🧠:[<think>Okay, so I need to find the area of triangle MQR given some side lengths. Let me start by visualizing the problem. We have triangle MNP with sides MN=4, MP=5, and NP=6. Points Q and R are on sides MN and MP respectively, such that MQ=3 and MR=4. The question is asking for the area of triangle MQR. Hmm, since we have some lengths given, maybe I can use coordinates or some ratio of areas. But Heron's formula is mentioned in the problem statement. Let me recall Heron's formula: the area of a triangle with sides a, b, c is sqrt[s(s-a)(s-b)(s-c)], where s is the semi-perimeter (a+b+c)/2. Maybe I can use Heron's formula on triangle MNP first to find its area and then use some ratio to find the area of MQR.Let me compute the area of triangle MNP first. The sides are MN=4, MP=5, NP=6. So, the semi-perimeter s = (4+5+6)/2 = 15/2 = 7.5. Then the area would be sqrt[7.5*(7.5-4)*(7.5-5)*(7.5-6)] = sqrt[7.5*3.5*2.5*1.5]. Let me compute that step by step. First, 7.5*1.5 = 11.25 and 3.5*2.5 = 8.75. Then multiply 11.25*8.75. Let me compute 11*8.75 = 96.25 and 0.25*8.75=2.1875. So total is 96.25 + 2.1875 = 98.4375. Therefore, the area is sqrt(98.4375). Let me check what that is. sqrt(98.4375) is approximately 9.921567. But maybe it's a whole number? Let me check if 98.4375 is a perfect square. 9.9^2 is 98.01, 9.92^2 is about 98.4064, 9.921567^2 is exactly 98.4375. So the area is exactly sqrt(98.4375). Let me write that as sqrt(98.4375). Alternatively, maybe convert 98.4375 to a fraction. 98.4375 = 98 + 0.4375 = 98 + 7/16 = (98*16 + 7)/16 = (1568 +7)/16 = 1575/16. Therefore, the area is sqrt(1575/16) = (sqrt(1575))/4. Simplify sqrt(1575). 1575 factors into 25*63, so sqrt(25*63) = 5*sqrt(63) = 5*sqrt(9*7) = 5*3*sqrt(7) = 15*sqrt(7). So area is 15*sqrt(7)/4. So triangle MNP has area 15√7 /4. Okay, now that I have the area of triangle MNP, maybe I can relate the area of triangle MQR to it using some ratio. Since points Q and R are on sides MN and MP respectively, with MQ=3 and MR=4. Let me note the lengths of the sides. MN is 4, so MQ=3, so Q divides MN into MQ=3 and QN=1. Similarly, MP is 5, so MR=4, so R divides MP into MR=4 and RP=1. So, triangles MQR and MNP share the angle at M. Therefore, the ratio of their areas could be the product of the ratios of the sides adjacent to the common angle. That is, if two triangles share an angle, the ratio of their areas is equal to the product of the ratios of the sides enclosing that angle. So, for triangle MQR and MNP, they share angle at M. The sides enclosing angle M in MQR are MQ and MR, which are 3 and 4. In triangle MNP, the sides enclosing angle M are MN=4 and MP=5. Therefore, the ratio of areas should be (3/4)*(4/5) = 12/20 = 3/5. Wait, but hold on, is that correct? Wait, actually, the formula for the ratio of areas when two triangles share a common angle is (side1/side1')*(side2/side2') where side1 and side2 are the sides enclosing the common angle in one triangle, and side1' and side2' are the corresponding sides in the other triangle. So in this case, triangle MQR and triangle MNP share angle at M. The sides around angle M in MQR are MQ=3 and MR=4, and in MNP they are MN=4 and MP=5. Therefore, the ratio is (3/4)*(4/5) = 12/20 = 3/5. So the area of MQR is 3/5 of the area of MNP? Wait, but that would be 3/5*(15√7 /4) = (45√7)/20 = 9√7 /4. But let me check if this is correct. Alternatively, maybe I need to use coordinates. Let me try that method to verify. Let's place point M at the origin (0,0). Let's let MN be along the x-axis. So point N is at (4,0). Point P is somewhere such that MP=5 and NP=6. Wait, coordinates might complicate things, but let's try. Let me assign coordinates. Let’s set point M at (0,0). Let’s set point N at (4,0). Now, we need to find coordinates of point P such that MP=5 and NP=6. So point P is somewhere in the plane. Let’s suppose P has coordinates (x,y). Then, distance from M to P is 5, so x² + y² = 25. Distance from N to P is 6, so (x-4)² + y² = 36. Subtract the first equation from the second: (x-4)² + y² - x² - y² = 36 -25 => (x² -8x +16) -x² = 11 => -8x +16 =11 => -8x = -5 => x = 5/8. Then, substituting back into x² + y² =25: (25/64) + y² =25 => y² =25 -25/64 = (1600/64 -25/64) =1575/64 => y= sqrt(1575)/8 = (15√7)/8. So coordinates of P are (5/8, 15√7/8). Now, points Q and R are on MN and MP. Q is on MN, which is from M(0,0) to N(4,0). Since MQ=3, Q is at (3,0). R is on MP, from M(0,0) to P(5/8,15√7/8). Since MR=4, and MP=5, the coordinates of R can be found by moving 4/5 of the way from M to P. So, parameter t=4/5. So R has coordinates ( (4/5)*(5/8), (4/5)*(15√7/8) ) = ( (4/5)*(5/8), (4/5)*(15√7/8) ) = ( (4/8), (60√7 /40) ) = (0.5, (3√7)/2 ). So coordinates:- M: (0,0)- Q: (3,0)- R: (0.5, (3√7)/2 )Now, compute area of triangle MQR. Using coordinates, the area can be found via determinant formula. The formula is (1/2)| (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) |. Let's assign points:M: (0,0)Q: (3,0)R: (0.5, (3√7)/2 )Plug into the formula:Area = (1/2)| 0*(0 - (3√7)/2) + 3*((3√7)/2 - 0) + 0.5*(0 - 0) |= (1/2)| 0 + 3*(3√7)/2 + 0 |= (1/2)*(9√7/2)= 9√7/4So that's 9√7 divided by 4. Which matches the ratio method. So this seems correct. Alternatively, using vectors or base-height, but the coordinate method gives the same answer as the ratio method. So I think that's the correct answer. Therefore, the area of triangle MQR is 9√7/4.Final AnswerThe area of triangle ( MQR ) is boxed{dfrac{9sqrt{7}}{4}}.</think>To find the area of triangle ( MQR ), we start by calculating the area of triangle ( MNP ) using Heron's formula. The sides of triangle ( MNP ) are ( MN = 4 ), ( MP = 5 ), and ( NP = 6 ). The semi-perimeter ( s ) is:[s = frac{4 + 5 + 6}{2} = 7.5]Using Heron's formula, the area ( A ) is:[A = sqrt{s(s - a)(s - b)(s - c)} = sqrt{7.5 times 3.5 times 2.5 times 1.5}]Converting to fractions:[7.5 = frac{15}{2}, quad 3.5 = frac{7}{2}, quad 2.5 = frac{5}{2}, quad 1.5 = frac{3}{2}]Thus,[A = sqrt{left(frac{15}{2}right) left(frac{7}{2}right) left(frac{5}{2}right) left(frac{3}{2}right)} = sqrt{frac{15 times 7 times 5 times 3}{16}} = sqrt{frac{1575}{16}} = frac{15sqrt{7}}{4}]Next, we use the ratio of areas for triangles sharing a common angle. Triangles ( MQR ) and ( MNP ) share angle ( M ). The sides enclosing angle ( M ) in ( MQR ) are ( MQ = 3 ) and ( MR = 4 ), while in ( MNP ) they are ( MN = 4 ) and ( MP = 5 ). The ratio of areas is:[left( frac{3}{4} right) left( frac{4}{5} right) = frac{3}{5}]Thus, the area of triangle ( MQR ) is:[frac{3}{5} times frac{15sqrt{7}}{4} = frac{45sqrt{7}}{20} = frac{9sqrt{7}}{4}]Verification using coordinates confirms this result. Placing points ( M ), ( N ), ( P ), ( Q ), and ( R ) in a coordinate system and using the determinant method for area calculation also yields the same result.The area of triangle ( MQR ) is boxed{dfrac{9sqrt{7}}{4}}.,<think>Okay, so I need to find the area of triangle MQR given some side lengths and positions of points Q and R on the sides of triangle MNP. Let me start by visualizing the problem. We have triangle MNP with sides MN=4, MP=5, and NP=6. Points Q and R are on sides MN and MP respectively, such that MQ=3 and MR=4. So, Q divides MN into MQ=3 and QN=1 (since MN=4). Similarly, R divides MP into MR=4 and RP=1 (since MP=5). The task is to find the area of triangle MQR.Hmm, I remember that when dealing with areas of triangles with a common angle, the ratio of areas can sometimes be found using the ratio of the sides. Since both triangles MQR and MNP share the common angle at M, maybe the area of MQR can be related to the area of MNP through the ratios of the adjacent sides.First, let me compute the area of triangle MNP. They give all three sides: MN=4, MP=5, NP=6. Since we have all three sides, Heron's formula would be the way to go here. Heron's formula states that the area of a triangle with sides a, b, c is sqrt[s(s-a)(s-b)(s-c)], where s is the semi-perimeter.Calculating the semi-perimeter (s) of triangle MNP:s = (MN + MP + NP)/2 = (4 + 5 + 6)/2 = 15/2 = 7.5Now, applying Heron's formula:Area = sqrt[s(s - MN)(s - MP)(s - NP)] = sqrt[7.5*(7.5 - 4)*(7.5 - 5)*(7.5 - 6)]Calculating each term inside the square root:7.5 - 4 = 3.57.5 - 5 = 2.57.5 - 6 = 1.5So, Area = sqrt[7.5 * 3.5 * 2.5 * 1.5]Hmm, this might be a bit tedious to compute. Let me compute this step by step.First, multiply 7.5 and 3.5:7.5 * 3.5 = 26.25Then multiply 2.5 and 1.5:2.5 * 1.5 = 3.75Now multiply those two results:26.25 * 3.75Let me compute that:26.25 * 3 = 78.7526.25 * 0.75 = 19.6875Adding those together: 78.75 + 19.6875 = 98.4375So, the area is sqrt(98.4375). Let me compute that.First, note that 98.4375 is equal to 98.4375. Let's see if this is a perfect square or can be simplified.Alternatively, maybe there's a mistake here? Let me double-check the Heron's formula calculation.Wait, perhaps I should compute the exact value using fractions instead of decimals to see if it simplifies.Original semi-perimeter: 15/2Area = sqrt[(15/2)*(15/2 - 4)*(15/2 - 5)*(15/2 - 6)]Convert all terms to fractions with denominator 2:15/2 - 4 = 15/2 - 8/2 = 7/215/2 - 5 = 15/2 - 10/2 = 5/215/2 - 6 = 15/2 - 12/2 = 3/2So, Area = sqrt[(15/2)*(7/2)*(5/2)*(3/2)] = sqrt[(15*7*5*3)/(2^4)]Compute numerator: 15*7=105; 105*5=525; 525*3=1575Denominator: 2^4=16So, Area = sqrt(1575/16) = sqrt(1575)/4Simplify sqrt(1575):1575 = 25*63 = 25*9*7 = 25*9*7Therefore, sqrt(1575) = 5*3*sqrt(7) = 15*sqrt(7)Hence, Area = (15*sqrt(7))/4So, the area of triangle MNP is (15√7)/4. Got that.Now, how to find the area of triangle MQR? Since both triangles MQR and MNP share angle at M, maybe the ratio of their areas is equal to the product of the ratios of the sides adjacent to angle M.In triangle MNP, sides MN=4 and MP=5. In triangle MQR, sides MQ=3 and MR=4. So, the ratio of MQ/MN = 3/4 and the ratio of MR/MP = 4/5. If the area ratio is the product of these, then Area MQR / Area MNP = (3/4)*(4/5) = 3/5. Wait, is that correct?Wait, that seems too straightforward, but let me think. The formula for the area ratio when two triangles share a common angle is (side1 * side2) / (side1' * side2'), where side1 and side2 are the sides forming the common angle. So, in this case, angle at M, sides MN and MP in triangle MNP, and sides MQ and MR in triangle MQR. So, the ratio of the areas would be (MQ * MR * sin(theta)) / (MN * MP * sin(theta)) ) = (MQ/MN)*(MR/MP). Since sin(theta) cancels out.Therefore, Area MQR / Area MNP = (MQ/MN)*(MR/MP) = (3/4)*(4/5) = 3/5.Therefore, Area MQR = (3/5)*Area MNP = (3/5)*(15√7)/4 = (45√7)/20 = (9√7)/4.Wait, that seems straightforward, but let me check if this is accurate.Alternatively, maybe using coordinates. Let me try setting coordinates to compute the area.Let me place point M at the origin (0,0). Then, let me assign coordinates to points N and P such that MN=4 and MP=5. However, since we know all sides of triangle MNP, we need to set coordinates properly.Let me set point M at (0,0). Let me set point N along the x-axis, so N is at (4,0). Now, point P is somewhere in the plane. We need to find coordinates of P such that MP=5 and NP=6.Given that M is (0,0), N is (4,0), and P is (x,y). Then, distance from M to P is 5, so sqrt(x^2 + y^2) = 5, which implies x^2 + y^2 = 25.Distance from N to P is 6, so sqrt((x - 4)^2 + y^2) = 6, so (x - 4)^2 + y^2 = 36.Subtracting the first equation from the second:(x - 4)^2 + y^2 - (x^2 + y^2) = 36 - 25 => x^2 - 8x + 16 + y^2 - x^2 - y^2 = 11 => -8x + 16 = 11 => -8x = -5 => x = 5/8.Then, substituting back into x^2 + y^2 = 25:(25/64) + y^2 = 25 => y^2 = 25 - 25/64 = (1600/64 - 25/64) = 1575/64 => y = ±(sqrt(1575)/8) = ±(15√7)/8.So, coordinates of P are (5/8, 15√7/8) or (5/8, -15√7/8). Since triangles are typically considered in positive y unless specified otherwise, we'll take (5/8, 15√7/8).So now, coordinates:M: (0,0)N: (4,0)P: (5/8, 15√7/8)Points Q and R are on MN and MP respectively.Point Q is on MN, which is from M(0,0) to N(4,0). Since MQ=3, Q is 3 units from M along MN. Since MN is along the x-axis from (0,0) to (4,0), moving 3 units along x-axis gives Q at (3,0).Point R is on MP, from M(0,0) to P(5/8, 15√7/8). MR=4. Since MP has length 5, moving 4 units from M towards P. Wait, but in coordinates, how to find the coordinates of R?Wait, the length from M to P is 5. So, to find the point R that is 4 units from M along MP. Since the total length is 5, the coordinates of R can be found by moving 4/5 of the way from M to P.Parametrize MP. The vector from M to P is (5/8, 15√7/8). So, the unit vector in the direction of MP is (5/8)/5, (15√7/8)/5) = (1/8, 3√7/8). Wait, scaling the vector (5/8, 15√7/8) by 1/5 gives (1/8, 3√7/8). Therefore, moving 4 units along MP would be scaling this unit vector by 4? Wait, no. Wait, the vector from M to P is (5/8, 15√7/8), which has length 5. So, to move 4 units along MP from M, we need to scale the vector by 4/5.So, coordinates of R = M + (4/5)*(vector MP) = (0,0) + (4/5)*(5/8, 15√7/8) = (4/5 * 5/8, 4/5 * 15√7/8) = ( (20/40), (60√7/40) ) = (1/2, (3√7)/2 )Therefore, coordinates of R are (1/2, (3√7)/2 )So now, points of triangle MQR are:M: (0,0)Q: (3,0)R: (1/2, (3√7)/2 )Now, let's compute the area of triangle MQR using coordinates. The formula for area given three points (x1,y1), (x2,y2), (x3,y3) is | (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2 |.Plugging in the coordinates:x1 = 0, y1 = 0x2 = 3, y2 = 0x3 = 1/2, y3 = (3√7)/2Compute the terms:0*(0 - (3√7)/2) + 3*((3√7)/2 - 0) + (1/2)*(0 - 0)Simplify:0 + 3*(3√7/2) + 0 = (9√7)/2Take absolute value and divide by 2:Area = | (9√7)/2 | / 2 = (9√7)/4So, the area is 9√7/4. That matches the earlier result from the ratio method. Therefore, the area of triangle MQR is 9√7 divided by 4.Hence, the answer is boxed{dfrac{9sqrt{7}}{4}}.Final Answerboxed{dfrac{9sqrt{7}}{4}}</think>To find the area of triangle ( MQR ), we start by calculating the area of triangle ( MNP ) using Heron's formula. The sides of triangle ( MNP ) are ( MN = 4 ), ( MP = 5 ), and ( NP = 6 ).1. Calculate the semi-perimeter ( s ) of triangle ( MNP ): [ s = frac{4 + 5 + 6}{2} = 7.5 ]2. Apply Heron's formula to find the area of triangle ( MNP ): [ text{Area} = sqrt{s(s - MN)(s - MP)(s - NP)} = sqrt{7.5 times (7.5 - 4) times (7.5 - 5) times (7.5 - 6)} ] [ = sqrt{7.5 times 3.5 times 2.5 times 1.5} ] Converting to fractions: [ = sqrt{frac{15}{2} times frac{7}{2} times frac{5}{2} times frac{3}{2}} = sqrt{frac{15 times 7 times 5 times 3}{16}} = sqrt{frac{1575}{16}} = frac{15sqrt{7}}{4} ]3. Determine the ratio of the areas of triangles ( MQR ) and ( MNP ). Since both triangles share the common angle at ( M ), the ratio of their areas is the product of the ratios of the sides adjacent to angle ( M ): [ frac{text{Area of } triangle MQR}{text{Area of } triangle MNP} = left( frac{MQ}{MN} right) left( frac{MR}{MP} right) = left( frac{3}{4} right) left( frac{4}{5} right) = frac{3}{5} ]4. Calculate the area of triangle ( MQR ): [ text{Area of } triangle MQR = frac{3}{5} times frac{15sqrt{7}}{4} = frac{45sqrt{7}}{20} = frac{9sqrt{7}}{4} ]Thus, the area of triangle ( MQR ) is (boxed{dfrac{9sqrt{7}}{4}}).]

🔑:s_{Delta mathrm{MQR}}=frac{M Q}{M N} cdot frac{M R}{M P} cdot s_{Delta mathrm{MNP}} cdot## Answerfrac{9 sqrt{7}}{4}.

🧠:[<think>Okay, let's try to solve this equation: √[(x + 3)/(3x - 5)] + 1 = 2√[(3x - 5)/(x + 3)]. Hmm, radicals in the equation. I need to be careful here. Maybe I can simplify it by substitution. Let me see.First, let me write the equation again:√[(x + 3)/(3x - 5)] + 1 = 2√[(3x - 5)/(x + 3)].Notice that the two radicals are reciprocals of each other. If I let y = √[(x + 3)/(3x - 5)], then the other radical would be √[(3x - 5)/(x + 3)] = 1/y. That might simplify the equation. Let's try that substitution.Let y = √[(x + 3)/(3x - 5)]. Then the equation becomes:y + 1 = 2*(1/y)Multiply both sides by y to eliminate the denominator:y(y + 1) = 2Which is:y² + y = 2Then bringing all terms to one side:y² + y - 2 = 0Now, factor this quadratic equation. Let's see, factors of -2 that add up to 1... 2 and -1? Yes:(y + 2)(y - 1) = 0So the solutions are y = -2 or y = 1.But wait, y was defined as a square root, which is always non-negative. So y can't be -2. Therefore, only y = 1 is valid.So y = 1 implies:√[(x + 3)/(3x - 5)] = 1Square both sides:(x + 3)/(3x - 5) = 1Multiply both sides by (3x - 5):x + 3 = 3x - 5Bring variables to left and constants to right:x - 3x = -5 - 3-2x = -8Divide both sides by -2:x = 4But I need to check if x = 4 is valid in the original equation, since sometimes squaring can introduce extraneous solutions.Let me plug x = 4 into the original equation:Left side: √[(4 + 3)/(3*4 - 5)] + 1 = √[7/7] + 1 = √1 + 1 = 1 + 1 = 2Right side: 2√[(3*4 - 5)/(4 + 3)] = 2√[(12 - 5)/7] = 2√[7/7] = 2√1 = 2*1 = 2Both sides equal 2, so x = 4 is a valid solution.Wait, but the problem says to solve the equation. Is there another solution? When we set y = 1, we got x =4, but what about y = -2? But since y is a square root, y can't be negative, so y = -2 is invalid. So x =4 is the only solution.But just to be thorough, let's check if there are any other possibilities. Maybe the substitution missed something. Let's try solving without substitution.Original equation:√[(x + 3)/(3x - 5)] + 1 = 2√[(3x - 5)/(x + 3)]Let me denote A = √[(x + 3)/(3x - 5)], then the equation becomes:A + 1 = 2*(1/A)Which is the same as before. So the substitution was correct.Alternatively, cross-multiplying or other methods.Alternatively, cross-multiplying the radicals. Let's see.Let me write the equation again:√[(x + 3)/(3x - 5)] + 1 = 2√[(3x - 5)/(x + 3)].Multiply both sides by √[(x + 3)/(3x - 5)] to eliminate denominators. Wait, but the denominators are inside the radicals. Alternatively, let me cross-multiply.But perhaps squaring both sides. Let me try that.Let me isolate one radical first. Let's move the 1 to the other side:√[(x + 3)/(3x - 5)] = 2√[(3x - 5)/(x + 3)] - 1Now square both sides:[(x + 3)/(3x - 5)] = [2√[(3x - 5)/(x + 3)} - 1]^2Expand the right side:= 4*(3x -5)/(x + 3) - 4√[(3x -5)/(x + 3)} + 1So, left side: (x + 3)/(3x -5)Right side: 4*(3x -5)/(x +3) - 4√[(3x -5)/(x +3)} +1Hmm, this looks complicated. Let me denote B = √[(3x -5)/(x +3)}, then the equation becomes:Left side: (x +3)/(3x -5) = 1/B²Right side: 4B² -4B +1So equation is:1/B² = 4B² -4B +1Multiply both sides by B²:1 = 4B^4 -4B^3 + B²Bring all terms to left side:4B^4 -4B^3 + B² -1 =0Hmm, quartic equation. This seems more complicated than the substitution method. Maybe the substitution method was better.Alternatively, maybe there is a different approach. Let me think.Wait, the original substitution gave us a valid solution. But perhaps I should check if there are any other solutions. Wait, maybe when we squared both sides, we introduced extraneous roots. But in the substitution method, since we only got y =1 as valid, which led to x=4, perhaps that's the only solution.Alternatively, maybe there are restrictions on x. Let's check the domain of the original equation.First, the expressions inside the radicals must be positive.So, (x +3)/(3x -5) > 0 and (3x -5)/(x +3) >0. Since they are reciprocals, if one is positive, the other is also positive. So we can just consider (x +3)/(3x -5) >0.Which implies that both numerator and denominator are positive or both negative.Case 1: x +3 >0 and 3x -5 >0.So x > -3 and x >5/3. So x >5/3.Case 2: x +3 <0 and 3x -5 <0.So x < -3 and x <5/3. So x < -3.Therefore, domain is x >5/3 or x < -3.But when we found x=4, which is greater than 5/3, so valid. If there were a solution less than -3, we need to check if that's possible.But in the substitution method, y is a square root, so y >=0, and solving y gave only y=1. So x=4 is the only solution. But let's verify if x < -3 can yield a solution. Let's test x = -4, which is in the domain.Plug x = -4 into original equation:Left side: √[(-4 +3)/(3*(-4) -5)] +1 = √[(-1)/(-17)] +1 = √(1/17) +1 ≈ 0.2425 +1 ≈1.2425Right side: 2√[(3*(-4) -5)/(-4 +3)] = 2√[(-17)/(-1)] = 2√17 ≈ 2*4.123 ≈8.246Not equal, so x=-4 isn't a solution. So maybe no solutions in x < -3. Therefore, only x=4.Alternatively, let's check during the substitution step: when we set y =1, which gives x=4, and y cannot be negative, so only y=1.Hence, x=4 is the only solution.But just to be thorough, let's check if in the equation after substitution y +1 = 2/y, even though y is non-negative, maybe when we square or do other steps, we might have missed something. Wait, but we solved it correctly:y +1 = 2/yMultiply by y:y² + y =2y² + y -2=0Solution y= [-1 ±√(1 +8)]/2= [-1 ±3]/2. So y=1 or y=-2. Since y is non-negative, only y=1. Correct.Thus, x=4 is the only solution. Checked in original equation, works. So answer is x=4.Final AnswerThe solution is boxed{4}.</think>To solve the equation (sqrt{frac{x+3}{3x-5}} + 1 = 2 sqrt{frac{3x-5}{x+3}}), we start by making a substitution. Let (y = sqrt{frac{x+3}{3x-5}}). Then the equation becomes:[y + 1 = 2 cdot frac{1}{y}]Multiplying both sides by (y) to clear the fraction:[y^2 + y = 2]Rearranging into a standard quadratic equation form:[y^2 + y - 2 = 0]Factoring the quadratic equation:[(y + 2)(y - 1) = 0]This gives the solutions (y = -2) or (y = 1). Since (y) is a square root and must be non-negative, we discard (y = -2). Thus, (y = 1).Substituting back (y = 1):[sqrt{frac{x+3}{3x-5}} = 1]Squaring both sides:[frac{x+3}{3x-5} = 1]Solving for (x):[x + 3 = 3x - 5][-2x = -8][x = 4]We verify (x = 4) in the original equation. Substituting (x = 4):Left side: (sqrt{frac{4+3}{3 cdot 4 - 5}} + 1 = sqrt{1} + 1 = 2)Right side: (2 sqrt{frac{3 cdot 4 - 5}{4 + 3}} = 2 sqrt{1} = 2)Both sides are equal, confirming (x = 4) is a valid solution. Checking the domain, (x = 4) is within the valid domain (x > frac{5}{3}). There are no solutions in (x < -3) as verified by testing values.Thus, the solution is (boxed{4}).,<think>Okay, let's see. I need to solve this equation: √[(x+3)/(3x−5)] + 1 = 2√[(3x−5)/(x+3)]. Hmm, radicals in denominators and numerators. Maybe I can simplify this by substitution. Let me think. If I let y = √[(x+3)/(3x−5)], then the equation becomes y + 1 = 2*(1/y), since √[(3x−5)/(x+3)] is the reciprocal of y. That might work. Let me try that.So, substituting y = √[(x+3)/(3x−5)], the equation becomes:y + 1 = 2*(1/y)Multiply both sides by y to eliminate the denominator:y(y) + y*1 = 2So that's y² + y = 2Bring all terms to one side:y² + y - 2 = 0Now, solve the quadratic equation. Let's factor it. Looking for two numbers that multiply to -2 and add to 1. Hmm, 2 and -1 work: (y + 2)(y - 1) = 0. Wait, no. Wait, (y + 2)(y - 1) = y² - y + 2y - 2 = y² + y - 2. Yes, that's right. So factors are (y + 2)(y - 1) = 0.So, y = -2 or y = 1. But y is a square root, which is always non-negative. So y cannot be -2. Therefore, the only possible solution is y = 1.So, set y = 1:√[(x+3)/(3x−5)] = 1Square both sides:(x + 3)/(3x - 5) = 1Multiply both sides by (3x - 5):x + 3 = 3x - 5Bring all terms to one side:3x - 5 - x - 3 = 0 → 2x - 8 = 0 → 2x = 8 → x = 4.Wait, but I need to check if this is a valid solution. Sometimes when squaring both sides, extraneous solutions can pop up, so I need to verify.Original equation: √[(4 + 3)/(3*4 - 5)] + 1 = √[7/7] + 1 = √1 + 1 = 1 + 1 = 2.Right-hand side: 2√[(3*4 -5)/(4 +3)] = 2√[(12 -5)/7] = 2√[7/7] = 2√1 = 2*1 = 2.Left side equals right side, so x = 4 is valid.But wait, let me check the domain. The expressions inside the radicals must be positive.First, (x + 3)/(3x - 5) must be positive. Also, (3x - 5)/(x + 3) must be positive. Actually, since they are reciprocals, if one is positive, the other is positive as well. So, both fractions must be positive.So, for (x + 3)/(3x - 5) > 0, numerator and denominator have same sign.Case 1: x + 3 > 0 and 3x -5 >0 → x > -3 and x > 5/3. So x > 5/3.Case 2: x +3 <0 and 3x -5 <0 → x < -3 and x < 5/3. Since 5/3 is approximately 1.666, so x < -3.Therefore, the domain is x > 5/3 or x < -3.But x = 4 is in x >5/3, so that's okay.But wait, maybe there was another solution from y = -2? But since y is a square root, it can't be negative. So we can discard y = -2. Hence, only x =4 is the solution.But let me check if maybe during the substitution we missed something. Let me try plugging in x = -4, which is less than -3, to see if maybe there's another solution.Wait, but how did we get only x=4? Let's see. Suppose x < -3. Let's take x = -4. Then:Left side: sqrt[(-4 +3)/(3*(-4) -5)] +1 = sqrt[(-1)/(-17)] +1 = sqrt[1/17] +1 ≈ 0.242 +1 ≈1.242Right side: 2*sqrt[(3*(-4)-5)/(-4 +3)] = 2*sqrt[(-17)/(-1)] = 2*sqrt[17] ≈ 2*4.123 ≈8.246Not equal. So x=-4 is not a solution. Maybe there's another solution in x < -3? Let me check.Wait, but when we solved for y, we only got y =1. So even if x < -3, unless there's another solution from y =1, which would still require (x +3)/(3x -5) =1. So same equation. So x=4 is the only solution. Let's check another value in x < -3. Suppose x = -5.Left side: sqrt[(-5 +3)/(3*(-5)-5)] +1 = sqrt[(-2)/(-20)] +1 = sqrt[1/10] +1 ≈0.316 +1 ≈1.316Right side: 2*sqrt[(3*(-5)-5)/(-5 +3)] = 2*sqrt[(-20)/(-2)] = 2*sqrt[10] ≈2*3.162≈6.324Not equal. So seems like x=4 is the only solution.Wait, but maybe I made a mistake in substitution. Let me try solving the original equation without substitution.Original equation: sqrt[(x+3)/(3x−5)] +1 = 2*sqrt[(3x−5)/(x+3)]Let me denote sqrt[(x+3)/(3x−5)] as A. Then the equation becomes A +1 = 2*(1/A), same as before.So, same steps. So the only solution is x=4.But just to be thorough, maybe I can cross multiply or something else.Multiply both sides by sqrt[(x+3)/(3x−5)]:Left side: [sqrt((x+3)/(3x−5))]^2 + sqrt((x+3)/(3x−5)) = (x+3)/(3x−5) + sqrt((x+3)/(3x−5))Right side: 2*sqrt[(3x−5)/(x+3)]*sqrt[(x+3)/(3x−5)}] = 2*sqrt{ [(3x−5)/(x+3)] * [(x+3)/(3x−5)] } = 2*sqrt{1} = 2*1 = 2So equation becomes:(x+3)/(3x−5) + sqrt[(x+3)/(3x−5)] = 2Let me set z = sqrt[(x+3)/(3x−5)], so the equation becomes z² + z = 2 → z² + z -2=0, which is same as before. So z=1 or z=-2. Again, z must be non-negative, so z=1.Therefore, same solution x=4. So indeed, x=4 is the only solution.Wait, but maybe when I squared the equation, another solution could come. But in this case, substitution led directly to same quadratic, and only one solution.Alternatively, let me try solving the equation step by step without substitution.Original equation:sqrt[(x+3)/(3x−5)] +1 = 2*sqrt[(3x−5)/(x+3)}Let me move 1 to the other side:sqrt[(x+3)/(3x−5)} = 2*sqrt[(3x−5)/(x+3)} -1Square both sides:(x+3)/(3x−5) = [2*sqrt{(3x−5)/(x+3)} -1]^2Expand the right side:[2*sqrt{(3x−5)/(x+3)}]^2 - 2*2*sqrt{(3x−5)/(x+3)}*1 +1^2=4*(3x−5)/(x+3) -4*sqrt{(3x−5)/(x+3)} +1So left side is (x+3)/(3x−5). So equation becomes:(x+3)/(3x−5) = 4*(3x−5)/(x+3) -4*sqrt{(3x−5)/(x+3)} +1This seems complicated, but maybe let's bring all terms to one side:(x+3)/(3x−5) -4*(3x−5)/(x+3) +4*sqrt{(3x−5)/(x+3)} -1 =0Hmm, this might not be helpful. Alternatively, maybe let me set t = sqrt{(3x−5)/(x+3)}, then t² = (3x−5)/(x+3). Then, sqrt{(x+3)/(3x−5)} = 1/t. So original equation becomes:1/t +1 =2tMultiply both sides by t:1 + t = 2t²Bring all terms to one side:2t² - t -1 =0Solve quadratic equation: t = [1 ± sqrt(1 +8)] /4 = [1 ±3]/4So t=(1+3)/4=1 or t=(1-3)/4=-0.5But t is sqrt{(3x−5)/(x+3)}, which is non-negative, so t=1. So t=-0.5 is rejected.Thus, t=1:sqrt{(3x−5)/(x+3)}=1Square both sides:(3x−5)/(x+3) =1Multiply both sides by (x+3):3x−5 =x +33x -x =3 +52x=8 →x=4Same solution. So regardless of substitution method, same answer. Therefore, x=4 is the only solution.But let me check if there's any other possible solutions. For example, perhaps when we squared both sides, did we introduce any extraneous roots?In the first substitution method, when we set y = sqrt[(x+3)/(3x-5)], then got y=1, which leads to x=4, which checks out.In the second method, squaring both sides, but since we only had t=1 as valid, leading to x=4.Alternatively, perhaps if I approach differently. Let me let u = sqrt[(x+3)/(3x−5)], then the original equation is u +1 = 2*(1/u). Multiply both sides by u: u² + u =2. So u² +u -2=0, same as before, leading to u=1 or u=-2. Discard u=-2, so u=1, leading to x=4.Same result. So it's consistent.Therefore, I think x=4 is the only solution. Let me just verify once again.Original equation with x=4:Left side: sqrt[(4+3)/(12-5)] +1 = sqrt[7/7] +1 =1 +1=2Right side:2*sqrt[(12-5)/7]=2*sqrt[7/7]=2*1=2Yes, equal. So correct.Is there a possibility that even though the domain is x>5/3 or x < -3, but x=4 is the only solution? Since in the substitution steps, we only found x=4, and testing x < -3 didn't work. So yes, x=4 is the only solution.Therefore, the answer is x=4.Final AnswerThe solution is boxed{4}.</think>To solve the equation (sqrt{frac{x+3}{3x-5}} + 1 = 2 sqrt{frac{3x-5}{x+3}}), we start by making a substitution. Let (y = sqrt{frac{x+3}{3x-5}}). This transforms the equation into:[y + 1 = 2 cdot frac{1}{y}]Multiplying both sides by (y) to clear the denominator, we get:[y^2 + y = 2]Rearranging terms, we have the quadratic equation:[y^2 + y - 2 = 0]Factoring, we get:[(y + 2)(y - 1) = 0]Since (y) is a square root and must be non-negative, we discard (y = -2). Thus, (y = 1). Substituting back, we have:[sqrt{frac{x+3}{3x-5}} = 1]Squaring both sides, we get:[frac{x+3}{3x-5} = 1]Solving for (x), we multiply both sides by (3x - 5):[x + 3 = 3x - 5]Rearranging terms, we find:[2x = 8 implies x = 4]We verify the solution by substituting (x = 4) back into the original equation. Both sides simplify to 2, confirming the solution is correct. Checking the domain, we ensure the expressions inside the radicals are positive, which they are for (x = 4).Thus, the solution is (boxed{4}).]

🔑:Solution:sqrt{frac{x+3}{3 x-5}}+1=2 sqrt{frac{3 x-5}{x+3}}frac{x+3}{3 x-5}>0t=frac{x+3}{3 x-5}, sqrt{t}+1=frac{2}{sqrt{t}}, frac{t+sqrt{t}-2}{sqrt{t}}=0, sqrt{t}=frac{-1 pm sqrt{1+8}}{2}=frac{-1 pm 3}{2}=left[begin{array}{l}-2 1end{array}right.t=1, frac{x+3}{3 x-5}=1,3 x-5=x+3,2 x=8, x=4. Verification: sqrt{frac{7}{7}}+1=2 sqrt{frac{7}{7}}. Answer: x=4.

🧠:[<think>Okay, let's see. So the problem says that from December to January, excluding the first and last day, each day's minimum temperature is the sum of the previous and next day's temperatures. We know that December 3rd was 5 degrees and January 31st was 2 degrees. We need to find the temperature on December 25th.Hmm, first, let me parse this. So the period in question is from December 1st to January 31st, right? Because December and January are the last months, and the period is December and January. Wait, but the problem says "the last months of December and January". Wait, depending on when this is, but maybe it's a span from December 1 to January 31. But the days are considered as a continuous period? So excluding the first day (December 1) and the last day (January 31), each day's temperature is the sum of the previous and next day's temperatures. Wait, but the first day is excluded, meaning December 1st is excluded, and the last day, January 31st, is excluded. So all the days in between, from December 2nd to January 30th, have temperatures that are the sum of the previous and next day's temperatures.But wait, we are given that on December 3rd the temperature was 5 degrees, and on January 31st it was 2 degrees. Wait, but January 31st is the last day, which is excluded from the condition. So maybe the condition applies from December 2nd to January 30th. So the first day is December 1st, which is excluded, and the last day is January 31st, which is excluded. So in between, December 2nd to January 30th, each day's temperature is the sum of the previous and next day's temperatures.But then the given data points are December 3rd (which is within the period) and January 31st (which is excluded). So maybe the temperatures on the excluded days (Dec 1 and Jan 31) are not governed by the condition, but we have some data points on days that are included (Dec 3) and excluded (Jan 31). Wait, the problem states: "on all the days of this period (excluding the first and last day), the minimum temperature was the sum of the minimum temperature of the previous day and the next day." So the period is from Dec 1 to Jan 31. But excluding the first (Dec 1) and last day (Jan 31), so the days from Dec 2 to Jan 30 inclusive. Each of these days (Dec 2 to Jan 30) has a temperature equal to the sum of the previous day and the next day.So given that, we have a recursive relationship for each day from Dec 2 to Jan 30: T_n = T_{n-1} + T_{n+1}, where T_n is the temperature on day n.But given that, we can rearrange the equation: T_{n+1} = T_n - T_{n-1}. So the temperature on day n+1 is equal to the temperature on day n minus the temperature on day n-1. That looks like a linear recurrence relation.Wait, if we have T_{n+1} = T_n - T_{n-1}, that's similar to the Fibonacci sequence, but with a subtraction. Let's see. If we define the recurrence relation as T_{n+1} = T_n - T_{n-1}, then perhaps we can model this as a linear recurrence and find a general solution.Given that, we can set up the characteristic equation. Let me recall that for linear recurrences of the form T_{n+1} = a*T_n + b*T_{n-1}, we use characteristic equations. In this case, the recurrence is T_{n+1} = T_n - T_{n-1}, so the characteristic equation would be r^2 = r - 1. Solving that: r^2 - r + 1 = 0. The roots would be [1 ± sqrt(1 - 4)] / 2 = [1 ± sqrt(-3)] / 2. So complex roots: (1 ± i√3)/2. These are complex numbers with magnitude 1 and angle π/3. So the solution would involve terms like cos(nπ/3) and sin(nπ/3), multiplied by constants.But since the roots are complex and have magnitude 1, the solution is periodic with period 6, because the angle is π/3, so 2π / (π/3) = 6. So the sequence should repeat every 6 days. Wait, that might be useful.Alternatively, since the recurrence is T_{n+1} = T_n - T_{n-1}, perhaps we can compute the terms step by step and see if there's a periodic pattern.Let me test this. Suppose we have T_1 and T_2. Then T_3 = T_2 - T_1, T_4 = T_3 - T_2 = (T_2 - T_1) - T_2 = -T_1, T_5 = T_4 - T_3 = (-T_1) - (T_2 - T_1) = -T_2, T_6 = T_5 - T_4 = (-T_2) - (-T_1) = -T_2 + T_1, T_7 = T_6 - T_5 = (-T_2 + T_1) - (-T_2) = T_1, T_8 = T_7 - T_6 = T_1 - (-T_2 + T_1) = T_1 + T_2 - T_1 = T_2. So after T_7 and T_8, we get back to T_1 and T_2. Therefore, the sequence is periodic with period 6. So every 6 days, the temperatures repeat.So if the recurrence relation leads to a period of 6, then the temperatures cycle every 6 days. Therefore, if we can figure out the starting temperatures, the entire sequence would repeat every 6 days.But in our problem, we are given two specific temperatures: December 3rd is 5 degrees, and January 31st is 2 degrees. However, January 31st is the last day of the period, which is excluded from the recurrence condition. Wait, so January 31st is outside the period where the temperatures follow the recurrence. Therefore, the temperature on January 31st is just given as 2 degrees, but it's not part of the sequence governed by the recurrence. Similarly, December 1st is excluded as the first day.Wait, perhaps we need to model the entire period from December 1st to January 31st, but with the middle days (Dec 2 to Jan 30) following the recurrence.But since we have a recurrence relation, perhaps we can model the temperatures from Dec 2 to Jan 30 as a sequence where each term is determined by the previous two terms. But we are given two specific temperatures: Dec 3 is 5 degrees, and Jan 31 is 2 degrees.Wait, but Jan 31 is the last day, which is excluded from the recurrence. So the temperature on Jan 31 is independent. However, the temperature on Jan 30 is part of the recurrence. Let me try to clarify.Let me list the days:- December 1: excluded (first day)- December 2: included (follows recurrence)- December 3: included, temperature 5°C...- January 30: included- January 31: excluded, temperature 2°CSo from Dec 2 to Jan 30, each day's temperature is the sum of the previous and next days. Wait, but according to the problem statement: "on all the days of this period (excluding the first and last day), the minimum temperature was the sum of the minimum temperature of the previous day and the next day."Wait, the wording is a bit confusing. It says: "the minimum temperature was the sum of the minimum temperature of the previous day and the next day." So for each day (excluding first and last), T_n = T_{n-1} + T_{n+1}. So rearranged, T_{n+1} = T_n - T_{n-1}. Which is the same recurrence as before.So the recurrence is T_{n+1} = T_n - T_{n-1}. Therefore, starting from some initial terms, the sequence progresses with this recurrence. However, the problem gives us T_dec3 = 5, and T_jan31 = 2. But T_jan31 is outside the recurrence period. So perhaps the recurrence runs from Dec 2 to Jan 30, and the temperatures on Dec 1 and Jan 31 are boundary conditions?But the problem states that Dec 1 and Jan 31 are excluded, so their temperatures are not determined by the recurrence. But we are given that on Dec 3, which is inside the recurrence period, the temperature is 5°, and on Jan 31, which is excluded, the temperature is 2°. So perhaps Jan 31's temperature is a boundary condition that affects Jan 30's temperature via the recurrence?Wait, let's think. If Jan 30 is part of the recurrence period, then according to the recurrence, T_jan30 = T_jan29 + T_jan31. But T_jan31 is given as 2°, so T_jan30 = T_jan29 + 2. But T_jan30 is also governed by the recurrence relation. Wait, no. Wait, the recurrence is for each day in the period (Dec 2 to Jan 30), their temperature is the sum of the previous and next day. Wait, hold on. If day n is in Dec 2 to Jan 30, then T_n = T_{n-1} + T_{n+1}. So, for example, T_dec2 = T_dec1 + T_dec3. But T_dec1 is excluded (the first day), so its temperature is unknown? Wait, this seems conflicting.Wait, perhaps the problem is worded such that for each day in the period (excluding first and last), the temperature is equal to the sum of the temperatures of the previous day and the next day. So, for day n (excluding first and last), T_n = T_{n-1} + T_{n+1}.Therefore, we have a system of equations where each T_n (from Dec 2 to Jan 30) is equal to T_{n-1} + T_{n+1}.This is a system that could be solved if we have boundary conditions. But we only have two temperatures given: Dec 3 is 5°, and Jan 31 is 2°. So Dec 3 is within the period, so T_dec3 = 5, and Jan 31 is outside, so T_jan31 = 2. But how does Jan 31 relate?Wait, let's consider the days:Total days from Dec 1 to Jan 31: December has 31 days, January has 31 days, so total days is 31 + 31 = 62 days. But the period is from Dec 1 to Jan 31, so 62 days. Excluding the first and last day, that's days 2 to 61. So days 2 to 61 (which correspond to Dec 2 to Jan 30) are governed by the recurrence T_n = T_{n-1} + T_{n+1}. Therefore, for each n from 2 to 61, T_n = T_{n-1} + T_{n+1}.But we need to convert the dates to day numbers. Let's consider day 1 as Dec 1, day 2 as Dec 2, ..., day 31 as Dec 31, day 32 as Jan 1, ..., day 62 as Jan 31.Given that, day 3 (Dec 3) has T_3 = 5, and day 62 (Jan 31) has T_62 = 2. But days 2 to 61 follow T_n = T_{n-1} + T_{n+1}, which rearranged is T_{n+1} = T_n - T_{n-1}. So starting from day 2, the recurrence is T_{n+1} = T_n - T_{n-1}.But we need to know T_2 and T_3 to compute T_4, but T_3 is given as 5. Wait, but T_3 is part of the recurrence. Let me see. For n=3: T_3 = T_2 + T_4. But T_3 is 5, so 5 = T_2 + T_4. But also, from the recurrence T_4 = T_3 - T_2. Wait, substituting T_4 from the recurrence into the equation: 5 = T_2 + (T_3 - T_2) => 5 = T_2 + 5 - T_2 => 5=5. That's a tautology. Hmm, so maybe the equations are dependent.Alternatively, perhaps we need to model this as a system of linear equations. For each day from 2 to 61, we have T_n = T_{n-1} + T_{n+1}. So for day 2: T_2 = T_1 + T_3. But T_1 is Dec 1, which is excluded, so we don't know T_1. Similarly, for day 61 (Jan 30): T_61 = T_60 + T_62. But T_62 is given as 2. So T_61 = T_60 + 2.However, since we don't know T_1 and T_62 is known, we might have a system with 60 equations (from day 2 to day 61) but with variables T_1 to T_61, but T_62 is known. But we are given T_3 = 5. So we have two known values: T_3 = 5 and T_62 = 2.This seems complicated. However, since the recurrence T_{n+1} = T_n - T_{n-1} is linear and homogeneous, the solution should be determined by the initial conditions. But since we have a boundary condition in the middle (T_3 = 5) and at the end (T_62 = 2), this might form a system that can be solved.Alternatively, maybe we can use the periodicity we noticed earlier. The recurrence T_{n+1} = T_n - T_{n-1} has a period of 6. Let's verify:Suppose we start with T_1 = a and T_2 = b.Then:T_3 = T_2 - T_1 = b - aT_4 = T_3 - T_2 = (b - a) - b = -aT_5 = T_4 - T_3 = (-a) - (b - a) = -bT_6 = T_5 - T_4 = (-b) - (-a) = -b + aT_7 = T_6 - T_5 = (-b + a) - (-b) = aT_8 = T_7 - T_6 = a - (-b + a) = bSo T_7 = a and T_8 = b, which are the same as T_1 and T_2. Therefore, the sequence is periodic with period 6.Therefore, the temperatures repeat every 6 days. Therefore, if we can figure out the temperatures for days 1 to 6, the rest of the days follow the same pattern.But in our problem, we have a fixed date: December 25th. Let's convert December 25th to the day number. Since December has 31 days, December 25th is day 25. Therefore, in the 62-day period, day 25 is December 25th.But since the sequence is periodic with period 6, the temperature on day n is equal to the temperature on day n mod 6. However, this is only true if the entire sequence is governed by the recurrence relation with the same initial conditions. But in our case, we have specific temperatures given on day 3 (5 degrees) and day 62 (2 degrees). Wait, day 62 is Jan 31, which is outside the recurrence period, so maybe it affects day 61 (Jan 30). Let's see.Given that, let's try to model this.First, note that the recurrence T_{n+1} = T_n - T_{n-1} leads to a periodicity of 6. Therefore, if the sequence is periodic every 6 days, then temperatures on days 1, 7, 13, etc., are the same. Similarly for other days. Therefore, if we can express day 3 and day 61 in terms of the periodicity, we can find a relationship.But let's see. Let's number the days from 1 to 62 as before. Days 2 to 61 follow the recurrence. We know T_3 = 5 and T_62 = 2. Also, T_61 = T_60 + T_62 (from the recurrence for day 61: T_61 = T_60 + T_62). Since T_62 = 2, T_61 = T_60 + 2.But from the recurrence relation, for day 61: T_61 = T_60 + T_62. But T_61 is also equal to T_61 = T_60 - T_59 (from the general recurrence T_{n} = T_{n-1} - T_{n-2} if we shift indices). Wait, no. Let me check.Wait, the recurrence is T_{n+1} = T_n - T_{n-1}. Therefore, for day 61 (n=61), T_{62} = T_{61} - T_{60}. But T_{62} is given as 2, so 2 = T_{61} - T_{60}. But from the equation for day 61 (from the problem's condition), T_{61} = T_{60} + T_{62}. Substituting T_{62} = 2, we have T_{61} = T_{60} + 2. But from the recurrence relation, T_{62} = T_{61} - T_{60} => 2 = T_{61} - T_{60}. Therefore, combining these two equations:From problem's condition: T_{61} = T_{60} + 2From recurrence relation: T_{61} - T_{60} = 2These are the same equations. So no new information. So that doesn't help.But perhaps we can model the entire sequence from day 1 to day 62, knowing that days 2 to 61 follow the recurrence, and days 1 and 62 are unknowns except for T_3 = 5 and T_62 = 2.But given the periodicity, perhaps we can write the temperature on day n as a function of the temperatures on days 1 and 2.Since the sequence is periodic every 6 days, the temperatures repeat every 6 days. Therefore, the temperature on day n is equal to the temperature on day n mod 6. However, this depends on the initial conditions. Let's assume that the temperatures cycle every 6 days. Therefore, if we can find the temperatures for days 1 to 6, the rest follow the cycle.But we have T_3 = 5. Let's suppose that the temperatures cycle every 6 days. Then, T_3 = T_9 = T_15 = ... = 5. Similarly, if day 3 is 5, then every 6 days after that is also 5. However, day 62 is Jan 31, which is day 62. Let's compute 62 mod 6: 6*10=60, so 62-60=2. So day 62 is equivalent to day 2 in the cycle. But day 62's temperature is 2, which would imply that day 2's temperature is also 2. But day 2 is part of the recurrence period. Wait, but day 62 is excluded from the recurrence period. Hmm, maybe this line of thinking isn't correct.Alternatively, maybe we can model this as a linear system. Let me try to see.Let me note that the recurrence T_{n+1} = T_n - T_{n-1} can be written in matrix form or using linear algebra. The general solution of such a linear recurrence can be expressed in terms of the roots of the characteristic equation, which we found to be complex with period 6. Therefore, the general solution is T_n = α * (cos(nπ/3) + i sin(nπ/3)) + β * (cos(nπ/3) - i sin(nπ/3)), which simplifies to T_n = C * cos(nπ/3) + D * sin(nπ/3). However, since the sequence must be real, we can write it in terms of cosine and sine with coefficients.But given that, the general solution is a combination of cos(nπ/3) and sin(nπ/3). However, this might be complicated. Alternatively, since we have a period of 6, we can express the temperature on day n as T_n = A * cos(2πn/6) + B * sin(2πn/6) = A * cos(πn/3) + B * sin(πn/3). But this is similar to what I had before.Alternatively, since the period is 6, the temperature on day n is equal to the temperature on day n + 6. Therefore, if we can figure out the temperatures for days 1 to 6, the rest of the days follow the same pattern.But we need to connect this to the given data: T_3 = 5 and T_62 = 2. Let's see. If the sequence is periodic with period 6, then T_62 = T_{62 mod 6} = T_{62 - 6*10} = T_2. Therefore, T_62 = T_2. But T_62 is given as 2, so T_2 = 2. Also, T_3 = 5. So in the periodic sequence, T_3 = 5. Therefore, in days 1 to 6:T_1 = ?T_2 = 2T_3 = 5T_4 = T_3 - T_2 = 5 - 2 = 3T_5 = T_4 - T_3 = 3 - 5 = -2T_6 = T_5 - T_4 = -2 - 3 = -5T_7 = T_6 - T_5 = -5 - (-2) = -3But wait, T_7 should be equal to T_1 if the period is 6. Wait, according to the earlier calculation, the sequence is periodic with period 6, so T_7 = T_1, T_8 = T_2, etc. But in this case, if T_1 is unknown, but from the recurrence, T_7 = T_6 - T_5 = -5 - (-2) = -3, so T_7 = -3. Therefore, if the period is 6, T_1 should be equal to T_7, so T_1 = -3. Then:T_1 = -3T_2 = 2T_3 = 5T_4 = 3T_5 = -2T_6 = -5T_7 = -3 (same as T_1)T_8 = 2 (same as T_2)So the period is indeed 6. Therefore, if T_2 = 2 and T_3 = 5, the rest of the temperatures follow as above, and the sequence repeats every 6 days. Then, since day 62 is equivalent to day 2 (62 mod 6 = 2), T_62 = T_2 = 2, which matches the given data. Therefore, this seems consistent.But wait, in this scenario, we have T_3 = 5 as given, and T_62 = 2, which is T_2 = 2. So this seems to fit. Therefore, the entire temperature sequence is periodic with period 6, starting from day 1 (T_1 = -3), day 2 (T_2 = 2), day 3 (5), etc.But the problem is to find the temperature on December 25th, which is day 25. Let's compute day 25 mod 6. 6*4=24, so 25-24=1. Therefore, day 25 is equivalent to day 1 in the cycle. Therefore, T_25 = T_1 = -3 degrees. But that seems odd; could the temperature be negative? The problem doesn't specify that temperatures are positive. However, the given temperatures are 5 and 2, which are positive. But perhaps in this hypothetical scenario, temperatures can be negative.But let's verify. Let's list the temperatures for the first few days:Day 1: -3 (Dec 1, excluded)Day 2: 2 (Dec 2)Day 3: 5 (Dec 3)Day 4: 3 (Dec 4)Day 5: -2 (Dec 5)Day 6: -5 (Dec 6)Day 7: -3 (Dec 7)Day 8: 2 (Dec 8)Day 9: 5 (Dec 9)Day 10: 3 (Dec 10)Day 11: -2 (Dec 11)Day 12: -5 (Dec 12)Day 13: -3 (Dec 13)Day 14: 2 (Dec 14)... and so on.So every 6 days, the pattern repeats. Then December 25th is day 25. Let's compute 25 mod 6. 6*4=24, 25-24=1. Therefore, day 25 corresponds to day 1 in the cycle. But day 1 is excluded, and its temperature is -3. However, day 25 is part of the recurrence period (Dec 25 is between Dec 2 and Jan 30). Wait, day 25 is December 25th, which is within the period Dec 2 to Jan 30 (day 2 to day 61). Therefore, day 25 is included in the recurrence, so according to our periodic model, T_25 = T_1 = -3. But T_1 is excluded, but in our model, the sequence is periodic, so day 25 would have the same temperature as day 1. However, day 1 is excluded, but the temperature on day 25 is part of the recurrence. Wait, but if the period is 6, then day 25 is day 1 in the cycle, but day 1's temperature is set to -3, which is not part of the recurrence. Wait, maybe there's confusion here.Wait, the problem states that the period is from Dec 1 to Jan 31, excluding the first and last day. Therefore, days 2 to 61 (Dec 2 to Jan 30) are included. So day 25 (Dec 25) is day 25 in the overall numbering, which is within the recurrence period. However, according to our periodic model, day 25 corresponds to day 1 in the cycle (since 25 mod 6 = 1). But day 1 is excluded, so how does that translate?Wait, maybe the cycle starts from day 2. Because days 2 to 61 follow the recurrence. If the period is 6, then starting from day 2, every 6 days, the temperatures repeat. Therefore, day 2, 8, 14, ..., 62 would be the same. But day 62 is excluded. Wait, perhaps we need to adjust the cycle.Alternatively, since the recurrence leads to a period of 6, and the temperatures from day 2 onward follow that period, then starting from day 2, the temperatures cycle every 6 days. Therefore, day 2, 8, 14, etc., have the same temperature. Similarly, day 3, 9, 15, etc., have the same temperature. So December 25th is day 25. Let's compute how many days after day 2 is day 25.Day 25 - day 2 = 23 days. 23 divided by 6 is 3 with a remainder of 5. So 6*3=18, 23-18=5. Therefore, day 25 is 5 days after day 2 + 18 days (which is day 20). Therefore, day 25 corresponds to position 5 in the cycle. Since day 2 is position 0 in the cycle:- day 2: position 0: temperature 2- day 3: position 1: temperature 5- day 4: position 2: temperature 3- day 5: position 3: temperature -2- day 6: position 4: temperature -5- day 7: position 5: temperature -3Then the cycle repeats:- day 8: position 0: temperature 2Therefore, day 25 is 23 days after day 2. 23 divided by 6 is 3 full cycles (18 days) plus 5 days. Therefore, starting from day 2, adding 18 days gets us to day 20 (day 2 + 18 = day 20). Then adding 5 days: day 21, 22, 23, 24, 25. So day 25 is 5 days into the cycle. Looking at the cycle positions:Position 0: day 2, 8, 14, 20, 26, ...Position 1: day 3, 9, 15, 21, 27, ...Position 2: day 4, 10, 16, 22, 28, ...Position 3: day 5, 11, 17, 23, 29, ...Position 4: day 6, 12, 18, 24, 30, ...Position 5: day 7, 13, 19, 25, 31, ...Therefore, day 25 is position 5 in the cycle, which corresponds to temperature -3 degrees. So the temperature on December 25th would be -3 degrees.But wait, this is the same as saying day 25 is equivalent to day 7, which is position 5, which has temperature -3. However, in our earlier list, day 7 is December 7th, which is part of the recurrence period and has temperature -3. Then day 13, 19, 25, etc., also have -3.But the problem didn't specify any other temperatures except December 3rd (5°) and January 31st (2°). According to this model, the temperature on January 31st (day 62) is 2°, which is the same as day 2 (Dec 2), which is 2°, so that checks out. December 3rd is day 3, which is 5°, which is position 1 in the cycle (5°), which also checks out. Therefore, this seems consistent.However, the answer of -3 degrees seems unexpected. Is there a way to confirm this?Alternatively, maybe the problem assumes that the temperatures are all positive, and there's a different approach. Let me think again.Wait, the recurrence T_{n+1} = T_n - T_{n-1} with T_3 = 5 and T_62 = 2. Let's try to traverse the sequence from both ends.Starting from day 3: T_3 = 5.We know T_{n+1} = T_n - T_{n-1}. So for day 3: T_4 = T_3 - T_2. But we don't know T_2 yet.Similarly, starting from the end: day 61 (Jan 30) is related to day 62 (Jan 31) by T_61 = T_60 + T_62. Since T_62 = 2, T_61 = T_60 + 2. Also, from the recurrence, T_{61} = T_{60} - T_{59}. So T_{60} - T_{59} = T_{60} + 2 => -T_{59} = 2 => T_{59} = -2.Similarly, T_{60} = T_{59} - T_{58} => T_{60} = -2 - T_{58}But T_{60} = T_{59} - T_{58} = -2 - T_{58}But we also have T_{60} + 2 = T_{61} = T_{60} - T_{59} = T_{60} - (-2) = T_{60} + 2, which gives us T_{61} = T_{60} + 2. Therefore, consistent.So proceeding backward:T_{62} = 2T_{61} = T_{60} + 2T_{61} = T_{60} - T_{59} => T_{60} + 2 = T_{60} - T_{59} => T_{59} = -2T_{59} = -2T_{59} = T_{58} - T_{57} => -2 = T_{58} - T_{57} => T_{58} = T_{57} - 2T_{58} = T_{57} - T_{56} => T_{57} - 2 = T_{57} - T_{56} => T_{56} = 2T_{56} = 2T_{56} = T_{55} - T_{54} => 2 = T_{55} - T_{54} => T_{55} = T_{54} + 2T_{55} = T_{54} - T_{53} => T_{54} + 2 = T_{54} - T_{53} => T_{53} = -2T_{53} = -2Hmm, so going backward, we see a pattern: T_{59} = -2, T_{56} = 2, T_{53} = -2, T_{50} = 2, etc., decreasing by 3 each time. Similarly, perhaps every 3 days, the temperature alternates between -2 and 2.Similarly, going forward from day 3:T_3 = 5T_4 = T_3 - T_2 = 5 - T_2T_5 = T_4 - T_3 = (5 - T_2) - 5 = -T_2T_6 = T_5 - T_4 = (-T_2) - (5 - T_2) = -5T_7 = T_6 - T_5 = (-5) - (-T_2) = -5 + T_2T_8 = T_7 - T_6 = (-5 + T_2) - (-5) = T_2T_9 = T_8 - T_7 = T_2 - (-5 + T_2) = 5So T_9 = 5, which is the same as T_3. Therefore, the sequence from day 3 onward has a period of 6 as well. Therefore, days 3, 9, 15, ..., 57, 63 would all be 5°, but day 63 is beyond our period.But since our period ends at day 62, let's see how many days are there from day 3 to day 61.Wait, days 3 to 61 inclusive is 59 days. 59 divided by 6 is 9 full cycles (54 days) with a remainder of 5 days. So starting from day 3, adding 54 days gets us to day 57. Then 5 more days is day 62, but day 62 is excluded. Therefore, day 61 is day 57 + 5 = day 62 -1 = day 61. Therefore, the temperature at day 61 would be the same as day 3 + 54 + 5 = day 3 + 59 days. Wait, this might not be helpful.Alternatively, since going backward from day 62, we found T_{59} = -2, T_{56} = 2, T_{53} = -2, etc., every 3 days alternating between -2 and 2. So T_{59}=-2, T_{56}=2, T_{53}=-2, T_{50}=2,..., until we reach day 3.Similarly, day 3 is 5°, so let's see how many steps we can go back from day 59 to day 3.From day 59 to day 3 is 59 - 3 = 56 days. Since we're stepping back 3 days each time (from 59 to 56 to 53, etc.), 56 divided by 3 is 18 with remainder 2. So 3*18=54, 56-54=2. So stepping back 18 times would get us to day 59 - 3*18 = day 59 - 54 = day 5. Then another step back would be day 5 -3 = day 2, which is included in the recurrence. But we have T_5 = -T_2 from earlier.But wait, from the backward pass, we have T_{59} = -2, T_{56}=2, T_{53}=-2,..., up to T_5 = -2 (since 59 - 3*18 = 59 -54=5). Therefore, T_5 = -2. But from the forward equations, T_5 = -T_2. Therefore, -T_2 = -2 => T_2 = 2. Which matches our earlier conclusion that T_2 = 2.Therefore, T_2 = 2. Then, from day 3 onward:T_3 =5T_4 = 5 - T_2 = 5 - 2 = 3T_5 = -T_2 = -2T_6 = -5T_7 = -5 + T_2 = -5 + 2 = -3T_8 = T_2 = 2T_9 =5So this loops every 6 days. Therefore, days 3,9,15,...,57,63: 5°, days 4,10,16,...,58: 3°, days 5,11,17,...,59: -2°, days 6,12,18,...,60: -5°, days 7,13,19,...,61: -3°, days 8,14,20,...,62:2°, but day 62 is given as 2°, which matches.So now, December 25th is day 25. Let's find the temperature on day 25.From the forward sequence:Day 3:5Day 4:3Day 5:-2Day 6:-5Day 7:-3Day 8:2Day 9:5Day 10:3Day 11:-2Day 12:-5Day 13:-3Day 14:2Day 15:5Day 16:3Day 17:-2Day 18:-5Day 19:-3Day 20:2Day 21:5Day 22:3Day 23:-2Day 24:-5Day 25:-3Therefore, day 25 is -3 degrees. So the minimum temperature on December 25th is -3 degrees.But the problem didn't specify that temperatures can be negative. However, logically, the recurrence allows for negative temperatures, and the given data doesn't restrict it. Therefore, the answer is -3 degrees.But to confirm, let's check day 25 in the backward pass.From the backward pass, we have T_{59}=-2, T_{56}=2, T_{53}=-2, T_{50}=2, T_{47}=-2, T_{44}=2, T_{41}=-2, T_{38}=2, T_{35}=-2, T_{32}=2, T_{29}=-2, T_{26}=2, T_{23}=-2, T_{20}=2, T_{17}=-2, T_{14}=2, T_{11}=-2, T_{8}=2, T_{5}=-2, T_{2}=2.So day 25: let's see. From day 59 to day 25 is 59 -25=34 days. Stepping back 3 days each time: 34 /3=11 with remainder 1. 3*11=33, 34-33=1. So day 25 = day 59 - 3*11 -1= day 59 -34= day 25. But this might not help. Alternatively, from day 25, moving forward 3 days each time:Day 25: ?Day 28: ?But this might not be helpful. Alternatively, from the forward sequence, day 25 is -3. Which is consistent with our earlier calculation.Therefore, the answer is boxed{-3}.Final Answerboxed{-3}</think>Given the problem, we need to determine the minimum temperature on December 25th, knowing that the temperature on each day (excluding the first and last day) is the sum of the temperatures of the previous and next days. We are given that the temperature on December 3rd is 5 degrees and on January 31st is 2 degrees.1. Understanding the Recurrence Relation: - The temperature on each day ( n ) (from December 2nd to January 30th) follows the relation ( T_n = T_{n-1} + T_{n+1} ). - Rearranging this, we get ( T_{n+1} = T_n - T_{n-1} ), which is a linear recurrence relation.2. Periodicity of the Recurrence Relation: - The recurrence relation ( T_{n+1} = T_n - T_{n-1} ) has a period of 6 days. This means the temperature sequence repeats every 6 days.3. Given Temperatures: - December 3rd (day 3) has a temperature of 5 degrees. - January 31st (day 62) has a temperature of 2 degrees. This day is excluded from the recurrence period, but it affects the temperature on January 30th (day 61).4. Determining the Periodicity: - The sequence of temperatures is periodic with a period of 6 days. We need to determine the temperatures for the first 6 days and then extend this periodicity to find the temperature on December 25th.5. Calculating Temperatures: - Using the given temperature on December 3rd (day 3) and the periodicity, we find the temperatures for the first 6 days: - ( T_2 = 2 ) - ( T_3 = 5 ) - ( T_4 = 3 ) - ( T_5 = -2 ) - ( T_6 = -5 ) - ( T_7 = -3 ) - ( T_8 = 2 ) (repeats the cycle)6. Finding the Temperature on December 25th: - December 25th is day 25. To find the equivalent day within the 6-day period: - ( 25 mod 6 = 1 ) - The temperature on day 1 (which corresponds to day 25 in the 6-day cycle) is -3 degrees.Thus, the minimum temperature on December 25th is (boxed{-3}).,<think>Okay, let me try to figure out this problem. So, Stefano noticed that from December to January, excluding the first and last day, each day's minimum temperature was the sum of the previous and next day's temperatures. We know that on December 3rd, it was 5 degrees, and on January 31st, it was 2 degrees. We need to find the temperature on December 25th. Hmm, interesting.First, let me make sure I understand the problem correctly. The period considered is from December 1st to January 31st. Wait, actually, the problem says "the last months of December and January," so maybe it's just December and January, so December 1st to January 31st. But then, excluding the first and last day, so December 1st and January 31st are excluded? Wait, but the problem says "excluding the first and last day" of the period. So maybe the period starts on December 1st and ends on January 31st, but those two days are excluded? Wait, but the problem mentions that on December 3rd the temperature was 5 degrees and on January 31st it was 2 degrees. Wait, January 31st is the last day, which was excluded? Wait, maybe I need to clarify.Wait, the problem states: "on all the days of this period (excluding the first and last day), the minimum temperature was the sum of the minimum temperature of the previous day and the next day." So the entire period considered is from December 1st to January 31st, but excluding December 1st and January 31st. So the days from December 2nd to January 30th. But wait, the problem gives temperatures on December 3rd (which is included) and on January 31st (which is excluded). Wait, but the temperature on January 31st is given as 2 degrees. Wait, maybe the period is December 2nd to January 30th? But then December 3rd is included, and January 31st is outside? But the problem says "the last months of December and January," so maybe the period is from December 1st to January 31st, but the first day (December 1st) and the last day (January 31st) are excluded. However, the problem gives the temperature on January 31st, which is outside the period. That seems conflicting.Wait, perhaps I need to parse the problem again. It says: "Observing the temperatures recorded in Cesenatico in the last months of December and January, Stefano noticed a strange coincidence: on all the days of this period (excluding the first and last day), the minimum temperature was the sum of the minimum temperature of the previous day and the next day."So the period is the last months of December and January. Wait, that's a bit unclear. Maybe the period is from December 1st to January 31st, and Stefano excluded the first day (December 1st) and the last day (January 31st) from his observation. So the days from December 2nd to January 30th have the property that each day's temperature is the sum of the previous and next day. But then, the problem gives us the temperature on December 3rd (which is within the period) and on January 31st (which is outside the period). So maybe the temperatures on December 3rd and January 31st are given, but the relation applies from December 2nd to January 30th.Wait, so perhaps the recurrence relation applies to days from December 2nd to January 30th. Then, we have a temperature given on December 3rd (inside the period) and on January 31st (outside the period). But how can we use the temperature on January 31st if it's outside the period where the recurrence applies?Alternatively, maybe the period is considered as the days from December 2nd to January 30th, and the temperatures on December 3rd and January 31st are given. Wait, but if the period is December 2nd to January 30th, then the first day is December 2nd and the last day is January 30th. But the problem mentions excluding the first and last day of the period. Then, the days from December 3rd to January 29th would have the property. But the problem gives temperatures on December 3rd and January 31st. Hmm, this is confusing. Maybe I need to clarify.Let me rephrase the problem:Stefano observed temperatures in Cesenatico during the last months of December and January. He noticed that, for all days in this period except the first and last day, the minimum temperature on each day was equal to the sum of the minimum temperatures of the previous day and the next day. We are told that on December 3rd, the temperature was 5 degrees, and on January 31st, it was 2 degrees. We need to find the temperature on December 25th.So the "period" is December and January, excluding the first day (December 1st?) and the last day (January 31st?). So the days from December 2nd to January 30th. Wait, but the problem mentions that the temperature on January 31st is 2 degrees. So even though January 31st is excluded from the recurrence relation, we know its temperature. Similarly, perhaps December 1st's temperature is unknown, but we know December 3rd's.So the recurrence applies from December 2nd to January 30th, inclusive. Each day's temperature (from December 2nd to January 30th) is equal to the sum of the previous day and next day. Wait, but that seems impossible because each temperature is dependent on two other days. Wait, let's model this.Let me denote the temperature on day ( n ) as ( T_n ). The problem states that for each day in the period (excluding first and last), ( T_n = T_{n-1} + T_{n+1} ).Wait, this is a recurrence relation. So rearranged, we have ( T_{n+1} = T_n - T_{n-1} ). That's a linear recurrence relation.Hmm, so for all days in the period (excluding first and last), the temperature satisfies ( T_{n+1} = T_n - T_{n-1} ). This is a second-order linear recurrence relation with constant coefficients.Let me check the characteristic equation. The recurrence is ( T_{n+1} - T_n + T_{n-1} = 0 ). The characteristic equation would be ( r^2 - r + 1 = 0 ). Solving this quadratic equation: discriminant ( D = (-1)^2 - 4*1*1 = 1 - 4 = -3 ). So the roots are complex: ( r = [1 pm isqrt{3}]/2 ). These are complex conjugates with magnitude 1 and angle 60 degrees. Therefore, the solution is periodic with period 6, because the angle is 60 degrees, which is 360/6. So the sequence should repeat every 6 days.Wait, that might be the key here. If the recurrence relation leads to a periodic sequence with period 6, then the temperatures repeat every 6 days. Therefore, if we can find the temperatures over a 6-day cycle, we can determine the temperature on any day, given two temperatures separated by a certain number of days.Given that, let's see. We have two known temperatures: December 3rd is 5 degrees, and January 31st is 2 degrees. We need to figure out how many days are between December 3rd and January 31st, then use the periodicity to find the relationship between these two temperatures, and then determine the temperature on December 25th.First, let's calculate the number of days between December 3rd and January 31st. December has 31 days, so from December 3rd to December 31st is 31 - 3 = 28 days. Then January has 31 days, so from December 31st to January 31st is 31 days. Therefore, total days from December 3rd to January 31st is 28 + 31 = 59 days. However, since both start and end dates are included, we need to check if we should add 1. Wait, December 3rd to December 4th is 1 day. So from December 3rd to December 31st inclusive is 31 - 3 + 1 = 29 days. Then from January 1st to January 31st is 31 days. So total days from December 3rd to January 31st inclusive is 29 + 31 = 60 days. But since January 31st is the end date, maybe the number of days between December 3rd and January 31st is 60 - 1 = 59 days? Wait, this can be confusing.Wait, let's use a concrete example. Suppose today is December 3rd (day 0). Then December 4th is day 1, ..., December 31st is day 28 (since 31-3=28). Then January has 31 days, so January 31st is day 28 + 31 = day 59. So from December 3rd to January 31st is 59 days. Therefore, there are 59 days between December 3rd and January 31st, inclusive of both endpoints. Wait, but if you count the days from December 3rd to January 31st, including both days, it's (31 - 3) + 31 + 1 = 28 + 31 +1 = 60 days? Wait, no. Wait, December has 31 days. From December 3rd to December 31st is 31 - 3 = 28 days, but including both the 3rd and 31st, it's 28 + 1 = 29 days. Then January has 31 days, so total days would be 29 + 31 = 60 days. Therefore, December 3rd to January 31st is 60 days, inclusive. Therefore, the number of days between them is 60 - 1 = 59 days. Hmm, this is conflicting.Alternatively, perhaps using a date calculator: December 3rd to January 31st. Let's count the days:December: 31 days. From December 3rd to December 31: 31 - 3 = 28 days (excluding December 3rd), but if including both start and end, it's 31 - 3 + 1 = 29 days. Then January 1st to January 31st is 31 days. So total days from December 3rd to January 31st inclusive is 29 + 31 = 60 days. Therefore, the number of days between December 3rd and January 31st is 59 days (since the number of days between day 1 and day 2 is 1 day, not 2). So December 3rd is day 0, then January 31st is day 59.But how does this relate to the problem? Well, if the sequence is periodic with period 6, then the temperature on day ( n ) is the same as day ( n + 6k ) for integer ( k ). Therefore, the temperature on January 31st (day 59) should be equal to the temperature on day ( 59 mod 6 ). Let's compute 59 divided by 6: 6*9=54, remainder 5. So 59 mod 6 is 5. Therefore, if the period is 6, then temperature on day 59 is equal to temperature on day 5. But temperature on day 0 (December 3rd) is 5 degrees. Then temperature on day 5 would be related to day 0 through the recurrence.But wait, let's see. The recurrence is ( T_{n+1} = T_n - T_{n-1} ). So starting from some initial terms, we can generate the sequence.Suppose we have ( T_0 = 5 ) (December 3rd). Let's try to compute the subsequent terms using the recurrence ( T_{n+1} = T_n - T_{n-1} ).But to compute this, we need two initial terms. Wait, the problem only gives us one temperature inside the period (December 3rd) and one outside (January 31st). Hmm. Wait, but maybe the periodicity allows us to relate them.Wait, the problem is that the recurrence relation ( T_{n+1} = T_n - T_{n-1} ) has a period of 6. Let me verify this.Suppose we start with T0 and T1, then compute T2 = T1 - T0, T3 = T2 - T1 = (T1 - T0) - T1 = -T0, T4 = T3 - T2 = (-T0) - (T1 - T0) = -T1, T5 = T4 - T3 = (-T1) - (-T0) = -T1 + T0, T6 = T5 - T4 = (-T1 + T0) - (-T1) = T0, T7 = T6 - T5 = T0 - (-T1 + T0) = T0 + T1 - T0 = T1.So the sequence is T0, T1, T1 - T0, -T0, -T1, -T1 + T0, T0, T1, ... So the period is indeed 6. Therefore, every 6 days, the sequence repeats. Therefore, Tn = T_{n+6}.Therefore, if we know that the temperature on December 3rd is T0 = 5, and the temperature on January 31st is 2 degrees, which is day 59 (assuming December 3rd is day 0), then T59 = T_{59 mod 6} = T5 = 2 degrees.From the earlier calculation, T5 = -T1 + T0. Therefore, -T1 + T0 = 2. Since T0 = 5, this becomes -T1 + 5 = 2 => -T1 = -3 => T1 = 3.Therefore, T1 = 3. Then we can compute all the terms in the sequence:T0 = 5T1 = 3T2 = T1 - T0 = 3 - 5 = -2T3 = T2 - T1 = -2 - 3 = -5T4 = T3 - T2 = -5 - (-2) = -5 + 2 = -3T5 = T4 - T3 = -3 - (-5) = -3 + 5 = 2T6 = T5 - T4 = 2 - (-3) = 2 + 3 = 5T7 = T6 - T5 = 5 - 2 = 3Which matches T0 and T1, so the period is indeed 6.Therefore, the temperatures repeat every 6 days. So the temperature on any day n is equal to T_{n mod 6}.Now, we need to find the temperature on December 25th. Let's figure out how many days December 25th is from December 3rd.December 3rd to December 25th: 25 - 3 = 22 days. So December 3rd is day 0, December 4th is day 1, ..., December 25th is day 22.Therefore, day number 22. Then, 22 divided by 6: 6*3=18, remainder 4. So 22 mod 6 = 4. Therefore, T22 = T4 = -3 degrees? Wait, but the problem is asking for the minimum temperature, and negative temperatures are possible? Well, the problem didn't specify that temperatures are positive, so maybe it's possible.Wait, but let's check. Wait, according to our earlier calculation, the sequence is:n : 0 1 2 3 4 5 6 7 8 9 10 11 ...T:5 3 -2 -5 -3 2 5 3 -2 -5 -3 2 ...So every 6 days, it repeats. Therefore, day 22: 22 divided by 6 is 3 with a remainder of 4 (since 6*3=18, 22-18=4). So T22 = T4 = -3 degrees. So the temperature on December 25th would be -3 degrees. But the problem says "minimum temperature," so maybe they are referring to the lowest temperature recorded, but here it's just the temperature value. The answer is -3 degrees. Wait, but the problem gives temperatures as 5 and 2 degrees, positive numbers, but maybe negative is allowed. However, maybe I made a mistake in the sign.Wait, let's check the recurrence again. The recurrence is ( T_{n} = T_{n-1} + T_{n+1} ). Wait, hold on! Wait, the problem states: "the minimum temperature was the sum of the minimum temperature of the previous day and the next day." So, ( T_n = T_{n-1} + T_{n+1} ). But in my earlier analysis, I rearranged this to ( T_{n+1} = T_n - T_{n-1} ). Is that correct?Wait, yes. If ( T_n = T_{n-1} + T_{n+1} ), then ( T_{n+1} = T_n - T_{n-1} ). That is correct. So the recurrence is correct. Therefore, the calculations for the sequence are correct. Therefore, if T0 = 5, T1 = 3, then T2 = -2, etc., as above.So, according to this, December 25th is day 22, which is T4 = -3 degrees. However, the problem states that the temperature on January 31st is 2 degrees, which in our sequence is T5 = 2, which occurs every 6 days. Therefore, that's consistent. So, the answer would be -3 degrees. But maybe there is a miscalculation in the day count.Wait, let's confirm the number of days between December 3rd and December 25th. December 3rd to December 25th: from the 3rd to the 25th, inclusive or exclusive? Wait, if December 3rd is day 0, then December 4th is day 1, ..., December 25th is day 22 (because 25 - 3 = 22 days later). So yes, 22 days. Therefore, 22 mod 6 is 4, so T4 = -3. Therefore, the answer is -3 degrees. But the problem might expect a positive number? Wait, maybe I messed up the direction of the recurrence.Wait, let's check the initial terms again. Given T0 = 5 (December 3rd), and T1 = 3 (from the calculation where T5 = 2 = -T1 + T0 => T1 = 3). Then T2 = T1 - T0 = 3 - 5 = -2. Then T3 = T2 - T1 = -2 -3 = -5. T4 = T3 - T2 = -5 - (-2) = -3. T5 = T4 - T3 = -3 - (-5) = 2. Then T6 = T5 - T4 = 2 - (-3) = 5, which brings us back to T0. So that's correct. Therefore, the temperature on day 4 is -3. Therefore, December 25th is -3 degrees. But is this possible? The problem didn't specify that temperatures are positive. It just said "minimum temperature," which could be negative. So maybe that's the answer.But let me cross-verify. Let's check the number of days again. From December 3rd (day 0) to December 25th (day 22). Wait, 22 days later. 22 divided by 6 is 3 cycles (18 days) with remainder 4. So day 22 is equivalent to day 4. T4 is -3. So the answer is -3. But let me see if there's another way to approach this.Alternatively, since the sequence is periodic with period 6, the temperature on December 3rd is 5, and every 6 days, it repeats. So December 3rd, 9th, 15th, 21st, 27th, January 2nd, 8th, 14th, 20th, 26th, February 1st, etc., all have the same temperature. Wait, but we are only dealing up to January 31st. But December 25th is how many days after December 3rd? 22 days, which is 3 weeks and 1 day, but in days, 22. 22 divided by 6 is 3*6=18, remainder 4. So 22 days later is 3 full cycles (18 days) plus 4 days. Therefore, temperature is T0 + 4 days: T4 = -3. So that's consistent.Alternatively, maybe Stefano's period is from December 2nd to January 30th, with the recurrence applying from December 2nd to January 30th. Wait, but the problem states "excluding the first and last day" of the period. If the period is December 1st to January 31st, then excluding those, it's December 2nd to January 30th. But the given temperatures are on December 3rd (included) and January 31st (excluded). Therefore, using December 3rd as a known temperature inside the period and January 31st as a temperature outside, but since the recurrence leads to periodicity, we can link them.But maybe my initial approach is correct. Therefore, the answer is -3 degrees. However, the problem might be expecting a positive temperature. Maybe I made an error in setting up the recurrence.Wait, let's check the original problem statement again: "the minimum temperature was the sum of the minimum temperature of the previous day and the next day." So, for each day (excluding the first and last), the temperature is equal to the sum of the previous day and the next day. Wait, hold on! Wait, if the temperature on day n is equal to the sum of the previous day and the next day, that is:( T_n = T_{n-1} + T_{n+1} )Which rearranged is:( T_{n+1} = T_n - T_{n-1} )Which is the same as before. So my recurrence is correct. Therefore, the calculations hold. So with T0 = 5, T1 = 3, the sequence proceeds as 5, 3, -2, -5, -3, 2, 5, 3, etc.Therefore, December 25th is 22 days after December 3rd, which is 3 full cycles (18 days) plus 4 days, so T4 = -3. Therefore, the answer is -3 degrees.But let's confirm with another approach. Suppose we consider the periodicity. Since the period is 6 days, then the temperature on day n is equal to the temperature on day n + 6. Therefore, the temperature on January 31st is 2 degrees. January 31st is 59 days after December 3rd. 59 divided by 6 is 9 with a remainder of 5 (since 6*9=54, 59-54=5). Therefore, the temperature on January 31st is equal to the temperature on day 5, which is 2 degrees. Therefore, T5 = 2. From the sequence, T5 = -T1 + T0 = 2. Since T0 = 5, then -T1 + 5 = 2 => T1 = 3. Therefore, T1 = 3, as before. Then, the temperature on December 25th is 22 days later, which is day 22, so T22 = T4 = -3. So this is consistent.Therefore, the minimum temperature on December 25th is -3 degrees. So the answer is boxed{-3}.Wait, but the problem says "minimum temperature," which is a bit confusing because it's the name of the measurement, not necessarily implying it's the lowest value. But in any case, the temperature on that day is -3. So I think that's the answer.But let me check once more. Suppose we list the temperatures:Starting from December 3rd as day 0: 5Day 1: 3Day 2: -2Day 3: -5Day 4: -3Day 5: 2Day 6: 5Day 7: 3Day 8: -2Day 9: -5Day 10: -3Day 11: 2Day 12: 5Day 13: 3Day 14: -2Day 15: -5Day 16: -3Day 17: 2Day 18: 5Day 19: 3Day 20: -2Day 21: -5Day 22: -3Yes, day 22 is -3. Therefore, December 25th is 22 days after December 3rd, so temperature is -3. Therefore, the answer is boxed{-3}.Final Answerboxed{-3}</think>Stefano observed that from December 2nd to January 30th (excluding the first and last day of the period), each day's minimum temperature was the sum of the minimum temperatures of the previous and next days. This gives the recurrence relation ( T_n = T_{n-1} + T_{n+1} ), which simplifies to ( T_{n+1} = T_n - T_{n-1} ). This recurrence relation has a periodic solution with period 6.Given:- The temperature on December 3rd (day 0) is 5 degrees.- The temperature on January 31st (day 59) is 2 degrees.We need to find the temperature on December 25th (day 22).1. Determine the periodicity: The sequence of temperatures repeats every 6 days. We verify this by solving the characteristic equation of the recurrence relation, leading to complex roots with a magnitude of 1 and an angle of 60 degrees, resulting in a period of 6.2. Calculate the number of days between December 3rd and January 31st: This is 59 days. Since the period is 6, the temperature on January 31st (day 59) is equivalent to the temperature on day ( 59 mod 6 = 5 ). Given ( T_5 = 2 ), we use the recurrence relation to find the initial terms: - ( T_5 = -T_1 + T_0 ) - ( 2 = -T_1 + 5 ) - ( T_1 = 3 )3. Generate the sequence: - ( T_0 = 5 ) - ( T_1 = 3 ) - ( T_2 = T_1 - T_0 = -2 ) - ( T_3 = T_2 - T_1 = -5 ) - ( T_4 = T_3 - T_2 = -3 ) - ( T_5 = T_4 - T_3 = 2 )4. Find the temperature on December 25th: December 25th is 22 days after December 3rd. Calculating ( 22 mod 6 = 4 ), we find ( T_4 = -3 ).Thus, the minimum temperature on December 25th is (boxed{-3}).]

🔑:## SOLUTION:The minimum temperature recorded on December 25th was -3 degrees.To demonstrate this, let x be the minimum temperature recorded on December 1st and y be the minimum temperature recorded on December 2nd. Using the relationship observed by Stefano, the minimum temperature on a given day can be derived from those of the two preceding days: in this way, the minimum temperatures for the first few days of December are as reported in the following table:| Day | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 || :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: || Min. Temp. | x | y | y-x | -x | -y | x-y | x | y | y-x | -x |It is therefore easily seen that the sequence of minimum temperatures repeats every 6 days. Consequently, the minimum temperature on December 3rd was y-x, while that on January 31st, which is the 62nd day of the period, coincides with that of the second, i.e., y. From the information, it is deduced that y-x=5 and y=2, from which x=-3.Now, the minimum temperature on December 25th coincides with that of the 1st, since 25-1 is a multiple of 6, and was therefore -3 degrees.

🧠:[<think>Alright, let's tackle this problem step by step. It's a bit complex, but if I break it down, maybe I can figure it out. So, the family has five people: Dad, Mom, Brother, Sister, and Xiao Yu. Each of them has to cook once and wash the dishes once over five days (1st to 5th). The goal is to find a five-digit number representing the days each person washes the dishes in the order Dad, Mom, Brother, Sister, Xiao Yu.First, let's list out the information given by each family member:1. Brother's Statement: - He washed the dishes yesterday (so if today is day X, he washed dishes on day X-1). - He has to cook tomorrow (so he will cook on day X+1). - Therefore, Brother's chores are: wash dishes on day X-1, cook on day X+1. So, between washing dishes and cooking, there's a gap of one day (day X). But he must have both chores on different days. Also, since everyone does each chore once, he can't do both on the same day.2. Sister's Statement: - She has chores both tomorrow and the day after. So, if today is day Y, she has chores on Y+1 and Y+2. Since each person does one chore per day, this means she is either cooking or washing dishes on both those days. But each person can only do one chore per day, so she must be doing different chores on those two days. Wait, but each person only cooks once and washes once, so she must cook on one day and wash on the other. So, Sister has chores on two consecutive days (but different chores each day). So, her two chores are on consecutive days.3. Xiao Yu's Statement: - Dad and Xiao Yu always do chores on the same day. So, whenever Dad cooks or washes dishes, Xiao Yu does the other chore on the same day. Wait, but each day there's one person cooking and one washing dishes. So, if Dad is cooking on a day, Xiao Yu must be washing dishes that day, and vice versa. Because they can't both cook or both wash on the same day. So, whenever Dad is assigned a chore, Xiao Yu does the other one on the same day. Therefore, Dad and Xiao Yu's chores are paired on the same days. So, their cooking and dishwashing days are the same dates, but with their roles swapped. Wait, but each person has to cook once and wash once. So, if Dad cooks on a day, Xiao Yu washes that day, and if Dad washes on another day, Xiao Yu cooks that day. So, they have their chores on the same days, but swapping roles. Therefore, Dad and Xiao Yu share the same dates for their chores, but with different tasks. Therefore, the days Dad cooks are the days Xiao Yu washes, and the days Dad washes are the days Xiao Yu cooks. But since each person has to do each chore once, they must each have two days assigned? Wait, no. Wait, each person only does each chore once. So, over five days, each person cooks once and washes once. So, Dad cooks on one day and washes on another day. Similarly, Xiao Yu cooks on one day and washes on another day. But according to Xiao Yu's statement, whenever Dad does a chore, Xiao Yu does the other chore on the same day. Therefore, whenever Dad is cooking, Xiao Yu is washing dishes that day, and whenever Dad is washing dishes, Xiao Yu is cooking that day. Therefore, Dad and Xiao Yu's chore days are the same. For example, if Dad cooks on day 3, Xiao Yu washes on day 3. If Dad washes on day 5, Xiao Yu cooks on day 5. Therefore, Dad has two chore days (one cooking, one washing), and Xiao Yu also has two chore days (opposite chores on the same days as Dad). But wait, each person is supposed to do each chore once. So, Dad cooks once and washes once, and Xiao Yu cooks once and washes once. Since they are paired on the same days, Dad's cooking day is Xiao Yu's washing day, and Dad's washing day is Xiao Yu's cooking day. Therefore, Dad and Xiao Yu share two days: one where Dad cooks and Xiao Yu washes, and another where Dad washes and Xiao Yu cooks. Therefore, those two days are shared between Dad and Xiao Yu. Then the other three days (since there are five days) must be covered by Mom, Brother, and Sister, each doing one cooking and one washing on different days. Wait, but each day has one cook and one washer. So, over five days, there are 5 cooks and 5 washers. Each person does one cook and one wash. Therefore, each of the five people has one cook day and one wash day.But according to Xiao Yu's statement, Dad and Xiao Yu do chores on the same days. So, their cook and wash days must coincide. Let me think again.If Dad cooks on day A, then Xiao Yu must wash on day A. If Dad washes on day B, then Xiao Yu must cook on day B. Therefore, Dad's two chores (cook on A, wash on B) correspond to Xiao Yu's two chores (wash on A, cook on B). Therefore, days A and B are the days when both Dad and Xiao Yu are doing chores. So, days A and B have Dad and Xiao Yu assigned. That leaves the other three days (C, D, E) to be covered by Mom, Brother, and Sister.But Mom, Brother, and Sister each need to cook once and wash once. So, in days C, D, E, each day needs a cook and a washer, but those three people can't repeat chores on the same day. Wait, no. Each person can only do one chore per day, but over the five days, each does one cook and one wash. So, in days 1-5, each person has one cook day and one wash day. But for days A and B, both chores are done by Dad and Xiao Yu. Then days C, D, E must have Mom, Brother, Sister doing the remaining three cooking and three washing duties, but each of them needs to do one cook and one wash. So, over three days, we have three cooks and three washes. Each of Mom, Brother, Sister must do one cook and one wash. Therefore, each of them will have one chore on two different days. So, in days C, D, E, each day has one cook and one wash, assigned to Mom, Brother, Sister. Each of them will do one cook and one wash on different days.Now, moving to Dad's statement:4. Dad's Statement: - After cooking, he can rest for two days before it's his turn to wash dishes again. So, if Dad cooks on day X, he washes dishes on day X + 3. Because after cooking, he rests for two days (days X+1 and X+2) and then washes on day X+3. Therefore, the gap between cooking and washing is three days. Since the days are from 1 to 5, if he cooks on day 1, he would wash on day 4 (1 + 3). If he cooks on day 2, he would wash on day 5. If he cooks on day 3, washing would be on day 6, which is outside the range. So, Dad can only cook on day 1 or 2, because cooking on day 3 would require washing on day 6, which doesn't exist. Therefore, Dad's cooking day is either day 1 or 2, and his washing day is day 4 or 5, respectively.But remember from Xiao Yu's statement, Dad's cooking day is the same as Xiao Yu's washing day, and Dad's washing day is the same as Xiao Yu's cooking day. So, if Dad cooks on day X, Xiao Yu washes on day X. If Dad washes on day Y, Xiao Yu cooks on day Y. And according to Dad's rest period, if he cooks on day X, he washes on day X + 3. Therefore, Y = X + 3. So, Xiao Yu's cooking day is Y = X + 3, which is the same as Dad's washing day.So, for example, if Dad cooks on day 1 (X=1), he washes on day 4 (Y=4). Then Xiao Yu washes on day 1 and cooks on day 4. Therefore, Xiao Yu's two chores are day 1 (wash) and day 4 (cook). Similarly, if Dad cooks on day 2 (X=2), he washes on day 5 (Y=5). Then Xiao Yu washes on day 2 and cooks on day 5.So, possible scenarios:Case 1: Dad cooks on day 1, washes on day 4.Then Xiao Yu washes on day 1, cooks on day 4.Case 2: Dad cooks on day 2, washes on day 5.Then Xiao Yu washes on day 2, cooks on day 5.These are the only possible cases because cooking on day 3 would make washing on day 6, which is invalid.Now, let's consider Brother's statement:Brother says: "I just washed the dishes yesterday, and I have to cook tomorrow, so tiring!"Let's parse this. Suppose the conversation is happening on day N. Then:- Yesterday (day N-1): Brother washed dishes.- Tomorrow (day N+1): Brother has to cook.Therefore, Brother's chores are:- Wash on day N-1.- Cook on day N+1.But since each person only does one chore per day, and the chores are spread over days 1 to 5, we need to figure out on which day this conversation is taking place.Wait, the problem says "One day they had the following conversation". So, the conversation is happening on a particular day, which is one of the days 1 to 5. But the problem doesn't specify which day. However, the statements refer to "yesterday", "tomorrow", and "the day after". So, we need to figure out the timeline.First, let's note that the conversation is happening on some day, let's call it day T. Then:- Brother washed dishes on day T-1.- Brother will cook on day T+1.Therefore, Brother's chores are on days T-1 (wash) and T+1 (cook). Since the days are 1 to 5, T must be such that T-1 and T+1 are within 1 to 5. Therefore, T-1 ≥ 1 and T+1 ≤5. So, T must be 2, 3, or 4.If T=2: Brother washed on day 1, cooks on day 3.If T=3: Brother washed on day 2, cooks on day 4.If T=4: Brother washed on day 3, cooks on day 5.So, these are the possibilities for Brother's chores.Now, Sister's statement: "I am even more tired, I have chores both tomorrow and the day after!"So, on the day of the conversation (day T), Sister is saying that she has chores on T+1 and T+2. Since each person does one chore per day, Sister must be doing one chore on T+1 and another on T+2. Since she can't do both chores on the same day, she must cook on one day and wash on the other. Therefore, Sister's chores are on two consecutive days: T+1 and T+2, with different chores each day.However, Sister can only do one cooking and one washing over the five days, so these two days (T+1 and T+2) must be her only two chore days, meaning she doesn't do any chores on other days. Wait, no. Wait, each person does one cooking and one washing on different days. So, Sister could have her two chores on days T+1 and T+2, but those are consecutive days. So, she must cook on one and wash on the other.So, Sister's chores are consecutive days. Therefore, Sister has two consecutive days where she does her chores, one cooking and one washing.Now, considering the conversation day T, Sister's chores are on T+1 and T+2. Therefore, the days T+1 and T+2 must be within 1 to 5, so T+2 ≤5. Hence, T can be at most 3. Therefore, possible T values for Sister's statement: T=1, 2, 3. But from Brother's statement, T must be 2,3,4. Therefore, overlapping possible T is 2 or 3.So, T could be 2 or 3.Let me summarize:Possible T (conversation day):From Brother's statement: T=2,3,4From Sister's statement: T=1,2,3Therefore, overlapping T=2,3So, T is either 2 or 3.Let's analyze both possibilities.First, let's consider T=2.If T=2:Brother's chores:- Washed dishes on day 1.- Cooks on day 3.Sister's chores:- Has chores on days 3 and 4.So, Sister must do one chore on day 3 and another on day 4. Since Brother is cooking on day 3, Sister can't cook on day 3 (each day only one cook). Therefore, Sister must wash dishes on day 3 and cook on day 4, or vice versa. Wait, but each person can only do one chore per day. So, if Brother is cooking on day 3, Sister can either cook or wash on day 3, but if Brother is cooking, Sister can't cook on the same day. So, if Brother is cooking on day 3, Sister must wash on day 3. Then, on day 4, Sister can cook or wash. But she has to do both chores once. So, if she washes on day 3, she needs to cook on another day. But according to her statement, her chores are on days 3 and 4 (T+1=3, T+2=4). Therefore, she must do one chore on day 3 and one on day 4. So, possible assignments:- Sister washes on day 3 and cooks on day 4.But on day 3, Brother is cooking, so Sister can wash on day 3. Then on day 4, she can cook, provided no one else is cooking that day.Alternatively, Sister cooks on day 3 and washes on day 4. But Brother is already cooking on day 3, so Sister can't cook on day 3. Therefore, Sister must wash on day 3 and cook on day 4.Thus, Sister's chores: wash on 3, cook on 4.Now, considering Dad's schedule. Let's recall the two cases for Dad:Case 1: Dad cooks on 1, washes on 4.Case 2: Dad cooks on 2, washes on 5.If T=2, which is the conversation day, let's see which case applies.First, in Case 1: Dad cooks on 1, washes on 4.Xiao Yu's chores: washes on 1, cooks on 4.So, on day 1: Dad cooks, Xiao Yu washes.On day 4: Dad washes, Xiao Yu cooks.In this case, let's look at day 3: Brother is cooking, Sister is washing.So, day 3: Cook = Brother, Wash = Sister.Then day 4: Cook = Xiao Yu, Wash = Dad.Then day 5: Who is left? The remaining people are Mom. Wait, Mom hasn't been assigned any chores yet. Wait, over the five days, each person must cook and wash once.Wait, let's check assignments so far (Case 1, T=2):- Dad: cooks 1, washes 4.- Xiao Yu: washes 1, cooks 4.- Brother: washes 1 (no, wait, Brother washes on day 1? Wait, in T=2, Brother washed on day 1 (T-1=1) and cooks on day 3.Wait, correction: In T=2, Brother's wash day is 1 (T-1=2-1=1) and cook day is 3 (T+1=3).Sister's chores: wash on 3 and cook on 4.Wait, but on day 3: Brother is cooking, Sister is washing.On day 4: Sister is cooking, Xiao Yu is cooking? Wait, no. Wait, day 4: Xiao Yu is cooking (from Dad's case 1: Xiao Yu cooks on 4), and Dad is washing. But Sister is also supposed to cook on day 4. That's a conflict because only one person can cook each day. Therefore, this is a problem.Therefore, Case 1 (Dad cooks 1, washes 4) with T=2 leads to a conflict on day 4: both Xiao Yu and Sister are cooking. That's impossible. Therefore, Case 1 is invalid if T=2.Now, check Case 2: Dad cooks on 2, washes on 5.Xiao Yu's chores: washes on 2, cooks on 5.So, day 2: Dad cooks, Xiao Yu washes.Day 5: Dad washes, Xiao Yu cooks.Now, with T=2:Brother's chores: wash on 1 (T-1=1), cook on 3 (T+1=3).Sister's chores: on days 3 and 4 (T+1=3, T+2=4). So, Sister must do one chore on 3 and one on 4.On day 3, Brother is cooking, so Sister can't cook. Therefore, Sister must wash on day 3 and cook on day 4.So:Day 1: Brother washes, who cooks? We need someone to cook on day 1. Since Dad is cooking on day 2, Xiao Yu is washing on day 2. On day 1, the cook must be someone else. The remaining people are Mom, Sister, and Xiao Yu. Wait, Xiao Yu's cooking day is day 5. So, day 1's cook must be Mom or Sister. Sister's cooking day is day 4. Therefore, day 1's cook is Mom.So:Day 1: Cook = Mom, Wash = Brother.Day 2: Cook = Dad, Wash = Xiao Yu.Day 3: Cook = Brother, Wash = Sister.Day 4: Cook = Sister, Wash = ?Day 5: Cook = Xiao Yu, Wash = Dad.On day 4, Sister is cooking, so the washer must be Mom, since the remaining people are Mom and Sister. But Sister is cooking, so Mom must wash on day 4.Wait, let's check assignments:- Mom: Cook on day 1, needs to wash once. So, Mom could wash on day 4.- Sister: Cook on day 4, wash on day 3.- Brother: Cook on day 3, wash on day 1.- Dad: Cook on day 2, wash on day 5.- Xiao Yu: Cook on day 5, wash on day 2.So, that works:Day 1: Mom cooks, Brother washes.Day 2: Dad cooks, Xiao Yu washes.Day 3: Brother cooks, Sister washes.Day 4: Sister cooks, Mom washes.Day 5: Xiao Yu cooks, Dad washes.Now, let's verify all conditions:1. Brother's Statement: He washed yesterday (day T-1 = day 1) and cooks tomorrow (day T+1 = day 3). Since T=2, this checks out.2. Sister's Statement: On day T=2, she says she has chores tomorrow and the day after (days 3 and 4). She washes on day 3 and cooks on day 4, which are consecutive days. That fits.3. Xiao Yu's Statement: Dad and Xiao Yu always do chores on the same day. Looking at the assignments: - Day 2: Dad cooks, Xiao Yu washes. - Day 5: Dad washes, Xiao Yu cooks. So, they are indeed paired on the same days, each doing opposite chores. Correct.4. Dad's Statement: After cooking, he rests for two days before washing. Dad cooks on day 2, rests on days 3 and 4, then washes on day 5. That's two days of rest (days 3 and 4) between cooking on day 2 and washing on day 5. Correct.Additionally, all five people have cooked once and washed once:- Dad: Cook 2, Wash 5.- Mom: Cook 1, Wash 4.- Brother: Cook 3, Wash 1.- Sister: Cook 4, Wash 3.- Xiao Yu: Cook 5, Wash 2.Each day has one cook and one washer:Day 1: Mom, Brother.Day 2: Dad, Xiao Yu.Day 3: Brother, Sister.Day 4: Sister, Mom.Day 5: Xiao Yu, Dad.No conflicts, everyone does each chore once. Looks good.Now, let's check if T=3 could also be possible.If T=3:Brother's chores:- Washed dishes on day 2 (T-1=2).- Cooks on day 4 (T+1=4).Sister's chores:- Has chores on days 4 and 5 (T+1=4, T+2=5).So, Sister must do one chore on day 4 and one on day 5. Since Brother is cooking on day 4, Sister can't cook on day 4, so she must wash on day 4 and cook on day 5. Alternatively, if Sister cooked on day 4, but Brother is already cooking, so she must wash on day 4 and cook on day 5.Now, let's consider Dad's two cases.Case 1: Dad cooks on day 1, washes on day 4.Then Xiao Yu washes on day 1, cooks on day 4.But in this scenario, on day 4, Xiao Yu is cooking and Sister is washing. However, Brother is also supposed to cook on day 4 (from Brother's statement when T=3). Wait, conflict: Brother is supposed to cook on day 4 (T+1=4), but Xiao Yu is also cooking on day 4 (from Dad's Case 1). That's impossible. Therefore, Case 1 with T=3 is invalid.Case 2: Dad cooks on day 2, washes on day 5.Xiao Yu washes on day 2, cooks on day 5.Now, with T=3:Brother's chores: wash on day 2, cook on day 4.Sister's chores: wash on day 4, cook on day 5.But let's assign chores step by step.Day 1: Who cooks and washes?Available people: Dad cooks on day 2, Xiao Yu washes on day 2. So, day 1 can be Mom, Sister, or Brother. But Brother is washing on day 2 and cooking on day 4. So, Brother's chores are day 2 (wash) and day 4 (cook). Therefore, Brother is free on day 1.So, day 1: Let's suppose Mom cooks and Sister washes. But we need to check all assignments.Wait, let's structure this:Dad: cooks on 2, washes on 5.Xiao Yu: washes on 2, cooks on 5.Brother: washes on 2, cooks on 4. Wait, but Xiao Yu is already washing on day 2. So, conflict: Brother can't wash on day 2 if Xiao Yu is washing there. Each day has only one washer. So, this is a problem.Wait, if T=3, Brother's wash day is T-1=2. But in Dad's Case 2, Xiao Yu is washing on day 2. Therefore, Brother and Xiao Yu both washing on day 2, which is impossible. Therefore, Case 2 with T=3 is invalid.Hence, T=3 leads to a conflict in both Dad's cases. Therefore, T=3 is invalid.Therefore, the only valid scenario is T=2 with Dad cooking on day 2 and washing on day 5 (Case 2), leading to the assignments:- Dad: Cook 2, Wash 5.- Xiao Yu: Wash 2, Cook 5.- Brother: Wash 1, Cook 3.- Sister: Wash 3, Cook 4.- Mom: Cook 1, Wash 4.Thus, the days each person washes dishes are:- Dad: 5- Mom: 4- Brother: 1- Sister: 3- Xiao Yu: 2So, in order (Dad, Mom, Brother, Sister, Xiao Yu), the washing days are 5, 4, 1, 3, 2. Therefore, the five-digit number is 54132.Wait, but let me confirm. The problem says "the five-digit number formed by the dates when Dad, Mom, Brother, Sister, and Xiao Yu wash the dishes in sequence". So, we need to list the days each person washes dishes, in the order Dad, Mom, Brother, Sister, Xiao Yu.From above:- Dad washes on day 5.- Mom washes on day 4.- Brother washes on day 1.- Sister washes on day 3.- Xiao Yu washes on day 2.Therefore, the sequence is 5 (Dad), 4 (Mom), 1 (Brother), 3 (Sister), 2 (Xiao Yu). So, the number is 54132.But let me cross-verify once more to ensure there's no mistake.Checking the assignment:Day 1:- Cook: Mom- Wash: BrotherDay 2:- Cook: Dad- Wash: Xiao YuDay 3:- Cook: Brother- Wash: SisterDay 4:- Cook: Sister- Wash: MomDay 5:- Cook: Xiao Yu- Wash: DadYes, this fits all conditions:- Each person cooks and washes once.- Dad and Xiao Yu are paired on days 2 and 5.- Brother's schedule: washes day 1, cooks day 3 (with conversation on day 2).- Sister's schedule: washes day 3, cooks day 4 (with conversation on day 2, chores on days 3 and 4).- Dad's rest: cooks day 2, rests days 3 and 4, washes day 5.Yes, everything checks out. Therefore, the five-digit number is 54132.Final Answerboxed{54132}</think>Xiao Yu's family consists of five people: Dad, Mom, Brother, Sister, and Xiao Yu. Each person must cook once and wash dishes once over five days (1st to 5th). The goal is to determine the five-digit number representing the days each person washes dishes in the order Dad, Mom, Brother, Sister, and Xiao Yu.1. Brother's Statement: - Washed dishes yesterday (day (T-1)). - Cooks tomorrow (day (T+1)). - Therefore, Brother's chores are on days (T-1) and (T+1).2. Sister's Statement: - Has chores tomorrow and the day after (days (T+1) and (T+2)). - Sister must cook and wash on these consecutive days.3. Xiao Yu's Statement: - Dad and Xiao Yu always do chores on the same day, swapping roles.4. Dad's Statement: - After cooking, rests for two days before washing dishes (cooks on day (X), washes on day (X+3)).Analyzing possible days for the conversation (day (T)):- Brother's chores require (T) to be 2, 3, or 4.- Sister's chores require (T) to be 1, 2, or 3.- Overlapping (T) values: 2 or 3.Case Analysis:- Case 1 (T=2): - Brother washes on day 1 and cooks on day 3. - Sister washes on day 3 and cooks on day 4. - Dad cooks on day 2 and washes on day 5 (rests days 3 and 4). - Xiao Yu washes on day 2 and cooks on day 5. - Mom cooks on day 1 and washes on day 4.Resulting Assignments:- Washing Days: - Dad: Day 5 - Mom: Day 4 - Brother: Day 1 - Sister: Day 3 - Xiao Yu: Day 2Thus, the five-digit number formed by the washing days in order (Dad, Mom, Brother, Sister, Xiao Yu) is (boxed{54132}).,<think>Okay, let's try to tackle this problem step by step. It's a logic puzzle involving scheduling chores for five family members over five days. Each person must cook once and wash dishes once, with each day having one cook and one dishwasher. The goal is to determine the five-digit number formed by the dates when each person (Dad, Mom, Brother, Sister, Xiao Yu) washes the dishes in order. First, let me summarize the given information:1. There are five family members: Dad, Mom, Brother, Sister, Xiao Yu.2. The period is from the 1st to the 5th of the month (5 days).3. Each day, one person cooks and another washes dishes.4. Each person cooks once and washes dishes once over the five days.5. Clues from the family members: - Brother: He washed dishes yesterday, has to cook tomorrow. So, if today is day X, he washed dishes on day X-1 and will cook on day X+1. - Sister: She has chores both tomorrow and the day after. So, if today is day X, she will cook or wash on days X+1 and X+2. - Xiao Yu: Dad and Xiao Yu always do chores on the same day. So, whenever Dad cooks or washes, Xiao Yu does the other chore that day. - Dad: After cooking, he can rest for two days before washing dishes again. So, if Dad cooks on day N, he washes dishes on day N + 3.Alright, let's break this down.First, let's note that each person has two chores: cook once and wash once. Therefore, each day has two different people assigned: one cooking, one washing.Key points to remember:- Xiao Yu and Dad are always on the same day. So, if Dad is cooking, Xiao Yu is washing dishes that day, and vice versa. So, their cooking and washing days are paired on the same dates.- Dad's schedule: After cooking on day N, he washes dishes on day N + 3. Since there are only 5 days, N must be such that N + 3 ≤ 5. So, N can be 1 or 2. Therefore, if Dad cooks on day 1, he washes on day 4; if he cooks on day 2, he washes on day 5.- Brother's statement: He washed dishes yesterday and has to cook tomorrow. Let's parse this. Suppose the conversation is happening on day X. Then: - Brother washed dishes on day X - 1. - Brother will cook on day X + 1. But we don't know which day the conversation is taking place. However, since all chores are from day 1 to day 5, the conversation must be on a day where X - 1 ≥ 1 and X + 1 ≤ 5. So, X must be 2, 3, or 4. Let's consider possible days for the conversation: If the conversation is on day 2: - Brother washed dishes on day 1. - Brother will cook on day 3. If the conversation is on day 3: - Brother washed dishes on day 2. - Brother will cook on day 4. If the conversation is on day 4: - Brother washed dishes on day 3. - Brother will cook on day 5. So, three possibilities for Brother's schedule.- Sister's statement: She has chores both tomorrow and the day after. Again, assuming the conversation is on day X, Sister is doing chores (either cooking or washing) on days X + 1 and X + 2. So, similar to Brother, X must be such that X + 2 ≤ 5, so X can be 1, 2, or 3. But Sister's statement is on the same day as Brother's. Since the conversation is one day when all these statements are made, X is the same for both Brother and Sister. From Brother's possible days (2, 3, 4), intersecting with Sister's possible days (1, 2, 3), the conversation must be on day 2 or 3. So, possible conversation days: 2 or 3. Let's check both possibilities.First, let's assume the conversation is on day 2.- Brother washed dishes on day 1, cooks on day 3.- Sister has chores on days 3 and 4.Now, Xiao Yu and Dad are always together. Let's consider Dad's schedule. If Dad cooks on day N, he washes on N + 3.Possible N for Dad: 1 or 2.If Dad cooks on day 1, he washes on day 4. Then Xiao Yu must be washing on day 1 if Dad is cooking, or cooking if Dad is washing. Wait: "Dad and I always do chores on the same day." So, whenever Dad does a chore (cook or wash), Xiao Yu does the other chore on the same day. So, if Dad cooks on day 1, Xiao Yu washes on day 1. If Dad washes on day 4, Xiao Yu cooks on day 4. So, Xiao Yu's chores are day 1 (wash) and day 4 (cook). But each person must cook once and wash once. So Xiao Yu must cook once and wash once. So, if Xiao Yu is washing on day 1 and cooking on day 4, that's correct.Alternatively, if Dad cooks on day 2, he washes on day 5. Then Xiao Yu would wash on day 2 (if Dad cooks) and cook on day 5 (if Dad washes). So Xiao Yu's chores: day 2 (wash) and day 5 (cook). That also works.So, two possibilities for Dad and Xiao Yu:1. Dad cooks day 1, washes day 4; Xiao Yu washes day 1, cooks day 4.2. Dad cooks day 2, washes day 5; Xiao Yu washes day 2, cooks day 5.Now, let's try the first case where conversation is on day 2.Case 1: Conversation day 2.Brother's schedule: washes day 1, cooks day 3.Sister's schedule: has chores on days 3 and 4.Dad and Xiao Yu: Dad cooks day 1, washes day 4; Xiao Yu washes day 1, cooks day 4.So, let's start filling a table.Days 1-5:Day 1:- Cook: Dad- Wash: Xiao YuBut Brother is washing on day 1 according to Brother's statement. Wait, conflict here. Because Brother is washing on day 1, but Xiao Yu is also washing on day 1? That can't be. Each day, only one person washes. So, conflict here.Therefore, this case is invalid. So if conversation is on day 2, and Brother is washing on day 1, but Xiao Yu is also washing on day 1 (due to Dad cooking day 1), which is impossible. Therefore, this combination is impossible.Therefore, the conversation cannot be on day 2. Thus, conversation must be on day 3.Let's check conversation day 3.Brother's schedule: washed dishes on day 2, cooks on day 4.Sister's schedule: has chores on days 4 and 5.Dad and Xiao Yu: Again, two possibilities.Possibility 1: Dad cooks day 1, washes day 4; Xiao Yu washes day 1, cooks day 4.Possibility 2: Dad cooks day 2, washes day 5; Xiao Yu washes day 2, cooks day 5.Let's check both.Case 1: Dad cooks day 1, washes day 4; Xiao Yu washes day 1, cooks day 4.So:Day 1:- Cook: Dad- Wash: Xiao YuDay 2:- Cook: ?- Wash: Brother (since Brother washed day 2? Wait, no. Wait, Brother washed dishes on day 2? Wait, if the conversation is on day 3, Brother washed yesterday (day 2) and is cooking tomorrow (day 4). So Brother's schedule:- Wash: day 2- Cook: day 4Therefore, on day 2, Brother is washing.But on day 2, someone else must be cooking.Day 2:- Cook: ?- Wash: BrotherDay 3:- Conversation day. We need to assign chores for day 3, 4, 5.Day 4:- Cook: Brother- Wash: Dad (since Dad washes day 4)- But Xiao Yu cooks day 4. Wait, conflict here.Wait, if Dad is washing on day 4, then Xiao Yu must be cooking on day 4. But Brother is also cooking on day 4. So two people cooking on day 4. Conflict. Therefore, invalid.Therefore, this case is invalid.Case 2: Dad cooks day 2, washes day 5; Xiao Yu washes day 2, cooks day 5.So:Day 1:- Cook: ?- Wash: ?Day 2:- Cook: Dad- Wash: Xiao YuBut Brother is washing on day 2. Wait, no, Brother's washing day is day 2 (since conversation is day 3, he washed yesterday, day 2). Therefore, Brother is washing on day 2, but Xiao Yu is also washing on day 2 (since Dad cooks day 2, Xiao Yu must wash day 2). Conflict again. Two people washing on day 2. Not allowed.Hence, this case is invalid. Therefore, even with conversation day 3, if we take Dad cooking day 2, it's invalid.Wait, maybe my initial assumption is wrong. Let me check again.If conversation is on day 3, Brother washed day 2 and will cook day 4. So Brother's assignments are:- Wash: day 2- Cook: day 4Sister has chores on days 4 and 5.Xiao Yu and Dad are paired. Let's try to see if Dad can cook on day 3? Wait, no, because Dad must cook on day N and wash on N + 3. If he cooks on day 3, he would need to wash on day 6, which is beyond the 5-day period. Therefore, Dad can only cook on day 1 or 2.But in the previous cases, both possibilities led to conflicts. So maybe there's another approach.Wait, perhaps the error is in assuming that the conversation day is 3. Let me check if there's an alternative.Alternatively, maybe the conversation is on day 4? But Brother's statement would require X + 1 ≤ 5, so X=4 would have Brother cooking on day 5, which is allowed, but Sister's statement would require X + 2 ≤5, so X=4 would have chores on days 5 and 6, which is invalid. So conversation can't be on day 4. Hence, only possible days are 2 or 3, but day 2 led to conflict, day 3 led to conflict. Hmm.Wait, maybe I made a mistake in interpreting the days. Let's verify.If conversation is on day 3:- Brother washed dishes yesterday (day 2) and cooks tomorrow (day 4).- Sister has chores tomorrow (day 4) and day after (day 5). So Sister has chores on days 4 and 5.So Sister must either cook or wash on days 4 and 5. Since each person does one cook and one wash, Sister has to do one chore on day 4 and one on day 5, but she might do cooking both days or washing both days? Wait, no. Each person must cook once and wash once. So Sister must do one cooking and one washing over the five days, but according to her statement, she has chores on days 4 and 5. So she must do one chore on day 4 and another on day 5. But she could be cooking one day and washing the other. So Sister's two chores are on days 4 and 5.Similarly, Brother's two chores are on days 2 and 4 (wash day 2, cook day 4).Now, Xiao Yu and Dad are always together. Let's list all possible Dad's cooking days.If Dad cooks on day 1, he washes on day 4. Then Xiao Yu would wash on day 1 and cook on day 4. But Brother is cooking on day 4, so Xiao Yu can't cook on day 4. Conflict.If Dad cooks on day 2, he washes on day 5. Then Xiao Yu would wash on day 2 and cook on day 5. However, Brother is washing on day 2, so Xiao Yu can't wash on day 2. Conflict.Wait, so both possibilities lead to conflicts. Hmm. That suggests that perhaps there's a misinterpretation of the clues.Wait, maybe the Dad's rest period is two days between cooking and washing. Let me check the original statement:Dad: After cooking, I can rest for two days before it's my turn to wash the dishes again.So, if Dad cooks on day N, he washes on day N + 3 (since after cooking on N, rest days are N+1 and N+2, then wash on N+3). So that part was correct.But if Dad can't cook on day 1 or 2 (due to conflicts), then there must be an error in the previous assumptions. Wait, but Dad must cook once, so he must cook on either day 1, 2, 3, 4, or 5. But with the rest period, cooking on day 3 would require washing on day 6, which is invalid. Hence, only days 1 and 2.But both days 1 and 2 lead to conflicts with Brother's and Xiao Yu's assignments. Hmm.Wait, perhaps the problem is that we haven't considered that the rest days are two full days, so cooking on day N, then rest days N+1 and N+2, then washing on N+3. So same as before.But given that, maybe the conflict arises because we need to assign Mom's chores as well. Let's try to map out all assignments step by step.Let's start over.We have five days (1-5). Each day has a cook and a washer. Each person does each chore once.Clues:1. Brother: washed yesterday, cooks tomorrow. So if today is X, then washed on X-1, cooks on X+1.2. Sister: has chores tomorrow and day after. So on X+1 and X+2.3. Xiao Yu and Dad are always on the same day: if Dad cooks, Xiao Yu washes, and vice versa.4. Dad: cooks on N, washes on N+3. So N=1 or 2.Let's consider possible conversation days (X). From Brother's statement, X can be 2,3,4. From Sister's, X can be 1,2,3. So overlapping days are 2,3.Let's try conversation day 3 first.Brother: washed on 2, cooks on 4.Sister: chores on 4 and 5.Dad: possible cooking days 1 or 2.Case 1: Dad cooks on 1, washes on 4. Then Xiao Yu washes on 1, cooks on 4.But Brother is cooking on 4, so conflict. Therefore, invalid.Case 2: Dad cooks on 2, washes on 5. Then Xiao Yu washes on 2, cooks on 5.But Brother is washing on 2, conflict. Therefore, invalid.So conversation day 3 leads to conflicts. Now check conversation day 2.Brother: washed on 1, cooks on 3.Sister: chores on 3 and 4.Dad: cooks on 1 or 2.Case 1: Dad cooks on 1, washes on 4. Xiao Yu washes on 1, cooks on 4.But Brother is washing on 1 (from Brother's schedule). Conflict. Can't have two washers on day 1.Case 2: Dad cooks on 2, washes on 5. Xiao Yu washes on 2, cooks on 5.But Brother's cooking day is 3, and Sister's chores are 3 and 4.Let's see:Day 1:- Cook: ?- Wash: Brother (since Brother washed on 1)But Dad is cooking on 2, so day 1's cook can't be Dad. It has to be someone else. Let's assign.Day 1:- Cook: Mom or Sister- Wash: BrotherDay 2:- Cook: Dad- Wash: Xiao YuDay 3:- Cook: Brother- Wash: ?Day 4:- Cook: ?- Wash: ?Day 5:- Cook: Xiao Yu- Wash: DadNow, Sister's chores are on days 3 and 4. So Sister must do one chore on day 3 and one on day 4.But let's see who's left. The people are Dad, Mom, Brother, Sister, Xiao Yu.Each person must cook and wash once. So far:- Dad: cooks 2, washes 5- Brother: washes 1, cooks 3- Xiao Yu: washes 2, cooks 5- Sister: needs to cook and wash on days 3 and 4- Mom: needs to cook and wash on remaining days.Let's check day 3:- Cook: Brother- Wash: ?Possible washers: Sister or Mom. If Sister washes on day 3, then she has to cook or wash on day 4. If Mom washes on day 3, then Sister has to do both chores on day 4, which is impossible (since each day only one cook and one washer). So Sister must wash on day 3 or cook on day 3. But day 3's cook is Brother, so Sister can't cook on day 3. Therefore, Sister must wash on day 3. Then her other chore is cooking or washing on day 4.Wait, Sister has chores on days 3 and 4. So she does one chore on day 3 and one on day 4. Since she's washing on day 3, she must do either cooking or washing on day 4. But she can't wash twice, so she must cook on day 4.Therefore:Day 3:- Cook: Brother- Wash: SisterDay 4:- Cook: Sister- Wash: ?Remaining washer on day 4: Mom.Then day 4:- Cook: Sister- Wash: MomThen day 5:- Cook: Xiao Yu- Wash: DadNow, let's check Mom's assignments. Mom needs to cook once and wash once. So far, Mom has washed on day 4. She needs to cook on one of the remaining days: day 1.Day 1's cook is still unassigned. Day 1:- Cook: Mom (since the only remaining person is Mom)- Wash: BrotherThus, Mom's assignments:- Cook: day 1- Wash: day 4Sister's assignments:- Wash: day 3- Cook: day 4Xiao Yu:- Wash: day 2- Cook: day 5Brother:- Wash: day 1- Cook: day 3Dad:- Cook: day 2- Wash: day 5Wait, but Dad's wash day is 5, which is N + 3 where N = 2 (cooking day). So 2 + 3 = 5, which is correct.Xiao Yu is paired with Dad: day 2 (Dad cooks, Xiao Yu washes) and day 5 (Dad washes, Xiao Yu cooks). Correct.Now, let's check if everyone has one cook and one wash:- Dad: cook 2, wash 5 ✔️- Mom: cook 1, wash 4 ✔️- Brother: wash 1, cook 3 ✔️- Sister: wash 3, cook 4 ✔️- Xiao Yu: wash 2, cook 5 ✔️Perfect. All assignments are valid. Now, the question is: the five-digit number formed by the dates when Dad, Mom, Brother, Sister, and Xiao Yu wash the dishes in sequence.So we need to list the days each person washes dishes, in the order Dad, Mom, Brother, Sister, Xiao Yu.From above:- Dad washes on day 5- Mom washes on day 4- Brother washes on day 1- Sister washes on day 3- Xiao Yu washes on day 2So the sequence is 5 (Dad), 4 (Mom), 1 (Brother), 3 (Sister), 2 (Xiao Yu). Therefore, the number is 54132.Let me verify again to ensure no mistakes.Day 1:- Cook: Mom- Wash: Brother ✔️Day 2:- Cook: Dad- Wash: Xiao Yu ✔️Day 3:- Cook: Brother- Wash: Sister ✔️Day 4:- Cook: Sister- Wash: Mom ✔️Day 5:- Cook: Xiao Yu- Wash: Dad ✔️Yes, all days have distinct cooks and washers, each person does one of each. Dad's schedule: cooks day 2, washes day 5 (2 + 3 = 5). Xiao Yu is paired with Dad on days 2 and 5. Sister has chores on days 3 and 4. Brother washed day 1, cooks day 3. The conversation day was day 2, but wait, in this scenario, the conversation would have to be on day 2, but Brother's statement would be "washed yesterday (day 1), cook tomorrow (day 3)". Sister's statement would be "chores tomorrow (day 3) and day after (day 4)", which matches her assignments. So the conversation is on day 2.Wait, but earlier we thought conversation day was day 2, but when we tried that, there was a conflict. Wait, how did that resolve?Wait, no, earlier when we tried conversation day 2, we assumed Dad cooking day 1, but in this correct case, Dad is cooking day 2. So perhaps the error was in the initial assumption. Let's recap.When the conversation is on day 2:- Brother says he washed yesterday (day 1) and cooks tomorrow (day 3). Correct, as Brother washes day 1 and cooks day 3.- Sister says she has chores tomorrow (day 3) and day after (day 4). Which is correct, Sister washes day 3 and cooks day 4.- Xiao Yu says Dad and I always do chores on the same day. Correct, as they are paired on day 2 (Dad cooks, Xiao Yu washes) and day 5 (Dad washes, Xiao Yu cooks).- Dad says after cooking, he rests two days before washing. Dad cooks day 2, rests days 3 and 4, washes day 5. Correct.Therefore, all statements are satisfied with the conversation happening on day 2.But initially, when I considered conversation day 2, I incorrectly assumed Dad was cooking day 1, leading to conflict. But actually, in the correct schedule, Dad is cooking day 2, which avoids the conflict.Therefore, the correct five-digit number is 54132.</think>The five-digit number formed by the dates when Dad, Mom, Brother, Sister, and Xiao Yu wash the dishes in sequence is determined by analyzing the given clues and scheduling their chores accordingly. Here's the breakdown:1. Dad's Schedule: - Cooks on day 2 and washes on day 5 (resting days 3 and 4 in between). 2. Xiao Yu's Schedule: - Washes on day 2 (paired with Dad's cooking) and cooks on day 5 (paired with Dad's washing).3. Brother's Schedule: - Washes on day 1 and cooks on day 3 (as stated in the conversation).4. Sister's Schedule: - Washes on day 3 and cooks on day 4 (chores on days 3 and 4 as per the conversation).5. Mom's Schedule: - Cooks on day 1 and washes on day 4 (remaining chores after assigning others).Washing Days Sequence:- Dad: Day 5- Mom: Day 4- Brother: Day 1- Sister: Day 3- Xiao Yu: Day 2Thus, the five-digit number is boxed{54132}.]

🔑:54132

🧠:[<think>Okay, so I need to find the length of the third median CM in triangle ABC, given that the other two medians AN and BP are 3 and 6 units long respectively, and the area of the triangle is 3√15. The options are given, but I should try to solve it step by step.First, I remember that in a triangle, the medians intersect at the centroid, which divides each median into a ratio of 2:1, with the longer part being closer to the vertex. But I'm not sure how that directly helps here. Maybe I need a formula that relates the lengths of the medians to the area of the triangle?Wait, there's a formula for the area of a triangle in terms of its medians. But I can't recall it exactly. Maybe it's similar to Heron's formula? Let me think. Heron's formula uses the lengths of the sides, but here we have medians. Alternatively, there's another approach using vectors or coordinates?Alternatively, I remember that if we have the lengths of all three medians, we can use a formula similar to Heron's formula to compute the area. The formula is called Apollonius's theorem, but maybe not exactly. Apollonius's theorem relates the length of the medians to the sides. Let me check.Apollonius's theorem states that the sum of the squares of any two sides of a triangle equals twice the square of the median to the third side plus twice the square of half the third side. For example, for median AN, which is from vertex A to side BC, the theorem states:AB² + AC² = 2AN² + 2(BC/2)²Similarly for the other medians. But if we don't know the sides, maybe this would lead to a system of equations?But with three medians, maybe we can relate them to the area. Alternatively, there's a formula that gives the area in terms of the three medians. Let me try to recall or derive it.I think the area can be calculated using the formula:Area = (4/3) * area formed by the medians.Wait, is that correct? If you form a triangle from the medians of the original triangle, then the area of the original triangle is (4/3) times the area of the median triangle. Hmm, that might be useful. Let me confirm.Suppose we have a triangle with medians m_a, m_b, m_c. Then the area of the original triangle is (4/3) times the area of the triangle formed by the medians. So if I can compute the area of the median triangle, multiply by 4/3, and set it equal to 3√15. But wait, in this problem, we have two medians and need to find the third. So if we can form a triangle with the three medians, compute its area using Heron's formula, and then multiply by 4/3 to get the original area. But since we don't have the third median, perhaps we can set up an equation.Wait, let's denote the medians as m_a = AN = 3, m_b = BP = 6, and m_c = CM = x (which we need to find). Then, the area of the original triangle is (4/3) times the area of the triangle formed by sides 3, 6, and x. According to the problem, the area is 3√15. So:(4/3) * Area_median_triangle = 3√15Therefore, Area_median_triangle = (3√15) * (3/4) = (9√15)/4So, the area of the triangle formed by the medians is (9√15)/4. Now, using Heron's formula on the median triangle with sides 3, 6, and x.First, compute the semi-perimeter: s = (3 + 6 + x)/2 = (9 + x)/2Then, the area is sqrt[s(s - 3)(s - 6)(s - x)] = (9√15)/4Therefore,sqrt[ ((9 + x)/2 ) * ((9 + x)/2 - 3) * ((9 + x)/2 - 6) * ((9 + x)/2 - x) ] = (9√15)/4Simplify each term:First term: ((9 + x)/2 )Second term: ((9 + x)/2 - 3) = (9 + x - 6)/2 = (3 + x)/2Third term: ((9 + x)/2 - 6) = (9 + x - 12)/2 = (x - 3)/2Fourth term: ((9 + x)/2 - x) = (9 + x - 2x)/2 = (9 - x)/2So, the product inside the square root becomes:[ (9 + x)/2 * (3 + x)/2 * (x - 3)/2 * (9 - x)/2 ] = [ (9 + x)(9 - x)(x + 3)(x - 3) ) ] / 16Notice that (9 + x)(9 - x) = 81 - x² and (x + 3)(x - 3) = x² - 9. Therefore, the product becomes:[ (81 - x²)(x² - 9) ] / 16So, putting back into the equation:sqrt[ (81 - x²)(x² - 9)/16 ] = (9√15)/4Multiply both sides by 4:sqrt[ (81 - x²)(x² - 9) ] = 9√15Square both sides:(81 - x²)(x² - 9) = 81² * 15Wait, let's compute the right side first. 9√15 squared is 81*15 = 1215. So:(81 - x²)(x² - 9) = 1215Let’s expand the left side:81x² - 729 - x⁴ + 9x² = (81x² + 9x²) - x⁴ - 729 = 90x² - x⁴ - 729So:-x⁴ + 90x² - 729 = 1215Bring all terms to left side:-x⁴ + 90x² - 729 - 1215 = 0 => -x⁴ + 90x² - 1944 = 0Multiply both sides by -1:x⁴ - 90x² + 1944 = 0Let’s set y = x², so equation becomes:y² - 90y + 1944 = 0Solving quadratic equation:y = [90 ± sqrt(90² - 4*1*1944)] / 2Compute discriminant:90² = 81004*1*1944 = 7776So discriminant is 8100 - 7776 = 324Therefore:y = [90 ± sqrt(324)] / 2 = [90 ± 18]/2Thus,y = (90 + 18)/2 = 108/2 = 54ory = (90 - 18)/2 = 72/2 = 36Therefore, x² = 54 or x² = 36Therefore, x = √54 = 3√6 or x = 6But x is the length of the median CM. So possible answers are 3√6 or 6. But the options include 6 (option A) and 3√6 (option C). So which one is correct?Wait, we need to check if both solutions are valid. Since medians must satisfy the triangle inequality in the median triangle. The medians must be able to form a triangle. Let's check both possibilities.First, if CM = 6, then the medians are 3, 6, 6. Check if 3, 6, 6 can form a triangle.3 + 6 > 6? 9 > 6, yes.6 + 6 > 3? 12 > 3, yes.3 + 6 > 6? Same as first, yes. So that's okay.Second, if CM = 3√6 ≈ 7.348, then the medians are 3, 6, 7.348. Check triangle inequality.3 + 6 > 7.348? 9 > 7.348, yes.6 + 7.348 > 3? 13.348 > 3, yes.3 + 7.348 > 6? 10.348 > 6, yes. So both are possible.But the original triangle has unequal sides. Wait, the problem states "triangle ABC with unequal sides". So the original triangle is scalene, so all medians should be of different lengths? Wait, but in the case where CM = 6, we have medians 3, 6, 6. So two medians are equal. Does that correspond to the original triangle being scalene?Wait, in a triangle, if two medians are equal, then the triangle is isosceles. Because if two medians are equal, the triangle must have two sides equal. Wait, is that true?Yes, actually, in a triangle, if two medians are equal, then the triangle is isosceles. Let me recall. For example, if the medians from A and B are equal, then the triangle is isosceles with AC = BC. Similarly, if two medians are equal, the triangle is isosceles. Therefore, if the original triangle has unequal sides (i.e., scalene), then all medians must be of different lengths. Therefore, CM cannot be 6, because that would make two medians equal (BP = CM = 6), implying the triangle is isosceles, which contradicts the given that the triangle has unequal sides.Therefore, the correct answer must be 3√6, which is option C.Let me verify this conclusion.If CM = 3√6 ≈ 7.348, then all medians are distinct: 3, 6, 7.348, which fits the triangle being scalene.Alternatively, if CM = 6, then two medians are equal (6 and 6), leading to an isosceles triangle, which is not allowed. Therefore, the answer must be 3√6.Hence, the correct option is C.Final Answerboxed{C}</think>Given the problem of finding the length of the median ( CM ) in triangle ( triangle ABC ) with unequal sides, where the medians ( AN = 3 ), ( BP = 6 ), and the area of the triangle is ( 3sqrt{15} ):1. Recall the formula relating the area of a triangle to its medians: The area of the original triangle is ( frac{4}{3} ) times the area of the triangle formed by its medians. Let the medians be ( m_a = 3 ), ( m_b = 6 ), and ( m_c = x ). The area of the original triangle is given as ( 3sqrt{15} ), so the area of the median triangle is ( frac{9sqrt{15}}{4} ).2. Apply Heron's formula to the median triangle: The semi-perimeter ( s ) of the median triangle is ( frac{3 + 6 + x}{2} ). Using Heron's formula, the area of the median triangle is: [ sqrt{s(s - 3)(s - 6)(s - x)} = frac{9sqrt{15}}{4} ]3. Simplify and solve the equation: Substituting ( s = frac{9 + x}{2} ) into the equation, we get: [ sqrt{left( frac{(9 + x)(3 + x)(x - 3)(9 - x)}{16} right)} = frac{9sqrt{15}}{4} ] Simplifying the product inside the square root: [ sqrt{frac{(81 - x^2)(x^2 - 9)}{16}} = frac{9sqrt{15}}{4} ] Squaring both sides and solving the resulting equation: [ (81 - x^2)(x^2 - 9) = 1215 ] This leads to the quartic equation: [ x^4 - 90x^2 + 1944 = 0 ] Solving this quadratic in terms of ( y = x^2 ), we find ( y = 54 ) or ( y = 36 ), giving ( x = 3sqrt{6} ) or ( x = 6 ).4. Check for triangle inequality and scalene condition: Since the original triangle has unequal sides, all medians must be distinct. If ( CM = 6 ), two medians would be equal, implying an isosceles triangle, which is a contradiction. Thus, the valid solution is ( x = 3sqrt{6} ).Thus, the length of the median ( CM ) is (boxed{C}).,<think>Okay, so I need to find the length of median CM in triangle ABC, given that the other two medians AN and BP are 3 and 6 units long, respectively, and the area of the triangle is 3√15. The answer choices are provided, so maybe I can use some formulas related to medians and area?First, I remember that in a triangle, the three medians intersect at the centroid, which divides each median into a ratio of 2:1. But how does that help here? Hmm. Maybe there's a formula that relates the lengths of the medians to the area of the triangle. Let me think.Wait, there's a formula called Apollonius's theorem which relates the lengths of the medians to the sides of the triangle. Apollonius's theorem states that the sum of the squares of any two sides of a triangle equals twice the square of the median to the third side plus twice the square of half the third side. But if I don't know the sides of the triangle, maybe that's not directly helpful here.Alternatively, I recall that there's a formula for the area of a triangle in terms of its medians. But I can't remember the exact formula. Maybe it's similar to Heron's formula but for medians? Let me check my thoughts.Yes, the formula for the area of a triangle given the lengths of its three medians is called formula using medians, and it is similar to Heron's formula. The formula is:Area = (4/3) * √[s(s - m1)(s - m2)(s - m3)]where m1, m2, m3 are the lengths of the medians, and s is the semi-sum of the medians: s = (m1 + m2 + m3)/2.But wait, in this problem, we have two medians and the area, and we need to find the third median. So maybe I can set up the equation using this formula and solve for the third median.Let me write down the given information:AN = 3, BP = 6, Area = 3√15, need to find CM.Let's denote the medians as m_a, m_b, m_c. Typically, m_a is the median from vertex A, m_b from B, and m_c from C. But need to confirm which median is which here.Given AN is the median from A to BC, so that would be m_a = 3. BP is the median from B to AC, so that's m_b = 6. CM is the median from C to AB, which is m_c, the one we need to find.So, using the formula:Area = (4/3) * √[s(s - m_a)(s - m_b)(s - m_c)]where s = (m_a + m_b + m_c)/2Plugging in the known values:3√15 = (4/3) * √[s(s - 3)(s - 6)(s - m_c)]But since we need to find m_c, this seems a bit complicated. Let's denote m_c as x for simplicity.So,3√15 = (4/3) * √[s(s - 3)(s - 6)(s - x)]where s = (3 + 6 + x)/2 = (9 + x)/2Therefore,3√15 = (4/3) * √[ ((9 + x)/2 ) * ( (9 + x)/2 - 3 ) * ( (9 + x)/2 - 6 ) * ( (9 + x)/2 - x ) ]Let me simplify each term inside the square root step by step.First, compute each factor:s = (9 + x)/2s - 3 = (9 + x)/2 - 6/2 = (9 + x - 6)/2 = (3 + x)/2s - 6 = (9 + x)/2 - 12/2 = (x - 3)/2s - x = (9 + x)/2 - 2x/2 = (9 - x)/2Therefore, the product inside the square root becomes:[(9 + x)/2] * [(3 + x)/2] * [(x - 3)/2] * [(9 - x)/2]Which can be written as:[(9 + x)(9 - x)(x + 3)(x - 3)] / (2^4)= [(81 - x²)(x² - 9)] / 16So, substituting back into the area formula:3√15 = (4/3) * √[ (81 - x²)(x² - 9)/16 ]Let me simplify the right-hand side step by step.First, square both sides to eliminate the square root:(3√15)^2 = [(4/3)^2 * (81 - x²)(x² - 9)/16 ]Calculate left-hand side:9 * 15 = 135Right-hand side:(16/9) * (81 - x²)(x² - 9)/16 = (1/9) * (81 - x²)(x² - 9)Thus:135 = (1/9)(81 - x²)(x² - 9)Multiply both sides by 9:1215 = (81 - x²)(x² - 9)Let me expand the right-hand side:(81 - x²)(x² - 9) = 81x² - 729 - x^4 + 9x² = (81x² + 9x²) - x^4 - 729 = 90x² - x^4 - 729So,1215 = -x^4 + 90x² - 729Bring all terms to left-hand side:x^4 - 90x² + 1215 + 729 = 0x^4 - 90x² + 1944 = 0Let me set y = x², then equation becomes:y² - 90y + 1944 = 0Solve quadratic equation:y = [90 ± √(90² - 4*1*1944)] / 2Calculate discriminant:90² = 81004*1*1944 = 7776Discriminant = 8100 - 7776 = 324Thus,y = [90 ± √324]/2 = [90 ± 18]/2So,y = (90 + 18)/2 = 108/2 = 54or y = (90 - 18)/2 = 72/2 = 36Therefore, x² = 54 or x² = 36Thus, x = √54 = 3√6 or x = 6 (since length can't be negative)But x is the length of median CM, so possible values are 3√6 or 6. Need to check which one is valid.Wait, but the answer choices include both 6 and 3√6. So which one is correct?Wait, maybe both solutions are possible, but we need to check which one is consistent with the triangle inequality for medians or something else.Alternatively, maybe only one of them gives a positive area when plugged back into the formula. Let me check.If x = 6, then let's compute s = (9 + 6)/2 = 15/2 = 7.5Then the product inside the square root becomes:[(81 - 6²)(6² - 9)] = (81 - 36)(36 - 9) = 45 * 27 = 1215Thus, Area = (4/3) * √(1215 / 16) = (4/3) * (√1215)/4 = (√1215)/3But √1215 = √(81*15) = 9√15, so Area = 9√15 / 3 = 3√15, which matches the given area. So x=6 is a valid solution.Similarly, if x = 3√6, which is approximately 7.348, then s = (9 + 3√6)/2Then compute (81 - x²)(x² - 9):x² = 54, so 81 - 54 = 27, and 54 - 9 = 45, so product is 27*45 = 1215 again.Therefore, Area would also be 3√15. So both x=6 and x=3√6 satisfy the equation. But this is a problem because the answer options have both 6 and 3√6. So how do we decide which one is correct?Wait, but in the problem statement, it's mentioned that the triangle has unequal sides. So if the medians are 3, 6, and 6, would that imply the triangle has equal sides? Wait, medians being equal would not necessarily mean the sides are equal. Let me think.Wait, if two medians are equal, does that imply the triangle is isoceles? Yes, in a triangle, if two medians are equal, then the triangle is isoceles with respect to the sides those medians are drawn from. For example, if medians from A and B are equal, then sides BC and AC are equal. But here, if two medians are 3 and 6, and the third is either 6 or 3√6.Wait, but in the problem statement, it says "triangle ABC with unequal sides", so the triangle is scalene. So if we have two medians equal, that would contradict the triangle being scalene. Therefore, if x=6, then the medians would be 3,6,6. But if two medians are equal, that would imply the triangle is isoceles, but the problem states it's a triangle with unequal sides. Therefore, x=6 is invalid, because that would make two medians equal, implying the triangle is isoceles, contradicting the given condition. Therefore, the correct answer must be x=3√6, which is option C.Therefore, the length of CM is 3√6.Final Answerboxed{C}</think>Given the problem, we need to find the length of median ( CM ) in triangle ( ABC ) with unequal sides, given that the other two medians ( AN = 3 ) and ( BP = 6 ), and the area of the triangle is ( 3sqrt{15} ).We use the formula for the area of a triangle in terms of its medians:[text{Area} = frac{4}{3} sqrt{s(s - m_a)(s - m_b)(s - m_c)}]where ( m_a, m_b, m_c ) are the lengths of the medians, and ( s = frac{m_a + m_b + m_c}{2} ).Given:- ( m_a = 3 )- ( m_b = 6 )- ( text{Area} = 3sqrt{15} )Let ( m_c = x ). Then, ( s = frac{3 + 6 + x}{2} = frac{9 + x}{2} ).Substituting into the area formula:[3sqrt{15} = frac{4}{3} sqrt{left( frac{9 + x}{2} right) left( frac{9 + x}{2} - 3 right) left( frac{9 + x}{2} - 6 right) left( frac{9 + x}{2} - x right)}]Simplifying each term inside the square root:[left( frac{9 + x}{2} right) left( frac{3 + x}{2} right) left( frac{x - 3}{2} right) left( frac{9 - x}{2} right)]This simplifies to:[frac{(81 - x^2)(x^2 - 9)}{16}]Substituting back into the area formula and squaring both sides:[(3sqrt{15})^2 = left( frac{4}{3} right)^2 cdot frac{(81 - x^2)(x^2 - 9)}{16}]Simplifying:[135 = frac{1}{9} left( (81 - x^2)(x^2 - 9) right)]Solving the quadratic equation:[x^4 - 90x^2 + 1944 = 0]Let ( y = x^2 ):[y^2 - 90y + 1944 = 0]Solving for ( y ):[y = frac{90 pm 18}{2} implies y = 54 text{ or } y = 36]Thus, ( x = sqrt{54} = 3sqrt{6} ) or ( x = 6 ). Since the triangle has unequal sides, the medians must also be unequal. Therefore, ( x = 6 ) is invalid as it would imply two equal medians, contradicting the triangle being scalene. Hence, the correct answer is ( 3sqrt{6} ).[boxed{C}]]

🔑:6. C.As shown in Figure 4, let the centroid of triangle ABC be G, and CM = 3x. Then CG = 2x, GM = x.Extend CM to point Q such that MQ = GM. Connect BQ and AQ. Then quadrilateral AQB G is a parallelogram.Thus, AQ = BG = 4.Also, S_{triangle AQC} = S_{triangle ABG} = frac{1}{3} S_{triangle ABC} = sqrt{15}.In triangle AQG, AG = 2, AQ = 4, QG = 2x, and the semi-perimeter p = frac{1}{2}(2 + 4 + 2x) = 3 + x.By Heron's formula,sqrt{15} = sqrt{(3 + x)(3 - x)(x - 1)(x + 1)}.Simplifying, we get x^4 - 10x^2 + 24 = 0.Solving, we find x^2 = 4 or x^2 = 6.Thus, x = 2 or x = sqrt{6}.When x = 2, CM = 6.Given BP = 6, it follows that BA = AC.This contradicts the given conditions.Therefore, x = sqrt{6}, and CM = 3sqrt{6}.

🧠:[<think>Okay, let's tackle this problem step by step. So, we have the function f(x) = x³ + 2f’(1)x² + 1. And we need to find the equation of the tangent line at the point (1, f(1)). The options are given as four different linear equations, so once we find the tangent line's equation, we can match it with the options provided.First, let me recall that the equation of a tangent line at a point (a, f(a)) is given by y = f(a) + f’(a)(x - a). In this case, a is 1, so the tangent line will be y = f(1) + f’(1)(x - 1). So, to find the equation, we need both f(1) and f’(1). But here's the catch: the function f(x) itself includes the term 2f’(1)x². That means f’(1) is part of the function's definition, so we can't directly compute f’(x) without knowing f’(1). This seems like we'll have to set up an equation to solve for f’(1) first. Let me write down the function again: f(x) = x³ + 2f’(1)x² + 1. Let's denote f’(1) as some constant, say c. Then the function becomes f(x) = x³ + 2c x² + 1. Now, if we can find the value of c, we can substitute it back into the function and compute both f(1) and f’(1).To find c, which is f’(1), we need to first compute the derivative f’(x) of the function f(x). Let's do that.f(x) = x³ + 2c x² + 1Taking the derivative term by term:The derivative of x³ is 3x².The derivative of 2c x² is 4c x (since 2c is a constant, multiplied by 2 from x² gives 4c x).The derivative of 1 is 0.So, f’(x) = 3x² + 4c x.Now, since c is f’(1), we can substitute x = 1 into the derivative to find c.Therefore:c = f’(1) = 3(1)² + 4c(1) = 3 + 4c.Wait, that gives us an equation where c is expressed in terms of itself. Let me write that again:c = 3 + 4c.Hmm, solving for c. Let's subtract 4c from both sides:c - 4c = 3-3c = 3Divide both sides by -3:c = -1.So, f’(1) is -1. That means the constant c in the original function is -1. Let me substitute that back into the original function.So, f(x) = x³ + 2*(-1)x² + 1 = x³ - 2x² + 1.Now, let's compute f(1):f(1) = (1)³ - 2*(1)² + 1 = 1 - 2 + 1 = 0.So, the point (1, f(1)) is (1, 0).Next, we already found that f’(1) = -1. Therefore, the slope of the tangent line at x = 1 is -1.Now, let's write the equation of the tangent line using the point-slope form. The general formula is:y - f(1) = f’(1)(x - 1)Substituting the known values:y - 0 = -1*(x - 1)Simplifying:y = -x + 1To write this in the standard form as given in the options, let's bring all terms to one side:x + y - 1 = 0Looking at the options provided:A: x + y - 1 = 0B: x - y + 1 = 0C: x - y - 1 = 0D: x + y + 1 = 0Therefore, the correct answer should be option A.But wait, let me double-check all steps to make sure I didn't make a mistake.First, we set f’(1) = c, then wrote f(x) as x³ + 2c x² + 1. Then, took the derivative to get f’(x) = 3x² + 4c x. Then substituted x = 1 into the derivative to get c = 3 + 4c. Solving gives c = -1. Then f(1) = 1 - 2 + 1 = 0. Then, the tangent line is y = -1(x - 1), which is y = -x + 1. Converting to standard form: x + y = 1 => x + y - 1 = 0. Yes, that's option A. Wait, but just to make sure, let me check if there is any miscalculation here. Let's verify the derivative again. The function f(x) is x³ + 2c x² + 1. The derivative should be 3x² + 4c x, because 2c times x squared derivative is 4c x. So, yes, 2c * 2x = 4c x. So, f’(x) = 3x² + 4c x. Then f’(1) is 3 + 4c, which equals c. So c = 3 + 4c. Then, solving for c gives c = -1. That seems correct.Then f(1) = 1 - 2 + 1 = 0. Correct.Equation of the tangent line: slope is -1, passing through (1,0). So, y = -x + 1. Converting to standard form: x + y - 1 = 0. That's option A.Alternatively, let me check if another option could be equivalent. For example, if we multiply through by -1, we might get -x - y + 1 = 0, but that's not one of the options. The options given are A: x + y - 1, B: x - y +1, C: x - y -1, D: x + y +1. So A is the only one matching. Therefore, the answer is A.But hold on, the original problem had f(x) = x³ + 2f’(1)x² + 1. So when we substituted f’(1) as c, we had 2c x². Then, derivative was 3x² + 4c x. Wait, is that correct? Let me check again. The derivative of 2c x² is 4c x? Wait, hold on. Wait, if you have 2c x², then the derivative should be 2c * 2x = 4c x. Yes, that's correct. So, f’(x) = 3x² + 4c x. Then, when x=1, that's 3*1 + 4c*1 = 3 + 4c. Which we set equal to c. So 3 + 4c = c. Then, 3 = -3c => c = -1. Correct.Alternatively, maybe the problem is with the standard form? Let me check the equation again. The tangent line is y = -x + 1. So, rearranged, x + y = 1, which is x + y - 1 = 0. That's exactly option A. So that's correct. Therefore, the answer should be A. But wait, the options in the problem are labeled A to D. Let me check again:A: x + y -1 =0B: x - y +1 =0C: x - y -1 =0D: x + y +1 =0Yes, A is x + y -1 =0, which is the same as x + y = 1. So that's the equation we derived. Therefore, the correct answer is A.Wait, but let me think again. Sometimes, people might miscalculate signs. Let me rederive the tangent line. At point (1,0), slope is -1. So point-slope form: y - 0 = -1(x - 1). So, y = -x +1. Adding x to both sides: x + y = 1. Subtract 1: x + y -1 =0. Yes, that's correct. So no mistake here. So option A is correct.Therefore, after all these steps, I confirm that the answer is option A.Final Answerboxed{A}</think>Given the function ( f(x) = x^3 + 2f'(1)x^2 + 1 ), we need to find the equation of the tangent line at the point ((1, f(1))).1. Express ( f(x) ) in terms of ( c ): Let ( c = f'(1) ). Then the function becomes: [ f(x) = x^3 + 2cx^2 + 1 ]2. Compute the derivative ( f'(x) ): [ f'(x) = 3x^2 + 4cx ]3. Solve for ( c ): Substitute ( x = 1 ) into the derivative: [ c = f'(1) = 3(1)^2 + 4c(1) implies c = 3 + 4c ] Solving for ( c ): [ c - 4c = 3 implies -3c = 3 implies c = -1 ]4. Substitute ( c = -1 ) back into the function: [ f(x) = x^3 - 2x^2 + 1 ]5. Calculate ( f(1) ): [ f(1) = 1^3 - 2(1)^2 + 1 = 1 - 2 + 1 = 0 ]6. Determine the slope of the tangent line: The slope at ( x = 1 ) is ( f'(1) = -1 ).7. Form the equation of the tangent line: Using the point-slope form ( y - f(1) = f'(1)(x - 1) ): [ y - 0 = -1(x - 1) implies y = -x + 1 ] Converting to standard form: [ x + y - 1 = 0 ]Thus, the equation of the tangent line is (boxed{A}).,<think>Okay, so I need to find the equation of the tangent line to the function f(x) at the point (1, f(1)). The function is given as f(x) = x³ + 2f’(1)x² + 1. The options are A through D, each with different equations. Let me think step by step how to approach this.First, I remember that the equation of the tangent line at a point (a, f(a)) is given by y = f(a) + f’(a)(x - a). In this case, the point is (1, f(1)), so a is 1. That means I need to find f(1) and f’(1) to plug into the tangent line formula.But wait, the function f(x) itself includes the term 2f’(1)x². That seems a bit circular because f’(1) is part of the function's definition. So I can’t just compute f’(x) directly without knowing f’(1). Hmm, this must mean I need to solve for f’(1) first.Let me write down the function again: f(x) = x³ + 2f’(1)x² + 1. So f(x) depends on f’(1), which is the derivative of f at x=1. Let me denote f’(1) as some constant, say c, to make it easier. Let’s let c = f’(1). Then the function becomes f(x) = x³ + 2c x² + 1.Now, if I can find the value of c, then I can fully define f(x) and compute both f(1) and f’(1). So the key here is to find the value of c. To do that, I need to take the derivative of f(x) and then evaluate it at x=1, which should give me c. Let me proceed step by step.First, compute f’(x). The derivative of f(x) with respect to x is:f’(x) = d/dx [x³ + 2c x² + 1] = 3x² + 4c x + 0 = 3x² + 4c x.Now, evaluate f’(1):f’(1) = 3(1)² + 4c(1) = 3 + 4c.But we defined c as f’(1), so:c = 3 + 4c.Wait, that equation is c = 3 + 4c. Let me solve for c:Subtract 4c from both sides:c - 4c = 3-3c = 3Divide both sides by -3:c = -1.Okay, so f’(1) is -1. That means the original function f(x) is:f(x) = x³ + 2*(-1)*x² + 1 = x³ - 2x² + 1.Now, let me verify that. If f(x) = x³ - 2x² + 1, then f’(x) would be 3x² - 4x. Then f’(1) is 3(1)² - 4(1) = 3 - 4 = -1, which matches the value we found for c. Great, that checks out.Now, let's compute f(1). Plugging x=1 into f(x):f(1) = (1)^3 - 2*(1)^2 + 1 = 1 - 2 + 1 = 0.So the point (1, f(1)) is (1, 0). Now, we need the equation of the tangent line at this point. We have f’(1) = -1, so the slope of the tangent line is -1.Using the point-slope form of a line: y - f(1) = f’(1)(x - 1). Plugging in the values:y - 0 = -1*(x - 1)Simplify:y = -x + 1.But the answer options are given in the form Ax + By + C = 0. Let me convert y = -x + 1 to that form.Bring all terms to the left side:x + y - 1 = 0.Looking at the options, that's option A: x + y - 1 = 0.Wait, but let me double-check everything to be sure.First, f(x) was correctly derived by substituting c = -1, so f(x) = x³ - 2x² + 1. Then f(1) is indeed 0. Then f’(x) = 3x² - 4x, so f’(1) = 3 - 4 = -1. So the slope is -1. Then the tangent line equation is y = -x + 1, which converts to x + y - 1 = 0, which is option A.But wait, let me check the answer options again:A: x + y -1 = 0B: x - y +1 =0C: x - y -1=0D: x + y +1=0Yes, A is x + y -1 =0. But hold on, I just want to make sure I didn't make a mistake in the algebra when converting y = -x +1 to the standard form.Starting from y = -x + 1.Add x to both sides: x + y = 1.Subtract 1 from both sides: x + y -1 =0. That's correct. So that is option A. So the answer should be A.But wait, let me check the other options just to be thorough.Option B: x - y +1=0. That would be x - y = -1, or y = x +1. Which has a slope of 1, but our slope is -1, so not B.Option C: x - y -1=0, which is x - y =1, so y = x -1. Slope 1, not our slope.Option D: x + y +1=0, which is y = -x -1. Slope -1, but the intercept is -1, but our equation is y = -x +1, so intercept +1, so D is different.So yes, A is correct. Therefore, the answer is A.Wait, but hold on. Let me check once more if the point (1,0) is indeed on the tangent line. For equation A: x + y -1 =0. Plugging in x=1, y=0: 1 +0 -1 =0. 0=0, which is true. So yes, the point is on the line. And the slope is -1. So that's correct.Alternatively, maybe I should write the tangent line equation in another way. Let me see:Slope is -1, point is (1,0). So using slope-intercept form: y = mx + b. Plugging in x=1, y=0:0 = -1*(1) + b => 0 = -1 + b => b=1. So equation is y = -x +1, which is same as x + y -1=0. So yes, that's correct. So answer A is correct. So the answer is A.Wait, but the options are given with letters, and in the problem statement, the options are labeled A to D. So according to my reasoning, the correct answer is A. So I should check the options again.Wait, the options given are:A: x + y -1 =0B: x - y +1=0C: x - y -1=0D: x + y +1=0Yes, so A is x + y -1=0, which is what I derived. So that's correct. Therefore, the answer is option A.But wait, hold on a second. Let me check once again if there's any miscalculation. Sometimes when problems involve f’(1) in the function itself, there could be a trick, but here I think it's straightforward.We had f(x) = x³ + 2f’(1)x² +1. Let me rederive f’(1) once again.Set c = f’(1). Then f(x) = x³ + 2c x² +1. Then f’(x) = 3x² + 4c x. Then f’(1) = 3(1) +4c(1)=3 +4c. But since c = f’(1), we have c =3 +4c. Solving gives c= -1. Then f(1)=1 + 2*(-1)*1 +1=1 -2 +1=0. Then f’(1)= -1, so tangent line is y= -1(x-1) +0= -x +1. So equation is x + y -1=0. So that's correct.Therefore, the answer is A.But wait, the problem is given as multiple-choice with options, so I need to make sure that in the problem statement, the function is indeed written as f(x) =x³ +2f’(1)x² +1. Let me check again.Original problem: "Let the function f(x) = x³ + 2f'(1)x² + 1. The equation of the tangent line to the function f(x) at the point (1, f(1)) is ( )"Yes, that's correct. So the function is defined in terms of its own derivative at x=1, which creates an equation we can solve for f’(1). So my approach was correct.Another way to think about it: if someone is confused about the f’(1) inside the function, but treating it as a constant coefficient makes sense because f’(1) is just a number (once computed), even though it's part of the function's definition. So solving for that constant first allows us to fully define f(x) before computing the tangent line.Therefore, all steps check out. So the answer is A.Final Answerboxed{A}</think>Given the function ( f(x) = x^3 + 2f'(1)x^2 + 1 ), we need to find the equation of the tangent line at the point ((1, f(1))).First, we denote ( f'(1) ) as ( c ). This transforms the function into:[ f(x) = x^3 + 2cx^2 + 1 ]Next, we compute the derivative ( f'(x) ):[ f'(x) = 3x^2 + 4cx ]Evaluating the derivative at ( x = 1 ):[ f'(1) = 3(1)^2 + 4c(1) = 3 + 4c ]Since ( c = f'(1) ), we substitute ( c ) into the equation:[ c = 3 + 4c ]Solving for ( c ):[ c - 4c = 3 ][ -3c = 3 ][ c = -1 ]Thus, ( f'(1) = -1 ). Substituting ( c = -1 ) back into the original function:[ f(x) = x^3 - 2x^2 + 1 ]We then compute ( f(1) ):[ f(1) = (1)^3 - 2(1)^2 + 1 = 1 - 2 + 1 = 0 ]The point is ((1, 0)) and the slope of the tangent line at this point is ( f'(1) = -1 ). Using the point-slope form of the tangent line equation:[ y - 0 = -1(x - 1) ][ y = -x + 1 ]Converting this to the standard form:[ x + y - 1 = 0 ]Thus, the equation of the tangent line is:[boxed{A}]]

🔑:First, we find the derivative of f(x), which is f'(x) = 3x^2 + 4f'(1)x. Setting x=1, we get f'(1) = 3 + 4f'(1), solving this equation yields f'(1) = -1.Therefore, the function becomes f(x) = x^3 - 2x^2 + 1, and its derivative is f'(x) = 3x^2 - 4x.Thus, f(1) = 0 and f'(1) = -1.Therefore, the equation of the tangent line to f(x) at the point (1, f(1)) is y = -(x-1), which simplifies to x + y - 1 = 0.Hence, the correct option is boxed{text{A}}.