answer：Okay, so I have this geometry problem here. Let me try to understand it step by step. We have a circle, and there's a chord AB. The smaller arc subtended by AB is α degrees. So, that means if I draw chord AB, the arc that's smaller between the two arcs created by AB is α degrees. Got that. Now, C is the midpoint of chord AB. So, AC equals BC. That makes sense because the midpoint divides the chord into two equal parts. Next, a chord DE is drawn through point C such that the ratio of DC to CE is 1:3. So, DE is another chord passing through C, and it's divided by C into two parts where DC is one part and CE is three times that part. So, DC:CE = 1:3. The problem asks for the acute angle ACD and the permissible values of α. So, I need to find the measure of angle ACD, which is the angle between chords AC and DC, and also determine the possible values of α for which this configuration is possible.Alright, let me start by drawing a diagram in my mind. There's a circle with chord AB. The midpoint of AB is C. Then, another chord DE passes through C, with DC:CE = 1:3. So, DE is split into four equal parts, with DC being one part and CE being three parts.I think I should assign some variables to make this concrete. Let me let DC = x, so CE = 3x, making DE = 4x. That seems straightforward.Now, since C is the midpoint of AB, AC = BC. Let me denote AC as y. So, AB = 2y.Since AB is a chord of the circle, the length of AB can be related to the radius of the circle and the angle α. The formula for the length of a chord is 2R sin(θ/2), where θ is the central angle subtended by the chord. So, AB = 2R sin(α/2). Therefore, 2y = 2R sin(α/2), which simplifies to y = R sin(α/2). So, AC = R sin(α/2).Similarly, DE is another chord, and its length is 4x. So, DE = 4x = 2R sin(β/2), where β is the central angle subtended by DE. So, 4x = 2R sin(β/2), which gives x = (R/2) sin(β/2). But I don't know β yet. Maybe I can relate it to α somehow.Since DE passes through C, which is the midpoint of AB, perhaps there's a relationship between the angles α and β. Maybe I can use some properties of circles here.I remember that the angle subtended by a chord at the center is twice the angle subtended at the circumference. So, maybe I can relate the angles at point C to the central angles.Also, since C is the midpoint of AB, the line OC (where O is the center of the circle) is perpendicular to AB. So, OC is the perpendicular bisector of AB. That might be useful.Let me try to visualize triangle ACD. Points A, C, D. I need to find angle ACD. So, angle at point C between points A and D.Since I have lengths AC and DC, maybe I can use the Law of Cosines in triangle ACD to find angle ACD. But for that, I need to know the length of AD or some other information.Wait, maybe I can find the length of AD using other properties. Since DE is a chord passing through C, and DC:CE = 1:3, so DE is divided into 1:3 ratio at C.I also remember that when two chords intersect, the products of the segments are equal. So, AC * CB = DC * CE. Since AC = CB = y, and DC = x, CE = 3x, so y * y = x * 3x, which gives y² = 3x². Therefore, y = x√3.So, AC = x√3. That's helpful because I can relate AC and DC.Earlier, I had AC = R sin(α/2) and x = (R/2) sin(β/2). So, substituting AC = x√3, we get R sin(α/2) = (R/2) sin(β/2) * √3. Simplifying, sin(α/2) = (√3/2) sin(β/2). Hmm, that relates α and β. Maybe I can find another relationship between α and β.Since DE is a chord passing through C, and OC is perpendicular to AB, maybe I can consider triangle OCD or something like that.Let me think about the coordinates. Maybe assigning coordinates can help. Let me place the circle with center at O(0,0). Let me place chord AB horizontally for simplicity. So, point A is (-R,0), point B is (R,0), so midpoint C is (0,0). Wait, but then OC would be the same as the center, which is (0,0). That might complicate things because then point C is the center. But in the problem, C is the midpoint of AB, which is not necessarily the center unless AB is a diameter.Wait, hold on. If AB is a diameter, then C would coincide with the center. But in the problem, AB is just a chord, not necessarily a diameter. So, I shouldn't assume that.Let me correct that. Let me place the center O somewhere above the chord AB. So, chord AB is horizontal, midpoint C is at some point below O. Let me assign coordinates accordingly.Let me set point C at (0,0). Then, chord AB is horizontal, so points A and B are (-y,0) and (y,0), respectively, where y = AC = BC. The center O is somewhere along the perpendicular bisector of AB, which is the y-axis. So, O is at (0, h) for some h.Now, DE is another chord passing through C(0,0), with DC:CE = 1:3. Let me denote point D as (d_x, d_y) and point E as (e_x, e_y). Since DE passes through C, which is (0,0), the line DE passes through the origin.Given DC:CE = 1:3, so the ratio of lengths from D to C to C to E is 1:3. So, in terms of vectors, vector DC = ( -d_x, -d_y ) and vector CE = ( e_x, e_y ). The ratio of their magnitudes is 1:3.But since DE is a straight line through the origin, points D and E are colinear with the origin, so they lie on a straight line passing through (0,0). So, the coordinates of D and E can be expressed as scalar multiples. Let me say that point D is (k, mk) and point E is (-3k, -3mk), so that DC:CE = 1:3. Because from D to C is vector (-k, -mk), and from C to E is vector (-3k, -3mk), so their magnitudes are in the ratio 1:3.But wait, actually, since DC:CE = 1:3, the length from D to C is one part, and from C to E is three parts. So, if I take D as (k, mk), then E would be (-3k, -3mk), so that the vector from D to C is (-k, -mk), and from C to E is (-3k, -3mk). The lengths would be sqrt(k² + (mk)²) and sqrt(9k² + 9m²k²) = 3 sqrt(k² + m²k²). So, the ratio is 1:3, as desired.So, points D and E are (k, mk) and (-3k, -3mk). Now, since D and E lie on the circle centered at (0, h) with radius R, their coordinates must satisfy the circle equation:For point D: (k - 0)² + (mk - h)² = R²For point E: (-3k - 0)² + (-3mk - h)² = R²So, both equations equal R². Let me write them out:1. k² + (mk - h)² = R²2. 9k² + (-3mk - h)² = R²Since both equal R², set them equal to each other:k² + (mk - h)² = 9k² + (-3mk - h)²Let me expand both sides:Left side: k² + (m²k² - 2mhk + h²)Right side: 9k² + (9m²k² + 6mhk + h²)So, left side: k² + m²k² - 2mhk + h²Right side: 9k² + 9m²k² + 6mhk + h²Subtract left side from right side:(9k² + 9m²k² + 6mhk + h²) - (k² + m²k² - 2mhk + h²) = 0Simplify:(9k² - k²) + (9m²k² - m²k²) + (6mhk + 2mhk) + (h² - h²) = 0Which is:8k² + 8m²k² + 8mhk = 0Factor out 8k:8k(k + m²k + mh) = 0So, 8k(k(1 + m²) + mh) = 0Since k ≠ 0 (otherwise, points D and E would coincide with C), we have:k(1 + m²) + mh = 0So, k = -mh / (1 + m²)So, now we have k expressed in terms of m and h.Now, let's substitute k back into one of the circle equations. Let's take the first one:k² + (mk - h)² = R²Substitute k = -mh / (1 + m²):[(-mh / (1 + m²))²] + [m*(-mh / (1 + m²)) - h]² = R²Let me compute each term:First term: (m²h²) / (1 + m²)²Second term: [ -m²h / (1 + m²) - h ] = [ (-m²h - h(1 + m²)) / (1 + m²) ] = [ (-m²h - h - m²h) / (1 + m²) ] = [ (-2m²h - h) / (1 + m²) ] = -h(2m² + 1) / (1 + m²)So, squaring the second term: [ h²(2m² + 1)² ] / (1 + m²)²So, putting it all together:(m²h²)/(1 + m²)² + [ h²(2m² + 1)² ] / (1 + m²)² = R²Combine the terms:[ m²h² + h²(2m² + 1)² ] / (1 + m²)² = R²Factor out h²:h²[ m² + (2m² + 1)² ] / (1 + m²)² = R²Let me compute the numerator inside the brackets:m² + (2m² + 1)² = m² + 4m⁴ + 4m² + 1 = 4m⁴ + 5m² + 1So, we have:h²(4m⁴ + 5m² + 1) / (1 + m²)² = R²Let me see if I can factor 4m⁴ + 5m² + 1. Let me treat it as a quadratic in m²:4m⁴ + 5m² + 1 = (4m² + a)(m² + b) = 4m⁴ + (a + 4b)m² + abComparing coefficients:a + 4b = 5ab = 1Looking for integers a and b. Let me try a=1, b=1: 1 + 4=5, which works. So,4m⁴ + 5m² + 1 = (4m² + 1)(m² + 1)So, numerator becomes (4m² + 1)(m² + 1)Denominator is (1 + m²)²So, h²(4m² + 1)(m² + 1) / (m² + 1)² = h²(4m² + 1)/(m² + 1) = R²Thus,h²(4m² + 1)/(m² + 1) = R²So, h² = R²(m² + 1)/(4m² + 1)Okay, so now I have h in terms of R and m.Now, let me recall that point A is (-y, 0), point B is (y, 0), and the center is (0, h). So, the distance from center O(0,h) to point A(-y,0) is R.So, OA² = (-y - 0)² + (0 - h)² = y² + h² = R²Therefore, y² + h² = R²But from above, h² = R²(m² + 1)/(4m² + 1)So,y² + R²(m² + 1)/(4m² + 1) = R²Therefore,y² = R² - R²(m² + 1)/(4m² + 1) = R²[1 - (m² + 1)/(4m² + 1)]Simplify the expression inside the brackets:1 - (m² + 1)/(4m² + 1) = [ (4m² + 1) - (m² + 1) ] / (4m² + 1) = (3m²) / (4m² + 1)Therefore,y² = R² * (3m²)/(4m² + 1)So, y = R * sqrt(3m²/(4m² + 1)) = R * (m√3)/sqrt(4m² + 1)But earlier, we had AC = y = x√3, and x = DC = sqrt(k² + (mk)²) = |k| sqrt(1 + m²)But k = -mh / (1 + m²), so |k| = mh / (1 + m²)Therefore, x = (mh / (1 + m²)) * sqrt(1 + m²) = mh / sqrt(1 + m²)So, x = mh / sqrt(1 + m²)But from above, y = R * (m√3)/sqrt(4m² + 1)And since y = x√3, we have:R * (m√3)/sqrt(4m² + 1) = (mh / sqrt(1 + m²)) * √3Simplify:R * m / sqrt(4m² + 1) = mh / sqrt(1 + m²)Cancel m from both sides (assuming m ≠ 0, which it isn't because otherwise DE would be vertical and DC:CE ratio wouldn't make sense):R / sqrt(4m² + 1) = h / sqrt(1 + m²)But from earlier, h² = R²(m² + 1)/(4m² + 1)So, h = R sqrt( (m² + 1)/(4m² + 1) )Therefore, substituting h into the equation:R / sqrt(4m² + 1) = [ R sqrt( (m² + 1)/(4m² + 1) ) ] / sqrt(1 + m²)Simplify the right side:R sqrt( (m² + 1)/(4m² + 1) ) / sqrt(1 + m²) = R sqrt(1/(4m² + 1)) = R / sqrt(4m² + 1)So, both sides are equal, which is consistent. So, that checks out.Now, going back, I need to find angle ACD. Let me think about triangle ACD. Points A(-y,0), C(0,0), D(k, mk). So, vectors CA = (-y, 0) and CD = (k, mk). The angle between vectors CA and CD is angle ACD.The formula for the angle between two vectors u and v is:cosθ = (u • v) / (|u| |v|)So, let's compute the dot product of CA and CD.Vector CA = (-y, 0)Vector CD = (k, mk)Dot product: (-y)(k) + 0*(mk) = -ykThe magnitudes:|CA| = sqrt( (-y)^2 + 0^2 ) = y|CD| = sqrt( k^2 + (mk)^2 ) = |k| sqrt(1 + m²) = x sqrt(1 + m²) (since x = |k| sqrt(1 + m²))Wait, but earlier, x = |k| sqrt(1 + m²). So, |CD| = x.Wait, but x was defined as DC, which is the length from D to C, which is |CD|. So, yes, |CD| = x.So, |CD| = x.Therefore, cosθ = (-yk)/(y * x) = -k / xBut k = -mh / (1 + m²), so:cosθ = - [ (-mh / (1 + m²)) ] / x = (mh / (1 + m²)) / xBut x = |k| sqrt(1 + m²) = (mh / (1 + m²)) sqrt(1 + m²) = mh / sqrt(1 + m²)So, x = mh / sqrt(1 + m²)Therefore, cosθ = (mh / (1 + m²)) / (mh / sqrt(1 + m²)) ) = [ mh / (1 + m²) ] * [ sqrt(1 + m²) / mh ] = 1 / sqrt(1 + m²)So, cosθ = 1 / sqrt(1 + m²)Therefore, θ = arccos(1 / sqrt(1 + m²)) = arcsin(m / sqrt(1 + m²))Wait, because sinθ = opposite/hypotenuse = m / sqrt(1 + m²). So, θ = arcsin(m / sqrt(1 + m²))But I need to express this in terms of α.Earlier, we had:From the chord AB, we had y = R sin(α/2)But y = R * (m√3)/sqrt(4m² + 1)So, R sin(α/2) = R * (m√3)/sqrt(4m² + 1)Cancel R:sin(α/2) = (m√3)/sqrt(4m² + 1)Let me square both sides:sin²(α/2) = (3m²)/(4m² + 1)Let me denote t = m². Then,sin²(α/2) = 3t / (4t + 1)Let me solve for t:sin²(α/2) = 3t / (4t + 1)Multiply both sides by (4t + 1):sin²(α/2)(4t + 1) = 3tExpand:4 sin²(α/2) t + sin²(α/2) = 3tBring all terms to one side:4 sin²(α/2) t - 3t + sin²(α/2) = 0Factor t:t(4 sin²(α/2) - 3) + sin²(α/2) = 0Solve for t:t = - sin²(α/2) / (4 sin²(α/2) - 3)But t = m² must be positive, so the numerator and denominator must have the same sign.So,- sin²(α/2) / (4 sin²(α/2) - 3) > 0Which implies that either both numerator and denominator are positive or both are negative.Case 1: Numerator positive, denominator positive.- sin²(α/2) > 0 implies sin²(α/2) < 0, which is impossible because squares are non-negative.Case 2: Numerator negative, denominator negative.- sin²(α/2) < 0 implies sin²(α/2) > 0, which is always true except when sin(α/2)=0, which would mean α=0, but α is the measure of an arc, so it must be greater than 0.Denominator negative:4 sin²(α/2) - 3 < 0So,4 sin²(α/2) < 3sin²(α/2) < 3/4sin(α/2) < sqrt(3)/2Which implies α/2 < 60°, so α < 120°Therefore, permissible values of α are α < 120°, since α is the measure of the smaller arc, it must be less than 180°, but here it's further restricted to less than 120°.So, α must be less than 120°.Now, going back to t:t = m² = - sin²(α/2) / (4 sin²(α/2) - 3)But since denominator is negative (from above), and numerator is negative (because of the negative sign), t is positive, as required.So,m² = [ - sin²(α/2) ] / [4 sin²(α/2) - 3] = sin²(α/2) / [3 - 4 sin²(α/2)]Therefore,m = sqrt[ sin²(α/2) / (3 - 4 sin²(α/2)) ] = sin(α/2) / sqrt(3 - 4 sin²(α/2))Now, recall that angle ACD = θ = arcsin(m / sqrt(1 + m²))Let me compute m / sqrt(1 + m²):m / sqrt(1 + m²) = [ sin(α/2) / sqrt(3 - 4 sin²(α/2)) ] / sqrt(1 + [ sin²(α/2) / (3 - 4 sin²(α/2)) ] )Simplify denominator:sqrt[ (3 - 4 sin²(α/2) + sin²(α/2)) / (3 - 4 sin²(α/2)) ) ] = sqrt[ (3 - 3 sin²(α/2)) / (3 - 4 sin²(α/2)) ) ] = sqrt[ 3(1 - sin²(α/2)) / (3 - 4 sin²(α/2)) ) ] = sqrt[ 3 cos²(α/2) / (3 - 4 sin²(α/2)) ) ] = sqrt(3) cos(α/2) / sqrt(3 - 4 sin²(α/2))Therefore,m / sqrt(1 + m²) = [ sin(α/2) / sqrt(3 - 4 sin²(α/2)) ] / [ sqrt(3) cos(α/2) / sqrt(3 - 4 sin²(α/2)) ) ] = sin(α/2) / (sqrt(3) cos(α/2)) = (1/√3) tan(α/2)So, angle ACD = arcsin( (1/√3) tan(α/2) )Therefore, the acute angle ACD is arcsin( tan(α/2) / √3 )And the permissible values of α are α < 120°, since α must be less than 120° for the configuration to be possible.So, summarizing:The acute angle ACD is arcsin( tan(α/2) / √3 ), and α must be less than 120 degrees.**Final Answer**The acute angle ( angle ACD ) is ( boxed{arcsinleft(frac{tanleft(frac{alpha}{2}right)}{sqrt{3}}right)} ) and the permissible values of ( alpha ) are ( boxed{alpha < 120^circ} ).

answer：Okay, so I have this problem where there are two spheres of one radius and two spheres of another radius arranged such that each sphere touches the other three and a plane. I need to find the ratio of the radius of the larger sphere to the smaller one. Hmm, okay, let me try to visualize this.First, I imagine four spheres: two larger ones and two smaller ones. Each sphere touches the other three and also rests on a plane. So, all four spheres are touching the same plane. That means the centers of these spheres are all at a height equal to their respective radii above the plane.Let me denote the radius of the larger spheres as ( R ) and the radius of the smaller spheres as ( r ). So, the centers of the larger spheres are ( R ) units above the plane, and the centers of the smaller spheres are ( r ) units above the plane.Now, since each sphere touches the other three, the distance between the centers of any two spheres should be equal to the sum of their radii. For example, the distance between the centers of a larger sphere and a smaller sphere should be ( R + r ). Similarly, the distance between the centers of two larger spheres should be ( 2R ), and the distance between the centers of two smaller spheres should be ( 2r ).But wait, all four spheres are touching each other. So, the centers of these spheres must form a regular tetrahedron? Or maybe not exactly, because two of them are larger and two are smaller. Hmm, maybe it's a different kind of tetrahedron where the edges have different lengths.Let me think. If I have two larger spheres and two smaller spheres, each larger sphere touches both smaller spheres and the other larger sphere. Similarly, each smaller sphere touches both larger spheres and the other smaller sphere. So, the centers form a tetrahedron where two edges are ( 2R ), two edges are ( 2r ), and the remaining edges are ( R + r ).But actually, in a tetrahedron, each pair of vertices is connected by an edge, so in this case, there are six edges. Among these six edges, how many are ( R + r ), ( 2R ), or ( 2r )?Wait, no. Each larger sphere touches the other larger sphere, so that's one edge of ( 2R ). Each smaller sphere touches the other smaller sphere, so that's another edge of ( 2r ). Then, each larger sphere touches each smaller sphere, so that's four edges each of length ( R + r ). So, in total, we have one edge of ( 2R ), one edge of ( 2r ), and four edges of ( R + r ).But a tetrahedron has six edges, so that doesn't add up. Wait, no, actually, each larger sphere touches both smaller spheres and the other larger sphere, so each larger sphere has three connections: one to the other larger sphere and two to the smaller spheres. Similarly, each smaller sphere has three connections: one to the other smaller sphere and two to the larger spheres.So, in total, the edges are: one ( 2R ), one ( 2r ), and four ( R + r ). That makes six edges, which is correct for a tetrahedron.But how can we model this? Maybe it's easier to think in terms of coordinates. Let me try to assign coordinates to the centers of the spheres.Let me place the plane as the xy-plane for simplicity. Then, the centers of the larger spheres will be at some points ( (x, y, R) ) and the centers of the smaller spheres will be at some points ( (u, v, r) ).Since each sphere touches the plane, their z-coordinates are equal to their radii.Now, the distance between the centers of the two larger spheres should be ( 2R ). Similarly, the distance between the centers of the two smaller spheres should be ( 2r ). The distance between the centers of a larger sphere and a smaller sphere should be ( R + r ).Let me denote the centers of the larger spheres as ( O_1 ) and ( O_2 ), and the centers of the smaller spheres as ( O_3 ) and ( O_4 ).So, ( O_1 ) is at ( (x_1, y_1, R) ), ( O_2 ) is at ( (x_2, y_2, R) ), ( O_3 ) is at ( (x_3, y_3, r) ), and ( O_4 ) is at ( (x_4, y_4, r) ).Now, the distance between ( O_1 ) and ( O_2 ) is ( 2R ), so:[sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (R - R)^2} = 2R]Simplifying, we get:[sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = 2R]Similarly, the distance between ( O_3 ) and ( O_4 ) is ( 2r ):[sqrt{(x_4 - x_3)^2 + (y_4 - y_3)^2 + (r - r)^2} = 2r]Which simplifies to:[sqrt{(x_4 - x_3)^2 + (y_4 - y_3)^2} = 2r]Now, the distance between ( O_1 ) and ( O_3 ) is ( R + r ):[sqrt{(x_3 - x_1)^2 + (y_3 - y_1)^2 + (r - R)^2} = R + r]Squaring both sides:[(x_3 - x_1)^2 + (y_3 - y_1)^2 + (R - r)^2 = (R + r)^2]Expanding both sides:Left side:[(x_3 - x_1)^2 + (y_3 - y_1)^2 + R^2 - 2Rr + r^2]Right side:[R^2 + 2Rr + r^2]Subtracting left side from right side:[(x_3 - x_1)^2 + (y_3 - y_1)^2 + R^2 - 2Rr + r^2 = R^2 + 2Rr + r^2]Subtracting ( R^2 + r^2 ) from both sides:[(x_3 - x_1)^2 + (y_3 - y_1)^2 - 2Rr = 2Rr]Adding ( 2Rr ) to both sides:[(x_3 - x_1)^2 + (y_3 - y_1)^2 = 4Rr]So, the distance between the projections of ( O_1 ) and ( O_3 ) onto the plane is ( 2sqrt{Rr} ).Similarly, the distance between ( O_1 ) and ( O_4 ) is also ( R + r ), so the same applies:[(x_4 - x_1)^2 + (y_4 - y_1)^2 = 4Rr]Similarly, the distance between ( O_2 ) and ( O_3 ):[(x_3 - x_2)^2 + (y_3 - y_2)^2 = 4Rr]And between ( O_2 ) and ( O_4 ):[(x_4 - x_2)^2 + (y_4 - y_2)^2 = 4Rr]So, all four projections of the smaller spheres onto the plane are at a distance of ( 2sqrt{Rr} ) from both projections of the larger spheres.Now, let's think about the configuration on the plane. The projections of the centers of the two larger spheres, ( O_1 ) and ( O_2 ), are separated by ( 2R ). The projections of the centers of the two smaller spheres, ( O_3 ) and ( O_4 ), are separated by ( 2r ).Moreover, each projection of a smaller sphere is at a distance of ( 2sqrt{Rr} ) from both projections of the larger spheres.So, if I consider the projections on the plane, we have two points ( A ) and ( B ) (projections of ( O_1 ) and ( O_2 )) separated by ( 2R ), and two other points ( C ) and ( D ) (projections of ( O_3 ) and ( O_4 )) separated by ( 2r ). Each of ( C ) and ( D ) is at a distance of ( 2sqrt{Rr} ) from both ( A ) and ( B ).This seems like a rhombus. Let me check. If ( AC = AD = BC = BD = 2sqrt{Rr} ), and ( AB = 2R ), ( CD = 2r ), then the figure ( ACBD ) is a rhombus with sides ( 2sqrt{Rr} ) and diagonals ( 2R ) and ( 2r ).Wait, in a rhombus, the diagonals are perpendicular bisectors of each other. So, if ( ACBD ) is a rhombus with diagonals ( AB = 2R ) and ( CD = 2r ), then the sides of the rhombus can be calculated using the formula for the side length in terms of the diagonals.The formula for the side length ( s ) of a rhombus with diagonals ( d_1 ) and ( d_2 ) is:[s = sqrt{left(frac{d_1}{2}right)^2 + left(frac{d_2}{2}right)^2}]So, in this case, the side length ( s ) is:[s = sqrt{left(frac{2R}{2}right)^2 + left(frac{2r}{2}right)^2} = sqrt{R^2 + r^2}]But earlier, we found that the side length is ( 2sqrt{Rr} ). Therefore, we have:[2sqrt{Rr} = sqrt{R^2 + r^2}]Let me square both sides to eliminate the square roots:[(2sqrt{Rr})^2 = (sqrt{R^2 + r^2})^2]Simplifying:[4Rr = R^2 + r^2]Let me rearrange this equation:[R^2 - 4Rr + r^2 = 0]This is a quadratic equation in terms of ( R ). Let me write it as:[R^2 - 4rR + r^2 = 0]Using the quadratic formula, ( R = frac{4r pm sqrt{(4r)^2 - 4 cdot 1 cdot r^2}}{2 cdot 1} )Calculating the discriminant:[(4r)^2 - 4 cdot 1 cdot r^2 = 16r^2 - 4r^2 = 12r^2]So,[R = frac{4r pm sqrt{12r^2}}{2} = frac{4r pm 2rsqrt{3}}{2} = 2r pm rsqrt{3}]Therefore, we have two solutions:1. ( R = 2r + rsqrt{3} = r(2 + sqrt{3}) )2. ( R = 2r - rsqrt{3} = r(2 - sqrt{3}) )Since ( R ) is the radius of the larger sphere, it must be greater than ( r ). Let's check the second solution:( 2 - sqrt{3} ) is approximately ( 2 - 1.732 = 0.267 ), which is positive, but ( R = r(2 - sqrt{3}) ) would be smaller than ( r ), which contradicts the assumption that ( R ) is larger. Therefore, we discard this solution.Thus, the valid solution is:[R = r(2 + sqrt{3})]Therefore, the ratio ( frac{R}{r} ) is ( 2 + sqrt{3} ).Let me just recap to make sure I haven't missed anything. We considered the projections of the centers onto the plane, recognized the rhombus formed by these projections, used the relationship between the diagonals and the sides of the rhombus, set up the equation, solved the quadratic, and discarded the invalid solution. It seems consistent.I think this is the correct ratio.**Final Answer**The ratio of the radius of the larger sphere to the radius of the smaller sphere is boxed{2 + sqrt{3}}.

answer：First, I need to determine the speed of the man in still water, given his upstream and downstream speeds.I know that when rowing upstream, the man's effective speed is reduced by the speed of the current, and when rowing downstream, his effective speed is increased by the speed of the current.Let’s denote:- ( S ) as the speed of the man in still water,- ( C ) as the speed of the current.From the problem, the upstream speed is 60 km/h, and the downstream speed is 90 km/h. This gives me two equations:1. ( S - C = 60 )2. ( S + C = 90 )To find ( S ), I can add these two equations together:( (S - C) + (S + C) = 60 + 90 )( 2S = 150 )( S = 75 )Therefore, the speed of the man in still water is 75 km/h.

answer：Okay, so I have this problem where I need to find the distance between two points, P and Q, which are given in polar coordinates. P is at (1, π/6) and Q is at (2, π/2). Hmm, polar coordinates, right? That means each point is defined by a radius and an angle. I remember that to find the distance between two points, it's often easier to work with Cartesian coordinates, so maybe I should convert these polar coordinates to Cartesian first.Alright, let's recall the conversion formulas. To convert from polar (r, θ) to Cartesian (x, y), the formulas are:x = r * cos(θ)y = r * sin(θ)So, let's start with point P(1, π/6). Plugging into the formulas:x_P = 1 * cos(π/6)y_P = 1 * sin(π/6)I remember that cos(π/6) is √3/2 and sin(π/6) is 1/2. So,x_P = √3/2y_P = 1/2So, point P in Cartesian coordinates is (√3/2, 1/2).Now, let's do the same for point Q(2, π/2).x_Q = 2 * cos(π/2)y_Q = 2 * sin(π/2)I know that cos(π/2) is 0 and sin(π/2) is 1. So,x_Q = 0y_Q = 2Therefore, point Q in Cartesian coordinates is (0, 2).Now that both points are in Cartesian coordinates, I can use the distance formula to find |PQ|. The distance formula between two points (x1, y1) and (x2, y2) is:distance = √[(x2 - x1)^2 + (y2 - y1)^2]Plugging in the coordinates of P and Q:distance = √[(0 - √3/2)^2 + (2 - 1/2)^2]Let's compute each part step by step. First, the x-component:(0 - √3/2)^2 = (√3/2)^2 = (3/4)Next, the y-component:(2 - 1/2)^2 = (3/2)^2 = 9/4Now, add these two results together:3/4 + 9/4 = 12/4 = 3So, the distance is √3.Wait, let me double-check my calculations to make sure I didn't make any mistakes. Starting with the conversion from polar to Cartesian, that seems correct. For point P, cos(π/6) is indeed √3/2 and sin(π/6) is 1/2. For point Q, cos(π/2) is 0 and sin(π/2) is 1, so that's correct too.Then, using the distance formula, I subtracted the x-coordinates and squared them, same with the y-coordinates. The x-component squared was (√3/2)^2 which is 3/4, and the y-component squared was (3/2)^2 which is 9/4. Adding those gives 12/4, which simplifies to 3. Taking the square root of 3 gives √3.Hmm, that seems right. I don't think I made any calculation errors. So, the distance between P and Q is √3.

answer：Alright, so I've got this problem here where I need to prove that for an odd number ( k geq 3 ), there exists a polynomial ( f(x) ) of degree ( k ) with non-integer coefficients that satisfies two specific properties. The first property is straightforward: ( f(0) = 0 ) and ( f(1) = 1 ). The second property is a bit more involved—it says that there are infinitely many positive integers ( n ) such that if the equation ( n = f(x_1) + cdots + f(x_s) ) has integer solutions ( x_1, ldots, x_s ), then the number of terms ( s ) must be at least ( 2^k - 1 ).Okay, so first, let's unpack what this means. We're dealing with a polynomial of odd degree ( k ), which is at least 3. The polynomial has non-integer coefficients, but it still maps integers to integers—so it's an integer-valued polynomial. That's interesting because usually, polynomials with integer coefficients are integer-valued, but here, the coefficients don't have to be integers, yet the polynomial still takes integer values at integer points.The first property is clear: when we plug in 0, we get 0, and when we plug in 1, we get 1. That seems like a normalization condition. The second property is about the representation of integers ( n ) as sums of values of this polynomial. Specifically, for infinitely many ( n ), you can't express ( n ) as a sum of fewer than ( 2^k - 1 ) values of ( f ). That's a pretty strong condition—it suggests that the polynomial has some kind of "sparsity" property when it comes to representing integers.I think the key here is to construct such a polynomial ( f(x) ) explicitly. The problem mentions that ( f(x) ) is an integer-valued polynomial, which usually means that it can be expressed in terms of binomial coefficients with integer coefficients. For example, polynomials like ( f(x) = binom{x}{1} = x ), ( f(x) = binom{x}{2} = frac{x(x-1)}{2} ), etc., are integer-valued because the binomial coefficients ensure that the result is always an integer when ( x ) is an integer.Given that ( f(x) ) is of degree ( k ), I might need to express it in terms of binomial coefficients up to ( binom{x}{k} ). Since the coefficients don't have to be integers, but the polynomial still needs to be integer-valued, this gives me some flexibility in choosing the coefficients.Let me think about how to ensure that ( f(0) = 0 ) and ( f(1) = 1 ). If I write ( f(x) ) as a linear combination of binomial coefficients:[f(x) = a_k binom{x}{k} + a_{k-1} binom{x}{k-1} + cdots + a_1 binom{x}{1}]Then, evaluating at ( x = 0 ) gives ( f(0) = 0 ) because all binomial coefficients ( binom{0}{i} = 0 ) for ( i geq 1 ). Evaluating at ( x = 1 ) gives:[f(1) = a_k binom{1}{k} + a_{k-1} binom{1}{k-1} + cdots + a_1 binom{1}{1}]But ( binom{1}{i} = 0 ) for ( i geq 2 ), so this simplifies to:[f(1) = a_1 binom{1}{1} = a_1]Since ( f(1) = 1 ), this tells me that ( a_1 = 1 ).So, I have:[f(x) = a_k binom{x}{k} + a_{k-1} binom{x}{k-1} + cdots + a_2 binom{x}{2} + binom{x}{1}]Now, I need to choose coefficients ( a_2, ldots, a_k ) such that ( f(x) ) has non-integer coefficients but is still integer-valued. Also, I need to ensure the second property about the representation of ( n ).I recall that integer-valued polynomials can have non-integer coefficients as long as the combination of terms results in integer outputs for integer inputs. For example, ( frac{x(x-1)}{2} ) is integer-valued because for any integer ( x ), ( x(x-1) ) is even, so dividing by 2 still gives an integer.So, perhaps I can choose coefficients ( a_i ) such that when combined with the binomial coefficients, the overall polynomial remains integer-valued. Since binomial coefficients have denominators that are factorials, choosing coefficients that are multiples of those factorials could help in keeping the polynomial integer-valued.Wait, but the problem states that the coefficients themselves are non-integer. So, I need to choose ( a_i ) such that, when multiplied by the binomial coefficients (which have denominators), the overall coefficients of ( f(x) ) are non-integer, but the polynomial still maps integers to integers.This seems a bit tricky. Maybe I can set up the coefficients ( a_i ) in such a way that the denominators in the binomial coefficients cancel out the denominators in ( a_i ), but the coefficients of the polynomial ( f(x) ) are not integers.Alternatively, perhaps I can use a basis of integer-valued polynomials with non-integer coefficients. I know that the set of integer-valued polynomials forms a ring, and they can be expressed in terms of binomial coefficients with integer coefficients, but here we need non-integer coefficients.Maybe I can use a scaling factor. For example, if I take a polynomial with integer coefficients and then scale it by a non-integer rational number, but in such a way that the resulting polynomial is still integer-valued. However, scaling by a non-integer would likely result in non-integer coefficients, but we have to ensure that the polynomial remains integer-valued.But scaling a polynomial with integer coefficients by a non-integer rational number would generally result in a non-integer-valued polynomial unless the scaling factor is chosen carefully. For example, if I have a polynomial ( P(x) ) with integer coefficients and I scale it by ( frac{1}{m} ), then ( frac{1}{m} P(x) ) is integer-valued only if ( m ) divides ( P(x) ) for all integer ( x ).So, perhaps I can construct ( f(x) ) as ( frac{1}{m} P(x) ), where ( P(x) ) is a polynomial with integer coefficients such that ( m ) divides ( P(x) ) for all integer ( x ). Then, ( f(x) ) would have non-integer coefficients (if ( m > 1 )) but still be integer-valued.This seems promising. So, how can I construct such a polynomial ( P(x) )?One way is to use the concept of integer-valued polynomials with denominators. For example, the binomial coefficients ( binom{x}{k} ) have denominators ( k! ), but they are integer-valued because the numerator is always divisible by ( k! ) when ( x ) is an integer.So, perhaps I can construct ( f(x) ) as a linear combination of such binomial coefficients with coefficients chosen such that the overall polynomial has non-integer coefficients but remains integer-valued.Given that ( f(x) ) is of degree ( k ), I can write it as:[f(x) = a_k binom{x}{k} + a_{k-1} binom{x}{k-1} + cdots + a_1 binom{x}{1}]We already established that ( a_1 = 1 ) to satisfy ( f(1) = 1 ). Now, I need to choose ( a_2, ldots, a_k ) such that the coefficients of ( f(x) ) are non-integer, but ( f(x) ) is still integer-valued.Let me think about the coefficients of ( f(x) ). The binomial coefficient ( binom{x}{i} ) can be written as:[binom{x}{i} = frac{x(x-1)(x-2)cdots(x-i+1)}{i!}]So, when expressed as a polynomial, it has coefficients that are fractions with denominators dividing ( i! ). Therefore, if I choose ( a_i ) such that when multiplied by the coefficients of ( binom{x}{i} ), the resulting coefficients of ( f(x) ) are non-integer, but the entire polynomial ( f(x) ) still evaluates to integers at integer points.This seems a bit abstract. Maybe I can look for a specific construction. Let's consider the case when ( k = 3 ) to get some intuition.For ( k = 3 ), we need a cubic polynomial ( f(x) ) with non-integer coefficients, such that ( f(0) = 0 ), ( f(1) = 1 ), and for infinitely many ( n ), any representation of ( n ) as a sum of ( f(x_i) ) requires at least ( 2^3 - 1 = 7 ) terms.So, let's try to construct such a polynomial for ( k = 3 ).We can write:[f(x) = a_3 binom{x}{3} + a_2 binom{x}{2} + binom{x}{1}]We need ( f(0) = 0 ), which is already satisfied. ( f(1) = 1 ) gives:[f(1) = a_3 binom{1}{3} + a_2 binom{1}{2} + binom{1}{1} = 0 + 0 + 1 = 1]So, that condition is satisfied regardless of ( a_3 ) and ( a_2 ). Now, we need to choose ( a_3 ) and ( a_2 ) such that the coefficients of ( f(x) ) are non-integer, but ( f(x) ) is still integer-valued.Let's express ( f(x) ) in standard polynomial form. Recall that:[binom{x}{3} = frac{x^3 - 3x^2 + 2x}{6}][binom{x}{2} = frac{x^2 - x}{2}][binom{x}{1} = x]So, substituting these into ( f(x) ):[f(x) = a_3 left( frac{x^3 - 3x^2 + 2x}{6} right) + a_2 left( frac{x^2 - x}{2} right) + x]Simplify:[f(x) = frac{a_3}{6} x^3 - frac{3a_3}{6} x^2 + frac{2a_3}{6} x + frac{a_2}{2} x^2 - frac{a_2}{2} x + x]Combine like terms:[f(x) = frac{a_3}{6} x^3 + left( -frac{a_3}{2} + frac{a_2}{2} right) x^2 + left( frac{a_3}{3} - frac{a_2}{2} + 1 right) x]Now, the coefficients are:- ( x^3 ): ( frac{a_3}{6} )- ( x^2 ): ( -frac{a_3}{2} + frac{a_2}{2} )- ( x ): ( frac{a_3}{3} - frac{a_2}{2} + 1 )We need these coefficients to be non-integer. Let's choose ( a_3 ) and ( a_2 ) such that the coefficients are fractions but the polynomial remains integer-valued.Suppose we choose ( a_3 = 2 ) and ( a_2 = 1 ). Then:- ( x^3 ): ( frac{2}{6} = frac{1}{3} ) (non-integer)- ( x^2 ): ( -frac{2}{2} + frac{1}{2} = -1 + 0.5 = -0.5 ) (non-integer)- ( x ): ( frac{2}{3} - frac{1}{2} + 1 = frac{4}{6} - frac{3}{6} + frac{6}{6} = frac{7}{6} ) (non-integer)So, all coefficients are non-integer. Now, is ( f(x) ) integer-valued?Let's check for some integer values:- ( x = 0 ): ( f(0) = 0 ) (good)- ( x = 1 ): ( f(1) = frac{1}{3} - 0.5 + frac{7}{6} = frac{2}{6} - frac{3}{6} + frac{7}{6} = frac{6}{6} = 1 ) (good)- ( x = 2 ): ( f(2) = frac{1}{3}(8) + (-0.5)(4) + frac{7}{6}(2) = frac{8}{3} - 2 + frac{14}{6} = frac{8}{3} - 2 + frac{7}{3} = frac{15}{3} - 2 = 5 - 2 = 3 ) (good)- ( x = 3 ): ( f(3) = frac{1}{3}(27) + (-0.5)(9) + frac{7}{6}(3) = 9 - 4.5 + 3.5 = 8 ) (good)Okay, so it seems to work for these values. Let's check ( x = -1 ):( f(-1) = frac{1}{3}(-1)^3 + (-0.5)(1) + frac{7}{6}(-1) = -frac{1}{3} - 0.5 - frac{7}{6} = -frac{2}{6} - frac{3}{6} - frac{7}{6} = -frac{12}{6} = -2 ) (integer)So, it seems that ( f(x) ) is indeed integer-valued even though the coefficients are non-integer. That's good.Now, the second property: there are infinitely many positive integers ( n ) such that if ( n = f(x_1) + cdots + f(x_s) ), then ( s geq 7 ).How can I ensure this? It seems like we need to construct ( f(x) ) such that for infinitely many ( n ), the minimal number of terms needed to express ( n ) as a sum of ( f(x_i) ) is at least ( 2^k - 1 ). For ( k = 3 ), that's 7.I think this relates to the concept of additive bases in number theory. Specifically, we want ( f(x) ) to be such that its image forms a basis of order ( 2^k - 1 ), meaning that every sufficiently large integer can be expressed as a sum of at most ( 2^k - 1 ) terms from the image of ( f ). However, in our case, it's slightly different—we need infinitely many ( n ) that cannot be expressed with fewer than ( 2^k - 1 ) terms.Perhaps we can use the fact that the polynomial ( f(x) ) has a certain growth rate or modular properties that make it difficult to represent certain numbers with fewer terms.Looking back at the lemma mentioned earlier, it talks about constructing a polynomial that is 0 modulo ( 2^k ) when ( x ) is even and 1 modulo ( 2^k ) when ( x ) is odd. This suggests that the polynomial has specific congruence properties that could be used to control the representations of ( n ).If ( f(x) ) is 0 modulo ( 2^k ) for even ( x ) and 1 modulo ( 2^k ) for odd ( x ), then when we sum multiple ( f(x_i) ), the total sum modulo ( 2^k ) will depend on the number of odd ( x_i ) terms. Specifically, if we have ( s ) terms, the sum modulo ( 2^k ) will be equal to the number of odd ( x_i ) modulo ( 2^k ).Now, if we choose ( n ) such that ( n equiv -1 mod 2^k ), then the sum ( f(x_1) + cdots + f(x_s) equiv -1 mod 2^k ). Since each odd ( x_i ) contributes 1 modulo ( 2^k ) and each even ( x_i ) contributes 0, the total sum modulo ( 2^k ) is equal to the number of odd ( x_i ) terms. Therefore, to get a total of ( -1 mod 2^k ), which is equivalent to ( 2^k - 1 mod 2^k ), we need exactly ( 2^k - 1 ) odd ( x_i ) terms. Hence, the number of terms ( s ) must be at least ( 2^k - 1 ).This seems to align with the second property we need to prove. So, if we can construct such a polynomial ( f(x) ) with these congruence properties, then for infinitely many ( n ) (specifically, those congruent to ( -1 mod 2^k )), the minimal number of terms ( s ) required to express ( n ) as a sum of ( f(x_i) ) is at least ( 2^k - 1 ).Therefore, the strategy is to construct ( f(x) ) such that:1. ( f(x) equiv 0 mod 2^k ) when ( x ) is even.2. ( f(x) equiv 1 mod 2^k ) when ( x ) is odd.This ensures that the sum ( f(x_1) + cdots + f(x_s) equiv t mod 2^k ), where ( t ) is the number of odd ( x_i ) terms. For ( n equiv -1 mod 2^k ), we need ( t equiv -1 mod 2^k ), which requires ( t = 2^k - 1 ), hence ( s geq 2^k - 1 ).Now, how do we construct such a polynomial ( f(x) )? The lemma suggests that we can use a polynomial expressed in terms of binomial coefficients with specific coefficients. For ( k = 3 ), we saw that choosing ( a_3 = 2 ) and ( a_2 = 1 ) worked, but we need to generalize this for any odd ( k geq 3 ).The general construction would involve setting up the coefficients ( a_i ) such that the polynomial ( f(x) ) satisfies the required congruence properties. Specifically, we need:[f(x) equiv begin{cases}0 mod 2^k & text{if } x text{ is even}, 1 mod 2^k & text{if } x text{ is odd}.end{cases}]To achieve this, we can use the fact that the binomial coefficients ( binom{x}{i} ) have known congruence properties. For example, ( binom{x}{i} ) is an integer for integer ( x ), and its behavior modulo powers of 2 can be controlled by choosing appropriate coefficients.The key idea is to set up a system of congruences for the coefficients ( a_i ) such that the resulting polynomial ( f(x) ) satisfies the desired properties. This likely involves solving a system of linear equations modulo ( 2^k ) to determine the coefficients ( a_i ).Once we have such a polynomial ( f(x) ), it will automatically satisfy the second property because of the congruence conditions. Specifically, for any ( n equiv -1 mod 2^k ), the minimal number of terms ( s ) required to express ( n ) as a sum of ( f(x_i) ) must be at least ( 2^k - 1 ), as each odd ( x_i ) contributes 1 modulo ( 2^k ) and we need a total of ( 2^k - 1 ) such contributions to reach ( -1 mod 2^k ).Therefore, the existence of such a polynomial ( f(x) ) is guaranteed by the lemma, and the second property follows from the congruence properties of ( f(x) ).In summary, the steps are:1. Recognize that ( f(x) ) must be an integer-valued polynomial of degree ( k ) with non-integer coefficients.2. Use the lemma to construct ( f(x) ) such that it satisfies the congruence properties modulo ( 2^k ).3. Show that these congruence properties imply the second property about the minimal number of terms ( s ).This approach should work for any odd ( k geq 3 ), ensuring the existence of the desired polynomial ( f(x) ).

answer：Okay, so I have this problem with three identical circles, each touching the other two. The circumference of each circle is given as 48. I need to find two things: the perimeter of the shaded region and the area of the triangle formed by joining the centers of the three circles.First, let me visualize the diagram. There are three circles, each touching the other two. That means the centers of these circles form a triangle, and each side of the triangle is equal to twice the radius of the circles because the circles touch each other exactly once. So, the distance between any two centers is twice the radius.Given the circumference of each circle is 48, I can find the radius. The formula for circumference is ( C = 2pi r ). So, if I solve for ( r ), I get:[ r = frac{C}{2pi} = frac{48}{2pi} = frac{24}{pi} ]Okay, so the radius is ( frac{24}{pi} ). That means the distance between any two centers is ( 2r = frac{48}{pi} ). So, the triangle formed by the centers is an equilateral triangle with each side equal to ( frac{48}{pi} ).Now, for the perimeter of the shaded region. Looking at the Asymptote code, it seems the shaded region is a sort of lens-shaped area formed by the intersection of the three circles. But actually, in the Asymptote code, it's a path that goes from one circle to another, creating a sort of Reuleaux triangle. A Reuleaux triangle has a perimeter made up of three circular arcs, each centered at a different circle's center.Each of these arcs is 60 degrees because the triangle is equilateral, so each angle is 60 degrees. Since a full circle is 360 degrees, each arc is ( frac{60}{360} = frac{1}{6} ) of the circumference of a circle.So, the length of each arc is ( frac{1}{6} times 48 = 8 ). Since there are three such arcs in the perimeter of the shaded region, the total perimeter is ( 3 times 8 = 24 ).Wait, let me double-check that. If each arc is 60 degrees, which is a sixth of the circle, then each arc length is ( frac{1}{6} times 2pi r ). But since the circumference is 48, each arc is ( frac{1}{6} times 48 = 8 ). So, yes, three arcs make 24. That seems right.Now, moving on to the area of the triangle formed by joining the centers. Since the triangle is equilateral, all sides are equal, and each side is ( frac{48}{pi} ). The formula for the area of an equilateral triangle is ( frac{sqrt{3}}{4} times text{side}^2 ).So, plugging in the side length:[ text{Area} = frac{sqrt{3}}{4} times left( frac{48}{pi} right)^2 ]Calculating that:First, square the side length:[ left( frac{48}{pi} right)^2 = frac{2304}{pi^2} ]Then multiply by ( frac{sqrt{3}}{4} ):[ frac{sqrt{3}}{4} times frac{2304}{pi^2} = frac{2304 sqrt{3}}{4pi^2} ]Simplify the fraction:[ frac{2304}{4} = 576 ]So, the area is:[ frac{576 sqrt{3}}{pi^2} ]Let me just make sure I didn't make any calculation errors. The side length is ( frac{48}{pi} ), squaring that gives ( frac{2304}{pi^2} ). Multiplying by ( frac{sqrt{3}}{4} ) gives ( frac{576 sqrt{3}}{pi^2} ). Yep, that looks correct.So, to recap:- The radius of each circle is ( frac{24}{pi} ).- The side length of the equilateral triangle is ( frac{48}{pi} ).- The perimeter of the shaded region is 24.- The area of the triangle is ( frac{576 sqrt{3}}{pi^2} ).I think that's all. I don't see any mistakes in my reasoning, so I feel confident about these answers.**Final Answer**The perimeter of the shaded region is boxed{24} and the area of the triangle is boxed{dfrac{576sqrt{3}}{pi^2}}.