these circles are concentric, meaning they share the same center.I need to figure out why these points lie on circles and why these circles are concentric. Maybe I can use some properties of circles and tangents to approach this.Let me recall that the tangent to a circle is perpendicular to the radius at the point of contact. So, if I have a tangent line touching a circle at a point, the radius to that point is perpendicular to the tangent. That seems important.Also, for two circles, the external and internal tangent lines can be constructed by drawing lines that touch both circles without crossing the line connecting the centers (for external) or crossing it (for internal). The points where these tangents touch the circles are called the points of contact or contact points.So, if I consider all the contact points for external tangents, they should lie on some circle. Similarly, the contact points for internal tangents lie on another circle. And the intersection points of external and internal tangents lie on a third circle.Wait, the problem says that these three sets of points lie on three different circles, and these circles are concentric. So, all three circles share the same center.I need to find the location of this common center. Maybe it's related to the line connecting the centers of the two original circles?Let me denote the two original circles as Circle 1 and Circle 2, with centers O1 and O2, and radii r1 and r2, respectively. The distance between O1 and O2 is d.For external tangents, the points of contact can be found by drawing lines that touch both circles without crossing the line O1O2. The internal tangents cross between the circles.I remember that the length of the external tangent between two circles is given by sqrt(d^2 - (r1 + r2)^2), and the length of the internal tangent is sqrt(d^2 - (r1 - r2)^2). But I'm not sure if that's directly helpful here.Maybe I should think about the locus of the contact points. For external tangents, as the tangent lines rotate around the circles, the contact points trace out some curve. Similarly for internal tangents.Wait, the problem says these contact points lie on circles. So, maybe each set of contact points (external and internal) lies on a circle. Then, the intersection points of external and internal tangents also lie on another circle.And all three circles are concentric. So, they share the same center.I need to find the center of these circles. Maybe it's the midpoint between O1 and O2? Or perhaps some other point related to the two circles.Alternatively, maybe it's the center of the circle that is the radical axis or something like that.Wait, the radical axis of two circles is the set of points with equal power concerning both circles. It's the line perpendicular to the line joining the centers. But I don't think that's directly the center we're looking for.Alternatively, maybe it's the center of homothety that maps one circle to the other. The external and internal homothety centers.Wait, homothety centers! That might be it.The external homothety center is the point from which the external tangents emanate, and the internal homothety center is the point from which the internal tangents emanate.But in this problem, we're dealing with contact points, not the points from which the tangents are drawn.Hmm, perhaps I need to think differently.Let me consider the contact points of the external tangents. For each external tangent, there are two contact points, one on each circle. Similarly, for internal tangents.If I can show that all these contact points lie on a circle, then I can find the center.Similarly, for the intersection points of external and internal tangents, I need to show they lie on another circle.But how?Maybe I can use inversion. Inversion is a powerful tool in geometry that can turn circles into lines and vice versa, preserving tangency.Alternatively, maybe coordinate geometry. Let me try setting up coordinates.Let me place the two circles on a coordinate plane. Let me set O1 at (0,0) and O2 at (d,0). So, the line connecting the centers is the x-axis.Let me denote Circle 1 as (x)^2 + (y)^2 = r1^2, and Circle 2 as (x - d)^2 + y^2 = r2^2.Now, I can find the equations of the external and internal tangents.The external tangent lines can be found by solving for lines that touch both circles without crossing the x-axis. Similarly, internal tangents cross the x-axis.The contact points on each circle for these tangents can be found, and then I can see if they lie on a circle.Similarly, the intersection points of external and internal tangents can be found, and then I can check if they lie on a circle.This might get complicated, but let me try.First, let's find the equations of the external tangents.The formula for the external tangent lines between two circles can be derived using similar triangles or by solving the system of equations.The slope of the external tangent can be found by ensuring that the distance from the center to the line is equal to the radius.Let me denote the equation of a line as y = mx + c.The distance from O1(0,0) to this line is |c| / sqrt(m^2 + 1) = r1.Similarly, the distance from O2(d,0) to this line is |m*d + c| / sqrt(m^2 + 1) = r2.So, we have:|c| = r1 * sqrt(m^2 + 1)and|m*d + c| = r2 * sqrt(m^2 + 1)Assuming the external tangent is above the x-axis, we can drop the absolute value:c = r1 * sqrt(m^2 + 1)andm*d + c = r2 * sqrt(m^2 + 1)Substituting c from the first equation into the second:m*d + r1 * sqrt(m^2 + 1) = r2 * sqrt(m^2 + 1)So,m*d = (r2 - r1) * sqrt(m^2 + 1)Let me square both sides to eliminate the square root:m^2 * d^2 = (r2 - r1)^2 * (m^2 + 1)Expanding:m^2 * d^2 = (r2 - r1)^2 * m^2 + (r2 - r1)^2Bring all terms to one side:m^2 * d^2 - (r2 - r1)^2 * m^2 - (r2 - r1)^2 = 0Factor m^2:m^2 [d^2 - (r2 - r1)^2] - (r2 - r1)^2 = 0So,m^2 = (r2 - r1)^2 / [d^2 - (r2 - r1)^2]Therefore,m = ± (r2 - r1) / sqrt(d^2 - (r2 - r1)^2)So, the slopes of the external tangents are m = ± (r2 - r1) / sqrt(d^2 - (r2 - r1)^2)Similarly, the y-intercept c can be found from c = r1 * sqrt(m^2 + 1)Let me compute sqrt(m^2 + 1):sqrt( [ (r2 - r1)^2 / (d^2 - (r2 - r1)^2) ] + 1 ) = sqrt( [ (r2 - r1)^2 + d^2 - (r2 - r1)^2 ] / (d^2 - (r2 - r1)^2) ) = sqrt(d^2 / (d^2 - (r2 - r1)^2)) = d / sqrt(d^2 - (r2 - r1)^2)Therefore, c = r1 * [ d / sqrt(d^2 - (r2 - r1)^2) ] = (r1 * d) / sqrt(d^2 - (r2 - r1)^2)So, the equations of the external tangents are:y = [ (r2 - r1) / sqrt(d^2 - (r2 - r1)^2) ] x + (r1 * d) / sqrt(d^2 - (r2 - r1)^2 )andy = [ - (r2 - r1) / sqrt(d^2 - (r2 - r1)^2) ] x + (r1 * d) / sqrt(d^2 - (r2 - r1)^2 )Similarly, for internal tangents, the process is similar, but the formula for the slope will be different.For internal tangents, the distance equations would be:|c| = r1 * sqrt(m^2 + 1)and|m*d + c| = r2 * sqrt(m^2 + 1)But since internal tangents cross between the circles, the signs might be different. Let me assume that the internal tangent is below the x-axis, so c is negative.But actually, the formula for internal tangents is similar, but with r1 + r2 instead of r2 - r1.Wait, let me think.For internal tangents, the formula for the slope is m = ± (r1 + r2) / sqrt(d^2 - (r1 + r2)^2 )But actually, the formula is similar, but the denominator is d^2 - (r1 + r2)^2.Wait, let me derive it.For internal tangents, the distance from O1 to the line is r1, and from O2 to the line is r2, but since the tangent is between the circles, the signs might be opposite.So, let's say the line is y = mx + c.Then, distance from O1(0,0) is |c| / sqrt(m^2 + 1) = r1Distance from O2(d,0) is |m*d + c| / sqrt(m^2 + 1) = r2Assuming the internal tangent is below the x-axis, c is negative, and m*d + c is also negative.So, we can write:c = - r1 * sqrt(m^2 + 1)andm*d + c = - r2 * sqrt(m^2 + 1)Substituting c:m*d - r1 * sqrt(m^2 + 1) = - r2 * sqrt(m^2 + 1)So,m*d = - (r2 - r1) * sqrt(m^2 + 1)Wait, that seems similar to the external tangent case, but with a negative sign.So, squaring both sides:m^2 * d^2 = (r2 - r1)^2 * (m^2 + 1)Which is the same equation as before.Wait, that can't be right. Maybe I made a mistake.Wait, no, for internal tangents, the formula should be different because the direction is different.Wait, perhaps I should consider that for internal tangents, the line is between the circles, so the distances have opposite signs.Let me think again.If the line is an internal tangent, then the distance from O1 is r1, and the distance from O2 is r2, but the line is between the circles, so the signs of the distances are opposite.Therefore, if I take the distance from O1 as positive, the distance from O2 would be negative, or vice versa.So, let's write:c / sqrt(m^2 + 1) = r1and(m*d + c) / sqrt(m^2 + 1) = - r2So,c = r1 * sqrt(m^2 + 1)andm*d + c = - r2 * sqrt(m^2 + 1)Substituting c:m*d + r1 * sqrt(m^2 + 1) = - r2 * sqrt(m^2 + 1)So,m*d = - (r1 + r2) * sqrt(m^2 + 1)Then, squaring both sides:m^2 * d^2 = (r1 + r2)^2 * (m^2 + 1)So,m^2 * d^2 = (r1 + r2)^2 * m^2 + (r1 + r2)^2Bring all terms to one side:m^2 * d^2 - (r1 + r2)^2 * m^2 - (r1 + r2)^2 = 0Factor m^2:m^2 [d^2 - (r1 + r2)^2] - (r1 + r2)^2 = 0So,m^2 = (r1 + r2)^2 / [d^2 - (r1 + r2)^2]Therefore,m = ± (r1 + r2) / sqrt(d^2 - (r1 + r2)^2 )So, the slopes of the internal tangents are m = ± (r1 + r2) / sqrt(d^2 - (r1 + r2)^2 )Similarly, the y-intercept c can be found from c = r1 * sqrt(m^2 + 1)Compute sqrt(m^2 + 1):sqrt( [ (r1 + r2)^2 / (d^2 - (r1 + r2)^2) ] + 1 ) = sqrt( [ (r1 + r2)^2 + d^2 - (r1 + r2)^2 ] / (d^2 - (r1 + r2)^2) ) = sqrt(d^2 / (d^2 - (r1 + r2)^2)) = d / sqrt(d^2 - (r1 + r2)^2 )Therefore, c = r1 * [ d / sqrt(d^2 - (r1 + r2)^2 ) ] = (r1 * d) / sqrt(d^2 - (r1 + r2)^2 )So, the equations of the internal tangents are:y = [ (r1 + r2) / sqrt(d^2 - (r1 + r2)^2 ) ] x + (r1 * d) / sqrt(d^2 - (r1 + r2)^2 )andy = [ - (r1 + r2) / sqrt(d^2 - (r1 + r2)^2 ) ] x + (r1 * d) / sqrt(d^2 - (r1 + r2)^2 )Okay, so now I have the equations of the external and internal tangents.Next, I need to find the contact points on each circle for these tangents.Let's start with the external tangents.For Circle 1 (O1 at (0,0)), the contact point can be found by moving along the radius perpendicular to the tangent line.Since the tangent line has slope m, the radius at the contact point has slope -1/m.So, the contact point on Circle 1 is along the line y = (-1/m) x, at a distance r1 from O1.Similarly, for Circle 2 (O2 at (d,0)), the contact point is along the line perpendicular to the tangent, which has slope -1/m, but from the point (d,0).Wait, actually, the direction of the radius depends on the side of the tangent.Wait, maybe it's better to parametrize the contact point.Alternatively, since the tangent line is y = m x + c, the contact point on Circle 1 can be found by solving the system:y = m x + candx^2 + y^2 = r1^2But since the line is tangent to the circle, this system has exactly one solution.So, substituting y into the circle equation:x^2 + (m x + c)^2 = r1^2Expanding:x^2 + m^2 x^2 + 2 m c x + c^2 - r1^2 = 0This is a quadratic in x:(1 + m^2) x^2 + 2 m c x + (c^2 - r1^2) = 0Since it's a tangent, the discriminant is zero:(2 m c)^2 - 4 (1 + m^2)(c^2 - r1^2) = 0Simplify:4 m^2 c^2 - 4 (1 + m^2)(c^2 - r1^2) = 0Divide by 4:m^2 c^2 - (1 + m^2)(c^2 - r1^2) = 0Expand:m^2 c^2 - c^2 + r1^2 - m^2 c^2 + m^2 r1^2 = 0Simplify:- c^2 + r1^2 + m^2 r1^2 = 0So,c^2 = r1^2 (1 + m^2 )Which is consistent with our earlier result, since c = r1 * sqrt(1 + m^2 )So, the contact point on Circle 1 is the solution to the system, which is the point where the tangent touches the circle.Since the discriminant is zero, the x-coordinate is x = - (2 m c) / [2 (1 + m^2)] = - (m c) / (1 + m^2 )Similarly, y = m x + c = m * [ - (m c) / (1 + m^2 ) ] + c = - m^2 c / (1 + m^2 ) + c = c (1 - m^2 / (1 + m^2 )) = c / (1 + m^2 )So, the contact point on Circle 1 is ( - m c / (1 + m^2 ), c / (1 + m^2 ) )Similarly, for Circle 2, the contact point can be found by solving the system:y = m x + cand(x - d)^2 + y^2 = r2^2Substitute y:(x - d)^2 + (m x + c)^2 = r2^2Expand:x^2 - 2 d x + d^2 + m^2 x^2 + 2 m c x + c^2 - r2^2 = 0Combine like terms:(1 + m^2) x^2 + (-2 d + 2 m c ) x + (d^2 + c^2 - r2^2 ) = 0Again, since it's a tangent, discriminant is zero:[ -2 d + 2 m c ]^2 - 4 (1 + m^2)(d^2 + c^2 - r2^2 ) = 0Simplify:4 (d - m c )^2 - 4 (1 + m^2)(d^2 + c^2 - r2^2 ) = 0Divide by 4:(d - m c )^2 - (1 + m^2)(d^2 + c^2 - r2^2 ) = 0Expand (d - m c )^2:d^2 - 2 d m c + m^2 c^2 - (1 + m^2)(d^2 + c^2 - r2^2 ) = 0Expand the second term:d^2 - 2 d m c + m^2 c^2 - d^2 - c^2 + r2^2 - m^2 d^2 - m^2 c^2 + m^2 r2^2 = 0Simplify:-2 d m c - c^2 + r2^2 - m^2 d^2 + m^2 r2^2 = 0Wait, this seems complicated. Maybe there's a better way.Alternatively, since the contact point on Circle 2 lies along the line perpendicular to the tangent from O2.So, the slope of the radius is -1/m, so the line from O2(d,0) to the contact point is y = (-1/m)(x - d)So, solving y = (-1/m)(x - d) and y = m x + cSet equal:(-1/m)(x - d) = m x + cMultiply both sides by m:- (x - d) = m^2 x + c mSo,- x + d = m^2 x + c mBring terms with x to one side:- x - m^2 x = c m - dFactor x:- x (1 + m^2 ) = c m - dSo,x = (d - c m ) / (1 + m^2 )Then, y = m x + c = m (d - c m ) / (1 + m^2 ) + c = (m d - m^2 c ) / (1 + m^2 ) + c = (m d - m^2 c + c (1 + m^2 )) / (1 + m^2 ) = (m d + c ) / (1 + m^2 )So, the contact point on Circle 2 is ( (d - c m ) / (1 + m^2 ), (m d + c ) / (1 + m^2 ) )Okay, so now I have expressions for the contact points on both circles for a given external tangent.Similarly, for internal tangents, the contact points can be found using the same method, but with different slopes and intercepts.Now, the problem states that all these contact points lie on a circle. So, I need to show that these points satisfy the equation of a circle.Let me denote the contact points on Circle 1 as P1 and P2 for the two external tangents, and similarly Q1 and Q2 for the internal tangents.Similarly, the contact points on Circle 2 are P1' and P2' for external tangents, and Q1' and Q2' for internal tangents.Then, I need to show that P1, P2, Q1, Q2 lie on a circle, and similarly for P1', P2', Q1', Q2', and also the intersection points of external and internal tangents lie on another circle.But this is getting quite involved. Maybe there's a better approach.Wait, the problem mentions that these three circles are concentric. So, if I can find the center of one of these circles, it should be the same for the others.Let me think about the external contact points. For each external tangent, the contact points on both circles lie on a circle. Similarly for internal tangents.Wait, actually, for each external tangent, there are two contact points, one on each circle. Similarly, for internal tangents.But the problem says that all contact points of external tangents lie on a circle, all contact points of internal tangents lie on another circle, and the intersection points of external and internal tangents lie on a third circle.So, in total, three circles.And all three are concentric.So, I need to find the center of these circles.Hmm.Wait, maybe the center is the midpoint between O1 and O2? Or perhaps the center is the external or internal homothety center.Wait, homothety centers are points from which the circles are scaled to each other. The external homothety center is along the line O1O2, beyond the circles, and the internal homothety center is between them.But in this problem, the contact points are on the circles, not the homothety centers.Wait, but maybe the center of the circle on which the contact points lie is related to the homothety center.Alternatively, perhaps the center is the midpoint of O1O2.Wait, let me test this with a simple case.Suppose the two circles are equal, so r1 = r2 = r, and the distance between centers is d.Then, the external tangents are symmetric with respect to the line O1O2, and the internal tangents are also symmetric.In this case, the contact points for external tangents would form a rectangle, and similarly for internal tangents.Wait, actually, for equal circles, the contact points of external tangents would lie on a circle centered at the midpoint of O1O2, with radius sqrt( (d/2)^2 + r^2 )Similarly, the contact points of internal tangents would lie on another circle centered at the midpoint.And the intersection points of external and internal tangents would also lie on a circle centered at the midpoint.So, in this special case, the common center is the midpoint of O1O2.Hmm, that's interesting.So, perhaps in the general case, the common center is the midpoint of O1O2.But wait, in the case where the circles are unequal, is this still true?Let me think.Suppose Circle 1 has radius r1, Circle 2 has radius r2, and the distance between centers is d.If I can show that the contact points lie on a circle centered at the midpoint of O1O2, then that would be the common center.Alternatively, maybe it's another point.Wait, let's consider the midpoint M of O1O2.If I can show that all contact points are equidistant from M, then M is the center.So, let's compute the distance from M to a contact point.Let me take a contact point on Circle 1 for an external tangent.From earlier, the contact point on Circle 1 is ( - m c / (1 + m^2 ), c / (1 + m^2 ) )Similarly, the midpoint M is at (d/2, 0)So, the distance squared from M to the contact point is:( - m c / (1 + m^2 ) - d/2 )^2 + ( c / (1 + m^2 ) - 0 )^2Let me compute this.First, let me denote m as the slope of the external tangent, which we found earlier as m = (r2 - r1)/sqrt(d^2 - (r2 - r1)^2 )And c = (r1 d)/sqrt(d^2 - (r2 - r1)^2 )Let me compute - m c / (1 + m^2 )First, compute m c:m c = [ (r2 - r1)/sqrt(d^2 - (r2 - r1)^2 ) ] * [ r1 d / sqrt(d^2 - (r2 - r1)^2 ) ] = r1 d (r2 - r1 ) / (d^2 - (r2 - r1)^2 )Similarly, 1 + m^2 = 1 + [ (r2 - r1)^2 / (d^2 - (r2 - r1)^2 ) ] = [ d^2 - (r2 - r1)^2 + (r2 - r1)^2 ] / (d^2 - (r2 - r1)^2 ) = d^2 / (d^2 - (r2 - r1)^2 )So, - m c / (1 + m^2 ) = - [ r1 d (r2 - r1 ) / (d^2 - (r2 - r1)^2 ) ] / [ d^2 / (d^2 - (r2 - r1)^2 ) ] = - r1 d (r2 - r1 ) / d^2 = - r1 (r2 - r1 ) / dSimilarly, c / (1 + m^2 ) = [ r1 d / sqrt(d^2 - (r2 - r1)^2 ) ] / [ d^2 / (d^2 - (r2 - r1)^2 ) ] = [ r1 d / sqrt(d^2 - (r2 - r1)^2 ) ] * [ (d^2 - (r2 - r1)^2 ) / d^2 ] = r1 sqrt(d^2 - (r2 - r1)^2 ) / dSo, the contact point on Circle 1 is ( - r1 (r2 - r1 ) / d, r1 sqrt(d^2 - (r2 - r1)^2 ) / d )Similarly, the midpoint M is at (d/2, 0)So, the distance squared from M to the contact point is:[ - r1 (r2 - r1 ) / d - d/2 ]^2 + [ r1 sqrt(d^2 - (r2 - r1)^2 ) / d ]^2Let me compute each term.First term:- r1 (r2 - r1 ) / d - d/2 = - [ r1 (r2 - r1 ) + d^2 / 2 ] / dWait, no, let's compute it step by step.Compute x-coordinate difference:- r1 (r2 - r1 ) / d - d/2 = - [ r1 (r2 - r1 ) + d^2 / 2 ] / dWait, actually, let me compute it as:= ( - r1 (r2 - r1 ) / d ) - ( d / 2 )= - [ r1 (r2 - r1 ) / d + d / 2 ]= - [ (2 r1 (r2 - r1 ) + d^2 ) / (2 d ) ]But actually, let me compute the square:[ - r1 (r2 - r1 ) / d - d/2 ]^2 = [ - ( r1 (r2 - r1 ) + d^2 / 2 ) / d ]^2 = [ ( r1 (r2 - r1 ) + d^2 / 2 ) / d ]^2 = [ r1 (r2 - r1 ) + d^2 / 2 ]^2 / d^2Similarly, the y-coordinate squared is:[ r1 sqrt(d^2 - (r2 - r1)^2 ) / d ]^2 = r1^2 (d^2 - (r2 - r1)^2 ) / d^2So, total distance squared is:[ r1 (r2 - r1 ) + d^2 / 2 ]^2 / d^2 + r1^2 (d^2 - (r2 - r1)^2 ) / d^2Let me compute this:= [ ( r1 (r2 - r1 ) + d^2 / 2 )^2 + r1^2 (d^2 - (r2 - r1)^2 ) ] / d^2Expand the first square:= [ r1^2 (r2 - r1 )^2 + r1 (r2 - r1 ) d^2 + d^4 / 4 + r1^2 (d^2 - (r2 - r1)^2 ) ] / d^2Simplify term by term:First term: r1^2 (r2 - r1 )^2Second term: r1 (r2 - r1 ) d^2Third term: d^4 / 4Fourth term: r1^2 d^2 - r1^2 (r2 - r1 )^2So, combine like terms:r1^2 (r2 - r1 )^2 - r1^2 (r2 - r1 )^2 = 0r1 (r2 - r1 ) d^2 + r1^2 d^2 = r1 d^2 (r2 - r1 + r1 ) = r1 d^2 r2d^4 / 4So, total numerator:r1 d^2 r2 + d^4 / 4Therefore, distance squared is:( r1 r2 d^2 + d^4 / 4 ) / d^2 = r1 r2 + d^2 / 4So, the distance from M to the contact point is sqrt(r1 r2 + d^2 / 4 )Similarly, let's compute the distance from M to the contact point on Circle 2.The contact point on Circle 2 is ( (d - c m ) / (1 + m^2 ), (m d + c ) / (1 + m^2 ) )From earlier, we have:c = (r1 d ) / sqrt(d^2 - (r2 - r1 )^2 )m = (r2 - r1 ) / sqrt(d^2 - (r2 - r1 )^2 )So, c m = [ r1 d (r2 - r1 ) ] / (d^2 - (r2 - r1 )^2 )Similarly, 1 + m^2 = d^2 / (d^2 - (r2 - r1 )^2 )So, (d - c m ) / (1 + m^2 ) = [ d - r1 d (r2 - r1 ) / (d^2 - (r2 - r1 )^2 ) ] / [ d^2 / (d^2 - (r2 - r1 )^2 ) ]= [ d (d^2 - (r2 - r1 )^2 ) - r1 d (r2 - r1 ) ] / d^2= [ d^3 - d (r2 - r1 )^2 - r1 d (r2 - r1 ) ] / d^2= [ d^2 - (r2 - r1 )^2 - r1 (r2 - r1 ) ] / dSimilarly, (m d + c ) / (1 + m^2 ) = [ m d + c ] / (1 + m^2 )Compute m d + c:= [ (r2 - r1 ) / sqrt(d^2 - (r2 - r1 )^2 ) ] d + [ r1 d / sqrt(d^2 - (r2 - r1 )^2 ) ]= [ (r2 - r1 ) d + r1 d ] / sqrt(d^2 - (r2 - r1 )^2 )= [ r2 d ] / sqrt(d^2 - (r2 - r1 )^2 )Then, divide by (1 + m^2 ) = d^2 / (d^2 - (r2 - r1 )^2 )So,= [ r2 d / sqrt(d^2 - (r2 - r1 )^2 ) ] / [ d^2 / (d^2 - (r2 - r1 )^2 ) ]= r2 d (d^2 - (r2 - r1 )^2 ) / ( d^2 sqrt(d^2 - (r2 - r1 )^2 ) )= r2 (d^2 - (r2 - r1 )^2 ) / ( d sqrt(d^2 - (r2 - r1 )^2 ) )= r2 sqrt(d^2 - (r2 - r1 )^2 ) / dSo, the contact point on Circle 2 is ( [ d^2 - (r2 - r1 )^2 - r1 (r2 - r1 ) ] / d , r2 sqrt(d^2 - (r2 - r1 )^2 ) / d )Now, compute the distance from M(d/2, 0) to this point.x-coordinate difference:[ d^2 - (r2 - r1 )^2 - r1 (r2 - r1 ) ] / d - d/2 = [ d^2 - (r2 - r1 )^2 - r1 (r2 - r1 ) - d^2 / 2 ] / d= [ d^2 / 2 - (r2 - r1 )^2 - r1 (r2 - r1 ) ] / dSimilarly, y-coordinate difference:r2 sqrt(d^2 - (r2 - r1 )^2 ) / d - 0 = r2 sqrt(d^2 - (r2 - r1 )^2 ) / dSo, distance squared is:[ ( d^2 / 2 - (r2 - r1 )^2 - r1 (r2 - r1 ) ) / d ]^2 + [ r2 sqrt(d^2 - (r2 - r1 )^2 ) / d ]^2Let me compute this.First term:= [ ( d^2 / 2 - (r2 - r1 )^2 - r1 (r2 - r1 ) ) / d ]^2= [ ( d^2 / 2 - r2^2 + 2 r1 r2 - r1^2 - r1 r2 + r1^2 ) / d ]^2Simplify inside the brackets:d^2 / 2 - r2^2 + 2 r1 r2 - r1^2 - r1 r2 + r1^2 = d^2 / 2 - r2^2 + r1 r2So, first term squared:= ( d^2 / 2 - r2^2 + r1 r2 )^2 / d^2Second term:= [ r2 sqrt(d^2 - (r2 - r1 )^2 ) / d ]^2 = r2^2 (d^2 - (r2 - r1 )^2 ) / d^2So, total distance squared:[ ( d^2 / 2 - r2^2 + r1 r2 )^2 + r2^2 (d^2 - (r2 - r1 )^2 ) ] / d^2Let me expand the first square:( d^2 / 2 - r2^2 + r1 r2 )^2 = (d^2 / 2)^2 + (- r2^2 + r1 r2 )^2 + 2 * d^2 / 2 * (- r2^2 + r1 r2 )= d^4 / 4 + r2^4 - 2 r1 r2^3 + r1^2 r2^2 + d^2 (- r2^2 + r1 r2 )So, expanding:= d^4 / 4 + r2^4 - 2 r1 r2^3 + r1^2 r2^2 - d^2 r2^2 + d^2 r1 r2Now, add the second term:+ r2^2 (d^2 - (r2 - r1 )^2 ) = r2^2 d^2 - r2^2 (r2^2 - 2 r1 r2 + r1^2 ) = r2^2 d^2 - r2^4 + 2 r1 r2^3 - r1^2 r2^2So, total numerator:d^4 / 4 + r2^4 - 2 r1 r2^3 + r1^2 r2^2 - d^2 r2^2 + d^2 r1 r2 + r2^2 d^2 - r2^4 + 2 r1 r2^3 - r1^2 r2^2Simplify term by term:d^4 / 4r2^4 - r2^4 = 0-2 r1 r2^3 + 2 r1 r2^3 = 0r1^2 r2^2 - r1^2 r2^2 = 0- d^2 r2^2 + r2^2 d^2 = 0d^2 r1 r2So, total numerator:d^4 / 4 + d^2 r1 r2Therefore, distance squared is:( d^4 / 4 + d^2 r1 r2 ) / d^2 = d^2 / 4 + r1 r2Which is the same as the distance squared from M to the contact point on Circle 1.So, both contact points on external tangents are equidistant from M, with distance sqrt( d^2 / 4 + r1 r2 )Therefore, all contact points of external tangents lie on a circle centered at M with radius sqrt( d^2 / 4 + r1 r2 )Similarly, for internal tangents, I can perform the same calculation.Let me consider the contact points for internal tangents.Using similar steps, but with m = (r1 + r2 ) / sqrt(d^2 - (r1 + r2 )^2 ) and c = (r1 d ) / sqrt(d^2 - (r1 + r2 )^2 )Following the same process, the contact points on Circle 1 and Circle 2 will be equidistant from M, with distance sqrt( d^2 / 4 + r1 r2 )Wait, is that the same radius?Wait, let me compute it.For internal tangents, the contact point on Circle 1 is ( - m c / (1 + m^2 ), c / (1 + m^2 ) )With m = (r1 + r2 ) / sqrt(d^2 - (r1 + r2 )^2 )And c = (r1 d ) / sqrt(d^2 - (r1 + r2 )^2 )So, compute - m c / (1 + m^2 )First, m c = [ (r1 + r2 ) / sqrt(d^2 - (r1 + r2 )^2 ) ] * [ r1 d / sqrt(d^2 - (r1 + r2 )^2 ) ] = r1 d (r1 + r2 ) / (d^2 - (r1 + r2 )^2 )1 + m^2 = 1 + [ (r1 + r2 )^2 / (d^2 - (r1 + r2 )^2 ) ] = [ d^2 - (r1 + r2 )^2 + (r1 + r2 )^2 ] / (d^2 - (r1 + r2 )^2 ) = d^2 / (d^2 - (r1 + r2 )^2 )So, - m c / (1 + m^2 ) = - [ r1 d (r1 + r2 ) / (d^2 - (r1 + r2 )^2 ) ] / [ d^2 / (d^2 - (r1 + r2 )^2 ) ] = - r1 d (r1 + r2 ) / d^2 = - r1 (r1 + r2 ) / dSimilarly, c / (1 + m^2 ) = [ r1 d / sqrt(d^2 - (r1 + r2 )^2 ) ] / [ d^2 / (d^2 - (r1 + r2 )^2 ) ] = r1 sqrt(d^2 - (r1 + r2 )^2 ) / dSo, the contact point on Circle 1 is ( - r1 (r1 + r2 ) / d, r1 sqrt(d^2 - (r1 + r2 )^2 ) / d )Similarly, compute the distance from M(d/2, 0):x-coordinate difference:- r1 (r1 + r2 ) / d - d/2 = - [ r1 (r1 + r2 ) + d^2 / 2 ] / dWait, let me compute it step by step.= ( - r1 (r1 + r2 ) / d ) - ( d / 2 )= - [ r1 (r1 + r2 ) / d + d / 2 ]= - [ (2 r1 (r1 + r2 ) + d^2 ) / (2 d ) ]But actually, let me compute the square:[ - r1 (r1 + r2 ) / d - d/2 ]^2 = [ - ( r1 (r1 + r2 ) + d^2 / 2 ) / d ]^2 = [ ( r1 (r1 + r2 ) + d^2 / 2 ) / d ]^2 = [ r1 (r1 + r2 ) + d^2 / 2 ]^2 / d^2Similarly, the y-coordinate squared is:[ r1 sqrt(d^2 - (r1 + r2 )^2 ) / d ]^2 = r1^2 (d^2 - (r1 + r2 )^2 ) / d^2So, total distance squared is:[ r1 (r1 + r2 ) + d^2 / 2 ]^2 / d^2 + r1^2 (d^2 - (r1 + r2 )^2 ) / d^2Let me compute this:= [ ( r1 (r1 + r2 ) + d^2 / 2 )^2 + r1^2 (d^2 - (r1 + r2 )^2 ) ] / d^2Expand the first square:= [ r1^2 (r1 + r2 )^2 + r1 (r1 + r2 ) d^2 + d^4 / 4 + r1^2 (d^2 - (r1 + r2 )^2 ) ] / d^2Simplify term by term:First term: r1^2 (r1 + r2 )^2Second term: r1 (r1 + r2 ) d^2Third term: d^4 / 4Fourth term: r1^2 d^2 - r1^2 (r1 + r2 )^2Combine like terms:r1^2 (r1 + r2 )^2 - r1^2 (r1 + r2 )^2 = 0r1 (r1 + r2 ) d^2 + r1^2 d^2 = r1 d^2 (r1 + r2 + r1 ) = r1 d^2 (2 r1 + r2 )Wait, no, let me compute:r1 (r1 + r2 ) d^2 + r1^2 d^2 = r1 d^2 (r1 + r2 + r1 ) = r1 d^2 (2 r1 + r2 )Wait, that doesn't seem right. Let me re-express:r1 (r1 + r2 ) d^2 + r1^2 d^2 = r1 d^2 (r1 + r2 + r1 ) = r1 d^2 (2 r1 + r2 )Wait, actually, no. It's r1 (r1 + r2 ) d^2 + r1^2 d^2 = r1 d^2 (r1 + r2 + r1 ) = r1 d^2 (2 r1 + r2 )But that seems incorrect because:r1 (r1 + r2 ) d^2 + r1^2 d^2 = r1^2 d^2 + r1 r2 d^2 + r1^2 d^2 = 2 r1^2 d^2 + r1 r2 d^2So, it's 2 r1^2 d^2 + r1 r2 d^2Similarly, the third term is d^4 / 4So, total numerator:2 r1^2 d^2 + r1 r2 d^2 + d^4 / 4Therefore, distance squared is:( 2 r1^2 d^2 + r1 r2 d^2 + d^4 / 4 ) / d^2 = 2 r1^2 + r1 r2 + d^2 / 4Wait, that's different from the external tangent case.Wait, in the external tangent case, the distance squared was r1 r2 + d^2 / 4But here, it's 2 r1^2 + r1 r2 + d^2 / 4Hmm, that suggests that the contact points for internal tangents are not equidistant from M.Wait, that contradicts the problem statement, which says that all contact points lie on a circle.Wait, maybe I made a mistake in the calculation.Wait, let me double-check.In the external tangent case, the distance squared was r1 r2 + d^2 / 4In the internal tangent case, I got 2 r1^2 + r1 r2 + d^2 / 4But that can't be, because the problem says that all contact points lie on a circle.Wait, perhaps I made a mistake in the calculation.Let me recompute the distance squared for the internal tangent contact point.The contact point on Circle 1 is ( - r1 (r1 + r2 ) / d, r1 sqrt(d^2 - (r1 + r2 )^2 ) / d )Midpoint M is at (d/2, 0)So, x-coordinate difference:- r1 (r1 + r2 ) / d - d/2 = - [ r1 (r1 + r2 ) + d^2 / 2 ] / dWait, no, let me compute it correctly.Compute x1 - xM:= ( - r1 (r1 + r2 ) / d ) - ( d / 2 )= - r1 (r1 + r2 ) / d - d / 2= - [ r1 (r1 + r2 ) + d^2 / 2 ] / dSimilarly, y1 - yM = r1 sqrt(d^2 - (r1 + r2 )^2 ) / d - 0 = r1 sqrt(d^2 - (r1 + r2 )^2 ) / dSo, distance squared is:[ - ( r1 (r1 + r2 ) + d^2 / 2 ) / d ]^2 + [ r1 sqrt(d^2 - (r1 + r2 )^2 ) / d ]^2= [ ( r1 (r1 + r2 ) + d^2 / 2 )^2 + r1^2 (d^2 - (r1 + r2 )^2 ) ] / d^2Expand the first square:= [ r1^2 (r1 + r2 )^2 + r1 (r1 + r2 ) d^2 + d^4 / 4 + r1^2 (d^2 - (r1 + r2 )^2 ) ] / d^2Simplify term by term:First term: r1^2 (r1 + r2 )^2Second term: r1 (r1 + r2 ) d^2Third term: d^4 / 4Fourth term: r1^2 d^2 - r1^2 (r1 + r2 )^2Combine like terms:r1^2 (r1 + r2 )^2 - r1^2 (r1 + r2 )^2 = 0r1 (r1 + r2 ) d^2 + r1^2 d^2 = r1 d^2 (r1 + r2 + r1 ) = r1 d^2 (2 r1 + r2 )Wait, no, actually:r1 (r1 + r2 ) d^2 + r1^2 d^2 = r1^2 d^2 + r1 r2 d^2 + r1^2 d^2 = 2 r1^2 d^2 + r1 r2 d^2So, total numerator:2 r1^2 d^2 + r1 r2 d^2 + d^4 / 4Therefore, distance squared:(2 r1^2 d^2 + r1 r2 d^2 + d^4 / 4 ) / d^2 = 2 r1^2 + r1 r2 + d^2 / 4Hmm, so it's different from the external tangent case.But the problem states that all contact points lie on a circle. So, perhaps I made a mistake in assuming that the internal tangent contact points lie on the same circle as external ones.Wait, no, the problem says that the contact points of external tangents lie on one circle, contact points of internal tangents lie on another circle, and the intersection points lie on a third circle, and all three circles are concentric.So, in this case, the external tangent contact points lie on a circle centered at M with radius sqrt(r1 r2 + d^2 / 4 )The internal tangent contact points lie on another circle centered at M with radius sqrt(2 r1^2 + r1 r2 + d^2 / 4 )Wait, but that would mean that the two circles are concentric but with different radii.Similarly, the intersection points of external and internal tangents would lie on a third circle centered at M.But according to the problem, all three circles are concentric, which is true if they all share the same center, which is M.So, the common center is the midpoint of O1O2.Therefore, the geometric location of their common center is the midpoint of the line segment joining the centers of the two original circles.So, the answer is the midpoint of O1O2.But let me verify this with another approach.Another way to think about this is to consider the midcircle or the circle of antisimilitude.Wait, the circle of antisimilitude is the locus of points from which the two circles are seen under equal angles. It's centered at the midpoint of O1O2.Wait, actually, the circle of antisimilitude is the locus of centers of circles orthogonal to both given circles. It's centered at the midpoint of O1O2.But in our case, the contact points lie on circles centered at the midpoint.Alternatively, perhaps the nine-point circle or something else.But given the calculations above, it seems that the midpoint M is the common center.Therefore, the geometric location of the common center is the midpoint of the line segment joining the centers of the two original circles.

💡Alright, so I have this problem about a sequence where the sum of the first n terms is given by a specific formula. Let me try to unpack this step by step.First, the problem is divided into two parts: part (I) asks for the first term ( a_1 ) and the general term ( a_n ). Part (II) introduces another sequence ( T_n ) and asks to prove that the sum of the first n terms of ( T_n ) is less than ( frac{3}{2} ).Starting with part (I). The sum of the first n terms is given by:[S_n = frac{4}{3}a_n - frac{1}{3} times 2^{n+1} + frac{2}{3}]I know that the sum of the first n terms ( S_n ) is related to the nth term ( a_n ) by the formula:[S_n = a_1 + a_2 + dots + a_n]And also, the nth term can be found by:[a_n = S_n - S_{n-1}]So, maybe I can use this relationship to find a recursive formula for ( a_n ).But first, let's find ( a_1 ). Since ( S_1 = a_1 ), plugging n=1 into the given formula:[S_1 = frac{4}{3}a_1 - frac{1}{3} times 2^{2} + frac{2}{3}]Simplify:[a_1 = frac{4}{3}a_1 - frac{4}{3} + frac{2}{3}]Combine the constants:[a_1 = frac{4}{3}a_1 - frac{2}{3}]Subtract ( frac{4}{3}a_1 ) from both sides:[a_1 - frac{4}{3}a_1 = -frac{2}{3}][-frac{1}{3}a_1 = -frac{2}{3}]Multiply both sides by -3:[a_1 = 2]Okay, so the first term is 2. That wasn't too bad.Now, let's find the general term ( a_n ). As I thought earlier, ( a_n = S_n - S_{n-1} ). Let's compute that.Given:[S_n = frac{4}{3}a_n - frac{1}{3} times 2^{n+1} + frac{2}{3}]So,[S_{n-1} = frac{4}{3}a_{n-1} - frac{1}{3} times 2^{n} + frac{2}{3}]Therefore,[a_n = S_n - S_{n-1} = left( frac{4}{3}a_n - frac{1}{3} times 2^{n+1} + frac{2}{3} right) - left( frac{4}{3}a_{n-1} - frac{1}{3} times 2^{n} + frac{2}{3} right)]Simplify term by term:- The ( frac{4}{3}a_n ) and ( -frac{4}{3}a_{n-1} ) terms stay.- The constants ( frac{2}{3} ) cancel out.- The other terms: ( -frac{1}{3} times 2^{n+1} + frac{1}{3} times 2^{n} )So,[a_n = frac{4}{3}a_n - frac{4}{3}a_{n-1} - frac{1}{3}(2^{n+1} - 2^n)]Simplify ( 2^{n+1} - 2^n = 2^n(2 - 1) = 2^n ). So,[a_n = frac{4}{3}a_n - frac{4}{3}a_{n-1} - frac{1}{3} times 2^n]Let's get all terms involving ( a_n ) on one side:[a_n - frac{4}{3}a_n = -frac{4}{3}a_{n-1} - frac{1}{3} times 2^n][-frac{1}{3}a_n = -frac{4}{3}a_{n-1} - frac{1}{3} times 2^n]Multiply both sides by -3:[a_n = 4a_{n-1} + 2^n]Hmm, so we have a recurrence relation:[a_n = 4a_{n-1} + 2^n]This is a linear nonhomogeneous recurrence relation. To solve this, I can find the homogeneous solution and a particular solution.First, the homogeneous equation:[a_n - 4a_{n-1} = 0]Characteristic equation:[r - 4 = 0 implies r = 4]So, the homogeneous solution is:[a_n^{(h)} = C times 4^n]Now, find a particular solution. The nonhomogeneous term is ( 2^n ). Since 2 is not a root of the characteristic equation, we can try a particular solution of the form ( a_n^{(p)} = D times 2^n ).Plug into the recurrence:[D times 2^n = 4 times D times 2^{n-1} + 2^n]Simplify:[D times 2^n = 4 times D times 2^{n-1} + 2^n][D times 2^n = 2D times 2^{n-1} times 2 + 2^n]Wait, actually, let's compute it step by step:[4 times D times 2^{n-1} = 4D times 2^{n-1} = 2D times 2^{n}]So,[D times 2^n = 2D times 2^n + 2^n]Bring all terms to one side:[D times 2^n - 2D times 2^n - 2^n = 0][(-D) times 2^n - 2^n = 0]Factor out ( 2^n ):[(-D - 1)2^n = 0]Since ( 2^n neq 0 ), we have:[-D - 1 = 0 implies D = -1]So, the particular solution is:[a_n^{(p)} = -2^n]Therefore, the general solution is:[a_n = a_n^{(h)} + a_n^{(p)} = C times 4^n - 2^n]Now, apply the initial condition to find C. We know ( a_1 = 2 ).For n=1:[a_1 = C times 4^1 - 2^1 = 4C - 2 = 2]Solve for C:[4C - 2 = 2 implies 4C = 4 implies C = 1]Thus, the general term is:[a_n = 4^n - 2^n]Let me verify this with n=1:[a_1 = 4^1 - 2^1 = 4 - 2 = 2]Which matches. Let's check n=2. First, compute ( S_2 ):[S_2 = frac{4}{3}a_2 - frac{1}{3} times 2^{3} + frac{2}{3}]Compute ( a_2 = 4^2 - 2^2 = 16 - 4 = 12 )So,[S_2 = frac{4}{3} times 12 - frac{8}{3} + frac{2}{3} = 16 - frac{6}{3} = 16 - 2 = 14]Alternatively, ( S_2 = a_1 + a_2 = 2 + 12 = 14 ). It matches. Good.So, part (I) is done: ( a_1 = 2 ) and ( a_n = 4^n - 2^n ).Moving on to part (II). We have:[T_n = frac{2^n}{S_n}]And we need to prove that:[sum_{i=1}^{n} T_i < frac{3}{2}]First, let's express ( S_n ) in terms of ( a_n ). From part (I), we have ( a_n = 4^n - 2^n ), so:[S_n = frac{4}{3}a_n - frac{1}{3} times 2^{n+1} + frac{2}{3}]Substitute ( a_n ):[S_n = frac{4}{3}(4^n - 2^n) - frac{1}{3} times 2^{n+1} + frac{2}{3}]Simplify term by term:[frac{4}{3} times 4^n = frac{4^{n+1}}{3}][frac{4}{3} times (-2^n) = -frac{4 times 2^n}{3}][-frac{1}{3} times 2^{n+1} = -frac{2 times 2^n}{3}]So, combine all terms:[S_n = frac{4^{n+1}}{3} - frac{4 times 2^n}{3} - frac{2 times 2^n}{3} + frac{2}{3}]Combine the ( 2^n ) terms:[- frac{4 times 2^n + 2 times 2^n}{3} = -frac{6 times 2^n}{3} = -2 times 2^n = -2^{n+1}]So,[S_n = frac{4^{n+1}}{3} - 2^{n+1} + frac{2}{3}]Wait, let me double-check that. The original expression after substitution was:[S_n = frac{4^{n+1}}{3} - frac{4 times 2^n}{3} - frac{2 times 2^n}{3} + frac{2}{3}]So, combining the middle terms:[- frac{4 times 2^n + 2 times 2^n}{3} = -frac{6 times 2^n}{3} = -2 times 2^n = -2^{n+1}]Therefore,[S_n = frac{4^{n+1}}{3} - 2^{n+1} + frac{2}{3}]Alternatively, factor out ( frac{1}{3} ):[S_n = frac{4^{n+1} - 3 times 2^{n+1} + 2}{3}]Hmm, maybe we can factor this expression further.Notice that ( 4^{n+1} = (2^2)^{n+1} = 2^{2n+2} ). So,[S_n = frac{2^{2n+2} - 3 times 2^{n+1} + 2}{3}]Let me see if the numerator can be factored. Let me set ( x = 2^{n+1} ). Then the numerator becomes:[x^2 - 3x + 2]Which factors as:[(x - 1)(x - 2)]So, substituting back:[(2^{n+1} - 1)(2^{n+1} - 2)]But ( 2^{n+1} - 2 = 2(2^n - 1) ). So,[(2^{n+1} - 1) times 2(2^n - 1) = 2(2^{n+1} - 1)(2^n - 1)]Therefore, the numerator is ( 2(2^{n+1} - 1)(2^n - 1) ), so:[S_n = frac{2(2^{n+1} - 1)(2^n - 1)}{3}]Thus,[T_n = frac{2^n}{S_n} = frac{2^n}{frac{2(2^{n+1} - 1)(2^n - 1)}{3}} = frac{3 times 2^n}{2(2^{n+1} - 1)(2^n - 1)}]Simplify:[T_n = frac{3}{2} times frac{2^n}{(2^{n+1} - 1)(2^n - 1)}]Hmm, this looks like it might telescope. Let me try to express ( T_n ) as a difference of fractions.Let me consider the expression ( frac{2^n}{(2^{n+1} - 1)(2^n - 1)} ). Maybe partial fractions can help.Let me set:[frac{2^n}{(2^{n+1} - 1)(2^n - 1)} = frac{A}{2^n - 1} + frac{B}{2^{n+1} - 1}]Multiply both sides by ( (2^{n+1} - 1)(2^n - 1) ):[2^n = A(2^{n+1} - 1) + B(2^n - 1)]Expand:[2^n = A times 2^{n+1} - A + B times 2^n - B]Group like terms:[2^n = (2A + B)2^n - (A + B)]This must hold for all n, so equate coefficients:- Coefficient of ( 2^n ): ( 1 = 2A + B )- Constant term: ( 0 = - (A + B) implies A + B = 0 )From the second equation: ( B = -A )Substitute into the first equation:[1 = 2A - A = A]Thus, ( A = 1 ), ( B = -1 )Therefore,[frac{2^n}{(2^{n+1} - 1)(2^n - 1)} = frac{1}{2^n - 1} - frac{1}{2^{n+1} - 1}]So, substituting back into ( T_n ):[T_n = frac{3}{2} left( frac{1}{2^n - 1} - frac{1}{2^{n+1} - 1} right )]This is a telescoping series! When we sum ( T_i ) from i=1 to n, most terms will cancel out.Let's write out the sum:[sum_{i=1}^{n} T_i = frac{3}{2} sum_{i=1}^{n} left( frac{1}{2^i - 1} - frac{1}{2^{i+1} - 1} right )]This telescopes to:[frac{3}{2} left( frac{1}{2^1 - 1} - frac{1}{2^{n+1} - 1} right )]Simplify:[frac{3}{2} left( frac{1}{1} - frac{1}{2^{n+1} - 1} right ) = frac{3}{2} left( 1 - frac{1}{2^{n+1} - 1} right )]Since ( 2^{n+1} - 1 > 1 ) for all ( n geq 1 ), the term ( frac{1}{2^{n+1} - 1} ) is positive and less than 1. Therefore:[1 - frac{1}{2^{n+1} - 1} < 1]Thus,[frac{3}{2} left( 1 - frac{1}{2^{n+1} - 1} right ) < frac{3}{2} times 1 = frac{3}{2}]Therefore,[sum_{i=1}^{n} T_i < frac{3}{2}]Which is what we needed to prove.Let me just verify this with n=1 and n=2 to ensure it holds.For n=1:[T_1 = frac{2^1}{S_1} = frac{2}{2} = 1]Sum is 1, which is less than 3/2.For n=2:Compute ( S_2 = 14 ) as before.( T_2 = frac{4}{14} = frac{2}{7} )Sum: 1 + 2/7 = 9/7 ≈ 1.2857 < 1.5For n=3:Compute ( a_3 = 4^3 - 2^3 = 64 - 8 = 56 )( S_3 = S_2 + a_3 = 14 + 56 = 70 )( T_3 = frac{8}{70} = frac{4}{35} )Sum: 1 + 2/7 + 4/35 = (35 + 10 + 4)/35 = 49/35 = 1.4 < 1.5Seems consistent. So, the proof holds.**Final Answer**Ⅰ. The first term is (boxed{2}) and the general term is (boxed{4^n - 2^n}).Ⅱ. The sum (sum_{i=1}^{n} T_i) is indeed less than (boxed{dfrac{3}{2}}).

💡Okay, so I've got this problem here where I need to find the largest possible value of ( d ) given that ( a + b + c + d = 10 ) and ( ab + ac + ad + bc + bd + cd = 20 ). Hmm, let me think about how to approach this.First, the problem gives me two equations involving four variables. I remember that when dealing with symmetric equations like this, sometimes it's helpful to consider the variables as parts of a set or maybe even think about them in terms of means or something like that. But I'm not sure yet. Let me write down the equations again:1. ( a + b + c + d = 10 )2. ( ab + ac + ad + bc + bd + cd = 20 )I notice that the second equation is the sum of all possible products of two variables. That makes me think of the expansion of ( (a + b + c + d)^2 ), which is ( a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd) ). Maybe I can use that to find another equation involving the squares of the variables.Let me compute ( (a + b + c + d)^2 ):( (a + b + c + d)^2 = 10^2 = 100 )And expanding it, we have:( a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd) = 100 )We know from the second equation that ( ab + ac + ad + bc + bd + cd = 20 ), so substituting that in:( a^2 + b^2 + c^2 + d^2 + 2(20) = 100 )Simplifying:( a^2 + b^2 + c^2 + d^2 + 40 = 100 )So,( a^2 + b^2 + c^2 + d^2 = 60 )Alright, so now I have the sum of the squares of the variables. I need to find the maximum value of ( d ). Maybe I can express ( a^2 + b^2 + c^2 ) in terms of ( d ) and then use some inequality to find the maximum.Let me denote ( S = a + b + c ). Since ( a + b + c + d = 10 ), we have ( S = 10 - d ).Similarly, the sum of the squares ( a^2 + b^2 + c^2 = 60 - d^2 ).Now, I recall that for real numbers, the Cauchy-Schwarz inequality can be useful. Specifically, for three variables ( a, b, c ), the inequality states that:( (a^2 + b^2 + c^2)(1^2 + 1^2 + 1^2) geq (a + b + c)^2 )This is because Cauchy-Schwarz says that ( (x_1^2 + x_2^2 + ... + x_n^2)(y_1^2 + y_2^2 + ... + y_n^2) geq (x_1y_1 + x_2y_2 + ... + x_ny_n)^2 ). In this case, if I set all ( y_i = 1 ), it simplifies nicely.So applying that here:( (a^2 + b^2 + c^2)(1 + 1 + 1) geq (a + b + c)^2 )Plugging in the expressions I have:( (60 - d^2)(3) geq (10 - d)^2 )Let me write that out:( 3(60 - d^2) geq (10 - d)^2 )Expanding both sides:Left side: ( 180 - 3d^2 )Right side: ( 100 - 20d + d^2 )So putting it all together:( 180 - 3d^2 geq 100 - 20d + d^2 )Let me bring all terms to one side to form a quadratic inequality:( 180 - 3d^2 - 100 + 20d - d^2 geq 0 )Simplify:( 80 + 20d - 4d^2 geq 0 )Let me rearrange it:( -4d^2 + 20d + 80 geq 0 )Hmm, it's usually easier to work with positive coefficients for ( d^2 ), so I'll multiply both sides by -1. Remember that multiplying an inequality by a negative number reverses the inequality sign.( 4d^2 - 20d - 80 leq 0 )Now, let's simplify this quadratic inequality. I can factor out a 4:( 4(d^2 - 5d - 20) leq 0 )Divide both sides by 4 (which is positive, so inequality sign remains the same):( d^2 - 5d - 20 leq 0 )Now, I need to solve this quadratic inequality. First, let's find the roots of the equation ( d^2 - 5d - 20 = 0 ).Using the quadratic formula:( d = frac{5 pm sqrt{25 + 80}}{2} )Because ( b^2 - 4ac = 25 + 80 = 105 ), so:( d = frac{5 pm sqrt{105}}{2} )So the roots are ( frac{5 + sqrt{105}}{2} ) and ( frac{5 - sqrt{105}}{2} ).Since the coefficient of ( d^2 ) is positive, the parabola opens upwards. Therefore, the inequality ( d^2 - 5d - 20 leq 0 ) holds between the two roots.Thus, the solution is:( frac{5 - sqrt{105}}{2} leq d leq frac{5 + sqrt{105}}{2} )But since we're looking for the largest possible value of ( d ), the maximum occurs at the upper bound, which is ( frac{5 + sqrt{105}}{2} ).Wait, let me just make sure I didn't make any mistakes in my steps. I started by squaring the sum, then used Cauchy-Schwarz on ( a, b, c ). That should be valid because Cauchy-Schwarz holds for any real numbers. Then I formed the inequality, solved it, and found the roots. The quadratic opens upwards, so the inequality is satisfied between the roots. Therefore, the maximum value of ( d ) is indeed ( frac{5 + sqrt{105}}{2} ).Just to double-check, let me compute ( sqrt{105} ) approximately. ( 10^2 = 100 ), so ( sqrt{105} ) is a bit more than 10, say approximately 10.246. So ( frac{5 + 10.246}{2} ) is about ( frac{15.246}{2} = 7.623 ). So ( d ) can be up to roughly 7.623.Let me see if that makes sense. If ( d ) is about 7.623, then ( a + b + c ) is about ( 10 - 7.623 = 2.377 ). The sum of their squares would be ( 60 - d^2 approx 60 - (7.623)^2 approx 60 - 58.11 approx 1.89 ). So ( a^2 + b^2 + c^2 approx 1.89 ). Then, using Cauchy-Schwarz, ( (a^2 + b^2 + c^2)(1 + 1 + 1) geq (a + b + c)^2 ), so ( 1.89 times 3 geq (2.377)^2 ). Calculating, ( 5.67 geq 5.65 ), which is approximately true. So that seems consistent.Therefore, I think my answer is correct.**Final Answer**The largest possible value of ( d ) is boxed{dfrac{5 + sqrt{105}}{2}}.

💡Okay, so I have this problem here with two inequalities, p and q, and I need to figure out the range of m based on the relationship between p and q. Let me try to break this down step by step.First, let's understand what p and q represent.p is given by |x - m| < 1. I know that absolute value inequalities like this can be rewritten without the absolute value. So, |x - m| < 1 means that x is within 1 unit of m. That translates to m - 1 < x < m + 1. So, p is the interval (m - 1, m + 1).Next, q is given by x² - 8x + 12 < 0. This is a quadratic inequality. To solve this, I should factor the quadratic if possible. Let me try factoring x² - 8x + 12.Looking for two numbers that multiply to 12 and add up to -8. Hmm, -6 and -2 work because (-6) * (-2) = 12 and (-6) + (-2) = -8. So, the quadratic factors as (x - 6)(x - 2) < 0.Now, to solve the inequality (x - 6)(x - 2) < 0, I can use a sign chart. The critical points are x = 2 and x = 6. These divide the number line into three intervals: (-∞, 2), (2, 6), and (6, ∞). I'll test each interval.- For x < 2, say x = 1: (1 - 6)(1 - 2) = (-5)(-1) = 5 > 0.- For 2 < x < 6, say x = 4: (4 - 6)(4 - 2) = (-2)(2) = -4 < 0.- For x > 6, say x = 7: (7 - 6)(7 - 2) = (1)(5) = 5 > 0.Since we're looking for where the product is less than zero, the solution is the interval where the product is negative, which is (2, 6). So, q is the interval (2, 6).Now, the problem states that q is a necessary but not sufficient condition for p. I need to recall what that means.In logic, if q is a necessary condition for p, it means that p can only be true if q is true. In other words, p implies q, or p → q. So, every x that satisfies p must also satisfy q. However, it's also stated that q is not a sufficient condition for p, meaning that q does not necessarily imply p. So, there are some x that satisfy q but do not satisfy p.Translating this into intervals: The interval for p must be entirely contained within the interval for q. But since q is not sufficient, the interval for p must be a proper subset of q, not equal to q.So, p: (m - 1, m + 1) must be entirely within q: (2, 6). That means the lower bound of p must be greater than or equal to 2, and the upper bound of p must be less than or equal to 6.Mathematically, that gives us two inequalities:1. m - 1 ≥ 22. m + 1 ≤ 6Let me solve these.Starting with the first inequality:m - 1 ≥ 2 Add 1 to both sides: m ≥ 3Second inequality:m + 1 ≤ 6 Subtract 1 from both sides: m ≤ 5So, combining these two results, m must satisfy both m ≥ 3 and m ≤ 5. Therefore, m is in the interval [3, 5].But wait, the problem says q is not a sufficient condition for p. That means p is not equivalent to q, so p must be a strict subset of q. If m were exactly 3 or 5, would p still be a strict subset?Let me check:If m = 3, then p is (2, 4). Is (2, 4) entirely within (2, 6)? Yes, but does q imply p? If x is in (2, 6), does it necessarily mean x is in (2, 4)? No, because x could be 5, which is in q but not in p. So, even if m = 3, q is not sufficient for p. Similarly, if m = 5, p is (4, 6). Again, x could be 3, which is in q but not in p. So, m can be 3 or 5, and q is still not sufficient for p.Therefore, the range of m is 3 ≤ m ≤ 5.Looking back at the options:A: 3 < m < 5 B: 3 ≤ m ≤ 5 C: m > 5 or m < 3 D: m ≥ 5 or m ≤ 3So, the correct answer is B: 3 ≤ m ≤ 5.I think that's it. I considered the necessary and sufficient conditions, translated them into inequalities, solved for m, and checked the endpoints to ensure they still satisfy the conditions. It all seems to fit.**Final Answer**boxed{B}

💡First, I need to determine the percentage of the budget allocated to transportation. I know the percentages for all other categories: salaries (39%), research and development (27%), utilities (14%), equipment (9%), supplies (5%), and advertising (3.5%). Adding these together gives a total of 97.5%. Since the total budget must account for 100%, the remaining percentage for transportation is 100% - 97.5% = 2.5%.Next, to find the number of degrees representing transportation in the circle graph, I use the fact that a full circle is 360 degrees. The degrees for transportation can be calculated by multiplying the percentage (2.5%) by 360 degrees. This gives 0.025 * 360 = 9 degrees.To convert degrees to radians, I use the conversion factor π radians = 180 degrees. Dividing the degrees by 180 and multiplying by π gives (9 / 180) * π = π/20 radians.Finally, to find the actual amount spent on transportation, I use the total budget of 1,200,000 and the percentage for transportation (2.5%). Multiplying these together gives 0.025 * 1,200,000 = 30,000.