🧠:Okay, so I need to find the values of A, B, and C such that the fraction on the left side is equal to the sum of the two fractions on the right side. This seems like a partial fraction decomposition problem. Let me recall how partial fractions work. Generally, when you have a rational function and you want to decompose it into simpler fractions, you factor the denominator as much as possible and then express the original fraction as a sum of fractions with those factors in the denominator. In this case, the denominator on the left is already factored into (x+2)(x² + x + 1). Since x² + x + 1 is a quadratic that doesn't factor further (its discriminant is 1 - 4 = -3, which is negative), we can express the original fraction as A/(x+2) + (Bx + C)/(x² + x + 1), which is exactly what's given here.So, to find A, B, and C, I need to combine the right-hand side fractions into a single fraction and then set the numerators equal to each other because the denominators are already the same. Let's write that out step by step.Starting with the right-hand side:A/(x+2) + (Bx + C)/(x² + x + 1)To combine these, we need a common denominator, which is (x+2)(x² + x + 1). So we multiply the first fraction by (x² + x + 1)/(x² + x + 1) and the second fraction by (x+2)/(x+2):[A(x² + x + 1) + (Bx + C)(x + 2)] / [(x+2)(x² + x + 1)]This should be equal to the left-hand side, which is:(x + 1)² / [(x+2)(x² + x + 1)]Since the denominators are the same, we can set the numerators equal:A(x² + x + 1) + (Bx + C)(x + 2) = (x + 1)²Now, expand both sides and collect like terms to equate coefficients.First, expand the left side:A(x² + x + 1) becomes A x² + A x + A(Bx + C)(x + 2) becomes Bx(x) + Bx(2) + C(x) + C(2) = Bx² + 2Bx + Cx + 2CCombine these two parts:A x² + A x + A + B x² + 2B x + C x + 2CNow combine like terms:x² terms: (A + B) x²x terms: (A + 2B + C) xconstant terms: A + 2CSo the left side is:(A + B)x² + (A + 2B + C)x + (A + 2C)The right side is (x + 1)², which expands to x² + 2x + 1Therefore, we can set up equations by equating the coefficients of corresponding powers of x:For x²: (A + B) = 1For x: (A + 2B + C) = 2For constant term: (A + 2C) = 1So now we have a system of three equations:1) A + B = 12) A + 2B + C = 23) A + 2C = 1We need to solve this system for A, B, C.Let me write down the equations again:Equation 1: A + B = 1Equation 2: A + 2B + C = 2Equation 3: A + 2C = 1Let me solve this step by step. Maybe express A from equation 1 and substitute into the others.From equation 1: A = 1 - BSubstitute A = 1 - B into equation 2 and equation 3.Equation 2 becomes:(1 - B) + 2B + C = 2Simplify:1 - B + 2B + C = 21 + B + C = 2So B + C = 1 (Equation 2a)Equation 3 becomes:(1 - B) + 2C = 1Simplify:1 - B + 2C = 1Subtract 1 from both sides:-B + 2C = 0So -B + 2C = 0 (Equation 3a)Now we have two equations:Equation 2a: B + C = 1Equation 3a: -B + 2C = 0Let me add these two equations to eliminate B.(B + C) + (-B + 2C) = 1 + 0Simplify:B - B + C + 2C = 10 + 3C = 1 => 3C = 1 => C = 1/3Now plug C = 1/3 into equation 2a: B + 1/3 = 1 => B = 1 - 1/3 = 2/3Then from equation 1: A = 1 - B = 1 - 2/3 = 1/3So A = 1/3, B = 2/3, C = 1/3Wait, let me check if these values satisfy all three equations.Check equation 1: A + B = 1/3 + 2/3 = 1. Correct.Check equation 2: A + 2B + C = 1/3 + 2*(2/3) + 1/3 = 1/3 + 4/3 + 1/3 = (1 + 4 + 1)/3 = 6/3 = 2. Correct.Check equation 3: A + 2C = 1/3 + 2*(1/3) = 1/3 + 2/3 = 1. Correct.So all equations are satisfied. Therefore, the solution is A = 1/3, B = 2/3, C = 1/3.Alternatively, maybe I can verify by plugging these values back into the original equation.Original equation:Left side: (x + 1)^2 / [(x + 2)(x² + x + 1)]Right side: A/(x + 2) + (Bx + C)/(x² + x + 1) = (1/3)/(x + 2) + ( (2/3)x + 1/3 )/(x² + x + 1)Let me combine these two fractions:[1/3 * (x² + x + 1) + (2/3 x + 1/3)(x + 2)] / [(x + 2)(x² + x + 1)]Compute numerator:1/3*(x² + x + 1) + (2/3 x + 1/3)*(x + 2)First term: 1/3 x² + 1/3 x + 1/3Second term: expand (2/3 x + 1/3)(x + 2)Multiply out:2/3 x * x = 2/3 x²2/3 x * 2 = 4/3 x1/3 * x = 1/3 x1/3 * 2 = 2/3So total second term: 2/3 x² + 4/3 x + 1/3 x + 2/3 = 2/3 x² + (4/3 + 1/3) x + 2/3 = 2/3 x² + 5/3 x + 2/3Now add first term and second term:1/3 x² + 1/3 x + 1/3 + 2/3 x² + 5/3 x + 2/3Combine like terms:x² terms: 1/3 + 2/3 = 1 x²x terms: 1/3 + 5/3 = 6/3 = 2 xconstant terms: 1/3 + 2/3 = 1So numerator is x² + 2x + 1, which is (x + 1)^2. That matches the left side's numerator. Therefore, the values of A, B, C are correct.Alternatively, another method to find A, B, C is to substitute specific values of x to create equations. For example, pick x values that simplify the equation.But since the denominator on the left is (x + 2)(x² + x + 1), we can choose x = -2 to eliminate the second term (since x² + x + 1 evaluated at x = -2 is (-2)^2 + (-2) + 1 = 4 -2 +1 = 3 ≠ 0, so plugging x = -2 won't cause division by zero in the original equation, but in the decomposed form, the term A/(x + 2) would be problematic if x = -2, but since in the original equation x = -2 is allowed (denominator becomes (0)(...)), so actually x = -2 is not in the domain. Wait, actually, x = -2 makes the denominator zero, so x = -2 is excluded from the domain. Therefore, substituting x = -2 into the equation would not be valid. However, when we set the numerators equal, which is valid for all x except x = -2 and roots of x² + x +1, but since we have an identity, the equation of the numerators must hold for all x. So even though x = -2 makes the denominator zero, the numerator equation A(x² + x +1) + (Bx + C)(x +2) = (x +1)^2 must hold for all x, including x = -2. So actually, even though x = -2 is not in the domain, the equality of polynomials must hold for all x, so we can substitute x = -2 into the numerator equation to find A.Let me try that. Let x = -2:Left side numerator: A((-2)^2 + (-2) +1) + (B*(-2) + C)((-2) + 2) = A(4 -2 +1) + (-2B + C)(0) = A(3) + 0 = 3ARight side numerator: (-2 +1)^2 = (-1)^2 = 1Therefore, 3A = 1 => A = 1/3. Which matches our previous result.Then, with A known as 1/3, we can substitute other x values to find B and C. For example, let's choose x = 0:Left side numerator: (0 +1)^2 = 1Right side numerator: A(0^2 +0 +1) + (B*0 + C)(0 +2) = A(1) + C*2 = A + 2CFrom A = 1/3, then A + 2C = 1/3 + 2C. Set equal to 1:1/3 + 2C = 1 => 2C = 2/3 => C = 1/3. Which matches earlier.Then, choose x =1:Left side numerator: (1 +1)^2 = 4Right side numerator: A(1 +1 +1) + (B*1 + C)(1 +2) = 3A + 3(B + C)Substitute A =1/3, C=1/3:3*(1/3) + 3*(B +1/3) = 1 + 3B +1 = 2 + 3BSet equal to 4:2 + 3B =4 => 3B=2 => B=2/3. Correct.Thus, all three methods: expanding and equating coefficients, verifying by substitution into the original equation, and using specific x values to create equations, lead to the same result. Therefore, the solution is confirmed.Final AnswerThe values are (A = boxed{dfrac{1}{3}}), (B = boxed{dfrac{2}{3}}), and (C = boxed{dfrac{1}{3}}).

✅:To determine the values of (A), (B), and (C) such thatfrac{(x+1)^{2}}{(x+2)left(x^{2}+x+1right)} equiv frac{A}{x+2}+frac{B x+C}{x^{2}+x+1},we start by expressing the right-hand side as a single fraction with a common denominator:frac{A(x^{2} + x + 1) + (Bx + C)(x + 2)}{(x+2)(x^{2}+x+1)}.Equating the numerators of both sides, we get:A(x^{2} + x + 1) + (Bx + C)(x + 2) = (x + 1)^2.Expanding and combining like terms on the left side:A x^{2} + A x + A + B x^{2} + 2B x + C x + 2C = (A + B)x^{2} + (A + 2B + C)x + (A + 2C).Expanding the right side:(x + 1)^2 = x^2 + 2x + 1.Equating coefficients of corresponding powers of (x):1. For (x^2): (A + B = 1)2. For (x): (A + 2B + C = 2)3. For the constant term: (A + 2C = 1)Solving this system of equations:1. From (A + B = 1), we get (A = 1 - B).2. Substituting (A = 1 - B) into (A + 2B + C = 2) gives (1 + B + C = 2) or (B + C = 1).3. Substituting (A = 1 - B) into (A + 2C = 1) gives (-B + 2C = 0).Solving the system of equations (B + C = 1) and (-B + 2C = 0):- Adding these equations: (3C = 1 implies C = frac{1}{3}).- Substituting (C = frac{1}{3}) into (B + C = 1) gives (B = frac{2}{3}).- Substituting (B = frac{2}{3}) into (A = 1 - B) gives (A = frac{1}{3}).Verification by substituting back into the original equation confirms the solution is correct. Final AnswerThe values are (A = boxed{dfrac{1}{3}}), (B = boxed{dfrac{2}{3}}), and (C = boxed{dfrac{1}{3}}).

🔑:To solve for (A), (B), and (C) in the given equation:[frac{(x + 1)^2}{(x + 2)(x^2 + x + 1)} equiv frac{A}{x + 2} + frac{Bx + C}{x^2 + x + 1}]we follow these steps:1. Remove the Fraction: Multiply both sides by ((x + 2)(x^2 + x + 1)) to clear the fractions:[(x + 1)^2 equiv A(x^2 + x + 1) + (Bx + C)(x + 2)]2. Expand and Simplify: Expand both sides of the equation:[(x + 1)^2 = x^2 + 2x + 1][A(x^2 + x + 1) + (Bx + C)(x + 2) = A(x^2 + x + 1) + Bx(x + 2) + C(x + 2)][= A(x^2 + x + 1) + Bx^2 + 2Bx + Cx + 2C][= Ax^2 + Ax + A + Bx^2 + 2Bx + Cx + 2C][= (A + B)x^2 + (A + 2B + C)x + A + 2C]3. Match Coefficients: Compare the coefficients of powers of (x) to find (A), (B), and (C):[x^2: quad A + B = 1][x: quad A + 2B + C = 2][text{constant term}: quad A + 2C = 1]4. Solve the System of Equations: - From (A + B = 1):[B = 1 - A]- Substitute (B = 1 - A) into (A + 2B + C = 2):[A + 2(1 - A) + C = 2][A + 2 - 2A + C = 2][-A + 2 + C = 2][-A + C = 0][C = A]- Substitute (C = A) into (A + 2C = 1):[A + 2A = 1][3A = 1][A = frac{1}{3}]- Substitute (A = frac{1}{3}) into (B = 1 - A):[B = 1 - frac{1}{3}][B = frac{2}{3}]- And (C = A):[C = frac{1}{3}]Thus, the values of (A), (B), and (C) are:[boxed{A = frac{1}{3}, B = frac{2}{3}, C = frac{1}{3}}]

🧠:Okay, so the problem is: How many two-digit numbers exist that increase by at least twice when their digits are rearranged. Hmm, let me try to figure this out step by step. First, let me make sure I understand the question correctly. We have a two-digit number. If we rearrange its digits (which means swapping the tens and units digits), the new number should be at least twice the original number. So, for example, if the original number is AB (where A is the tens digit and B is the units digit), then the rearranged number is BA, and we need BA ≥ 2 * AB. Wait, but when we rearrange the digits, the new number could be either larger or smaller depending on the digits. But the problem specifies that the number should "increase by at least twice," so I think that means the rearranged number (BA) must be at least double the original number (AB). So BA ≥ 2 * AB. That seems right.So, we need to find all two-digit numbers AB (where A is from 1 to 9 and B is from 0 to 9, and A ≠ B because if A = B, rearranging doesn't change the number) such that when you swap the digits to get BA, BA is at least twice AB. Let me formalize this. Let the original number be 10A + B, and the rearranged number is 10B + A. The condition is 10B + A ≥ 2*(10A + B). Let me write that as an inequality:10B + A ≥ 20A + 2BThen, moving all terms to one side:10B + A - 20A - 2B ≥ 0Simplify:(10B - 2B) + (A - 20A) ≥ 08B - 19A ≥ 0So, 8B ≥ 19ATherefore, B ≥ (19/8)ABut B and A are digits. A is from 1 to 9, and B is from 0 to 9. But since B must be a digit, B can only be an integer from 0 to 9. Similarly, A is an integer from 1 to 9. Therefore, the inequality 8B ≥ 19A must hold with B and A integers.So, we can rephrase the inequality as:B ≥ (19/8)ABut since B must be an integer, we can write B ≥ ceiling(19A/8). Because if 19A/8 is not an integer, B has to be the next integer greater than that. However, ceiling function might complicate things. Alternatively, since 8B must be greater than or equal to 19A, we can write 8B ≥ 19A. Let's see:For each A from 1 to 9, we need to find B (0-9, integer) such that 8B ≥ 19A. But B must also be a digit different from A (since the digits are rearranged; if B = A, the number remains the same, but the problem says "rearranged," so maybe they need to be different digits? Wait, the problem doesn't specify that the digits must be different. Hmm. Let me check the original problem again: "which increase by at least twice when their digits are rearranged". If the digits are the same, rearranging them doesn't change the number, so the number would not increase. Therefore, such numbers (with same digits) can be excluded because they don't satisfy the condition. Therefore, A ≠ B is required.Therefore, we can proceed under the assumption that A ≠ B, so B can be 0-9 except A.So, going back to 8B ≥ 19A. Let's solve for B:B ≥ (19A)/8Since B must be an integer, B must be at least the ceiling of (19A)/8. Let's compute (19A)/8 for each A from 1 to 9 and find the minimum B required.Let's make a table:For A=1: 19*1/8 = 2.375 → ceiling is 3. So B ≥ 3. But B must be a digit (0-9), different from A=1. So possible B: 3,4,5,6,7,8,9. So 7 values.Wait, but B can be 0? Let's check for A=1: 8B ≥ 19*1 → 8B ≥19 → B ≥19/8≈2.375. So B must be at least 3. So for A=1, B can be 3-9 (excluding 1 if necessary). Since A=1 and B=1 is not allowed (digits must be different), but B is already ≥3, so all B=3-9 are allowed (since 3-9 don't include 1). So 7 numbers: 13,14,15,16,17,18,19. Wait, but wait: original number is 10A + B, which is 13,14,...,19. Then when you swap digits, you get 31,41,51,61,71,81,91. Then check if 31 ≥ 2*13=26, which is true (31-26=5). Similarly, 41 ≥ 28, yes. All of these will satisfy BA ≥ 2*AB. So for A=1, there are 7 numbers.For A=2: 19*2/8=38/8=4.75 → ceiling is 5. So B must be ≥5. Also, B ≠2. So B can be 5,6,7,8,9. So 5 values. The original numbers are 25,26,27,28,29. Swapped: 52,62,72,82,92. Check 52 ≥ 2*25=50 → 52 ≥50, yes. 62 ≥50, yes. All satisfy. So 5 numbers.A=3: 19*3/8=57/8=7.125 → ceiling is 8. So B ≥8. Also, B≠3. So B=8,9. So two numbers: 38,39. Swapped:83,93. Check 83 ≥ 2*38=76 →83≥76, yes. 93≥76, yes. So two numbers.A=4: 19*4/8=76/8=9.5 → ceiling is 10. But B is a digit (0-9), so ceiling is 10, but maximum B is 9. Therefore, no solution for A=4. So no numbers here.A=5: 19*5/8=95/8=11.875. B must be ≥12, which is impossible since B is a digit. So no solutions.Similarly, for A=6 to 9, 19*A/8 will be even larger, so B would need to be higher than 9, which is impossible. Therefore, no solutions for A=4 to 9.Therefore, the total numbers are for A=1:7, A=2:5, A=3:2. Total 7+5+2=14. But wait, let me check again.Wait, when A=1, B can be 3-9 (7 numbers). For A=2, B=5-9 (5 numbers). For A=3, B=8,9 (2 numbers). So 7+5+2=14. So the answer is 14.But wait, let's verify each of these numbers to make sure they actually satisfy the condition.Take A=1, B=3: original number 13, swapped 31. 31 is 2*13=26. 31 is more than 26, so yes. Similarly, 14 becomes 41, which is 41 vs 28, yes. All the way up to 19→91, 91 vs 38, yes.A=2, B=5: 25→52. 52 vs 50. 52 is exactly twice 26, wait. Wait, 2*25=50. 52 is greater than 50. So yes, satisfies BA ≥2*AB. Similarly, 26→62, 62≥52, yes. 27→72≥54, yes. 28→82≥56, yes. 29→92≥58, yes.A=3, B=8:38→83. 83 vs 76 (2*38=76). 83≥76. 39→93 vs 78. 93≥78. Correct.So all these numbers satisfy BA ≥2*AB. Therefore, total 14 numbers.Wait, but let me check if there are any other numbers that we might have missed. For example, what about numbers where B is less than A? For example, if original number is 51, swapping gives 15, which is less. But since we need the number to increase, we only consider cases where B > A, right? Wait, because when you swap the digits, BA is larger than AB only if B > A. Otherwise, if B < A, swapping would make the number smaller.Therefore, perhaps another approach is that we only need to consider numbers where B > A, because otherwise BA would be smaller than AB. Therefore, the condition BA ≥2*AB can only hold if BA > AB, which requires B > A. Therefore, in our earlier analysis, when we considered A from 1 to 9 and B ≥ ceiling(19A/8), but since B must be greater than A (to make BA > AB), we should also impose B > A. Wait, but in our previous calculation, for A=1, B≥3. Since B must be ≥3 and B≠1, but since A=1, B can be 3-9, which are all greater than A=1. Similarly, for A=2, B≥5, which are all greater than 2. For A=3, B≥8, which are greater than 3. So in those cases, B is already greater than A. Therefore, the previous count already includes only B > A. Therefore, 14 numbers.But let me verify with some edge cases. For example, take number 19: 19 swapped is 91. 91 is 4.789... times 19. So that's more than twice. What about number 13: 31 is 2.384... times 13, which is more than twice. Similarly, 25→52, which is exactly 2.08 times. Wait, 52 is exactly 2.08 times 25. So 52 = 2.08*25. Therefore, it's at least twice. So that's acceptable.Is there a number where BA is exactly twice AB? For example, 25 and 50. Wait, 25 swapped is 52. 52 is 2.08 times 25. So slightly more than twice. To get exactly twice, we would need 10B + A = 2*(10A + B). Let's solve:10B + A = 20A + 2B10B - 2B = 20A - A8B = 19ASo 8B =19A. Since A and B are digits, 19A must be divisible by 8. So 19A ≡ 0 mod8. Since 19 ≡3 mod8, so 3A ≡0 mod8. So 3A must be divisible by 8. Therefore, A must be such that 3A is divisible by 8. Since 3 and 8 are coprime, A must be divisible by 8. But A is a digit from1-9, so the only possibility is A=8. Then, 3*8=24≡0 mod8. Then, 8B=19*8 → B= (19*8)/8=19. But B is a digit (0-9), so impossible. Therefore, there's no two-digit number where swapping the digits gives exactly double. So all solutions satisfy BA >2*AB. Therefore, the count remains 14.Therefore, the answer is 14. But let me just check once more.List all numbers:For A=1, B=3:13→31, 31/13≈2.384B=4:14→41, 41/14≈2.928B=5:15→51, 51/15=3.4B=6:16→61, 61/16≈3.8125B=7:17→71≈4.176B=8:18→81≈4.5B=9:19→91≈4.789Total 7 numbers.For A=2:B=5:25→52≈2.08B=6:26→62≈2.384B=7:27→72≈2.666B=8:28→82≈2.928B=9:29→92≈3.172Total 5 numbers.For A=3:B=8:38→83≈2.184B=9:39→93≈2.384Total 2 numbers.So 7+5+2=14 numbers.Yes, each of these ratios is above 2. So all 14 numbers satisfy the condition. Therefore, the answer is 14.Final AnswerThe number of such two-digit numbers is boxed{14}.

✅:To determine how many two-digit numbers exist that increase by at least twice when their digits are rearranged, we start by considering a two-digit number (10A + B) (where (A) and (B) are digits, (A neq B)). When the digits are rearranged, the new number is (10B + A). We need this rearranged number to be at least twice the original number:[10B + A geq 2(10A + B)]Simplifying this inequality:[10B + A geq 20A + 2B implies 8B geq 19A]We need to find all pairs of digits (A) and (B) (with (A neq B)) such that (8B geq 19A). We analyze each possible value of (A) (from 1 to 9) and determine the minimum (B) required:- For (A = 1): (8B geq 19 implies B geq lceil 19/8 rceil = 3). Valid (B) values: 3, 4, 5, 6, 7, 8, 9 (7 values).- For (A = 2): (8B geq 38 implies B geq lceil 38/8 rceil = 5). Valid (B) values: 5, 6, 7, 8, 9 (5 values).- For (A = 3): (8B geq 57 implies B geq lceil 57/8 rceil = 8). Valid (B) values: 8, 9 (2 values).- For (A geq 4): The required (B) exceeds 9, which is not possible.Summarizing the valid numbers:- (A = 1): 13, 14, 15, 16, 17, 18, 19 (7 numbers)- (A = 2): 25, 26, 27, 28, 29 (5 numbers)- (A = 3): 38, 39 (2 numbers)Adding these, we get a total of (7 + 5 + 2 = 14) numbers.Thus, the number of two-digit numbers that increase by at least twice when their digits are rearranged is (boxed{14}).

🔑:Let's analyze the problem step by step to find out how many two-digit numbers satisfy the given condition.1. Definition of the Condition: To determine the number of two-digit numbers that increase by at least twice when their digits are swapped, let's denote the original two-digit number as ( overline{ab} ), where ( a ) and ( b ) are its digits and ( a neq 0 ).2. Forming the Equation: The original number ( overline{ab} ) is represented as ( 10a + b ). After swapping the digits, the new number will be ( overline{ba} ), represented as ( 10b + a ).3. Setting Up the Inequality: According to the problem, after swapping the digits, the new number should be at least twice the original number. Therefore, the inequality can be set up as: [ 10b + a geq 2(10a + b) ]4. Simplifying the Inequality: Rearrange and simplify the inequality step-by-step: [ 10b + a geq 20a + 2b ] [ 10b - 2b geq 20a - a ] [ 8b geq 19a ] [ b geq frac{19a}{8} ]5. Finding Valid Pairs (a, b): [ b geq frac{19a}{8} ] Since ( a ) and ( b ) must be digits (0 through 9, with ( a neq 0 )), we need to check for each value of ( a ) from 1 to 9 and see if there exists a corresponding ( b ) that satisfies the inequality: For ( a = 1 ): [ b geq frac{19 cdot 1}{8} = frac{19}{8} approx 2.375 ] Thus, ( b geq 3 ). Therefore, valid ( b ) values are 3, 4, 5, 6, 7, 8, 9, giving 7 possibilities. For ( a = 2 ): [ b geq frac{19 cdot 2}{8} = frac{38}{8} = 4.75 ] Thus, ( b geq 5 ). Therefore, valid ( b ) values are 5, 6, 7, 8, 9, giving 5 possibilities. For ( a = 3 ): [ b geq frac{19 cdot 3}{8} = frac{57}{8} approx 7.125 ] Thus, ( b geq 8 ). Therefore, valid ( b ) values are 8, 9, giving 2 possibilities. For ( a = 4 ): [ b geq frac{19 cdot 4}{8} = frac{76}{8} = 9.5 ] Thus, no valid value for ( b ) since ( b ) must be a digit from 0-9. For ( a geq 5 ): [ b geq frac{19 cdot a}{8} ] This value exceeds 10, which is not possible for ( b ) to be a single digit. Hence, no valid values for ( b ) exist when ( a geq 5 ).6. Summing Up the Valid Numbers: By adding the valid pairs: - For ( a = 1 ): 7 pairs. - For ( a = 2 ): 5 pairs. - For ( a = 3 ): 2 pairs. Therefore, the total number of two-digit numbers that meet the condition is: [ 7 + 5 + 2 = 14 ] Conclusion:The number of two-digit numbers that increase by at least twice when their digits are swapped is (boxed{14}).

🧠:Okay, let's see. The problem is about finding the largest possible number in a set of five different positive integers with an average of 11. The options are given from A to E. Hmm. So first, I need to recall that the average is the sum divided by the number of elements. Since there are five numbers, the total sum must be 5 times 11, which is 55. So the sum of all five numbers is 55.Now, the goal is to maximize the largest number in the set. To do that, I think I need to minimize the other four numbers as much as possible because if the other numbers are as small as they can be, then the remaining value for the largest number will be as big as possible. But they have to be different positive integers. So that means the four smallest numbers should be 1, 2, 3, 4, right? Let me check.Wait, if the numbers are different positive integers, the smallest possible four numbers would be 1, 2, 3, and 4. Let me add those up: 1 + 2 + 3 + 4 = 10. Then the fifth number would be 55 - 10 = 45. Hmm, 45 is one of the options, option A. But wait, the answer given here might not be correct because sometimes you have to check if that's possible. Wait, but 45 is in the options. Wait, but let me think again. If the four smallest numbers are 1, 2, 3, 4, then the largest number is 45. But 1, 2, 3, 4, 45 sum up to 55. All are different positive integers. So that seems okay. So why are the other options here? Maybe there's a mistake in my reasoning.Wait, wait. Let me check again. The sum of the five numbers is 55. If the four smallest numbers are 1, 2, 3, 4, sum to 10. Then the fifth number is 55 - 10 = 45. So that's possible. But maybe there's a restriction I'm not considering. Let me see. The numbers have to be different positive integers, which they are. So why is 45 an option but there's a higher number like 44 or 46? Wait, option E is 46. Hmm. Wait, maybe my approach is wrong. Let me think again.Wait, if the four smallest numbers are 1, 2, 3, 4, which sum to 10, then the fifth number is 45. But maybe the numbers can't be that spread out? Wait, no. There's no restriction except they have to be different positive integers. So 45 should be possible. But 46 is an option. Let me see. If the fifth number is 46, then the sum of the other four numbers would be 55 - 46 = 9. So the other four numbers would need to sum to 9. But they have to be different positive integers. Let me try to find four different positive integers that add up to 9. The smallest possible four numbers would be 1, 2, 3, 4, which sum to 10. So 9 is even less than that. So that's impossible. Therefore, 46 can't be achieved. So 45 is possible. Therefore, the answer should be 45, which is option A.But wait, the options also include 44 and 40 and 35. Maybe I'm missing something here. Let me check again. Let's see, if the largest number is 45, then the other numbers are 1, 2, 3, 4. That works. But maybe the problem requires that the numbers are in some order or have another constraint? The problem says "different positive integers," so that's all. So why is 46 an option? Let me see. If I try to get 46 as the largest number, the other four numbers must sum to 55 - 46 = 9. The minimal sum for four different positive integers is 1+2+3+4=10, so 9 is not possible. Therefore, 46 is impossible. So 45 is possible. Therefore, answer A. But why is 44 an option? Maybe there's another scenario where the numbers can't be 1, 2, 3, 4, 45 for some reason? Let me check again.Wait, perhaps the problem is that in some cases, the numbers must be distinct, but maybe if you use 1, 2, 3, 4, you can't use 45? No, that's not a problem. All numbers are distinct. So 1, 2, 3, 4, 45 are all different. So that should be acceptable. Therefore, the maximum possible number is 45. So answer A.Wait, but maybe there's a mistake here. Let me see. If the sum of the five numbers is 55, and we need five distinct positive integers. To maximize one number, minimize the others. The minimal possible sum for four distinct positive integers is 1+2+3+4=10, so the fifth number is 45. So that's correct. So 45 is possible. Then why is 46 not possible? Because 55 - 46 = 9, which is less than the minimal sum of 10. So 46 is impossible. Therefore, the answer is 45. So option A.But let me check the answer options again. The options are (A) 45, (B) 40, (C)35, (D)44, (E)46. So 45 is an option. But wait, perhaps there is a mistake in my reasoning. Let me check once more.Suppose we try to have the largest number as 44. Then the sum of the other four numbers would be 55 - 44 = 11. Can we find four different positive integers that sum to 11? Let's see. The minimal sum is 1+2+3+4=10. So 11 is possible. For example, 1, 2, 3, 5: 1+2+3+5=11. So the numbers would be 1,2,3,5,44. That's valid. So the largest number could be 44. But wait, but 45 is higher than 44. So why would 44 be a possible answer? Because maybe when we take 1,2,3,4,45, that's valid. But maybe there's a mistake here. Wait, no. If 1,2,3,4,45 is valid, then 45 is possible. So then why is 44 an option?Wait, maybe the problem is that when you use 1,2,3,4,45, the numbers are distinct and sum to 55. So that's correct. Therefore, 45 is achievable, so 45 is the maximum. But then why is option D (44) even there? Maybe there's a trick I'm missing.Wait, let me check again. Let's see, if the four numbers sum to 10, then the fifth is 45. But maybe there's a requirement that the numbers have to be in a certain way? Or perhaps the numbers can't have gaps that are too large? No, the problem only states different positive integers, so any distinct positive integers are allowed. So 1,2,3,4,45 is acceptable.Alternatively, maybe the problem is that I'm considering the numbers to be 1,2,3,4, but perhaps they need to be consecutive? No, the problem doesn't say that. They just need to be different. So 1,2,3,4 is fine. So then the answer should be 45. Therefore, option A.But maybe the answer is 44. Let me think again. Suppose that the four smallest numbers are not 1,2,3,4. Maybe there's another combination where the sum is higher, making the largest number smaller. But why would that happen? Wait, to maximize the largest number, you need to minimize the sum of the others. So if you can make the sum of the other four numbers as small as possible, then the largest number is as big as possible. So the minimal sum for four different positive integers is 1+2+3+4=10. Therefore, the maximum possible number is 55 - 10 = 45. Therefore, 45 is achievable. So answer A.But maybe there's an error in this line of reasoning. Let me test with another example. Suppose there are 5 numbers with average 10, so sum 50. Then the maximum number would be 50 - (1+2+3+4) = 50 -10=40. So the numbers would be 1,2,3,4,40. That's correct. Similarly, in this problem, it's average 11, sum 55, so maximum number is 55 -10=45. That seems consistent.Therefore, I think the correct answer is A)45. But the options include D)44 and E)46, so why are those present? Let me check again. For example, if the four numbers can't include 1? But the problem says positive integers, which start at 1. So 1 is allowed. Unless there's an unstated constraint, but the problem doesn't mention that. Therefore, I think the answer should be 45. So option A.

✅:To determine the largest possible number in a set of five different positive integers with an average of 11, we start by calculating the total sum of the set: [text{Total sum} = 5 times 11 = 55.]To maximize the largest number, the other four numbers must be as small as possible while remaining distinct. The smallest four distinct positive integers are (1, 2, 3, 4), which sum to: [1 + 2 + 3 + 4 = 10.]Subtracting this from the total sum gives the largest number: [55 - 10 = 45.]Verification: - The set (1, 2, 3, 4, 45) contains distinct positive integers. - Their sum is (1 + 2 + 3 + 4 + 45 = 55), confirming the average of 11. Rejecting Larger Options: - If the largest number were (46), the remaining four numbers would need to sum to (55 - 46 = 9). However, the smallest four distinct positive integers sum to (10), making (9) impossible. - Similarly, (44) would require the other four numbers to sum to (11), which is feasible (e.g., (1, 2, 3, 5)), but (45) is larger and valid. Thus, the largest possible number is 45.Answer: (boxed{A})

🔑:1. Let the set of five different positive integers be denoted by ( a_1, a_2, a_3, a_4, a_5 ). Since the integers are different and positive, we can assume ( a_1 < a_2 < a_3 < a_4 < a_5 ).2. Given that the average of these five integers is 11, we have: [ frac{a_1 + a_2 + a_3 + a_4 + a_5}{5} = 11 ]3. Multiplying both sides of the equation by 5 to eliminate the fraction, we get: [ a_1 + a_2 + a_3 + a_4 + a_5 = 55 ]4. To maximize ( a_5 ) (the largest number in the set), we need to minimize the sum of the other four integers: ( a_1, a_2, a_3, ) and ( a_4 ).5. The smallest four different positive integers are 1, 2, 3, and 4. Therefore, we set: [ a_1 = 1, quad a_2 = 2, quad a_3 = 3, quad a_4 = 4 ]6. Substitute these values into the equation: [ 1 + 2 + 3 + 4 + a_5 = 55 ]7. Calculate the sum of the four smallest integers: [ 1 + 2 + 3 + 4 = 10 ]8. Now, solve for ( a_5 ) by isolating it on one side of the equation: [ 10 + a_5 = 55 implies a_5 = 55 - 10 implies a_5 = 45 ]9. Thus, the largest possible number in the set, given that the sum of the integers must be 55 and using the four smallest integers (1, 2, 3, and 4), is 45.Conclusion:[boxed{45}]

🧠:Okay, so I need to figure out if three planes, each passing through an edge of a trihedral angle and the bisector of the opposite face, intersect at a single line. Hmm. Let me start by recalling some basic concepts. A trihedral angle is formed by three planes intersecting at a common point, creating three edges. Each pair of edges defines a face, and the angle between each pair of edges is a face angle. The problem mentions a bisector of the opposite face. Wait, so for each edge, there's an opposite face. The bisector of that face would be the angle bisector of the face angle opposite the edge. So, for each edge, we take the angle bisector of the face that's not adjacent to that edge, right? Then, the plane in question passes through the edge and this bisector. The question is whether these three planes intersect along a single line.Let me try to visualize this. Imagine a trihedral angle with vertex at point O, formed by three edges OA, OB, and OC. Each face is a plane: OAB, OBC, and OCA. The face angles are at each of these faces. For edge OA, the opposite face is OBC, so the bisector would be the bisector of angle BOC. Similarly, for edge OB, the opposite face is OAC, so the bisector is the bisector of angle AOC, and for edge OC, the opposite face is OAB, so the bisector is the bisector of angle AOB.Therefore, each plane is constructed by taking an edge and the bisector of the opposite face angle. For example, the plane for edge OA would contain OA and the bisector of angle BOC. Similarly for the others. The question is: do these three planes intersect along a single line?Hmm. Let's consider the properties of these planes. Since each plane contains an edge of the trihedral angle, all three planes pass through the vertex O. Therefore, all three planes intersect at O. But the question is whether they intersect along a common line, not just at a point. So, if the three planes intersect along a single line, that line must pass through O. So, perhaps we need to determine if the three planes share a common line through O.Alternatively, maybe the intersection is more complicated. Let me think. If three planes all pass through a common point, they can intersect along three different lines, each pair of planes intersecting along a line through O. But for all three planes to intersect along a single line, their lines of intersection must coincide. That would require that each pair of planes intersects along the same line, which would mean that all three planes are coincident, but that's not the case here because each plane is defined by a different edge and bisector.Wait, no. Each plane is different because each is associated with a different edge. So, for instance, plane P1 contains edge OA and the bisector of angle BOC, plane P2 contains edge OB and the bisector of angle AOC, and plane P3 contains edge OC and the bisector of angle AOB. Since each plane contains a different edge, the planes are distinct. Therefore, each pair of planes should intersect along a line. If all three planes share a common line, that line must contain O and lie in all three planes.Alternatively, maybe there's a line through O that lies in all three planes. To check this, maybe we can use coordinate geometry. Let me try to model the trihedral angle in a coordinate system.Let's place the vertex O at the origin. Let’s assume the edges OA, OB, and OC are along the coordinate axes for simplicity. Wait, but if OA, OB, OC are along the coordinate axes, then the face angles are all right angles if the trihedral is orthogonal. But the problem states an arbitrary trihedral angle, so the edges can form any angles. However, maybe for simplicity, I can start with an orthogonal trihedral angle (like the positive x, y, z axes) and see if the statement holds, and then consider a more general case.Wait, but in an orthogonal trihedral, the face angles are all 90 degrees. The bisector of a 90-degree angle is the line that splits it into two 45-degree angles. For example, in the OBC face (which is the y-z plane), the bisector of angle BOC (which is the 90-degree angle between the y and z axes) would lie along the line y = z in the y-z plane. Similarly, in the OAC face (x-z plane), the bisector of angle AOC would be x = z, and in the OAB face (x-y plane), the bisector of angle AOB would be x = y.So, for edge OA (which is along the x-axis), the plane containing OA and the bisector of angle BOC (which is y = z). So, this plane would be the x-axis and the line y = z in the y-z plane. Wait, how do these two define a plane? The x-axis is (x, 0, 0), and the line y = z in the y-z plane is (0, t, t). So, the plane containing these two lines. Let me find the equation of this plane.To find the plane equation, we can use two direction vectors. One direction vector is along the x-axis, say (1, 0, 0). Another direction vector is along the line y = z in the y-z plane, which is (0, 1, 1). The normal vector to the plane is the cross product of these two vectors:(1, 0, 0) × (0, 1, 1) = (0*1 - 0*1, 0*0 - 1*1, 1*1 - 0*0) = (0 - 0, 0 - 1, 1 - 0) = (0, -1, 1).Therefore, the plane equation is 0(x) -1(y) +1(z) = d. Since the plane passes through the origin (because both lines pass through the origin), d=0. So, the plane equation is -y + z = 0, or z = y.Similarly, for edge OB (along the y-axis), the plane contains OB and the bisector of angle AOC, which is x = z in the x-z plane. The plane containing the y-axis and the line x = z in the x-z plane.Direction vectors: along the y-axis (0, 1, 0), and along x = z in x-z plane (1, 0, 1). Cross product:(0, 1, 0) × (1, 0, 1) = (1*1 - 0*0, 0*1 - 0*1, 0*0 - 1*1) = (1 - 0, 0 - 0, 0 - 1) = (1, 0, -1).Plane equation: 1(x) + 0(y) -1(z) = 0, so x - z = 0, or x = z.Third plane, for edge OC (along z-axis), contains OC and the bisector of angle AOB, which is x = y in the x-y plane. The plane containing the z-axis and the line x = y in the x-y plane.Direction vectors: along z-axis (0, 0, 1) and along x = y in x-y plane (1, 1, 0). Cross product:(0, 0, 1) × (1, 1, 0) = (0*0 - 1*1, 1*1 - 0*0, 0*1 - 0*1) = (0 - 1, 1 - 0, 0 - 0) = (-1, 1, 0).Plane equation: -1(x) +1(y) +0(z) = 0, so -x + y = 0, or y = x.So, in the orthogonal case, the three planes are:1. z = y2. x = z3. y = xDo these three planes intersect along a single line? Let's see. The intersection of the first two planes: z = y and x = z. Substituting x = z into z = y gives x = y = z. So, the intersection line is x = y = z. Then, the third plane is y = x, which is satisfied by x = y = z. Therefore, all three planes intersect along the line x = y = z. So, in the orthogonal trihedral case, they do intersect along a single line.But the problem states an arbitrary trihedral angle. So, does this hold in general? Let's test with a non-orthogonal trihedral angle.Suppose we have a trihedral angle where the edges OA, OB, OC are not orthogonal. Let's assign coordinates such that OA is along the x-axis, OB is in the xy-plane making an angle α with OA, and OC is somewhere out of the xy-plane, making angles β with OA and γ with OB. Hmm, this might get complicated, but maybe we can use vector analysis.Alternatively, perhaps we can use symmetry or properties of bisectors. Wait, in 3D geometry, the bisector of a face angle is a line in that face which divides the angle into two equal parts. If each plane is constructed by an edge and the bisector of the opposite face, then these planes might have a common line of intersection if certain symmetry exists.Alternatively, perhaps this is related to the concept of the internal bisector planes of a trihedral angle. In 3D, the bisector planes are those that divide the dihedral angles between pairs of faces. However, here we are dealing with planes that contain an edge and the bisector of the opposite face angle, which is different.Wait, in the orthogonal case, the line x=y=z is the common line. Maybe in the general case, the three planes intersect along the line that is the internal bisector of the trihedral angle. But does such a line exist? For a trihedral angle, there isn't a unique 'bisector' line, but there are bisector planes. However, perhaps the three planes in question intersect along a line that equally divides the trihedral angles in some way.Alternatively, perhaps we can parameterize the problem. Let me try to assign coordinates to a general trihedral angle.Let’s place vertex O at the origin. Let’s define three unit vectors along the edges OA, OB, OC as vectors a, b, c. The face opposite OA is the face formed by OB and OC, so the angle bisector of this face is the bisector of angle BOC. Similarly for others.In vector terms, the bisector of angle BOC would be a vector in the plane of OB and OC that is the angle bisector between b and c. The direction of this bisector can be given by the normalized sum of the unit vectors b and c, i.e., (b + c)/|b + c|. Similarly, the bisector of angle AOC is (a + c)/|a + c|, and the bisector of angle AOB is (a + b)/|a + b|.Therefore, each plane is defined by an edge (say OA, direction a) and the bisector of the opposite face (direction (b + c)/|b + c|). The plane containing OA and the bisector of angle BOC is the plane spanned by vectors a and (b + c). Similarly for the other planes.To determine if these three planes intersect along a single line, we need to check if there exists a line that lies in all three planes. Since all three planes pass through the origin, such a line would pass through the origin and have a direction vector d that lies in each of the three planes.So, d must be orthogonal to the normal vectors of each of the three planes. Let’s compute the normal vectors of each plane.For the plane containing OA and the bisector of BOC: this plane is spanned by a and (b + c). Therefore, the normal vector n1 is a × (b + c).Similarly, the normal vector n2 for the plane containing OB and the bisector of AOC is b × (a + c).The normal vector n3 for the plane containing OC and the bisector of AOB is c × (a + b).For a line with direction d to lie in all three planes, d must be orthogonal to n1, n2, and n3. That is:d • (a × (b + c)) = 0d • (b × (a + c)) = 0d • (c × (a + b)) = 0Alternatively, using scalar triple product properties, d must be orthogonal to each of the cross products, meaning d is in the intersection of the three planes.Alternatively, if such a d exists, then the three planes intersect along the line through O with direction d.Alternatively, if the three normal vectors n1, n2, n3 are coplanar, then their cross products would satisfy some linear relation, implying that there's a common direction d orthogonal to all three.Alternatively, perhaps there's a symmetry here. Let me consider the case where a, b, c are arbitrary vectors. Let's compute the scalar triple products.But this might get complicated. Maybe there's a better approach. Let's think in terms of linear algebra. If the three planes intersect along a common line, then the system of equations defined by the three planes has a one-dimensional solution space (i.e., a line). Since all three planes pass through the origin, this is equivalent to the intersection of their solution spaces being a line. The dimension of the intersection is given by dim(P1 ∩ P2 ∩ P3). For three planes in three-dimensional space, the intersection can be a line if the three planes are not all distinct and their normals are not linearly independent. Wait, but in three-dimensional space, three planes usually intersect at a point unless their normals are coplanar. If the normals are coplanar, then the planes intersect along a line.So, if the normals n1, n2, n3 are linearly dependent (i.e., they lie in the same plane), then the three planes intersect along a line. Therefore, we need to check if n1, n2, n3 are linearly dependent.Let’s compute n1, n2, n3:n1 = a × (b + c) = a × b + a × cn2 = b × (a + c) = b × a + b × c = -a × b + b × cn3 = c × (a + b) = c × a + c × b = -a × c - b × cSo, n1 = a×b + a×cn2 = -a×b + b×cn3 = -a×c - b×cLet’s see if these three vectors are linearly dependent. Let’s attempt to find scalars λ1, λ2, λ3, not all zero, such that λ1n1 + λ2n2 + λ3n3 = 0.Substitute the expressions:λ1(a×b + a×c) + λ2(-a×b + b×c) + λ3(-a×c - b×c) = 0Let’s expand this:(λ1a×b - λ2a×b) + (λ1a×c - λ3a×c) + (λ2b×c - λ3b×c) = 0Factor the terms:[(λ1 - λ2)a×b] + [(λ1 - λ3)a×c] + [(λ2 - λ3)b×c] = 0For this equation to hold, the coefficients of each cross product must be zero, because a×b, a×c, b×c are linearly independent in three-dimensional space (unless the vectors a, b, c are coplanar, but since they form a trihedral angle, they are not coplanar). Therefore, the coefficients must each be zero:1. λ1 - λ2 = 02. λ1 - λ3 = 03. λ2 - λ3 = 0From equation 1: λ1 = λ2From equation 2: λ1 = λ3From equation 3: λ2 = λ3Thus, all λ's are equal: λ1 = λ2 = λ3. Let’s denote this common value as λ.Therefore, the equation becomes:λ(n1 + n2 + n3) = 0But let’s compute n1 + n2 + n3:(a×b + a×c) + (-a×b + b×c) + (-a×c - b×c)Simplify term by term:a×b - a×b cancels out.a×c - a×c cancels out.b×c - b×c cancels out.So, n1 + n2 + n3 = 0.Therefore, the equation reduces to λ*0 = 0, which is always true. Therefore, the three normal vectors are linearly dependent with coefficients λ1 = λ2 = λ3. Hence, the three normals lie in a plane, so the three planes intersect along a common line.Therefore, regardless of the trihedral angle, the three planes intersect along a single line.Wait, but this seems too general. The conclusion is that the three normals are linearly dependent, which implies that the three planes intersect along a line. Therefore, the answer is yes, the three planes intersect at a single line.But let me verify this with an example where the trihedral angle is not orthogonal. Suppose OA, OB, OC are edges with angles between them not equal to 90 degrees. Let's take a specific example.Let’s say in the trihedral angle, OA is along the x-axis, OB is in the xy-plane making 60 degrees with OA, and OC is in the xz-plane making 60 degrees with OA. So, the unit vectors would be:a = (1, 0, 0)b = (cos60°, sin60°, 0) = (0.5, √3/2, 0)c = (cos60°, 0, sin60°) = (0.5, 0, √3/2)Now, compute the bisectors of the opposite faces.For edge OA, opposite face is BOC. The bisector of angle BOC. Since vectors b and c are:b = (0.5, √3/2, 0)c = (0.5, 0, √3/2)The angle between b and c can be computed using the dot product:b • c = 0.5*0.5 + (√3/2)*0 + 0*(√3/2) = 0.25The magnitude of b and c is 1 (unit vectors). So, the angle θ between b and c satisfies cosθ = 0.25, so θ ≈ 75.5 degrees. The bisector direction would be (b + c) normalized.b + c = (0.5 + 0.5, √3/2 + 0, 0 + √3/2) = (1, √3/2, √3/2)Normalize this vector:|b + c| = sqrt(1^2 + ( (√3/2)^2 + (√3/2)^2 )) = sqrt(1 + (3/4 + 3/4)) = sqrt(1 + 1.5) = sqrt(2.5) ≈ 1.5811So, the bisector direction is (1, √3/2, √3/2) / sqrt(2.5)Similarly, the plane for edge OA is spanned by a and this bisector. Let's compute the normal vector n1 = a × (b + c) = (1, 0, 0) × (1, √3/2, √3/2) = (0*√3/2 - 0*√3/2, 0*1 - 1*√3/2, 1*√3/2 - 0*1) = (0, -√3/2, √3/2)Similarly, compute n2 for edge OB. The opposite face is AOC, bisector is (a + c) normalized.a + c = (1 + 0.5, 0 + 0, 0 + √3/2) = (1.5, 0, √3/2)Normalize this vector:|a + c| = sqrt(1.5^2 + 0 + (√3/2)^2) = sqrt(2.25 + 0.75) = sqrt(3) ≈ 1.732Bisector direction: (1.5, 0, √3/2)/sqrt(3) = ( (3/2)/sqrt(3), 0, (√3/2)/sqrt(3) ) = ( sqrt(3)/2, 0, 1/2 )The plane for edge OB is spanned by b and this bisector. Compute normal vector n2 = b × (a + c) = (0.5, √3/2, 0) × (1.5, 0, √3/2)Compute cross product:i component: (√3/2 * √3/2 - 0*0) = (3/4 - 0) = 3/4j component: -(0.5 * √3/2 - 0*1.5) = -( (√3/4) - 0 ) = -√3/4k component: (0.5*0 - √3/2*1.5) = (0 - (3√3)/4 ) = -3√3/4So, n2 = (3/4, -√3/4, -3√3/4 )Similarly, compute n3 for edge OC. The opposite face is AOB, bisector is (a + b) normalized.a + b = (1 + 0.5, 0 + √3/2, 0 + 0) = (1.5, √3/2, 0)Normalize this vector:|a + b| = sqrt(1.5^2 + (√3/2)^2 + 0) = sqrt(2.25 + 0.75) = sqrt(3) ≈ 1.732Bisector direction: (1.5, √3/2, 0)/sqrt(3) = (sqrt(3)/2, 1/2, 0 )Plane for edge OC is spanned by c and this bisector. Compute normal vector n3 = c × (a + b) = (0.5, 0, √3/2) × (1.5, √3/2, 0)Compute cross product:i component: 0*0 - √3/2*√3/2 = 0 - 3/4 = -3/4j component: -(0.5*0 - √3/2*1.5) = -(0 - (3√3)/4 ) = 3√3/4k component: 0.5*√3/2 - 0*1.5 = (√3)/4 - 0 = √3/4So, n3 = (-3/4, 3√3/4, √3/4 )Now, we have n1, n2, n3:n1 = (0, -√3/2, √3/2 )n2 = (3/4, -√3/4, -3√3/4 )n3 = (-3/4, 3√3/4, √3/4 )We need to check if these vectors are linearly dependent. Let's see if there exist scalars λ1, λ2, λ3, not all zero, such that λ1n1 + λ2n2 + λ3n3 = 0.Set up the equations:For the i component:0*λ1 + (3/4)λ2 + (-3/4)λ3 = 0 → (3/4)λ2 - (3/4)λ3 = 0 → λ2 = λ3For the j component:(-√3/2)λ1 + (-√3/4)λ2 + (3√3/4)λ3 = 0For the k component:(√3/2)λ1 + (-3√3/4)λ2 + (√3/4)λ3 = 0Since λ2 = λ3, substitute λ3 = λ2 into the j and k equations.j component:(-√3/2)λ1 + (-√3/4)λ2 + (3√3/4)λ2 = 0Simplify:(-√3/2)λ1 + ( (-√3/4 + 3√3/4 ) )λ2 = 0(-√3/2)λ1 + (2√3/4)λ2 = 0 → (-√3/2)λ1 + (√3/2)λ2 = 0Divide both sides by √3/2:-λ1 + λ2 = 0 → λ2 = λ1k component:(√3/2)λ1 + (-3√3/4)λ2 + (√3/4)λ2 = 0Simplify:(√3/2)λ1 + ( (-3√3/4 + √3/4 ) )λ2 = 0(√3/2)λ1 + (-2√3/4)λ2 = 0 → (√3/2)λ1 + (-√3/2)λ2 = 0Divide by √3/2:λ1 - λ2 = 0 → λ1 = λ2So, from j and k components, we get λ1 = λ2 = λ3.Therefore, the vectors are linearly dependent, with λ1 = λ2 = λ3. Hence, the normals are linearly dependent, so the three planes intersect along a line.Thus, even in this non-orthogonal trihedral angle, the three planes intersect along a single line. Therefore, the answer should be yes, it is true that these three planes intersect at a single line.Another way to think about it: since the sum of the normal vectors n1 + n2 + n3 = 0, as shown earlier, this implies that the three planes intersect along a line. Because if the normals sum to zero, they are linearly dependent, confirming the previous conclusion.Hence, regardless of the trihedral angle, the three planes intersect along a single line.

✅:The three planes in question, each passing through an edge of a trihedral angle and the bisector of the opposite face, indeed intersect along a single line. Here's the systematic breakdown:1. Trihedral Angle Structure: A trihedral angle consists of three edges meeting at a vertex ( O ), forming three faces. For each edge, the opposite face is the one not containing that edge.2. Bisector Planes: Each plane is defined by an edge and the angle bisector of the opposite face. The angle bisector in a face lies in that face and splits the face's angle into two equal parts. 3. Normals to Planes: Using vector analysis, the normal vectors to these planes are derived from cross products of the edge vectors and the bisectors of the opposite face angles. For edges ( OA, OB, OC ) with unit vectors ( mathbf{a}, mathbf{b}, mathbf{c} ), the normals are: - ( mathbf{n_1} = mathbf{a} times (mathbf{b} + mathbf{c}) ) - ( mathbf{n_2} = mathbf{b} times (mathbf{a} + mathbf{c}) ) - ( mathbf{n_3} = mathbf{c} times (mathbf{a} + mathbf{b}) )4. Linear Dependence of Normals: By expanding these expressions, it is shown that ( mathbf{n_1} + mathbf{n_2} + mathbf{n_3} = 0 ). This linear dependence confirms that the three normal vectors lie in the same plane, implying the corresponding planes intersect along a common line.5. Verification with Examples: Both orthogonal and non-orthogonal trihedral angles were analyzed. In all cases, the normals were linearly dependent, and the planes intersected along a single line through the vertex ( O ).Conclusion: Regardless of the trihedral angle's specific geometry, the three planes intersect at a single line. boxed{text{Yes}}

🔑:1. Setup the scenario: Consider an arbitrary trihedral angle with vertex S. We denote the three edges emerging from S as SA, SB, and SC.2. Bisectors through opposing faces: We introduce the three planes: - The plane passing through edge SA and the angle bisector of the opposing face, denoted as SM. - The plane passing through edge SB and the angle bisector of the opposing face, denoted as SN. - The plane passing through edge SC and the angle bisector of the opposing face, denoted as SK.3. Choose equal segments on the edges: Select equal segments on the edges of the trihedral angle such that SA = SB = SC.4. Form triangles: From A, B, and C, construct triangles SAB, SBC, and SCA, connecting the points as follows: - SM is the bisector of angle ASB. - SN is the bisector of angle BSC. - SK is the bisector of angle ASC.5. Identify triangles' collateral properties: Since SM, SN, and SK are bisectors in their respective triangle SAB, triangle SBC, and triangle SCA, it means that they also serve as the medians in these isosceles triangles.6. Medial triangle intersection point: Medians of a triangle intersect at a single point known as the centroid. For each triangle formed by these bisectors, their counterparts in the opposing face must meet at this centroid.7. Intersection of planes on common line: Consequently, each of the planes contains a median of triangle ABC, intersecting at a single point of the medians, O, which forms a common intersection line SO from the vertex S.Conclusion: Thus, the three planes intersect along the same line passing through S and the centroid O of the triangle ABC.boxed{text{Yes}}

🧠:Okay, so I need to figure out why King Arthur has a winning strategy in this game where he and Merlin take turns placing coins on a round table. The coins are identical, can't overlap, and the first person who can't place a coin loses. Arthur goes first. Hmm. Let me start by breaking down the problem.First, the table is round. That symmetry might be important. Since Arthur starts first, maybe he can use some kind of symmetry strategy. Often in these kinds of games, the first player can mirror the second player's moves, but since the table is round, maybe Arthur can place the first coin in the exact center. Then, whatever Merlin does, Arthur can mirror it on the opposite side of the center. But wait, coins are placed on the table, so if the table is a perfect circle, placing a coin in the center would allow for rotational symmetry. Wait, but if the table is a circle, the center is a unique point. So if Arthur places the first coin in the center, then Merlin has to place a coin somewhere else. Then Arthur can mirror Merlin's moves with respect to the center. For every coin Merlin places at some radius and angle, Arthur can place one at the same radius but 180 degrees opposite. That way, if Merlin has a spot to play, Arthur always has the mirrored spot. Since Arthur started with the center, which can't be mirrored, Merlin can never mirror Arthur's moves. Wait, but in this case, since the center is unique, once Arthur takes it, Merlin can't mirror. Then Arthur can mirror all subsequent moves. So the key is that the first move occupies the center, which breaks the symmetry but gives Arthur control over the mirroring. Let me check.But wait, is the center considered part of the table? If the table is a circle, then the center is a valid position. If Arthur places the first coin at the center, then the remaining area is an annulus (a ring) around it. Now, whenever Merlin places a coin somewhere in the annulus, Arthur can place a coin diametrically opposite to Merlin's coin. Since the table is rotationally symmetric, as long as Arthur can mirror each of Merlin's moves, Arthur will always have a move after Merlin. But since the center was taken by Arthur first, Merlin can never take away Arthur's ability to mirror. Therefore, Arthur can always respond to Merlin's moves, ensuring that he is the one who makes the last move. Therefore, Arthur can win.But let me think again. Is there a way Merlin can disrupt this mirroring? For example, if Merlin places a coin such that its diametrically opposite position is already occupied. But since Arthur is mirroring, he only needs to mirror the moves after his own. Wait, Arthur places the first coin in the center. Then Merlin places a coin somewhere. Then Arthur mirrors it. Then Merlin places another coin, and Arthur mirrors that. Since all of Merlin's moves are mirrored, the only way Merlin could block Arthur's mirroring is if Merlin places a coin in a position that's already mirrored. But since the center is already occupied, and the mirrored positions are unique as long as Merlin doesn't place a coin in a position that's diametrically opposite to an existing one. But Merlin can choose any position, so he might try to place a coin in a spot where the mirror image is already occupied. But if Arthur is mirroring, then Arthur would have already placed that mirrored coin. Wait, let's clarify.Suppose Arthur places the first coin at the center. Then Merlin places a coin at point A. Arthur then places a coin at point A', which is diametrically opposite to A. Then Merlin places a coin at point B. Arthur places at B'. And so on. Each time Merlin places a coin, Arthur places its mirror image. Since the center is already occupied, Merlin can never place a coin in such a way that its mirror image is already occupied, because Arthur is always responding by mirroring the previous move. Therefore, as long as there is space for Merlin to place a coin, there must be space for Arthur to place the mirror image. This is because the table's rotational symmetry ensures that if a spot is available, its mirror image is also available unless the spot is on the axis through the center and the original coin. Wait, but if Merlin places a coin on the axis, then the mirror image would be the same spot. But since coins can't overlap, Merlin can't place a coin on the axis after Arthur has placed one in the center. Wait, actually, the axis is a line through the center, but coins are placed at points. If Arthur places a coin at the center, then the diameter through that center is just a line. Merlin can place a coin along that diameter but not overlapping the center. If Merlin places a coin along the diameter at some distance from the center, then the mirror image would be on the opposite side along the same diameter. But since the center is already occupied, those two spots (along the diameter on either side of the center) are distinct. So if Merlin places a coin at (r, 0) in polar coordinates, Arthur can place one at (r, π). Similarly, if Merlin places at (r, θ), Arthur places at (r, θ + π). Since the center is already taken, these mirrored positions are unique and not overlapping.Therefore, Arthur's strategy is to take the center first, then mirror every move Merlin makes. Since Arthur starts, he can always mirror, and because the total number of available positions is even (if we consider pairs of mirrored positions), except for the center. But since the center is taken first by Arthur, every subsequent move by Merlin can be mirrored by Arthur. Therefore, Merlin will eventually be the one who can't make a move, because Arthur always has a response. Hence, Arthur has a winning strategy.Wait, but another way to think about it: the game is similar to the classic game of placing tiles where the first player can take the center and mirror. For example, in a game on a chessboard where players alternate placing pieces, the first player can take the center and mirror. However, in that case, the board has even or odd dimensions, which affects the strategy. Here, the table is a circle, which is symmetric in all directions. So the key is the first player takes the center, which is the only point with full rotational symmetry, and then mirrors all other moves. Since every move after the center can be mirrored, Arthur can always respond, and thus will make the last move. Therefore, Arthur can force a win.But let me test this with a simple case. Suppose the table is very small, just big enough for one coin in the center. Then Arthur places the center, and Merlin can't place any, so Arthur wins. If the table is slightly larger, so that after the center, there's space for two coins opposite each other. Arthur places center, Merlin places one somewhere, Arthur mirrors, then Merlin can't place anymore. Wait, depends on the size. Let's say the table's radius is just a bit more than the coin's radius. So after center, there's space for two coins around it. Arthur places center, Merlin places one at (r, 0), Arthur places one at (r, π). Now the table is full, Merlin can't place any more coins. So Arthur wins. If the table is even larger, allowing more coins, same logic applies. Each time Merlin places a coin, Arthur places the mirror, so Merlin runs out of moves first.Another angle: this is similar to the concept of "strategy stealing". If the second player had a winning strategy, the first player could "steal" it by making an arbitrary first move and then mirroring. But in this case, the mirroring is possible because of the center. Alternatively, the key is that the starting move occupies a unique position that allows for symmetry.Wait, but another thought: what if the table isn't a perfect circle? But the problem states it's a round table, so we can assume rotational symmetry. Therefore, Arthur's strategy is valid.So conclusion: Arthur places the first coin at the center, then mirrors every subsequent move by Merlin. This ensures Arthur always has a move, so Merlin will be the first unable to place a coin. Therefore, Arthur has a winning strategy.

✅:Arthur has a winning strategy by leveraging the symmetry of the round table. Here's the step-by-step reasoning:1. First Move: Occupy the Center Arthur places the first coin at the exact center of the table. This central position is unique and breaks the table's rotational symmetry in a way that benefits Arthur.2. Mirroring Strategy After Merlin places a coin anywhere on the table, Arthur responds by placing a coin diametrically opposite to Merlin’s placement. For example, if Merlin places a coin at polar coordinates ((r, theta)), Arthur places one at ((r, theta + pi)). This ensures that every valid move by Merlin can be mirrored by Arthur.3. Symmetry Preservation The central coin does not interfere with the mirroring strategy because it occupies a unique point. All subsequent coins placed by Merlin will have distinct mirrored positions, which Arthur can always utilize. Since the table is circular and infinite coins are available, the symmetry guarantees that if Merlin has space to place a coin, Arthur will always have the corresponding mirrored space.4. Forcing Merlin to Lose Since Arthur starts first and mirrors every move, he ensures that he always has the last move. Merlin will eventually face a situation where no space remains for a new coin, while Arthur’s mirrored responses exhaust all available positions.Conclusion: By occupying the center first and systematically mirroring Merlin’s moves, Arthur guarantees that he cannot be the first player unable to place a coin. Thus, Arthur has a winning strategy.

🔑:1. Initial Move by Arthur: - Arthur places his first coin at the exact center of the round table. By taking the center, Arthur ensures that this initial placement is symmetric. 2. Response Strategy: - For each subsequent move made by Merlin, Arthur places his coin diametrically opposite to Merlin’s coin. This strategy works because: 1. The round table is symmetric around the center. 2. If Merlin can place a coin at a certain position, by symmetry, the position diametrically opposite to it must be unoccupied and fitting for another coin of the same size.3. Symmetry Argument: - The symmetry of the strategy ensures that every move Merlin can make, Arthur can also make a corresponding move. - If Merlin places his coin at coordinates ( (x, y) ) relative to the center of the table, Arthur will place his coin at ( (-x, -y) ).4. Winning Condition: - Arthur can always place his coin if Merlin can place his, due to the symmetric arrangement. - Hence, Merlin will be the first to run out of valid moves because every valid move by Merlin will correspond to a valid move by Arthur. Conclusion:Arthur’s strategy of starting from the center and following with placements diametrically opposite to Merlin’s ensures he will always have a move if Merlin has one. Therefore, Arthur has a guaranteed winning strategy.boxed{text{Arthur has a winning strategy}}

🧠:Okay, let's try to tackle this physics problem. It seems a bit complex at first glance, but I'll take it step by step. First, I need to understand the setup.So, there are two masses, m₁ = 1 kg and m₂ = 10 kg, placed on two horizontal shelves. They're connected by a weightless and inextensible string over a smooth, weightless pulley. The coefficients of friction are μ₁ = 0.3 for the first mass and μ₂ = 0.1 for the second. There's a horizontal force F = 20 N applied to the pulley's center. We need to find the accelerations of each mass and the acceleration of the center of mass of the system. Gravity is 10 m/s².Hmm, let me visualize this. The pulley is smooth and weightless, so the tension should be the same on both sides? Wait, but the pulley is being pulled horizontally with a force F. How does that affect the system? Maybe the pulley's position affects the angles of the string? But since the shelves are horizontal and the pulley is weightless, maybe the strings are horizontal too? Wait, no, because if the pulley is in between the two shelves, the strings might be at an angle. Wait, but the problem says the shelves are horizontal and the pulley is connected by a weightless, inextensible thread. Hmm, maybe both masses are on the same horizontal level, connected over a pulley, and the pulley is being pulled horizontally. But I need to be careful here.Wait, the problem says "two horizontal shelves" – maybe the two masses are on different shelves, one higher than the other? But then the pulley would need to be between them. Alternatively, maybe both masses are on separate horizontal planes, connected by a string over a pulley, and the pulley is being pulled sideways. Hmm.Alternatively, maybe the pulley is positioned such that when the force F is applied horizontally, it changes the direction of the tension. For example, if the pulley is being pulled to the side, the strings from the pulley to the masses would form an angle. But since the pulley is weightless and smooth, the tension should still be equal on both sides. But how does the horizontal force F affect the tensions?Wait, the problem states that the pulley is weightless and smooth. So, the tension on either side of the pulley must be the same. However, the pulley itself is being pulled with a horizontal force F. In typical pulley problems, if you have a pulley fixed in place, the tension forces are balanced. But here, the pulley is being moved by force F. So, perhaps the application of force F on the pulley introduces a different scenario where the pulley is accelerating, and the motion of the masses is affected by that.Wait, but if the pulley is massless (weightless), then the net force on it should be zero. But here, force F is applied to it. Hmm, that seems contradictory. If the pulley is massless, then even if a force is applied, it should have infinite acceleration, but that doesn't make sense. Wait, maybe the pulley is considered part of the system, and the force F is an external force applied to the system. Alternatively, maybe the problem is set up such that the pulley's movement affects the motion of the masses.This is getting a bit confusing. Let me think again. Maybe the pulley is being pulled horizontally, which causes the strings to be at an angle, thereby changing the effective tension on each mass. For example, if the pulley is accelerated to the right with force F, the strings would make some angle with the horizontal, and the horizontal component of the tension would contribute to the forces on the masses.Wait, perhaps we need to model the system with the pulley moving, which would cause the strings to not be vertical but at an angle. Then, the tension in the string would have both vertical and horizontal components. However, since the pulley is massless, the net force on it must be zero. But force F is applied to it, so the tension forces must balance F.Wait, here's an approach: Since the pulley is massless, the sum of the forces on it must be zero. The applied force F is 20 N. The tensions from both sides of the pulley would have horizontal components adding up to F. Since the pulley is smooth, the tension T is the same on both sides. If the strings make an angle θ with the horizontal, then the horizontal components from each side would be T cos θ each. Since there are two tensions (left and right), the total horizontal force from the tensions is 2T cos θ. This must equal the applied force F.So, 2T cos θ = F = 20 N. Therefore, T cos θ = 10 N. But we also need to consider the vertical components of the tension, which would affect the normal forces on the masses, and hence the frictional forces. Wait, but the problem states that the shelves are horizontal. So the masses are on horizontal surfaces, connected by a string over a pulley, and the pulley is being pulled horizontally. Therefore, the strings must be at an angle to the horizontal.Let me try to draw a mental picture. The pulley is being pulled to the right with force F. The two masses are on the left and right sides of the pulley, each on their own horizontal shelf. The string from each mass goes over the pulley. If the pulley is moved to the right, the strings would slope towards the pulley from each mass. So each mass is being pulled towards the pulley at an angle. The angle would depend on the position of the pulley relative to the masses. But since the pulley is moving, this angle might be changing, making the problem dynamic.Wait, but the problem doesn't mention any constraints on the movement of the pulley. It just says a horizontally directed force is applied. So maybe we can assume that the pulley is moving with some acceleration, and the angle of the string is such that the horizontal components of the tension provide the necessary force for the pulley's acceleration, while the vertical components affect the normal force on the masses.But this is getting complicated. Let me break it down.First, let's consider the pulley. It is massless, so the net force on it must be zero. However, an external force F is applied. The only way this is possible is if the tensions in the string provide a reaction force. Since the pulley is massless, the sum of forces on it must be zero. So, the horizontal components of the tensions from both sides of the pulley must add up to F.Assuming the strings make an angle θ with the horizontal on both sides, then each tension T has a horizontal component T cos θ. Since there are two tensions (from each side), the total horizontal force from the string is 2T cos θ. Therefore:2T cos θ = F => T cos θ = F/2 = 10 N.But we don't know θ. However, the vertical component of the tension would be T sin θ, which would affect the normal force on each mass. The normal force, in turn, affects the friction.Wait, but the masses are on horizontal shelves. So, for each mass, the vertical component of the tension would either add to or subtract from the normal force. For example, if the string is pulling upward on a mass, the normal force would decrease, reducing the friction. If the string is pulling downward, the normal force would increase, increasing the friction. However, since the pulley is being pulled horizontally, the angle of the string would cause the tension to have both horizontal and vertical components.But how do we find θ? The angle θ depends on the geometry of the system. If the pulley moves horizontally, the length of the string would have to adjust, which affects the angle. However, since the string is inextensible, the movement of the pulley and the masses must be coordinated such that the total length of the string remains constant.This seems like a problem that involves constraints from the string length. Let me consider the positions of the masses and the pulley.Let’s denote the position of the pulley as x_p(t), moving to the right due to force F. The two masses have positions x₁(t) and x₂(t) on their respective shelves. Since the string is inextensible, the length of the string from each mass to the pulley must remain constant. However, since the pulley is moving, this introduces a relationship between the accelerations of the pulley and the masses.Wait, but how exactly? Let's model this. Suppose the pulley is at position x_p(t), and each mass is connected to the pulley. The total length of the string from mass 1 to mass 2 via the pulley is L = L1 + L2, where L1 is the length from mass 1 to the pulley, and L2 is from the pulley to mass 2. Since the string is inextensible, L must remain constant.If the pulley moves to the right by Δx_p, then the length L1 would decrease by the horizontal component of the movement, but since the string is also moving around the pulley, perhaps the vertical component also plays a role? Wait, this is getting too vague. Maybe we need to consider the geometry more carefully.Alternatively, maybe we can consider that when the pulley is displaced horizontally, the vertical position of the masses must adjust accordingly. But since the shelves are horizontal, the masses can only move horizontally as well. Therefore, any vertical component in the string would require the masses to move vertically, but they can't because they're constrained to the shelves. Hmm, this is a contradiction. Wait, that can't be. So perhaps the strings are horizontal? But then, if the pulley is pulled horizontally, the strings would have to be horizontal, meaning that the angle θ is zero, but then cos θ = 1, and the horizontal component of tension would be T. Then 2T = F => T = 10 N. But in that case, the tension would be 10 N, and there's no vertical component. But the problem states that the pulley is smooth and weightless. If the strings are horizontal, then the pulley being pulled with force F = 20 N would have two tensions of 10 N each, balancing the force. Then, the masses are being pulled with tension T = 10 N each. But wait, the pulley is massless, so the net force must be zero. So if F = 20 N is applied, then 2T = 20 N => T = 10 N. That makes sense.But then, if the strings are horizontal, there's no vertical component, so the normal forces on the masses would just be m₁g and m₂g. Therefore, the frictional forces would be μ₁m₁g and μ₂m₂g.But in that case, we can model each mass being pulled by a tension T = 10 N, with friction opposing the motion.Wait, but if the pulley is being pulled to the right, then the string would be moving to the right, so the masses would be pulled towards the pulley. If the pulley is moving, does that induce motion in the masses? Or are the masses moving relative to the shelves?Wait, this is confusing. Let's clarify. If the pulley is being pulled to the right, and the string is inextensible, then as the pulley moves right, the length of the string on both sides must adjust. However, since the masses are on the shelves, their movement is constrained by the string. If the pulley moves right, the left side of the string would get longer, requiring the mass m₁ to move left, and the right side would get shorter, requiring the mass m₂ to move right. But since the string is inextensible, the total length must remain the same. Therefore, if the pulley moves Δx_p to the right, then mass m₁ must move Δx₁ to the left, and mass m₂ must move Δx₂ to the right, such that Δx_p = (Δx₁ + Δx₂)/2. Wait, no. Let me think again. If the pulley moves to the right by Δx_p, then the length from the pulley to mass m₁ increases by Δx_p (if the mass doesn't move), but since the string is inextensible, mass m₁ must move left by Δx_p to compensate. Similarly, the length from the pulley to mass m₂ decreases by Δx_p, so mass m₂ must move right by Δx_p. Therefore, the displacement of each mass is equal and opposite to the pulley's displacement? Wait, that can't be. Let me model it.Suppose the pulley moves to the right by distance x_p. If the string is inextensible, the total length of the string is constant. Let’s assume the original length from m₁ to pulley is L₁ and from pulley to m₂ is L₂, so total length L = L₁ + L₂. When the pulley moves right by x_p, the new length from m₁ to pulley is L₁ + x_p (if m₁ stays still), and from pulley to m₂ is L₂ - x_p. But since L must remain the same, (L₁ + x_p) + (L₂ - x_p) = L₁ + L₂ = L. So that's okay. Wait, but that would mean the masses don't have to move? That doesn't make sense. Wait, maybe I'm missing something here.Alternatively, if the pulley moves right by x_p, and the masses move left and right by x₁ and x₂ respectively, then the change in length on the left side is x_p - x₁ (if m₁ moves left by x₁) and on the right side is x_p + x₂ (if m₂ moves right by x₂). Since the total length must remain the same, the sum of these changes should be zero:(x_p - x₁) + (x_p + x₂) = 0 => 2x_p + (x₂ - x₁) = 0.But unless x₂ - x₁ = -2x_p, which would mean that the displacement of the masses relative to the pulley is such that their total displacement compensates for twice the pulley's displacement. This is getting too convoluted. Maybe we need to relate the accelerations instead.If the pulley has an acceleration a_p to the right, then the accelerations of the masses relative to the pulley must be such that the total length of the string remains constant. This is similar to a pulley system where the acceleration of the pulley affects the accelerations of the masses.In such cases, the acceleration of each mass relative to the ground is a combination of the pulley's acceleration and their own acceleration relative to the pulley. For example, if the pulley accelerates to the right with a_p, and the mass m₁ accelerates to the left relative to the pulley with a'_1, then its absolute acceleration is a_p - a'_1. Similarly, if mass m₂ accelerates to the right relative to the pulley with a'_2, its absolute acceleration is a_p + a'_2. But since the string is inextensible, the accelerations relative to the pulley must be equal in magnitude and opposite in direction, so a'_1 = a'_2 = a'. Therefore, the absolute accelerations are a₁ = a_p - a' and a₂ = a_p + a'.But we have two unknowns here: a_p and a'. However, we also have the dynamics of the pulley and the masses. Since the pulley is massless, the net force on it must be zero. The external force F is applied to the pulley, and the tensions pull on it. So, for the pulley:F - 2T = 0 => T = F/2 = 10 N.Wait, but earlier we thought T is 10 N. However, the tensions also relate to the forces on the masses. So let's model the forces on each mass.For mass m₁: The horizontal force is T - f₁ = m₁a₁, where f₁ is the frictional force. Similarly, for mass m₂: T - f₂ = m₂a₂. But wait, the direction of the friction depends on the direction of motion. If the mass is moving to the left, friction acts to the right, and vice versa.But since we don't know the direction of acceleration yet, we need to assume a direction and check consistency.Let’s assume that both masses are moving in the direction of the tension, i.e., m₁ is being pulled towards the pulley (to the right) and m₂ is being pulled towards the pulley (to the left). Wait, but the pulley is being pulled to the right. If the pulley moves right, then the string on the left side (m₁) would need to lengthen, so m₁ would have to move left relative to the ground, while the string on the right side (m₂) shortens, so m₂ moves right relative to the ground. Wait, this is confusing.Alternatively, if the pulley is moving to the right, then relative to the pulley, the masses must be moving towards it. So, m₁ is moving to the right relative to the pulley, which would mean moving to the right faster than the pulley if the pulley itself is moving right. But in absolute terms, the acceleration of m₁ would be a_p + a' (if a' is the acceleration relative to the pulley). Similarly, m₂ would be accelerating to the left relative to the pulley, so its absolute acceleration would be a_p - a'. But this depends on the coordinate system.This is getting too tangled. Maybe a better approach is to use coordinates. Let's set up coordinate systems for each mass and the pulley.Let’s assume all motions are along the x-axis. Let’s take the positive x-direction to the right. Let’s denote the position of the pulley as x_p(t), mass m₁ as x₁(t), and mass m₂ as x₂(t). The string is inextensible, so the total length of the string is constant.Assuming the pulley is being pulled to the right, the length from m₁ to the pulley is x_p - x₁, and from the pulley to m₂ is x₂ - x_p. The total length L = (x_p - x₁) + (x₂ - x_p) = x₂ - x₁. So, L must be constant. Therefore, x₂ - x₁ = constant. Differentiating both sides with respect to time, we get v₂ - v₁ = 0 => v₂ = v₁. Differentiating again, a₂ = a₁. Wait, that's strange. If the total length is x₂ - x₁, then for it to be constant, the velocities of both masses must be equal, and accelerations equal. But this would mean that both masses move in the same direction with the same acceleration, which contradicts the pulley being pulled to the right.This suggests that my initial assumption about the string length is incorrect. Maybe the pulley is not between the masses but arranged such that the string goes over it in a way that allows the masses to move in opposite directions. Wait, maybe the problem is similar to a traditional pulley system where the two masses are on opposite sides of the pulley, but here, the pulley is on a horizontal plane.Wait, perhaps the problem is analogous to a pulley system on a table. Imagine both masses are on a horizontal table, connected by a string over a pulley fixed to the edge of the table. But in this case, the pulley isn't fixed; instead, it's being pulled with a horizontal force. However, the masses are on the table, so they can only move horizontally. The pulley, being pulled sideways, would cause the string to have an angle, affecting the tension.Alternatively, think of the pulley as being on a cart that is being pulled with force F. The two masses are on the cart's platform, connected over the pulley. As the cart is pulled, the masses may move relative to the cart. But I need to model this properly.Alternatively, since the problem mentions two horizontal shelves, maybe each mass is on a separate shelf at different heights, and the pulley is between them. But the applied force is horizontal. Hmmm.This is getting too confusing without a clear diagram. Let me try to approach it differently.Given that the pulley is massless and smooth, the tension in the string on both sides is equal. A force F is applied horizontally to the pulley. The masses are on horizontal shelves with friction.The key here is that applying force F to the pulley affects the tension in the string, which in turn affects the forces on the masses. The motion of the pulley will influence the accelerations of the masses through the string's constraints.Let me consider the forces on the pulley. Since it's massless, the net force must be zero. Therefore, the horizontal components of the tensions must balance the applied force F. If the string makes an angle θ with the horizontal on both sides, then each tension T contributes T cos θ horizontally. Hence, 2T cos θ = F.But we need to relate θ to the motion of the masses. If the pulley moves to the right with acceleration a_p, and the masses have accelerations a₁ and a₂, then the angle θ is determined by the relative accelerations.Alternatively, if the pulley is moving, the angle θ might change over time, making this a problem with varying θ. However, perhaps we can assume that the system reaches a steady state where θ is constant, but that might not be the case.Alternatively, use the instantaneous angle θ, considering that the accelerations are related through the string's constraint.Wait, perhaps using constraints. The key idea is that the acceleration of the masses relative to the pulley must be equal in magnitude but opposite in direction. Because as the pulley moves, the string's length on either side must change in a way that keeps the total length constant.If the pulley moves to the right with acceleration a_p, then the mass m₁ must move to the left relative to the pulley with some acceleration a', and mass m₂ must move to the right relative to the pulley with the same acceleration a'. Therefore, the absolute accelerations of the masses would be:a₁ = a_p - a' (since the pulley is moving right and the mass is moving left relative to it)a₂ = a_p + a' (since the pulley is moving right and the mass is moving right relative to it)But this is under the assumption that the relative acceleration a' is the same for both masses. Is this valid? Let's check.Since the string is inextensible, the rate at which the string is lengthened on one side must equal the rate it's shortened on the other. If the pulley moves right by Δx_p, then the left side of the string must lengthen by Δx_p, requiring mass m₁ to move left by Δx_p. Similarly, the right side shortens by Δx_p, requiring mass m₂ to move right by Δx_p. Therefore, the displacements of the masses are equal in magnitude and opposite in direction relative to the pulley. Hence, their accelerations relative to the pulley are equal in magnitude (a') and opposite in direction. Therefore, the absolute accelerations are:a₁ = a_p - a'a₂ = a_p + a'But actually, if the pulley accelerates to the right at a_p, and the masses have accelerations a₁ and a₂ relative to the ground, then the accelerations relative to the pulley are a₁' = a₁ - a_p (for mass m₁) and a₂' = a₂ - a_p (for mass m₂). Since the string is inextensible, these relative accelerations must satisfy a₁' = -a₂', so:a₁ - a_p = -(a₂ - a_p)=> a₁ - a_p = -a₂ + a_p=> a₁ + a₂ = 2a_pThat's an important constraint: the sum of the accelerations of the masses is twice the acceleration of the pulley.So, we have that equation: a₁ + a₂ = 2a_p.Now, we need to find expressions for a₁ and a₂ using Newton's laws.For each mass, the net force is equal to mass times acceleration. The forces on each mass are tension T, friction f, and any other forces.Let's first find the direction of friction. Assume that the pulley is accelerating to the right, causing the string to pull mass m₁ to the right and mass m₂ to the left (since if the pulley moves right, the left side of the string would pull m₁ right, and the right side would pull m₂ left). Wait, but based on previous analysis, if the pulley moves right, then m₁ is moving left relative to the ground (a₁ = a_p - a') and m₂ is moving right relative to the ground (a₂ = a_p + a'). Wait, this is conflicting. Maybe I need to re-examine the direction.Alternatively, perhaps when the pulley is pulled right, the string on the left side becomes longer, so m₁ has to move left, and the string on the right side becomes shorter, so m₂ has to move right. Therefore, the accelerations of the masses are a₁ to the left (negative) and a₂ to the right (positive). But according to the constraint equation a₁ + a₂ = 2a_p, if a_p is positive (to the right), then a₁ + a₂ must be positive. If a₁ is negative (left) and a₂ is positive (right), their sum could be positive or negative depending on magnitudes.But we need to model the forces.Let’s proceed step by step.First, consider mass m₁. It's on a horizontal shelf with friction μ₁ = 0.3. The forces acting on it are:- Tension T pulling to the right (assuming the string is trying to pull it towards the pulley, which is moving right; but if the pulley is moving right, then relative to the ground, the tension might be pulling m₁ to the right? Wait, no. If the pulley is moving right and the string is inextensible, then if the pulley moves right, the length of the string from m₁ to the pulley would increase unless m₁ moves left. Therefore, the tension would be pulling m₁ to the left? Wait, this is getting confusing.Wait, let's clarify. If the pulley is moving to the right, and the string is inextensible, then for the string to stay connected, m₁ must be moving towards the pulley. If the pulley moves right, then m₁ has to move left to keep the string from going slack. Similarly, m₂ has to move right to follow the pulley. Wait, but if the pulley moves right, the distance from m₁ to the pulley increases, so m₁ must move left to reduce that distance. Wait, no. If the pulley moves right, and the string is inextensible, then the only way to keep the string tight is if m₁ moves right as well, but maybe at a different rate. This is why the constraint equation is important.Given the constraint a₁ + a₂ = 2a_p, if we assume all accelerations are to the right, then:If a_p is the pulley's acceleration to the right, and a₁ and a₂ are the accelerations of the masses to the right, then the equation holds. However, depending on the forces, the accelerations could be in different directions.But let's tentatively assume that all accelerations are to the right. Then, for mass m₁, the tension T is pulling to the right, and friction is opposing the motion, so friction is to the left. Similarly, for mass m₂, tension T is pulling to the left (since it's on the other side of the pulley?), wait, no. Wait, the pulley is being pulled to the right, so the tension on both sides is directed towards the pulley. If the pulley is to the right of both masses, then both masses would be pulled towards the pulley. If the pulley is between the masses, then one mass is pulled to the right and the other to the left. Wait, this is unclear.Perhaps we need to consider the direction of the tension force on each mass. If the pulley is being pulled to the right, then the string on the left side of the pulley is going towards the pulley from the left, so the tension on m₁ is pulling it to the right. The string on the right side is going towards the pulley from the right, so the tension on m₂ is pulling it to the left. Wait, that makes sense. So, mass m₁ is pulled to the right by tension T, and mass m₂ is pulled to the left by tension T. Then, the friction forces will oppose the relative motion.Therefore, for mass m₁:Net force to the right: T - f₁ = m₁a₁Where f₁ = μ₁N₁. Since it's on a horizontal shelf, N₁ = m₁g. So, f₁ = μ₁m₁g.Similarly, for mass m₂:Net force to the left: T - f₂ = m₂a₂But wait, if the pulley is pulling m₂ to the left, then the friction force on m₂ would be to the right, opposing the motion. So:T - f₂ = m₂a₂, but we need to be careful with signs.Let's define positive direction to the right for all. Then, for mass m₁:T (to the right) - f₁ (to the left) = m₁a₁=> T - μ₁m₁g = m₁a₁For mass m₂:- T (to the left) - f₂ (to the left) = m₂a₂Wait, no. If positive is to the right, then tension on m₂ is to the left (-T), and friction, which opposes the motion, if the mass is moving to the left, friction would be to the right. But we don't know the direction of motion yet.This is the crux. We need to assume directions for the accelerations and check for consistency.Let’s assume that both masses are moving to the right. Then, friction would oppose that motion, so:For m₁: T - f₁ = m₁a₁For m₂: The tension is pulling it to the left, so the net force is (-T) - f₂ = m₂a₂ (since friction is to the left if the mass is moving right, but wait, if mass m₂ is moving right, friction is to the left. Tension is to the left, friction is to the left, so total force is (-T - f₂) = m₂a₂. But if a₂ is positive (to the right), then this would imply that the net force is to the left, causing a negative acceleration. Contradiction. Therefore, our assumption about the direction of motion might be wrong.Alternatively, suppose mass m₂ is moving to the left. Then, friction would be to the right. Tension is to the left, friction to the right:Net force on m₂: -T + f₂ = m₂a₂But if a₂ is negative (moving left), then:-T + f₂ = m₂a₂But we need to be consistent with signs. Let's define the positive direction for each mass:For mass m₁, positive to the right.For mass m₂, positive to the left.But this complicates things. Alternatively, let's take positive to the right for both, and let accelerations be positive or negative accordingly.If the pulley is pulled to the right, the tension on m₁ is to the right, and on m₂ is to the left. Assume that m₁ accelerates to the right (a₁ > 0) and m₂ accelerates to the left (a₂ < 0). But according to the constraint, a₁ + a₂ = 2a_p. If a_p is positive (pulley moving right), then a₁ + a₂ must equal 2a_p, which is positive. If a₁ is positive and a₂ is negative, their sum could be positive or negative depending on magnitudes.Alternatively, maybe both masses are accelerating to the right but with different accelerations. Let's proceed with equations.For mass m₁:T - f₁ = m₁a₁For mass m₂:-T - f₂ = m₂a₂But friction forces:f₁ = μ₁m₁g (opposing the motion, direction depends on the direction of motion)Wait, no. The friction force is always opposing the relative motion. So if m₁ is moving to the right, friction is to the left. If m₁ is moving to the left, friction is to the right. But we don't know the direction yet.This is why we need to assume directions and check.Assume that the pulley acceleration a_p is to the right. Assume that m₁ is moving to the right (a₁ > 0) and m₂ is moving to the left (a₂ < 0).But from the constraint equation:a₁ + a₂ = 2a_pIf a_p is positive, and a₂ is negative, their sum could still be positive if a₁ > |a₂|.But let's write the equations with the assumed directions.For m₁ (moving right):T - f₁ = m₁a₁f₁ = μ₁m₁g (to the left)For m₂ (moving left):-T + f₂ = m₂a₂Because friction opposes the motion, so if m₂ is moving left, friction is to the right (positive direction). Tension is to the left (-T), friction is to the right (+f₂). So:-T + f₂ = m₂a₂But a₂ is negative (moving left), so:-T + f₂ = m₂a₂Now, we also have the pulley equation: 2T cos θ = F. But we still need to relate θ to the accelerations.Wait, θ is the angle of the string with the horizontal. If the pulley is accelerating, the angle θ will change over time, making this a variable in the problem. However, perhaps we can relate θ to the accelerations through the kinematic constraint.The key idea is that the vertical component of the tension affects the normal force, which in turn affects friction. If the string is at an angle θ, then the tension has a vertical component T sin θ, which would add to or subtract from the normal force.Wait, this is crucial. I missed this earlier. If the string is at an angle θ, then the vertical component of tension would be T sin θ, changing the normal force.For mass m₁, the normal force N₁ would be m₁g - T sin θ (if the vertical component is upwards) or m₁g + T sin θ (if downwards). Similarly for mass m₂.But in which direction is the vertical component? If the pulley is above the masses, the string would pull upwards on the masses, reducing the normal force. If the pulley is below, it would pull downwards, increasing the normal force. However, the problem states that the shelves are horizontal, and the pulley is connected by a string. So, the pulley must be positioned such that the string is horizontal, or at an angle.Wait, if the pulley is on the same horizontal level as the shelves, then the string would also be horizontal, making θ = 0. But then the vertical component is zero, and the normal forces are just m₁g and m₂g. But in that case, the previous analysis applies, and the tension would be T = F/2 = 10 N.But the problem mentions that the pulley is weightless and smooth, and the string is weightless and inextensible. If the pulley is pulled horizontally, maybe the strings are horizontal, and the angle θ is zero. Then, the tension is T = 10 N, and we can compute the accelerations of the masses based on T, friction, and their masses.But let's check this scenario.If θ = 0, then the tension is horizontal, and there is no vertical component. Therefore, the normal forces are N₁ = m₁g and N₂ = m₂g.For mass m₁:T - f₁ = m₁a₁f₁ = μ₁N₁ = μ₁m₁g = 0.3 * 1 * 10 = 3 NSo, 10 N - 3 N = 1 kg * a₁ => a₁ = 7 m/s² to the right.For mass m₂:The tension is pulling to the left, so:-T - f₂ = m₂a₂Wait, but if the pulley is pulled to the right, the tension on m₂ is to the left. The friction force f₂ opposes the motion. If the mass m₂ is moving to the left, friction would be to the right. But if the mass is moving to the right, friction is to the left. But we don't know the direction yet.Wait, if T = 10 N, f₂ = μ₂m₂g = 0.1 * 10 * 10 = 10 N.So for mass m₂:-T - f₂ = m₂a₂But if tension is to the left (-10 N) and friction is opposing the motion. If the mass is moving to the left, friction is to the right (+10 N). If moving to the right, friction is to the left (-10 N). But we don't know direction.Assume that the mass m₂ is moving to the left (due to tension), then friction is to the right:-T + f₂ = m₂a₂=> -10 N + 10 N = 10 kg * a₂ => 0 = 10a₂ => a₂ = 0.But this contradicts the constraint equation a₁ + a₂ = 2a_p. If a₁ = 7 m/s² and a₂ = 0, then 7 = 2a_p => a_p = 3.5 m/s². But what is causing the pulley to accelerate? The applied force F = 20 N. According to the pulley equation, 2T = F => T = 10 N. But the pulley's acceleration is determined by the motion of the masses? Wait, but the pulley is massless, so it can't have inertia. Therefore, its acceleration is determined by the system's dynamics.But if we assume the pulley is accelerating at 3.5 m/s², but there's no mass to it, how does that happen? It seems like there's a contradiction here.Alternatively, maybe this approach is incorrect because we neglected the angle θ. If θ is not zero, then the vertical component of tension affects the normal force, changing the friction. Therefore, we can't assume θ = 0. The angle θ must be considered.So, we need to relate the angle θ to the accelerations. To do this, we can consider the vertical components of the tension affecting the normal forces.For mass m₁:Vertical forces: N₁ + T sin θ = m₁g (if tension is pulling upward)But if the string is pulling downward, it would be N₁ - T sin θ = m₁g.Wait, the direction of the vertical component depends on the position of the pulley. If the pulley is above the mass, the tension would pull upward, reducing the normal force. If below, it would pull downward, increasing the normal force. But since the shelves are horizontal and the pulley is being pulled horizontally, the pulley must be at the same height as the masses, making the string horizontal, or pulled to the side, making the string at an angle.But without knowing the pulley's position, it's difficult to determine. However, since the problem states that the pulley is placed between two horizontal shelves, perhaps the pulley is located such that the strings to each mass are horizontal. But if a horizontal force is applied to the pulley, it would move horizontally, and the strings would remain horizontal. However, this would require the masses to move in such a way that the string length remains constant. But if the pulley moves right, the left mass must move left and the right mass move right, keeping the string length the same. But since the string is inextensible, this isn't possible unless the masses move accordingly. Therefore, the movement of the pulley and masses are connected.This suggests that the angle θ is actually zero, and the previous analysis holds, but with the contradiction in accelerations.Alternatively, maybe the pulley is being pulled horizontally, causing the strings to slack or tighten, but since they are inextensible, the movement of the pulley must be directly related to the movement of the masses.Wait, let's think of it as a system where the pulley's movement is directly connected to the masses' movements. If the pulley moves right by Δx, then mass m₁ must move left by Δx, and mass m₂ must move right by Δx. Therefore, the displacement of the pulley is equal to the average displacement of the masses. Therefore, the acceleration of the pulley is the average of the accelerations of the masses.Wait, this is similar to a system where two masses are connected over a pulley, and the pulley is being accelerated. In such cases, the constraint is a_p = (a₁ + a₂)/2. Which matches our previous constraint a₁ + a₂ = 2a_p.So, if we can write the equations for each mass considering the tension, friction, and normal forces, and include the effect of the angle θ on the normal force, we can solve for the accelerations.But to find θ, we need to relate it to the accelerations. The angle θ depends on the relative acceleration between the pulley and the masses. If the pulley is moving with acceleration a_p, and the masses are moving with accelerations a₁ and a₂, then the angle θ is determined by the ratio of the vertical and horizontal components of the string's displacement.Wait, but if the pulley is moving horizontally and the masses are moving horizontally, where does the vertical component come from? If everything is horizontal, there is no vertical component. This suggests that θ = 0, making the tension purely horizontal. Therefore, the earlier assumption was correct, but then we faced a contradiction.Alternatively, perhaps the problem is designed with the pulley being pulled horizontally, but the strings have a vertical component because the pulley is above the masses. For instance, if the pulley is positioned higher than the shelves, the strings would slope down to the masses, creating an angle. In this case, the tension would have both horizontal and vertical components. The applied horizontal force on the pulley would then affect the horizontal component of the tension.Let's assume the pulley is positioned above the two masses, which are on the same horizontal level. The strings from the masses to the pulley make an angle θ below the horizontal. When a horizontal force F is applied to the pulley, it affects the tension in the strings.In this case, the horizontal component of the tension from each string is T cos θ, and the vertical component is T sin θ downward. Since there are two strings, the total horizontal force from the tensions is 2T cos θ, which must balance the applied force F. Therefore:2T cos θ = F => T = F / (2 cos θ)The vertical component of the tension on each mass is T sin θ downward. Therefore, the normal force on each mass is:For m₁: N₁ = m₁g + T sin θFor m₂: N₂ = m₂g + T sin θThe friction force for each mass is:f₁ = μ₁N₁ = μ₁(m₁g + T sin θ)f₂ = μ₂N₂ = μ₂(m₂g + T sin θ)The horizontal forces on each mass are:For m₁: T cos θ - f₁ = m₁a₁For m₂: T cos θ - f₂ = m₂a₂But we also have the constraint from the pulley's motion:a₁ + a₂ = 2a_pBut the pulley's acceleration a_p is related to the applied force F. However, the pulley is massless, so the net force on it must be zero. Therefore, 2T cos θ = F, which we've already used.But we now have a system of equations with variables T, θ, a₁, a₂, and a_p.Let's try to write all equations:1. 2T cos θ = F => T = F / (2 cos θ)2. a₁ + a₂ = 2a_p3. For m₁: T cos θ - μ₁(m₁g + T sin θ) = m₁a₁4. For m₂: T cos θ - μ₂(m₂g + T sin θ) = m₂a₂We have four equations and four unknowns: T, θ, a₁, a₂. (a_p can be expressed via equation 2)Let's substitute T from equation 1 into equations 3 and 4.From equation 1: T = F / (2 cos θ)Substitute into equation 3:(F / (2 cos θ)) cos θ - μ₁(m₁g + (F / (2 cos θ)) sin θ) = m₁a₁Simplify:F/2 - μ₁(m₁g + (F/2) tan θ) = m₁a₁Similarly, for equation 4:(F / (2 cos θ)) cos θ - μ₂(m₂g + (F / (2 cos θ)) sin θ) = m₂a₂Simplify:F/2 - μ₂(m₂g + (F/2) tan θ) = m₂a₂So now, we have:a₁ = [F/2 - μ₁(m₁g + (F/2) tan θ)] / m₁a₂ = [F/2 - μ₂(m₂g + (F/2) tan θ)] / m₂But we also have equation 2: a₁ + a₂ = 2a_pHowever, we need to find θ. To do that, we need another relation. The angle θ is related to the geometry of the system. If the pulley is at a certain height above the masses, the angle θ is determined by the horizontal distance between the pulley and the masses. However, since the problem doesn't provide any geometrical information, we might need to assume that the angle θ is such that the vertical components of the tension balance some forces, but it's not clear.Alternatively, maybe the angle θ is determined dynamically based on the accelerations. For instance, if the pulley is accelerating, the angle θ changes to accommodate the motion. However, without additional information, it's impossible to determine θ geometrically. Therefore, there must be another way to relate θ to the accelerations.Wait, considering that the pulley is massless and the net force is zero, we already used that in equation 1. The other equations involve the masses' accelerations and the angle θ.This seems like a dead end unless we can find another relation. Perhaps, if the vertical acceleration of the masses is zero (since they are on horizontal shelves), the vertical component of the tension must be balanced by the normal force.Wait, yes. The masses cannot accelerate vertically because they are constrained to move horizontally. Therefore, the vertical forces must balance.For mass m₁:N₁ = m₁g + T sin θSimilarly for mass m₂:N₂ = m₂g + T sin θBut since they can't move vertically, these equations hold true, and the friction forces depend on these normal forces.But how does this help us? We already used these in equations 3 and 4.Hmm. We have three equations:1. 2T cos θ = F2. a₁ = [F/2 - μ₁(m₁g + (F/2) tan θ)] / m₁3. a₂ = [F/2 - μ₂(m₂g + (F/2) tan θ)] / m₂4. a₁ + a₂ = 2a_pBut we still have two unknowns: θ and a_p. Wait, actually, θ is another unknown. So we have four equations but five unknowns (T, θ, a₁, a₂, a_p). However, T can be expressed in terms of θ via equation 1, so effectively, we have four equations with four unknowns (θ, a₁, a₂, a_p). Let's see if we can solve for θ.Let’s try to express a₁ and a₂ in terms of θ, then use the constraint a₁ + a₂ = 2a_p. But since a_p is the pulley's acceleration, which is related to how the system moves, perhaps we need another relation. However, without more information, it's challenging.Alternatively, maybe θ is constant, which would imply that the ratio of vertical to horizontal components of the tension is constant. But since the pulley is accelerating, θ would change over time, making this a time-dependent problem. However, the question asks for the accelerations of the masses, which are instantaneous. So perhaps we can consider the initial acceleration when the system starts moving, assuming the angle θ is such that the pulley just begins to move.Alternatively, maybe the angle θ is determined by the balance of forces in the vertical direction for the masses. Since the vertical acceleration is zero, the normal force equals the weight plus the vertical component of tension. But that’s already considered in the expressions for N₁ and N₂.Wait, let's plug the expression for T from equation 1 into the equations for a₁ and a₂.From equation 1:T = F / (2 cos θ)Therefore, T sin θ = (F / (2 cos θ)) sin θ = (F/2) tan θSo, substituting back into equations 2 and 3:a₁ = [F/2 - μ₁(m₁g + (F/2) tan θ)] / m₁a₂ = [F/2 - μ₂(m₂g + (F/2) tan θ)] / m₂But we need to express tan θ in terms of something else.Alternatively, let's assume that tan θ is a variable, say k = tan θ. Then, we can write:a₁ = [F/2 - μ₁(m₁g + (F/2)k)] / m₁a₂ = [F/2 - μ₂(m₂g + (F/2)k)] / m₂And the constraint equation:a₁ + a₂ = 2a_pBut we need another equation to relate k to the accelerations. However, without geometric constraints, we might not have enough information. This suggests that the problem might require an assumption that the angle θ is small or that the vertical component is negligible, but the problem doesn't state this.Alternatively, perhaps there's a different approach. Let me think.If we consider the entire system (both masses and the pulley), the external forces are the applied force F, the frictional forces f₁ and f₂, and the weights of the masses, which are balanced by the normal forces. The net external force in the horizontal direction is F - f₁ - f₂. This net force should equal the total mass times the acceleration of the center of mass.Wait, but the problem asks for the accelerations of the masses and the center of mass. Maybe we can first find the accelerations of the masses, then compute the center of mass acceleration.Alternatively, using the system approach:The total horizontal force on the system is F - f₁ - f₂. The total mass is m₁ + m₂. However, the system includes the pulley, which is massless. So, the total mass is m₁ + m₂ = 11 kg.The acceleration of the center of mass is given by a_cm = (F - f₁ - f₂) / (m₁ + m₂). But this is only true if all parts of the system have the same acceleration, which they don't. The center of mass acceleration is the weighted average of the accelerations of the individual masses.So, a_cm = (m₁a₁ + m₂a₂) / (m₁ + m₂)But we need to find a₁ and a₂ first.Alternatively, if we can find a_p, a₁, and a₂ from the previous equations, we can compute a_cm.But this brings us back to the original problem of solving for a₁ and a₂, which requires dealing with the angle θ.Wait, perhaps there's a different way to relate the accelerations. Since the angle θ affects the normal force and thus the friction, which in turn affects the accelerations, maybe we can express everything in terms of θ and solve for it.Let’s try substituting the expressions for a₁ and a₂ into the constraint equation a₁ + a₂ = 2a_p.But what is a_p in terms of the other variables? If the pulley is part of the system, its acceleration is governed by the applied force and the tensions. However, since the pulley is massless, it's acceleration is determined by the motion of the masses.But perhaps we can express a_p in terms of a₁ and a₂ via the constraint equation: a_p = (a₁ + a₂)/2Therefore, substituting into the previous equations:a₁ = [F/2 - μ₁(m₁g + (F/2) tan θ)] / m₁a₂ = [F/2 - μ₂(m₂g + (F/2) tan θ)] / m₂Then:a₁ + a₂ = [F/2 - μ₁(m₁g + (F/2) tan θ)] / m₁ + [F/2 - μ₂(m₂g + (F/2) tan θ)] / m₂But from the pulley equation, 2T cos θ = F, and T = F/(2 cos θ)We also have from the vertical balance:For mass m₁: N₁ = m₁g + T sin θ = m₁g + (F/2) tan θSimilarly for m₂: N₂ = m₂g + (F/2) tan θBut I still don't see how to eliminate θ.Let’s plug in the given values:F = 20 N, m₁ = 1 kg, m₂ = 10 kg, μ₁ = 0.3, μ₂ = 0.1, g = 10 m/s²So:For a₁:a₁ = [10 - 0.3(1*10 + 10 tan θ)] / 1= 10 - 0.3(10 + 10 tan θ)= 10 - 3 - 3 tan θ= 7 - 3 tan θFor a₂:a₂ = [10 - 0.1(10*10 + 10 tan θ)] / 10= [10 - 0.1(100 + 10 tan θ)] / 10= [10 - 10 - tan θ] / 10= (-tan θ)/10Now, the constraint equation:a₁ + a₂ = 2a_pBut we don't have an expression for a_p yet. However, since the pulley is massless, the net force on it must be zero. The applied force F is balanced by the horizontal components of the tensions. As before, 2T cos θ = F. With T = F/(2 cos θ), which we've already used.But perhaps we need to express a_p in terms of the movement of the pulley. However, since the pulley's acceleration is related to the masses' accelerations via a_p = (a₁ + a₂)/2, we can substitute:a_p = (a₁ + a₂)/2But from the expressions above:a₁ = 7 - 3 tan θa₂ = - tan θ / 10Therefore:a_p = (7 - 3 tan θ - tan θ / 10)/2= (7 - (30 tan θ + tan θ)/10)/2= (7 - (31 tan θ)/10)/2= 7/2 - (31 tan θ)/20But how does this help us? We still have tan θ as an unknown.Wait, perhaps we can find another equation involving tan θ. Let's think about the vertical motion. The masses don't move vertically, so the normal forces are balanced.For mass m₁:N₁ = m₁g + T sin θ = 10 + T sin θBut T sin θ = (F / (2 cos θ)) sin θ = (F/2) tan θ = 10 tan θTherefore, N₁ = 10 + 10 tan θSimilarly for m₂:N₂ = 10*10 + 10 tan θ = 100 + 10 tan θBut the friction forces are:f₁ = μ₁N₁ = 0.3(10 + 10 tan θ) = 3 + 3 tan θf₂ = μ₂N₂ = 0.1(100 + 10 tan θ) = 10 + tan θFrom the horizontal equations:For m₁: T cos θ - f₁ = m₁a₁But T cos θ = F/2 = 10 NSo:10 - (3 + 3 tan θ) = 1*a₁ => a₁ = 7 - 3 tan θWhich matches our previous result.For m₂: T cos θ - f₂ = m₂a₂10 - (10 + tan θ) = 10*a₂ => a₂ = (10 - 10 - tan θ)/10 = - tan θ / 10Also matches.Now, we need another equation to solve for tan θ. But we've exhausted all the equations. The only remaining relation is the constraint a_p = (a₁ + a₂)/2, which we've already used.But this gives us an expression for a_p in terms of tan θ, but we don't have a value for a_p. However, the pulley's acceleration is determined by the applied force F and the system's dynamics. But since the pulley is massless, F is balanced by the tensions' horizontal components, which we've already accounted for.This suggests that there's an infinite number of solutions depending on θ, which can't be right. The problem must have a unique solution, so I must have missed something.Wait, perhaps the angle θ is determined by the requirement that the string remains taut and inextensible. This means that the accelerations a₁ and a₂ must be related not just by the pulley's acceleration but also by the fact that the angle θ is consistent with the geometry of the system.But without knowing the initial geometry (like the height of the pulley or the length of the string), we can't determine θ. This implies that the problem might have an assumption that the strings are horizontal, making θ = 0. If θ = 0, then tan θ = 0, and the equations simplify:For θ = 0:a₁ = 7 - 3*0 = 7 m/s²a₂ = -0/10 = 0 m/s²Then, a_p = (7 + 0)/2 = 3.5 m/s²But then, checking if this is consistent with the pulley's force balance:2T cos θ = FIf θ = 0, then cos θ = 1, so T = F/2 = 10 N. This is consistent with the earlier calculation.But for mass m₂:T cos θ - f₂ = 10*0 => 10 - f₂ = 0 => f₂ = 10 NWhich matches the calculation for θ = 0:f₂ = μ₂N₂ = 0.1*(m₂g + T sin θ) = 0.1*(100 + 0) = 10 NSo, this is consistent.However, this results in mass m₂ not accelerating, which seems counterintuitive. Why would the 10 kg mass not accelerate when a force is applied?But according to the equations, with θ = 0, the tension is 10 N, which for m₂ is balanced by friction of 10 N, resulting in zero acceleration. Meanwhile, m₁ experiences 10 N tension minus 3 N friction, resulting in 7 N net force and 7 m/s² acceleration.The constraint equation gives a_p = (7 + 0)/2 = 3.5 m/s².But the pulley is being pulled with 20 N force, and since it's massless, this requires 2T = F => T = 10 N. This is satisfied.So, despite seeming odd, this solution is consistent. The reason m₂ doesn't accelerate is that the tension equals the friction force for m₂.Therefore, the accelerations would be:a₁ = 7 m/s² to the righta₂ = 0 m/s²a_p = 3.5 m/s² to the rightThen, the acceleration of the center of mass is:a_cm = (m₁a₁ + m₂a₂)/(m₁ + m₂) = (1*7 + 10*0)/11 = 7/11 ≈ 0.636 m/s² to the right.But wait, the problem states that the pulley is weightless and the force is applied to it. However, in this solution, the pulley is accelerating at 3.5 m/s², which is different from the center of mass acceleration.But this seems to be the only consistent solution if we assume θ = 0, i.e., the strings are horizontal. However, the problem didn't specify the pulley's position, so this assumption might be valid.Alternatively, maybe the angle θ is not zero, and we need to find its value. Let's attempt to solve for θ using the equations we have.We have:a₁ = 7 - 3 tan θa₂ = - tan θ / 10From the constraint:a₁ + a₂ = 2a_pBut a_p is related to the pulley's motion. However, since the pulley is massless, its acceleration is determined by the system. Without another equation, we can’t directly relate a_p to θ. However, perhaps we can use the fact that the tension T must also satisfy the vertical force balance for the masses.Wait, the vertical forces are already balanced by the normal forces, which we considered. We don't have any vertical motion, so those equations are already satisfied.Alternatively, perhaps there's an inconsistency in assuming that the angle θ is non-zero. If θ is not zero, then there must be a vertical component to the tension, which changes the normal force and thus the friction. But without knowing θ, we can't proceed.But wait, we can write the equations in terms of tan θ and solve for it.Let’s denote k = tan θFrom the expressions for a₁ and a₂:a₁ = 7 - 3ka₂ = -k/10Then, the constraint equation:a₁ + a₂ = 2a_pBut a_p is the acceleration of the pulley, which is also connected to the applied force F. However, since the pulley is massless, F is balanced by the horizontal components of the tensions. But we already used that to get T = 10 N. So, unless there's another relation involving k, we can't solve for it.But wait, in this case, the accelerations a₁ and a₂ depend on k, which is tan θ. However, there's no other equation to determine k. This suggests that the system is underdetermined, which can't be true for a physics problem. Therefore, there must be an implicit assumption in the problem.Re-reading the problem: "Two masses, m₁=1 kg and m₂=10 kg, are placed on two horizontal shelves and are connected by a weightless and inextensible thread passing over a weightless smooth pulley. The coefficients of friction between the masses and the shelves are μ₁=0,3 and μ₂=0,1. A horizontally directed force F=20 N is applied to the center of the pulley. Determine the accelerations of the masses and the center of mass of this system. The acceleration due to gravity is g=10 m/s²."The key detail I might have missed: the shelves are horizontal, and the pulley is also on a horizontal plane. Therefore, the string must be horizontal as well. Hence, angle θ = 0. Therefore, the previous solution is correct: a₁ = 7 m/s², a₂ = 0, a_cm = 7/11 m/s².But let's confirm this. If the pulley is on the same horizontal level as the masses, the string is horizontal. When a horizontal force is applied to the pulley, it tensions the string horizontally. The masses experience horizontal tension and friction. Therefore, the angle θ is zero.Therefore, the solution with θ = 0 is valid, and the accelerations are as calculated.However, the problem seems to imply that both masses would accelerate, but in this solution, m₂ doesn't move. Let's verify the numbers.For m₂:T = 10 Nf₂ = μ₂m₂g = 0.1*10*10 = 10 NNet force on m₂: T - f₂ = 10 - 10 = 0 NTherefore, a₂ = 0.This makes sense. The friction force exactly balances the tension for m₂, so it doesn't move. Meanwhile, m₁ has a net force of 10 - 3 = 7 N, so it accelerates at 7 m/s².The pulley's acceleration is (7 + 0)/2 = 3.5 m/s², which is consistent with the applied force F = 20 N and the tension T = 10 N (2T = 20 N).Therefore, the final answers are:a₁ = 7 m/s² to the righta₂ = 0 m/s²Acceleration of the center of mass, a_cm = (1*7 + 10*0)/11 = 7/11 ≈ 0.636 m/s² to the right.But let me check if there's any other possible interpretation. Suppose the angle θ is not zero, but the problem's configuration allows solving for it.Assume that the pulley is positioned such that the strings make an angle θ with the horizontal. Then, using the equations:From before:a₁ = 7 - 3ka₂ = -k/10And a₁ + a₂ = 2a_pBut a_p is also related to the system's dynamics. However, since the pulley is massless, there's no additional equation. Hence, the system is underdetermined unless θ = 0.Therefore, the only physically consistent solution is θ = 0, leading to the accelerations found earlier.Hence, the accelerations are 7 m/s² for m₁, 0 for m₂, and the center of mass accelerates at 7/11 m/s².

✅:To solve the problem, we analyze the forces and accelerations considering the pulley's motion and the constraints of the system.Step 1: Tension and Pulley Force BalanceThe pulley is massless, so the horizontal force applied ( F = 20 , text{N} ) balances the horizontal components of the tensions:[ 2T cos theta = F ]Assuming the strings are horizontal (( theta = 0 )), we get:[ 2T = 20 implies T = 10 , text{N} ]Step 2: Frictional ForcesFor each mass, friction opposes motion:- ( f_1 = mu_1 m_1 g = 0.3 times 1 times 10 = 3 , text{N} )- ( f_2 = mu_2 m_2 g = 0.1 times 10 times 10 = 10 , text{N} )Step 3: Net Force and Acceleration for Each Mass- Mass ( m_1 ): [ T - f_1 = m_1 a_1 implies 10 - 3 = 1 times a_1 implies a_1 = 7 , text{m/s}^2 ]- Mass ( m_2 ): [ T - f_2 = m_2 a_2 implies 10 - 10 = 10 times a_2 implies a_2 = 0 , text{m/s}^2 ]Step 4: Pulley AccelerationFrom the constraint ( a_1 + a_2 = 2a_p ):[ 7 + 0 = 2a_p implies a_p = 3.5 , text{m/s}^2 ]Step 5: Center of Mass AccelerationThe center of mass acceleration ( a_{text{cm}} ) is:[ a_{text{cm}} = frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} = frac{1 times 7 + 10 times 0}{11} = frac{7}{11} approx 0.636 , text{m/s}^2 ]Final Answers:- Acceleration of ( m_1 ): ( boxed{7 , text{m/s}^2} )- Acceleration of ( m_2 ): ( boxed{0 , text{m/s}^2} )- Acceleration of the center of mass: ( boxed{dfrac{7}{11} , text{m/s}^2} )

🔑:1. Determining the Tension in the String: Since the block is massless and frictionless, the tension T in the string is the same on both sides. Given the horizontally applied force F = 20 , mathrm{N}, we know that: [ T = frac{F}{2} = frac{20}{2} = 10 , mathrm{N} ]2. Calculating the Friction Forces: The friction force F_{text{tr}} for each mass can be determined using the formula: [ F_{text{tr}} = mu N ] For the first mass (m_1): [ F_{text{tr 1}} = mu_1 N_1 = mu_1 m_1 g = 0.3 cdot 1 cdot 10 = 3 , mathrm{N} ] For the second mass (m_2): [ F_{text{tr 2}} = mu_2 N_2 = mu_2 m_2 g = 0.1 cdot 10 cdot 10 = 10 , mathrm{N} ]3. Determining Acceleration of the Masses: Since the friction force F_{text{tr 2}} = 10 , mathrm{N} for the second mass is exactly equal to the tension T = 10 , mathrm{N}, the second mass does not move: [ a_2 = 0 , mathrm{m/s^2} ] For the first mass, apply Newton's Second Law: [ T - F_{text{tr 1}} = m_1 a_1 ] [ 10 - 3 = 1 cdot a_1 ] [ a_1 = frac{7}{1} = 7 , mathrm{m/s^2} ]4. Calculating the Acceleration of the Center of Mass: The acceleration of the center of mass for the system is given by: [ a = frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} ] Substitute the values: [ a = frac{1 cdot 7 + 10 cdot 0}{1 + 10} = frac{7 + 0}{11} = frac{7}{11} approx 0.64 , mathrm{m/s^2} ] Conclusion:The calculated accelerations are:- For the first mass: ( a_1 = 7 , mathrm{m/s^2} )- For the second mass: ( a_2 = 0 , mathrm{m/s^2} )- For the center of mass: ( a = 0.64 , mathrm{m/s^2} )These results can be summarized as follows:[boxed{a_1 = 7 , mathrm{m/s^2}, quad a_2 = 0 , mathrm{m/s^2}, quad a = 0.64 , mathrm{m/s^2}}]