💡Okay, so I have this problem about an ellipse, and it's divided into two parts, (Ⅰ) and (Ⅱ). Let me try to figure out how to approach each part step by step.Starting with part (Ⅰ):We are given an ellipse ( C: frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ) where ( a > b > 0 ). The length of the major axis is ( frac{3sqrt{5}}{5} ) times the length of the minor axis. There's a point ( D(-1, frac{2sqrt{10}}{3}) ) on the ellipse, and we need to find the maximum value of ( |NF| ), where ( F ) is the right focus of the ellipse.First, I remember that the major axis length is ( 2a ) and the minor axis length is ( 2b ). The problem states that ( 2a = frac{3sqrt{5}}{5} cdot 2b ). Simplifying this, we can divide both sides by 2 to get ( a = frac{3sqrt{5}}{5} b ).Next, since point ( D(-1, frac{2sqrt{10}}{3}) ) lies on the ellipse, it must satisfy the ellipse equation. Plugging in the coordinates of ( D ):[frac{(-1)^2}{a^2} + frac{left( frac{2sqrt{10}}{3} right)^2}{b^2} = 1][frac{1}{a^2} + frac{frac{40}{9}}{b^2} = 1]So now we have two equations:1. ( a = frac{3sqrt{5}}{5} b )2. ( frac{1}{a^2} + frac{40}{9b^2} = 1 )Let me substitute ( a ) from the first equation into the second equation. Let's express ( a^2 ) in terms of ( b^2 ):( a = frac{3sqrt{5}}{5} b ) implies ( a^2 = left( frac{3sqrt{5}}{5} right)^2 b^2 = frac{9 times 5}{25} b^2 = frac{45}{25} b^2 = frac{9}{5} b^2 )So, ( a^2 = frac{9}{5} b^2 ). Now, substitute ( a^2 ) into the second equation:[frac{1}{frac{9}{5} b^2} + frac{40}{9b^2} = 1][frac{5}{9b^2} + frac{40}{9b^2} = 1][frac{45}{9b^2} = 1][frac{5}{b^2} = 1][b^2 = 5]So, ( b = sqrt{5} ) since ( b > 0 ).Now, using ( a^2 = frac{9}{5} b^2 ), we can find ( a^2 ):[a^2 = frac{9}{5} times 5 = 9]Thus, ( a = 3 ) because ( a > 0 ).Now, knowing ( a ) and ( b ), we can find the distance ( c ) from the center to each focus. For an ellipse, ( c^2 = a^2 - b^2 ):[c^2 = 9 - 5 = 4][c = 2]So, the foci are located at ( (pm 2, 0) ). Since ( F ) is the right focus, its coordinates are ( (2, 0) ).Now, we need to find the maximum value of ( |NF| ), where ( N ) is a point on the ellipse. The maximum distance from a point on the ellipse to a focus occurs at the vertex farthest from that focus. Since ( F ) is at ( (2, 0) ), the farthest point on the ellipse from ( F ) would be the left vertex ( A ) at ( (-a, 0) = (-3, 0) ).Wait, but is that correct? Actually, the maximum distance from a focus is not necessarily the vertex. Let me think again.In an ellipse, the sum of distances from any point on the ellipse to the two foci is constant and equal to ( 2a ). So, for any point ( N ) on the ellipse, ( |NF_1| + |NF_2| = 2a = 6 ). But we are looking for the maximum of ( |NF| ), which is ( |NF_2| ) since ( F ) is ( F_2 ).To maximize ( |NF| ), we need to minimize ( |NF_1| ). The minimum distance from ( N ) to ( F_1 ) is ( |a - c| = |3 - 2| = 1 ). Therefore, the maximum ( |NF| ) would be ( 2a - |NF_1|_{text{min}} = 6 - 1 = 5 ).Alternatively, considering the ellipse's geometry, the maximum distance from ( F ) would be when ( N ) is at the farthest point from ( F ). Since the ellipse is symmetric, the farthest point from ( F ) on the ellipse would be the left vertex ( A(-3, 0) ). The distance from ( F(2, 0) ) to ( A(-3, 0) ) is ( |2 - (-3)| = 5 ). So, yes, the maximum value of ( |NF| ) is 5.Moving on to part (Ⅱ):We are told that ( overrightarrow{OM} = 2overrightarrow{AN} ), where ( O ) is the origin. We need to find the slope of line ( AN ).First, let's parse this vector equation. ( overrightarrow{OM} = 2overrightarrow{AN} ) implies that vector ( OM ) is twice vector ( AN ). So, if we denote ( M ) as ( (x_0, y_0) ) and ( N ) as ( (x_1, y_1) ), then:[overrightarrow{OM} = (x_0, y_0) = 2overrightarrow{AN}]Vector ( AN ) is ( overrightarrow{N} - overrightarrow{A} ). Since ( A ) is the left vertex, ( A = (-3, 0) ). So,[overrightarrow{AN} = (x_1 - (-3), y_1 - 0) = (x_1 + 3, y_1)]Therefore,[(x_0, y_0) = 2(x_1 + 3, y_1) = (2x_1 + 6, 2y_1)]So, we have:[x_0 = 2x_1 + 6 quad text{and} quad y_0 = 2y_1]Since both ( M ) and ( N ) are on the ellipse ( frac{x^2}{9} + frac{y^2}{5} = 1 ), their coordinates satisfy the ellipse equation.So, substituting ( M(x_0, y_0) ):[frac{x_0^2}{9} + frac{y_0^2}{5} = 1]And substituting ( N(x_1, y_1) ):[frac{x_1^2}{9} + frac{y_1^2}{5} = 1]Now, let's express ( x_0 ) and ( y_0 ) in terms of ( x_1 ) and ( y_1 ):From above, ( x_0 = 2x_1 + 6 ) and ( y_0 = 2y_1 ). Substitute these into the ellipse equation for ( M ):[frac{(2x_1 + 6)^2}{9} + frac{(2y_1)^2}{5} = 1][frac{4x_1^2 + 24x_1 + 36}{9} + frac{4y_1^2}{5} = 1]Simplify each term:[frac{4x_1^2}{9} + frac{24x_1}{9} + frac{36}{9} + frac{4y_1^2}{5} = 1][frac{4x_1^2}{9} + frac{8x_1}{3} + 4 + frac{4y_1^2}{5} = 1]Now, let's subtract 4 from both sides:[frac{4x_1^2}{9} + frac{8x_1}{3} + frac{4y_1^2}{5} = -3]Wait, that can't be right because the left side is a sum of squares and linear terms, which can't be negative. I must have made a mistake in the algebra.Let me double-check the substitution:Starting from ( frac{(2x_1 + 6)^2}{9} + frac{(2y_1)^2}{5} = 1 ):Expanding ( (2x_1 + 6)^2 ):( 4x_1^2 + 24x_1 + 36 )So,[frac{4x_1^2 + 24x_1 + 36}{9} + frac{4y_1^2}{5} = 1]Breaking it down:[frac{4x_1^2}{9} + frac{24x_1}{9} + frac{36}{9} + frac{4y_1^2}{5} = 1]Simplify each term:[frac{4x_1^2}{9} + frac{8x_1}{3} + 4 + frac{4y_1^2}{5} = 1]Subtract 1 from both sides:[frac{4x_1^2}{9} + frac{8x_1}{3} + 3 + frac{4y_1^2}{5} = 0]Hmm, this still results in a negative constant term. That doesn't make sense because the left side is a sum of squares and linear terms, which can't be negative. I must have messed up the substitution or the initial vector equation.Wait, perhaps I misinterpreted the vector equation. Let me double-check:( overrightarrow{OM} = 2overrightarrow{AN} )Vector ( AN ) is ( N - A ), so ( overrightarrow{AN} = (x_1 - (-3), y_1 - 0) = (x_1 + 3, y_1) ). Therefore, ( 2overrightarrow{AN} = (2x_1 + 6, 2y_1) ). So, ( overrightarrow{OM} = (x_0, y_0) = (2x_1 + 6, 2y_1) ). So, that part seems correct.So, substituting into the ellipse equation for ( M ):[frac{(2x_1 + 6)^2}{9} + frac{(2y_1)^2}{5} = 1]Which expands to:[frac{4x_1^2 + 24x_1 + 36}{9} + frac{4y_1^2}{5} = 1][frac{4x_1^2}{9} + frac{24x_1}{9} + frac{36}{9} + frac{4y_1^2}{5} = 1][frac{4x_1^2}{9} + frac{8x_1}{3} + 4 + frac{4y_1^2}{5} = 1]Subtract 1:[frac{4x_1^2}{9} + frac{8x_1}{3} + 3 + frac{4y_1^2}{5} = 0]This is problematic because the left side is a sum of positive terms (since ( x_1^2 ) and ( y_1^2 ) are non-negative, and the linear term ( frac{8x_1}{3} ) can be positive or negative depending on ( x_1 )), but the right side is zero. This suggests that perhaps there's no solution unless all terms are zero, which isn't possible because ( x_1 ) and ( y_1 ) can't satisfy all these simultaneously.Wait, maybe I made a mistake in the vector equation interpretation. Let me think again.The vector equation is ( overrightarrow{OM} = 2overrightarrow{AN} ). So, ( overrightarrow{OM} = 2(overrightarrow{N} - overrightarrow{A}) ). Since ( A = (-3, 0) ), ( overrightarrow{A} = (-3, 0) ). Therefore,[overrightarrow{OM} = 2(overrightarrow{N} - (-3, 0)) = 2(overrightarrow{N} + (3, 0)) = 2overrightarrow{N} + (6, 0)]So, ( overrightarrow{OM} = 2overrightarrow{N} + (6, 0) ). Therefore, ( M = 2N + (6, 0) ). So, in coordinates:[x_0 = 2x_1 + 6][y_0 = 2y_1]Which is what I had before. So, that part seems correct.But when substituting into the ellipse equation, I end up with a negative constant term, which is impossible. Maybe I need to approach this differently.Alternatively, perhaps I can express ( N ) in terms of ( M ). Since ( M = 2N + (6, 0) ), then ( N = frac{M - (6, 0)}{2} ). So, ( N = left( frac{x_0 - 6}{2}, frac{y_0}{2} right) ).Since ( N ) is on the ellipse, substituting into the ellipse equation:[frac{left( frac{x_0 - 6}{2} right)^2}{9} + frac{left( frac{y_0}{2} right)^2}{5} = 1]Simplify:[frac{(x_0 - 6)^2}{4 times 9} + frac{y_0^2}{4 times 5} = 1][frac{(x_0 - 6)^2}{36} + frac{y_0^2}{20} = 1]But we also know that ( M(x_0, y_0) ) is on the ellipse ( frac{x_0^2}{9} + frac{y_0^2}{5} = 1 ).So now we have two equations:1. ( frac{x_0^2}{9} + frac{y_0^2}{5} = 1 )2. ( frac{(x_0 - 6)^2}{36} + frac{y_0^2}{20} = 1 )Let me write them down:Equation (1): ( frac{x_0^2}{9} + frac{y_0^2}{5} = 1 )Equation (2): ( frac{(x_0 - 6)^2}{36} + frac{y_0^2}{20} = 1 )Let me try to solve these two equations simultaneously.First, let's express both equations with the same denominators to make it easier to subtract or manipulate them.Equation (1) can be written as:[frac{x_0^2}{9} + frac{y_0^2}{5} = 1]Multiply both sides by 180 (the least common multiple of 9 and 5) to eliminate denominators:[20x_0^2 + 36y_0^2 = 180]Equation (2):[frac{(x_0 - 6)^2}{36} + frac{y_0^2}{20} = 1]Multiply both sides by 180:[5(x_0 - 6)^2 + 9y_0^2 = 180]Now, we have:1. ( 20x_0^2 + 36y_0^2 = 180 ) (Equation 1a)2. ( 5(x_0 - 6)^2 + 9y_0^2 = 180 ) (Equation 2a)Let me expand Equation 2a:( 5(x_0^2 - 12x_0 + 36) + 9y_0^2 = 180 )Simplify:( 5x_0^2 - 60x_0 + 180 + 9y_0^2 = 180 )Subtract 180 from both sides:( 5x_0^2 - 60x_0 + 9y_0^2 = 0 ) (Equation 2b)Now, let's look at Equation 1a:( 20x_0^2 + 36y_0^2 = 180 )Let me try to express this in terms that can be subtracted from Equation 2b.First, notice that Equation 2b has coefficients 5, -60, 9, and Equation 1a has 20, 0, 36.Let me multiply Equation 2b by 4 to make the coefficients of ( x_0^2 ) and ( y_0^2 ) compatible with Equation 1a:Multiply Equation 2b by 4:( 20x_0^2 - 240x_0 + 36y_0^2 = 0 ) (Equation 2c)Now, subtract Equation 1a from Equation 2c:( (20x_0^2 - 240x_0 + 36y_0^2) - (20x_0^2 + 36y_0^2) = 0 - 180 )Simplify:( -240x_0 = -180 )Divide both sides by -60:( 4x_0 = 3 )So,( x_0 = frac{3}{4} )Now that we have ( x_0 ), we can substitute back into one of the equations to find ( y_0 ). Let's use Equation 1a:( 20x_0^2 + 36y_0^2 = 180 )Substitute ( x_0 = frac{3}{4} ):( 20 left( frac{9}{16} right) + 36y_0^2 = 180 )Calculate:( frac{180}{16} + 36y_0^2 = 180 )Simplify ( frac{180}{16} ) to ( frac{45}{4} ):( frac{45}{4} + 36y_0^2 = 180 )Subtract ( frac{45}{4} ) from both sides:( 36y_0^2 = 180 - frac{45}{4} = frac{720}{4} - frac{45}{4} = frac{675}{4} )Divide both sides by 36:( y_0^2 = frac{675}{4 times 36} = frac{675}{144} = frac{225}{48} = frac{75}{16} )So,( y_0 = sqrt{frac{75}{16}} = frac{5sqrt{3}}{4} ) (since ( y_0 > 0 ))Now, we have ( M(x_0, y_0) = left( frac{3}{4}, frac{5sqrt{3}}{4} right) ).Recall that ( N = left( frac{x_0 - 6}{2}, frac{y_0}{2} right) ). Let's compute ( N ):( x_1 = frac{frac{3}{4} - 6}{2} = frac{frac{3}{4} - frac{24}{4}}{2} = frac{-frac{21}{4}}{2} = -frac{21}{8} )( y_1 = frac{frac{5sqrt{3}}{4}}{2} = frac{5sqrt{3}}{8} )So, ( N = left( -frac{21}{8}, frac{5sqrt{3}}{8} right) )Now, we need to find the slope of line ( AN ). Point ( A ) is ( (-3, 0) ) and point ( N ) is ( left( -frac{21}{8}, frac{5sqrt{3}}{8} right) ).The slope ( m ) is given by:[m = frac{y_1 - y_A}{x_1 - x_A} = frac{frac{5sqrt{3}}{8} - 0}{-frac{21}{8} - (-3)} = frac{frac{5sqrt{3}}{8}}{-frac{21}{8} + frac{24}{8}} = frac{frac{5sqrt{3}}{8}}{frac{3}{8}} = frac{5sqrt{3}}{8} times frac{8}{3} = frac{5sqrt{3}}{3}]So, the slope of line ( AN ) is ( frac{5sqrt{3}}{3} ).Wait, but let me double-check the calculation for ( x_1 ):( x_1 = frac{x_0 - 6}{2} = frac{frac{3}{4} - 6}{2} = frac{frac{3}{4} - frac{24}{4}}{2} = frac{-frac{21}{4}}{2} = -frac{21}{8} ). That seems correct.And ( y_1 = frac{y_0}{2} = frac{5sqrt{3}/4}{2} = frac{5sqrt{3}}{8} ). Correct.Then, the slope calculation:( y_1 - y_A = frac{5sqrt{3}}{8} - 0 = frac{5sqrt{3}}{8} )( x_1 - x_A = -frac{21}{8} - (-3) = -frac{21}{8} + frac{24}{8} = frac{3}{8} )So, slope ( m = frac{5sqrt{3}/8}{3/8} = frac{5sqrt{3}}{3} ). Correct.Therefore, the slope of line ( AN ) is ( frac{5sqrt{3}}{3} ).**Final Answer**Ⅰ. The maximum value of ( |NF| ) is boxed{5}.Ⅱ. The slope of line ( AN ) is boxed{dfrac{5sqrt{3}}{3}}.

💡Alright, so I have this problem about a clock with an hour hand of length 3 and a minute hand of length 4. I need to find how many times the distance between the tips of these two hands is an integer from 1:00 am to 1:00 pm of the same day. Hmm, okay, let me try to break this down step by step.First, I know that the distance between two points can be found using the distance formula. Since the clock hands are moving, the distance between their tips will change over time. The hour hand is shorter than the minute hand, so their relative positions will vary as they move around the clock.I remember that the angle between the hour and minute hands can be calculated, and from there, I can use the law of cosines to find the distance between their tips. The formula for the distance ( d ) between two points with lengths ( r_1 ) and ( r_2 ) and angle ( theta ) between them is:[d = sqrt{r_1^2 + r_2^2 - 2r_1r_2cos(theta)}]In this case, ( r_1 = 3 ) and ( r_2 = 4 ), so plugging those in:[d = sqrt{3^2 + 4^2 - 2 cdot 3 cdot 4 cos(theta)} = sqrt{9 + 16 - 24cos(theta)} = sqrt{25 - 24cos(theta)}]So, ( d = sqrt{25 - 24cos(theta)} ). I need this distance ( d ) to be an integer. Let's think about what integer values ( d ) can take.Since the hour hand is 3 units and the minute hand is 4 units, the maximum possible distance between their tips is when they are opposite each other, which would be ( 3 + 4 = 7 ) units. The minimum distance is when they overlap, which would be ( |4 - 3| = 1 ) unit. So, ( d ) can be 1, 2, 3, 4, 5, 6, or 7.But wait, actually, when the hands overlap, the distance is 0, not 1. Hmm, so maybe the minimum distance is 0, but since we're looking for integer distances, and 0 is an integer, but the problem might be considering only positive integers. I should check that.Looking back at the problem, it says "the distance between the tips of the two hands is an integer." It doesn't specify positive integers, but in the context of distance, it's usually non-negative. So, 0 is a possible distance, but in reality, the hands only overlap at specific times, and the distance is 0 only when they overlap. However, in this case, since the hands are of different lengths, they might not actually overlap exactly, but get very close. Hmm, maybe I should consider 0 as a possible distance.But let's check. If the distance is 0, that would mean the tips are at the same point, which would require both hands to be at the same angle. But since the hour hand is shorter, it's only when the minute hand catches up to the hour hand. So, actually, the distance can be 0, but only at specific times when the hands overlap.But in the problem, it's from 1:00 am to 1:00 pm, which is 12 hours. In 12 hours, the hands overlap 11 times. So, the distance is 0 at 11 times. But the problem is asking for integer distances, so 0 is an integer, but maybe the problem is considering positive integers. I need to clarify that.Looking back at the problem statement: "the distance between the tips of the two hands is an integer." It doesn't specify positive, so 0 is included. However, in the context of clock hands, the distance being 0 is a special case when they overlap, which happens 11 times in 12 hours. So, that's 11 occurrences where ( d = 0 ).But let's see if the problem counts 0 as a valid distance. If it does, then we have 11 occurrences. But I think the problem is more likely looking for positive integer distances, so 1, 2, 3, 4, 5, 6, 7. So, maybe 0 is excluded. I'll keep that in mind and proceed, but I'll note that 0 is a possible distance.Now, let's consider the possible integer distances: 1, 2, 3, 4, 5, 6, 7. For each of these, I need to find how many times the distance ( d ) equals that integer between 1:00 am and 1:00 pm.Given that the distance formula is ( d = sqrt{25 - 24cos(theta)} ), I can set ( d ) to each integer and solve for ( theta ).Let's start with ( d = 1 ):[1 = sqrt{25 - 24cos(theta)}][1^2 = 25 - 24cos(theta)][1 = 25 - 24cos(theta)][24cos(theta) = 25 - 1 = 24][cos(theta) = 1][theta = 0 text{ radians}]So, ( theta = 0 ) radians, which means the hands are overlapping. As we discussed earlier, this happens 11 times in 12 hours. So, ( d = 1 ) is not possible because when ( theta = 0 ), the distance is 0, not 1. Wait, that's confusing. If ( theta = 0 ), the distance is 0, but according to the equation, ( d = 1 ) would require ( cos(theta) = 1 ), which gives ( theta = 0 ), but that results in ( d = 0 ). So, there's a contradiction here. That means there is no solution for ( d = 1 ). Hmm, interesting.Let me check my calculations again. If ( d = 1 ):[1 = sqrt{25 - 24cos(theta)}][1 = 25 - 24cos(theta)][24cos(theta) = 24][cos(theta) = 1][theta = 0]But when ( theta = 0 ), the distance is ( sqrt{3^2 + 4^2 - 2 cdot 3 cdot 4 cos(0)} = sqrt{9 + 16 - 24} = sqrt{1} = 1 ). Wait, that's not right. Wait, no, if ( theta = 0 ), the distance should be ( |4 - 3| = 1 ). Oh, right! Because when the hands overlap, the distance is the difference in their lengths, which is 1 unit. So, actually, ( d = 1 ) occurs when the hands overlap, which is 11 times. So, my initial thought was wrong. The distance is 1 when they overlap, not 0. Wait, that doesn't make sense because when they overlap, the tips are at the same point, so the distance should be 0. Hmm, I'm confused now.Wait, let's think about it. If both hands are pointing in the same direction, the tip of the hour hand is at a certain point, and the tip of the minute hand is 4 units away from the center, while the hour hand is only 3 units away. So, the distance between the tips is ( 4 - 3 = 1 ) unit. So, actually, when the hands overlap, the distance between the tips is 1 unit, not 0. That makes sense because they don't have the same length. So, the distance is 1 when they overlap, and 0 would only occur if they were the same length and overlapping. So, in this case, the minimum distance is 1, not 0.Therefore, ( d = 1 ) occurs 11 times in 12 hours. So, that's one integer distance accounted for.Now, let's try ( d = 2 ):[2 = sqrt{25 - 24cos(theta)}][4 = 25 - 24cos(theta)][24cos(theta) = 25 - 4 = 21][cos(theta) = frac{21}{24} = frac{7}{8}][theta = arccosleft(frac{7}{8}right)]So, ( theta ) is approximately ( arccos(0.875) ), which is about 28.955 degrees. Since the cosine function is positive in the first and fourth quadrants, there are two angles in the range ( [0, 2pi) ) that satisfy this equation: one in the first quadrant and one in the fourth quadrant. However, since the angle between clock hands is always between 0 and ( pi ) radians (0 to 180 degrees), we only consider the first quadrant solution. Wait, no, actually, the angle between the hands can be more than 180 degrees, but the smaller angle is usually considered. So, for each occurrence, there are two possible angles: one clockwise and one counterclockwise from the hour hand. But since the clock is circular, these two angles are supplementary, meaning they add up to ( 2pi ). So, for each ( theta ), there are two times when the angle between the hands is ( theta ) or ( 2pi - theta ).But in the context of clock hands, the angle between them is typically considered as the smallest angle, so between 0 and ( pi ). Therefore, for each ( theta ) in ( [0, pi] ), there are two times when the angle is ( theta ) or ( 2pi - theta ), but since ( 2pi - theta ) is equivalent to ( -theta ), which is the same as ( theta ) in the opposite direction, but on the clock, it's just the same angle measured the other way. So, in terms of occurrences, each ( theta ) corresponds to two times in a 12-hour period, except when ( theta = 0 ) or ( pi ), which only occur once.Wait, but in our case, ( theta = arccos(7/8) ) is approximately 28.955 degrees, which is less than 180 degrees, so it's in the first quadrant. Therefore, for each such ( theta ), there are two times in 12 hours when the angle between the hands is ( theta ). So, for ( d = 2 ), we have two occurrences per 12 hours.But wait, actually, in a 12-hour period, the hands overlap 11 times, and between each overlap, the minute hand gains on the hour hand, creating angles that increase and then decrease. So, for each angle ( theta ) between 0 and ( pi ), there are two times when the angle is ( theta ): once when the minute hand is moving towards overlapping, and once when it's moving away. Therefore, for each ( theta ) in ( (0, pi) ), there are two times in 12 hours when the angle is ( theta ). So, for ( d = 2 ), which corresponds to ( theta = arccos(7/8) ), there are two occurrences in 12 hours.Similarly, for ( d = 3 ):[3 = sqrt{25 - 24cos(theta)}][9 = 25 - 24cos(theta)][24cos(theta) = 25 - 9 = 16][cos(theta) = frac{16}{24} = frac{2}{3}][theta = arccosleft(frac{2}{3}right)]Which is approximately 48.19 degrees. Again, this is less than 180 degrees, so there are two occurrences in 12 hours.Continuing this way, let's check for ( d = 4 ):[4 = sqrt{25 - 24cos(theta)}][16 = 25 - 24cos(theta)][24cos(theta) = 25 - 16 = 9][cos(theta) = frac{9}{24} = frac{3}{8}][theta = arccosleft(frac{3}{8}right)]Approximately 72.54 degrees. Again, two occurrences in 12 hours.For ( d = 5 ):[5 = sqrt{25 - 24cos(theta)}][25 = 25 - 24cos(theta)][24cos(theta) = 0][cos(theta) = 0][theta = frac{pi}{2} text{ radians} = 90 text{ degrees}]So, ( theta = 90 ) degrees. This occurs twice in 12 hours as well.For ( d = 6 ):[6 = sqrt{25 - 24cos(theta)}][36 = 25 - 24cos(theta)][24cos(theta) = 25 - 36 = -11][cos(theta) = -frac{11}{24}][theta = arccosleft(-frac{11}{24}right)]Which is approximately 117.28 degrees. Since this is greater than 90 degrees but less than 180 degrees, it's in the second quadrant. However, as we discussed earlier, for each ( theta ) in ( (0, pi) ), there are two occurrences in 12 hours. So, even though this angle is in the second quadrant, it still corresponds to two times when the distance is 6.Finally, for ( d = 7 ):[7 = sqrt{25 - 24cos(theta)}][49 = 25 - 24cos(theta)][24cos(theta) = 25 - 49 = -24][cos(theta) = -1][theta = pi text{ radians} = 180 text{ degrees}]So, ( theta = 180 ) degrees. This occurs when the hands are directly opposite each other. In a 12-hour period, how many times does this happen? Well, the hands are opposite each other 11 times as well, similar to overlaps, but let me think. Actually, in 12 hours, the hands are opposite each other 11 times as well, but wait, is that correct?Wait, no. The hands overlap 11 times in 12 hours because the minute hand catches up to the hour hand 11 times. Similarly, the hands are opposite each other 11 times as well because the minute hand laps the hour hand 11 times, and each time it laps, it passes through the opposite position once. So, yes, ( d = 7 ) occurs 11 times in 12 hours.Wait, but earlier, for other distances, we had two occurrences per 12 hours. But for ( d = 7 ), it's 11 times. That seems inconsistent. Let me check.Actually, no. For each ( theta ) in ( (0, pi) ), there are two times in 12 hours when the angle is ( theta ). However, when ( theta = pi ), it's a special case because it's the maximum angle, and it occurs only once per lap. Wait, but in 12 hours, the minute hand goes around 12 times, and the hour hand goes around once. So, the number of times the hands are opposite each other is 11 times, similar to overlaps.Wait, let me think about it differently. The hands overlap 11 times in 12 hours because the minute hand gains 360 degrees on the hour hand every 12/11 hours. Similarly, the hands are opposite each other when the minute hand is 180 degrees ahead of the hour hand. So, the time between each opposition is also 12/11 hours, leading to 11 oppositions in 12 hours.Therefore, ( d = 7 ) occurs 11 times in 12 hours.So, summarizing:- ( d = 1 ): 11 times- ( d = 2 ): 2 times- ( d = 3 ): 2 times- ( d = 4 ): 2 times- ( d = 5 ): 2 times- ( d = 6 ): 2 times- ( d = 7 ): 11 timesWait, but that doesn't add up correctly. If ( d = 1 ) and ( d = 7 ) each occur 11 times, and the others occur 2 times each, then the total number of occurrences would be:11 (for d=1) + 2*5 (for d=2 to d=6) + 11 (for d=7) = 11 + 10 + 11 = 32.But that seems low. I think I'm making a mistake here.Wait, no. Actually, for each ( d ) from 2 to 6, there are two times in 12 hours when the distance is that integer. So, for each of these distances, there are two occurrences. But in reality, for each ( theta ) that gives a particular ( d ), there are two times in 12 hours when that ( theta ) occurs. However, since the hands move continuously, each ( theta ) (except for 0 and ( pi )) occurs twice in 12 hours.But wait, in the case of ( d = 1 ), which corresponds to ( theta = 0 ), it occurs 11 times, not twice. Similarly, ( d = 7 ) corresponds to ( theta = pi ), which occurs 11 times.So, perhaps the correct way to count is:- For ( d = 1 ) and ( d = 7 ): 11 times each- For ( d = 2, 3, 4, 5, 6 ): 2 times eachBut that would be:11 + 11 + 2*5 = 22 + 10 = 32.But I think this is incorrect because in reality, for each ( d ) from 2 to 6, there are two times in 12 hours when the distance is that integer. However, since the hands overlap 11 times, and between each overlap, there are two times when the distance is each of these integers. So, perhaps for each integer distance from 2 to 6, there are 22 occurrences in 12 hours? Wait, that can't be right because that would be too high.Wait, let me think differently. The hands overlap 11 times in 12 hours. Between each overlap, the minute hand goes around the clock once relative to the hour hand. So, in each interval between overlaps, the minute hand makes a full circle relative to the hour hand, meaning that for each integer distance from 2 to 6, there are two times when the distance is that integer. Therefore, for each of these distances, there are 2 occurrences per overlap interval.Since there are 11 overlap intervals in 12 hours, each distance from 2 to 6 occurs 2*11 = 22 times. But wait, that would be 22 times for each distance, which is 5 distances, so 110 times. Plus the 11 times for ( d = 1 ) and 11 times for ( d = 7 ), totaling 110 + 11 + 11 = 132.Wait, that seems high, but let me check the logic.Each time the minute hand laps the hour hand (which happens 11 times in 12 hours), between each lap, the minute hand goes around the clock once relative to the hour hand. During this relative movement, for each integer distance from 2 to 6, there are two times when the distance is that integer: once when the minute hand is approaching the hour hand and once when it's moving away. Therefore, for each of these distances, there are 2 occurrences per lap, leading to 2*11 = 22 occurrences per distance.Since there are 5 such distances (2,3,4,5,6), that's 5*22 = 110 occurrences.Additionally, the distance ( d = 1 ) occurs 11 times (when the hands overlap), and ( d = 7 ) occurs 11 times (when the hands are opposite each other). So, total occurrences are 110 + 11 + 11 = 132.But wait, does ( d = 7 ) occur 11 times? Because when the hands are opposite, that's another set of 11 times, similar to overlaps. So, yes, that makes sense.Therefore, the total number of occurrences when the distance between the tips of the two hands is an integer is 132.But let me verify this with another approach. The formula for the distance is ( d = sqrt{25 - 24cos(theta)} ). For ( d ) to be an integer, ( 25 - 24cos(theta) ) must be a perfect square.So, let's set ( 25 - 24cos(theta) = k^2 ), where ( k ) is an integer between 1 and 7.So, ( 24cos(theta) = 25 - k^2 ).Therefore, ( cos(theta) = frac{25 - k^2}{24} ).For ( cos(theta) ) to be valid, ( frac{25 - k^2}{24} ) must be between -1 and 1.So, let's check for each ( k ):- ( k = 1 ): ( cos(theta) = frac{25 - 1}{24} = 1 ). Valid.- ( k = 2 ): ( cos(theta) = frac{25 - 4}{24} = frac{21}{24} = 0.875 ). Valid.- ( k = 3 ): ( cos(theta) = frac{25 - 9}{24} = frac{16}{24} = 0.666... ). Valid.- ( k = 4 ): ( cos(theta) = frac{25 - 16}{24} = frac{9}{24} = 0.375 ). Valid.- ( k = 5 ): ( cos(theta) = frac{25 - 25}{24} = 0 ). Valid.- ( k = 6 ): ( cos(theta) = frac{25 - 36}{24} = frac{-11}{24} approx -0.4583 ). Valid.- ( k = 7 ): ( cos(theta) = frac{25 - 49}{24} = frac{-24}{24} = -1 ). Valid.So, all ( k ) from 1 to 7 are valid, meaning ( d ) can be 1,2,3,4,5,6,7.Now, for each ( k ), how many times does ( cos(theta) = frac{25 - k^2}{24} ) occur in 12 hours?As we discussed earlier, for ( k = 1 ) and ( k = 7 ), ( cos(theta) = 1 ) and ( cos(theta) = -1 ), which correspond to ( theta = 0 ) and ( theta = pi ), respectively. These occur 11 times each in 12 hours.For the other values of ( k ) (2,3,4,5,6), each corresponds to two angles ( theta ) in the range ( [0, 2pi) ), but since we're considering the smallest angle between the hands, each ( theta ) corresponds to two times in 12 hours when the distance is ( k ). Therefore, for each of these ( k ), there are 2*11 = 22 occurrences in 12 hours.Wait, no. Actually, for each ( k ) from 2 to 6, there are two times in each 12-hour period when the distance is ( k ). But since the hands overlap 11 times, and between each overlap, the minute hand goes around once relative to the hour hand, creating two opportunities for each ( k ) from 2 to 6. Therefore, for each ( k ) from 2 to 6, there are 2*11 = 22 occurrences.But wait, that would mean 5 distances (2,3,4,5,6) each occurring 22 times, which is 110, plus 11 for ( k=1 ) and 11 for ( k=7 ), totaling 132.Yes, that seems consistent with the earlier calculation.Therefore, the total number of occurrences is 132.

💡First, I need to determine the possible lengths of the third side of the triangle. Given two sides of 24 units and 51 units, the third side must satisfy the triangle inequality theorem.According to the triangle inequality theorem, the sum of any two sides of a triangle must be greater than the third side. Therefore, the third side must be greater than the difference of the two given sides and less than the sum of the two given sides.Calculating the difference:51 - 24 = 27Calculating the sum:51 + 24 = 75So, the third side must be greater than 27 units and less than 75 units.Since the third side must be an integer, the smallest possible integer greater than 27 is 28.Now, I can calculate the perimeter of the triangle by adding all three sides:24 + 51 + 28 = 103Therefore, the least possible perimeter of the triangle is 103 units.

💡Alright, so I've got this problem here about a 10x10 board with 91 white checkers. The painter is going to take one checker at a time, paint it black, and place it back on the board. He does this until all the checkers are black. The question is asking me to prove that at some point during this process, there will be two checkers of different colors on adjacent cells. Hmm, okay.Let me start by understanding the setup. The board is 10x10, so that's 100 cells in total. There are 91 white checkers, which means there are 9 empty cells initially. The painter is going to paint one checker black each time and place it on any free cell. So, each time he does this, the number of white checkers decreases by one, and the number of black checkers increases by one. Eventually, after 91 steps, all checkers will be black.Now, the key point is that at some moment, there must be two checkers of different colors on adjacent cells. So, I need to show that it's impossible for the painter to paint all the checkers black without ever having two differently colored adjacent checkers at some point.Let me think about this. If I assume the opposite, that is, suppose that at every step, all the checkers are either all white or all black, or arranged in such a way that no two adjacent cells have different colors. Is that possible?Wait, but the board starts with 91 white checkers and 9 empty cells. So, initially, all the checkers are white, and the empty cells are just empty. When the painter starts painting, he takes a white checker, paints it black, and places it on any free cell. So, the first black checker is placed somewhere on the board.Now, if I assume that after each painting step, there are no two adjacent cells with different colors, that would mean that the black checker must be placed in such a way that it doesn't create a conflict with its neighbors. But since all the other checkers are still white, the black checker would have white neighbors, which are different colors. So, that would violate the assumption.Wait, so maybe my initial assumption is wrong. Let me rephrase. If I assume that at every step, there are no two adjacent cells with different colors, then when the painter places the first black checker, it must be placed in a way that it doesn't have any white neighbors. But since all the checkers are white initially, the only way to place a black checker without having white neighbors is to place it in an empty cell that is surrounded by empty cells. But there are only 9 empty cells, and they are scattered on the board.Hmm, this is getting a bit confusing. Maybe I need to think about it differently. Let's consider the concept of a checkerboard pattern. If the board is colored like a chessboard, with alternating black and white squares, then no two adjacent squares have the same color. But in this case, the checkers themselves are being painted, not the squares.Wait, maybe that's not directly applicable. Let me think about the number of checkers and the number of empty cells. There are 91 checkers and 9 empty cells. So, as the painter paints each checker black, he's effectively moving a black checker into an empty cell, right? Because he takes a white checker, paints it black, and places it on a free cell.So, each time he does this, he's moving a black checker from one cell to another, leaving the original cell empty. So, the number of empty cells remains 9 throughout the process, but their positions change.Now, if I think about the movement of the black checkers, they have to move through the board, potentially crossing paths with other checkers. But since all checkers are initially white, the first black checker is just placed somewhere. The next black checker could be placed somewhere else, but it's possible that it's adjacent to the first black checker.Wait, but if it's adjacent, then we have two black checkers next to each other, which is fine because they're the same color. The problem is when a black checker is adjacent to a white checker. So, if the painter places a black checker next to a white checker, then we have two different colors adjacent, which is what we're supposed to prove happens at some point.But the question is, can the painter avoid this? Is there a way for the painter to place all the black checkers in such a way that no black checker is ever adjacent to a white checker until all are black?Hmm, that seems unlikely because as the number of black checkers increases, the number of white checkers decreases. At some point, the black checkers will have to be placed in positions that are adjacent to white checkers.Let me try to think about it more formally. Maybe I can use the pigeonhole principle or something like that.The board has 100 cells. Initially, 91 are white, 9 are empty. After each step, one white becomes black, and one empty cell is filled with a black checker. So, after k steps, there are 91 - k white checkers and k black checkers, plus 9 - k empty cells? Wait, no, the number of empty cells remains 9 because each time a checker is painted black, it's placed on an empty cell, so the total number of empty cells stays the same.Wait, actually, no. Initially, there are 91 checkers and 9 empty cells. When the painter takes a white checker, paints it black, and places it on an empty cell, he's effectively moving the checker from one cell to another, so the number of checkers remains 91, and the number of empty cells remains 9. So, the number of black checkers increases by one each time, and the number of white checkers decreases by one.So, after k steps, there are k black checkers and 91 - k white checkers, with 9 empty cells.Now, the question is, can the painter arrange the black checkers in such a way that no two black and white checkers are adjacent until all are black?I think not, because as the number of black checkers increases, they have to be placed in positions that are adjacent to white checkers.Let me think about the maximum number of black checkers that can be placed without being adjacent to any white checkers. If the painter wants to avoid having black and white checkers adjacent, he needs to place all black checkers in a region that is completely surrounded by other black checkers or empty cells.But since the board is 10x10, and the number of black checkers is increasing, at some point, the black checkers will have to expand into areas that are adjacent to white checkers.Alternatively, maybe I can think about the problem in terms of connected regions. If the painter wants to keep all black checkers isolated from white checkers, he has to keep the black checkers in separate regions that don't touch the white regions.But as the number of black checkers increases, it's impossible to keep them all isolated because the number of empty cells is limited.Wait, there are only 9 empty cells. So, the painter can only move the black checkers into these 9 cells. But as he paints more checkers black, he has to place them somewhere, and eventually, he'll have to place them next to white checkers.Hmm, I'm not sure if I'm on the right track here. Maybe I need to think about it differently.Let me consider the process step by step. Initially, all checkers are white. The first black checker is placed somewhere. Now, there's one black checker and 90 white checkers. The next black checker is placed somewhere else. If it's placed adjacent to the first black checker, that's fine, but if it's placed somewhere else, it might be adjacent to a white checker.Wait, but the painter can choose where to place the black checker. So, maybe he can always place the new black checker in a position that's not adjacent to any white checkers.But as the number of black checkers increases, the number of available positions that are not adjacent to any white checkers decreases.At some point, the painter will have to place a black checker next to a white checker because there are no more positions available that are not adjacent to white checkers.That seems like a plausible argument. So, maybe I can formalize that.Let me think about the number of cells that are adjacent to white checkers. Initially, all cells are white, so every cell is adjacent to white checkers. As the painter places black checkers, some cells become black, and their adjacent cells are no longer adjacent to white checkers.But the number of cells adjacent to white checkers decreases as more black checkers are placed. However, the number of black checkers increases, so the number of cells that are adjacent to white checkers also increases because each black checker can potentially block some white checkers from being adjacent to other cells.Wait, this is getting a bit tangled. Maybe I need to think about it in terms of the maximum number of black checkers that can be placed without being adjacent to any white checkers.If the painter wants to place black checkers in such a way that none are adjacent to white checkers, he needs to place them in a region that is completely surrounded by other black checkers or empty cells.But since the board is finite, and the number of empty cells is limited, the painter cannot keep expanding the black region indefinitely without eventually having to place a black checker adjacent to a white checker.Alternatively, maybe I can think about the problem in terms of graph theory. The board can be represented as a graph where each cell is a vertex, and edges connect adjacent cells. The problem then reduces to showing that during the process of turning vertices from white to black, there must be a step where a black vertex is adjacent to a white vertex.But I'm not sure if that's helpful.Wait, maybe I can use the concept of a checkerboard coloring. If I color the board in a checkerboard pattern, with alternating black and white squares, then no two adjacent squares have the same color. But in this problem, the checkers themselves are being colored, not the squares.Hmm, maybe that's not directly applicable.Let me try a different approach. Suppose that at every step, the painter places the new black checker in such a way that it is not adjacent to any white checkers. Is this possible?Initially, all checkers are white, so the first black checker can be placed anywhere. Let's say it's placed in the center of the board. Now, the next black checker needs to be placed somewhere that is not adjacent to any white checkers. But since all other checkers are white, the only way to place a black checker without being adjacent to white checkers is to place it in a cell that is surrounded by other black checkers or empty cells.But since there's only one black checker so far, the next black checker would have to be placed adjacent to it, which would make it adjacent to white checkers. Wait, no, because the first black checker is in the center, and the next black checker could be placed in a cell that is not adjacent to any white checkers.Wait, but all other cells are white, so unless the painter places the second black checker adjacent to the first one, it will be adjacent to white checkers. So, if he places it adjacent to the first one, then the two black checkers are next to each other, which is fine, but the surrounding cells are still white.Wait, but the problem is that as the number of black checkers increases, the number of cells that are adjacent to white checkers decreases. But the number of black checkers increases, so the number of cells that are adjacent to white checkers also increases because each black checker can potentially block some white checkers from being adjacent to other cells.I'm getting stuck here. Maybe I need to think about it in terms of the number of edges between black and white checkers.Initially, there are no black checkers, so the number of edges between black and white checkers is zero. As the painter places black checkers, the number of edges between black and white checkers increases.At some point, the number of edges between black and white checkers must be at least one, which would mean that there are two adjacent checkers of different colors.But I'm not sure if that's rigorous enough.Wait, maybe I can use the fact that the total number of edges on the board is fixed. For a 10x10 board, each cell has up to four edges, but the total number of edges is 10*9*2 = 180.Now, as the painter places black checkers, the number of edges between black and white checkers can be calculated. Initially, it's zero. After placing the first black checker, it has four edges, but since all its neighbors are white, the number of edges between black and white checkers is four.After placing the second black checker, if it's adjacent to the first one, then the number of edges between black and white checkers increases by three (since one edge is now between two black checkers). If it's not adjacent, it increases by four.Wait, so each time a black checker is placed, the number of edges between black and white checkers increases by at least three or four, depending on its position.But the total number of edges is 180, so eventually, the number of edges between black and white checkers will have to reach a point where it's impossible to avoid having at least one edge between black and white checkers.But I'm not sure if this is the right way to think about it.Maybe I need to consider the maximum number of black checkers that can be placed without having any adjacent to white checkers. If the painter can place all 91 black checkers without any being adjacent to white checkers, then the statement is false. But I don't think that's possible.Wait, but the board is 10x10, and the number of empty cells is 9. So, the painter can only move the black checkers into these 9 cells. But as he paints more checkers black, he has to place them somewhere, and eventually, he'll have to place them next to white checkers.I think the key idea is that since there are only 9 empty cells, the painter cannot keep all the black checkers isolated from white checkers because he has to place them in the 9 empty cells, which are limited.Wait, but the painter is not limited to placing black checkers only in the initial empty cells. He can move the black checkers around by painting white checkers black and placing them in any empty cell.So, as he paints more checkers black, he can move them into different positions, potentially creating a cluster of black checkers that are surrounded by other black checkers or empty cells.But even so, as the number of black checkers increases, the number of empty cells remains 9, so the painter has to place the black checkers in such a way that they don't have to be adjacent to white checkers.But I think that's impossible because the number of black checkers is too large compared to the number of empty cells.Wait, let me think about it more carefully. The board has 100 cells, 91 of which are checkers. So, 9 are empty. The painter is going to paint all 91 checkers black, one by one, each time placing the newly painted black checker into an empty cell.So, effectively, the painter is moving the black checkers around the board, using the 9 empty cells as temporary storage.Now, if the painter wants to avoid having any black and white checkers adjacent, he needs to arrange the black checkers in such a way that they are all clustered together, surrounded by other black checkers or empty cells.But since the number of empty cells is limited, the painter cannot keep expanding the black cluster indefinitely without eventually having to place a black checker adjacent to a white checker.In other words, the painter cannot keep all the black checkers isolated from white checkers because the number of empty cells is too small to allow for such isolation as the number of black checkers increases.Therefore, at some point, the painter will have to place a black checker adjacent to a white checker, resulting in two differently colored checkers on adjacent cells.I think that's the gist of the argument. So, to summarize, the limited number of empty cells forces the painter to eventually place a black checker next to a white checker as the number of black checkers increases.I hope that makes sense. I feel like I'm going in circles a bit, but I think the key idea is the limited number of empty cells and the increasing number of black checkers, which eventually forces an adjacency between black and white checkers.

💡First, I need to determine the height of the second building, which is twice the height of the first building. Given that the first building is 600 feet tall, the second building would be 600 multiplied by 2, resulting in 1200 feet.Next, I calculate the combined height of the first and second buildings by adding their individual heights: 600 feet plus 1200 feet equals 1800 feet.Then, the third building is three times as tall as the combined height of the first and second buildings. Multiplying 1800 feet by 3 gives the height of the third building as 5400 feet.Finally, to find the total height of all three buildings, I add the heights of the first, second, and third buildings together: 600 feet plus 1200 feet plus 5400 feet equals 7200 feet.

💡First, I need to determine the coordinates of point ( G ), which is the midpoint of segment ( overline{PQ} ). Given that ( P ) is at ( (0, 10) ) and ( Q ) is at ( (0, 0) ), I can calculate the midpoint by averaging the x-coordinates and the y-coordinates of ( P ) and ( Q ).Next, I will find the slope of the line passing through points ( R ) and ( G ). Point ( R ) is at ( (10, 0) ), and point ( G ) is at ( (0, 5) ). The slope ( m ) can be calculated using the formula ( m = frac{y_2 - y_1}{x_2 - x_1} ).After determining the slope, I will identify the y-intercept of the line. Since the line passes through point ( G ) at ( (0, 5) ), the y-intercept is simply the y-coordinate of ( G ).Finally, I will add the slope and the y-intercept together to find the required sum.