🧠:[<think>Okay, let's tackle this problem step by step. So we have points O, A, B, and P with given coordinates, and point X is moving along the line OP. We need to find the coordinates of X when the dot product of vectors XA and XB is minimized. Then, using that X, we have to find the cosine of the angle AXB. Starting with part (1). First, let me visualize the setup. O is the origin, right? So OA is the vector from the origin to point A, which is (1,7). Similarly, OB is (5,1), so point B is at (5,1). Point P is at (2,1), so OP is the line from the origin to P. X is a point moving along OP. That means the coordinates of X can be parameterized as some scalar multiple of OP. Since OP is from (0,0) to (2,1), any point X on OP can be written as t*(2,1), where t is a real number. So OX would be (2t, t) for some t.So, we can represent X as (2t, t). Then, vectors XA and XB can be found by subtracting the coordinates of X from A and B, respectively. Let's compute those.Vector XA is OA - OX, which is (1,7) - (2t, t) = (1 - 2t, 7 - t).Similarly, vector XB is OB - OX, which is (5,1) - (2t, t) = (5 - 2t, 1 - t).Now, the dot product of XA and XB is [(1 - 2t)(5 - 2t) + (7 - t)(1 - t)]. Let's compute this expression.First, expand the terms:(1 - 2t)(5 - 2t) = 1*5 + 1*(-2t) -2t*5 -2t*(-2t) = 5 - 2t -10t + 4t² = 5 -12t +4t²(7 - t)(1 - t) = 7*1 +7*(-t) -t*1 -t*(-t) = 7 -7t -t +t² = 7 -8t +t²Adding these together: (5 -12t +4t²) + (7 -8t +t²) = 12 -20t +5t²So the dot product XA · XB is 5t² -20t +12. Now, this is a quadratic function in terms of t. Since the coefficient of t² is positive (5), the parabola opens upwards, so its minimum occurs at the vertex.The vertex of a quadratic at² + bt + c occurs at t = -b/(2a). Here, a =5, b= -20, so t = -(-20)/(2*5) = 20/10 = 2. So t=2.Wait, but hold on. If t=2, then the coordinates of X would be (4,2). But point P is at (2,1). Wait, that would mean X is beyond point P on the line OP. But OP is from the origin to (2,1). So if t is a scalar multiple, when t=1, we reach P, and for t>1, we go beyond P. But the problem says "point X is a moving point on the line OP". Hmm, the line OP – does that refer to the line segment OP or the entire line through OP?Wait, the problem says "the line OP". In geometry, a line is usually infinite in both directions, unless specified as a segment. However, in some contexts, especially in vectors, "line OP" might refer to the line segment from O to P. Hmm. Wait, the problem says "point X is a moving point on the line OP". So if it's the entire line, then t can be any real number, so X can be anywhere on the line OP extended infinitely in both directions. But if it's the line segment OP, then t is between 0 and 1. So this is an important point. Let me check.But in the original problem statement, it's written as "the line OP". In standard terminology, "line OP" would be the infinite line passing through points O and P, whereas "segment OP" would be the finite part between O and P. So here, since it says "line OP", X can be anywhere on that infinite line. So t can be any real number. Therefore, t=2 would be allowed, even though it's beyond point P. Therefore, the minimum occurs at t=2, so X is at (4,2). But let me confirm.Wait, but let's check if this makes sense. Let me compute the dot product at t=2: XA · XB would be 5*(2)^2 -20*2 +12 = 20 -40 +12 = -8. Is that the minimum? Let me check another value. Let's say t=1 (point P). Then the dot product is 5*1 -20*1 +12 = 5 -20 +12 = -3. Which is larger than -8. So t=2 gives a smaller value. If we take t=3, it would be 5*9 -20*3 +12 = 45 -60 +12 = -3 again. Wait, that's strange. So at t=2, the dot product is -8, and as t increases beyond 2, it starts increasing again? Wait, but according to the quadratic, which is 5t² -20t +12, the minimum is indeed at t=2, and as t approaches ±infinity, the quadratic tends to positive infinity, so it's a parabola opening upwards. Therefore, the minimum is at t=2, which is on the line OP extended beyond P. So even though X is beyond P, it's still on the line OP, so that's acceptable. Therefore, the answer for part (1) is (4,2).But wait, let me verify this again. Let's compute XA and XB at t=2.At t=2, X is (4,2). Then, vector XA is OA - OX = (1,7)-(4,2) = (-3,5). Vector XB is OB - OX = (5,1)-(4,2) = (1,-1). Then their dot product is (-3)(1) + (5)(-1) = -3 -5 = -8, which matches the earlier calculation. So that's correct.But just to be thorough, let's check another point. Suppose t=0 (point O). Then X is (0,0). Then XA is (1,7), XB is (5,1). The dot product is 1*5 +7*1=5+7=12, which is higher. So indeed, the minimum is at t=2.Therefore, the coordinates of OX when the dot product is minimized is (4,2). So that's part (1) done.Now moving to part (2): When X is at (4,2), we need to find cos(angle AXB). So angle AXB is the angle at point X between points A and B. To find the cosine of this angle, we can use the dot product formula. Remember that cosθ = (XA · XB)/( |XA| |XB| ). We already have XA · XB = -8 from part (1). Now we need |XA| and |XB|.First, compute |XA|. XA is (-3,5). The magnitude is sqrt((-3)^2 +5^2)=sqrt(9+25)=sqrt(34).Next, compute |XB|. XB is (1,-1). The magnitude is sqrt(1^2 + (-1)^2)=sqrt(1+1)=sqrt(2).Therefore, cosθ = (-8)/(sqrt(34)*sqrt(2)) = -8/(sqrt(68)) = -8/(2*sqrt(17)) = -4/sqrt(17). Rationalizing the denominator, that would be -4*sqrt(17)/17. However, since cosine can be negative, but the question just asks for the value, so we can leave it as -4/sqrt(17) or rationalized.But let's verify the calculation once more.XA vector: from X(4,2) to A(1,7): (1-4,7-2)=(-3,5). Correct.XB vector: from X(4,2) to B(5,1): (5-4,1-2)=(1,-1). Correct.Dot product: (-3)(1) + (5)(-1) = -3 -5 = -8. Correct.|XA|: sqrt(9 +25)=sqrt(34). Correct.|XB|: sqrt(1 +1)=sqrt(2). Correct.Therefore, cosθ = -8 / (sqrt(34)*sqrt(2)) = -8 / sqrt(68) = -8 / (2*sqrt(17)) = -4 / sqrt(17) = -4√17 /17. So the value is -4√17 /17.But let me check if the angle is acute or obtuse. Since the cosine is negative, the angle is obtuse (greater than 90 degrees). That makes sense because the dot product is negative, which implies the angle between XA and XB is obtuse.So, the answer for part (2) is -4√17⁄17.Therefore, summarizing:1) The coordinates of OX when the dot product is minimized is (4,2).2) The cosine of angle AXB is -4√17⁄17.I think that's it. Let me just check if all steps make sense. For part (1), parameterizing X as (2t, t) along OP, then computing vectors XA and XB, their dot product, simplifying to a quadratic in t, finding the vertex, checking the coordinates, and verifying the dot product. Then for part (2), using the dot product formula to find the cosine. All steps seem correct. So I think this is the solution.Final Answer(1) boxed{(4,2)}(2) boxed{-dfrac{4sqrt{17}}{17}}</think>Given vectors (overrightarrow{OA} = (1, 7)), (overrightarrow{OB} = (5, 1)), (overrightarrow{OP} = (2, 1)), and point (X) moving on the line (OP):1. Coordinates of (overrightarrow{OX}) when (overrightarrow{XA} cdot overrightarrow{XB}) is minimized: - Parameterize point (X) on line (OP) as (X = (2t, t)). - Vectors (overrightarrow{XA}) and (overrightarrow{XB}) are calculated as: [ overrightarrow{XA} = (1 - 2t, 7 - t) ] [ overrightarrow{XB} = (5 - 2t, 1 - t) ] - The dot product (overrightarrow{XA} cdot overrightarrow{XB}) is: [ (1 - 2t)(5 - 2t) + (7 - t)(1 - t) = 5t^2 - 20t + 12 ] - This quadratic function in (t) has its minimum at (t = 2). Thus, the coordinates of (overrightarrow{OX}) are ((4, 2)).2. Value of (cos angle AXB) when (X) is at ((4, 2)): - Vectors (overrightarrow{XA}) and (overrightarrow{XB}) at (X = (4, 2)) are: [ overrightarrow{XA} = (-3, 5) ] [ overrightarrow{XB} = (1, -1) ] - Magnitudes of the vectors: [ |overrightarrow{XA}| = sqrt{(-3)^2 + 5^2} = sqrt{34} ] [ |overrightarrow{XB}| = sqrt{1^2 + (-1)^2} = sqrt{2} ] - Dot product (overrightarrow{XA} cdot overrightarrow{XB} = -8). - Using the dot product formula: [ cos angle AXB = frac{overrightarrow{XA} cdot overrightarrow{XB}}{|overrightarrow{XA}| |overrightarrow{XB}|} = frac{-8}{sqrt{34} cdot sqrt{2}} = frac{-8}{2sqrt{17}} = frac{-4sqrt{17}}{17} ] Final Answers(1) (boxed{(4,2)})(2) (boxed{-dfrac{4sqrt{17}}{17}}),<think>Okay, so I have this problem here about vectors in the plane. Let me try to parse it step by step. There are four points mentioned: O, A, B, and P. Their coordinates are given as vectors from O. Specifically, OA is (1,7), OB is (5,1), OP is (2,1), and point X is moving along the line OP. The first part asks me to find the coordinates of vector OX when the dot product of vectors XA and XB reaches its minimum value. Then, part two wants me to find the cosine of the angle AXB when X is at that position.Hmm. Let's start with part (1). Since X is moving along the line OP, I can parametrize its position. The line OP goes from the origin O to point P(2,1). So any point X on OP can be written as a scalar multiple of the vector OP. Let me denote the parameter as t, so OX = t*(2,1) = (2t, t), where t is a real number. So the coordinates of X are (2t, t).Now, I need to find the value of t that minimizes the dot product XA ⋅ XB. First, let's find the vectors XA and XB. Vector XA is OA - OX, right? Because XA goes from X to A, so it's A - X. Similarly, XB is B - X. Let me compute these.OA is (1,7), so XA = OA - OX = (1 - 2t, 7 - t).Similarly, OB is (5,1), so XB = OB - OX = (5 - 2t, 1 - t).Now, the dot product XA ⋅ XB is the product of their x-components plus the product of their y-components. Let me compute that:XA ⋅ XB = (1 - 2t)(5 - 2t) + (7 - t)(1 - t).Let me expand each term:First term: (1)(5) + (1)(-2t) + (-2t)(5) + (-2t)(-2t) = 5 - 2t - 10t + 4t² = 5 - 12t + 4t².Second term: (7)(1) + (7)(-t) + (-t)(1) + (-t)(-t) = 7 - 7t - t + t² = 7 - 8t + t².Adding both terms together: (5 - 12t + 4t²) + (7 - 8t + t²) = 12 - 20t + 5t².So the dot product simplifies to 5t² - 20t + 12. Wait, that's a quadratic function in terms of t. To find its minimum, since the coefficient of t² is positive (5), the parabola opens upwards, so the minimum occurs at the vertex. The vertex of a quadratic at² + bt + c is at t = -b/(2a). Here, a = 5, b = -20. So t = -(-20)/(2*5) = 20/10 = 2. Therefore, the minimum occurs at t = 2. So the coordinates of OX are (2*2, 1*2) = (4,2). But wait, hold on. Let me check my calculations again because sometimes when expanding terms, errors can happen. Let me verify the expansion of XA ⋅ XB.First, (1 - 2t)(5 - 2t):Multiply 1*5 = 5.1*(-2t) = -2t.(-2t)*5 = -10t.(-2t)*(-2t) = 4t².So adding these: 5 - 2t -10t +4t² = 5 -12t +4t². Correct.Then, (7 - t)(1 - t):7*1 =7.7*(-t) = -7t.(-t)*1 = -t.(-t)*(-t) = t².Adding these: 7 -7t -t + t² =7 -8t +t². Correct.So total dot product is 5 -12t +4t² +7 -8t +t² = (5+7) + (-12t -8t) + (4t² + t²) = 12 -20t +5t². Correct.So quadratic is 5t² -20t +12. Then the vertex is at t = 20/(2*5) = 2. So t=2. So X is at (4,2). Wait, but OP is the line from O to P(2,1). If t=2, then OX = (4,2). But OP is the line from (0,0) to (2,1), so parametrizing it as (2t, t), when t=1, we reach P(2,1). If t>1, then X is beyond point P on the extended line OP. But the problem says "point X is a moving point on the line OP". Depending on interpretation, "on the line OP" could mean just the segment OP, but in mathematics, a line is usually infinitely extended unless specified as a segment. So since it's just the line, t can be any real number. Therefore, even though t=2 would be beyond point P, it's still on the line OP. Therefore, the answer is (4,2).But let me check again if there's an error in the parametrization. The parametrization OX = t*(2,1) where t ∈ ℝ. So (2t, t). When t=0, it's O; when t=1, it's P. For t>1, it's beyond P, and t<0, it's behind O. So yes, that's correct.So, conclusion: the coordinates of OX when the dot product is minimized are (4,2).But wait, maybe I should verify by computing the derivative? Since the dot product is a quadratic function, the vertex is indeed the minimum, but just to be thorough, let me compute the derivative with respect to t.Let f(t) = 5t² -20t +12.df/dt = 10t -20.Setting derivative to zero: 10t -20 =0 → t=2. Yep, same result. So t=2, so X=(4,2). So part (1) answer is (4,2).Now part (2): When X is at (4,2), find the value of cos∠AXB.So angle AXB is the angle at point X between points A, X, B. So vectors XA and XB emanate from X, so the angle between XA and XB is ∠AXB.To find the cosine of that angle, we can use the dot product formula:cosθ = (XA ⋅ XB) / (|XA| |XB|).We already computed XA ⋅ XB when X is at (4,2). Wait, but at t=2, which gives X=(4,2), the dot product is f(2) =5*(4) -20*(2) +12=20 -40 +12= -8.Wait, let's compute f(2): 5*(2)^2 -20*(2) +12 =5*4 -40 +12=20 -40 +12= -8. Correct.So XA ⋅ XB = -8.Now we need |XA| and |XB|.First, compute XA and XB when X=(4,2). XA is OA - OX = (1,7) - (4,2) = (-3,5). XB is OB - OX = (5,1) - (4,2) = (1,-1).So vector XA is (-3,5), vector XB is (1,-1).Compute |XA|: sqrt((-3)^2 +5^2) = sqrt(9 +25)=sqrt(34).Compute |XB|: sqrt(1^2 + (-1)^2)=sqrt(1 +1)=sqrt(2).Therefore, cosθ = (-8)/(sqrt(34)*sqrt(2)) = -8/sqrt(68) = -8/(2*sqrt(17)) = -4/sqrt(17).Rationalizing the denominator: -4*sqrt(17)/17.So cos∠AXB is -4√17 /17.But let me confirm the vectors again. X is at (4,2). So coordinates of X are (4,2). Then vector XA is A - X: (1 -4,7 -2)=(-3,5). Correct. Vector XB is B - X: (5 -4,1 -2)=(1,-1). Correct.Dot product: (-3)(1) + (5)(-1) = -3 -5 = -8. Correct.Magnitudes:|XA| = sqrt(9 +25)=sqrt(34). |XB| = sqrt(1 +1)=sqrt(2). So product is sqrt(34*2)=sqrt(68)=2*sqrt(17). So -8/(2*sqrt(17))= -4/sqrt(17)= -4√17/17. Correct.So the value of cos∠AXB is -4√17/17.But let me check if the angle is acute or obtuse. Since the cosine is negative, the angle is obtuse (between 90° and 180°). Makes sense since the dot product is negative.Therefore, part (2) answer is -4√17⁄17.Wait, but the problem says "find the value of cos∠AXB", so just present it as a numerical value. So that's -4√17/17.So summarizing:1) The coordinates of OX are (4,2).2) cos∠AXB is -4√17⁄17.But let me check once again for part (1). When t=2, X=(4,2). But OP is the line from O to P(2,1). So when t=2, it's twice the vector OP. So it's on the line OP extended beyond P. So if the problem allows X to be anywhere on the line OP (not just the segment OP), then that's correct. Since it says "moving point on the line OP", then yes, t can be any real number, so (4,2) is valid.Alternatively, if there was a restriction to the segment OP, the minimum might be at an endpoint. But in this case, the quadratic achieves its minimum at t=2, which is outside the segment OP (since OP is t from 0 to1). So if we had been restricted to the segment, the minimum would be at t=1. Let me check the value at t=1.At t=1, X=(2,1). Then XA=(1-2,7-1)=(-1,6), XB=(5-2,1-1)=(3,0). The dot product is (-1)(3) +6*0= -3. But at t=2, the dot product is -8, which is smaller. So since X is allowed on the entire line, the minimum is at t=2. So yes, answer (4,2) is correct.Another way to approach part (1) is using calculus or projections. Let me try another method for verification.Let’s consider the line OP parametrized as X(t) = (2t, t). Then, express vectors XA and XB in terms of t, compute their dot product as a function of t, find its minimum.But that's essentially what I did before. Alternatively, since XA ⋅ XB is a quadratic function, which is convex, so the minimum is at the vertex. So same result.Alternatively, perhaps using vector calculus. Let me think.Let’s denote vector OX = t*(2,1). Then, vectors XA = OA - OX = (1,7) - (2t, t) = (1 -2t,7 -t).Similarly, XB = OB - OX = (5,1) - (2t, t) = (5 -2t,1 -t).The dot product is (1 -2t)(5 -2t) + (7 -t)(1 -t). Which we expanded to 5t² -20t +12.So the same result. So calculus and algebraic methods both lead to t=2.Alternatively, another approach: to minimize XA ⋅ XB, which is equal to (A - X) ⋅ (B - X).Expanding this, we get A ⋅ B - A ⋅ X - B ⋅ X + X ⋅ X.Since A ⋅ B is a constant, the expression to minimize is (-A -B) ⋅ X + |X|².But since X is on line OP, we can write X = tP, where P=(2,1). So substitute X = tP, then the expression becomes (-A -B) ⋅ (tP) + |tP|².Which is -t(A + B) ⋅ P + t²|P|².Compute coefficients:First, A + B = (1+5,7+1)=(6,8). Then, (A + B) ⋅ P = (6,8) ⋅ (2,1) =12 +8=20.|P|² = (2)^2 +1^2=4 +1=5.So the expression becomes -20t +5t². To minimize this quadratic in t: 5t² -20t. Wait, but in previous calculation, the constant term was 12, but here, it seems different. Wait, perhaps because here, we separated A ⋅ B as a constant. Let me check.Original expression: (A - X) ⋅ (B - X) = A ⋅ B - A ⋅ X - B ⋅ X + X ⋅ X.So if we write this as A ⋅ B - (A + B) ⋅ X + |X|².So when we set X = tP, then:Expression becomes A ⋅ B - (A + B) ⋅ (tP) + |tP|².So A ⋅ B is a constant. The rest is -t(A + B) ⋅ P + t²|P|².Thus, the quadratic in t is |P|² t² - (A + B) ⋅ P t + (A ⋅ B). Wait, but in the previous calculation, we had 5t² -20t +12. Let me check:Compute A ⋅ B: (1,7) ⋅ (5,1)=1*5 +7*1=5 +7=12.So indeed, the expression is 5t² -20t +12. Therefore, this alternative approach gives the same quadratic. Therefore, the minimum is at t=2, same as before. So the result is consistent.Therefore, part (1) is definitely (4,2). Part (2) is -4√17⁄17.So the answers are:(1) boxed{(4,2)}(2) boxed{-dfrac{4sqrt{17}}{17}}Final Answer(1) boxed{(4, 2)}(2) boxed{-dfrac{4sqrt{17}}{17}}</think>Given vectors (overrightarrow{OA} = (1, 7)), (overrightarrow{OB} = (5, 1)), (overrightarrow{OP} = (2, 1)), and point (X) moving on the line (OP):1. To find the coordinates of (overrightarrow{OX}) when (overrightarrow{XA} cdot overrightarrow{XB}) reaches its minimum value: - Parametrize point (X) on line (OP) as (overrightarrow{OX} = t(2, 1) = (2t, t)). - Calculate vectors (overrightarrow{XA}) and (overrightarrow{XB}): [ overrightarrow{XA} = (1 - 2t, 7 - t) ] [ overrightarrow{XB} = (5 - 2t, 1 - t) ] - Compute the dot product (overrightarrow{XA} cdot overrightarrow{XB}): [ (1 - 2t)(5 - 2t) + (7 - t)(1 - t) = 5t^2 - 20t + 12 ] - The quadratic function (5t^2 - 20t + 12) reaches its minimum at (t = 2) (vertex of the parabola). - Therefore, (overrightarrow{OX} = (2 cdot 2, 2) = (4, 2)).2. To find (cos angle AXB) when (X) is at ((4, 2)): - Calculate vectors (overrightarrow{XA}) and (overrightarrow{XB}) at (X = (4, 2)): [ overrightarrow{XA} = (1 - 4, 7 - 2) = (-3, 5) ] [ overrightarrow{XB} = (5 - 4, 1 - 2) = (1, -1) ] - Compute the dot product (overrightarrow{XA} cdot overrightarrow{XB} = -3 cdot 1 + 5 cdot (-1) = -8). - Compute magnitudes (|overrightarrow{XA}|) and (|overrightarrow{XB}|): [ |overrightarrow{XA}| = sqrt{(-3)^2 + 5^2} = sqrt{34} ] [ |overrightarrow{XB}| = sqrt{1^2 + (-1)^2} = sqrt{2} ] - Calculate (cos angle AXB): [ cos angle AXB = frac{-8}{sqrt{34} cdot sqrt{2}} = frac{-8}{2sqrt{17}} = frac{-4sqrt{17}}{17} ]Final answers:(1) (boxed{(4, 2)})(2) (boxed{-dfrac{4sqrt{17}}{17}})]

🔑:(1) Let overrightarrow{OX}=(x,y).Since point X is on the line OP, vector overrightarrow{OX} is collinear with overrightarrow{OP}.Given overrightarrow{OP}=(2,1), we have x-2y=0, i.e., x=2y.Thus, overrightarrow{OX}=(2y,y). Also, overrightarrow{XA}=overrightarrow{OA}-overrightarrow{OX}, and overrightarrow{OA}=(1,7),so overrightarrow{XA}=(1-2y,7-y).Similarly, overrightarrow{XB}=overrightarrow{OB}-overrightarrow{OX}=(5-2y,1-y).Hence, overrightarrow{XA} cdot overrightarrow{XB}=(1-2y)(5-2y)+(7-y)(1-y)=5y^2-20y+12=5(y-2)^2-8.Therefore, when y=2, overrightarrow{XA} cdot overrightarrow{XB} attains its minimum value (-8), and at this time, overrightarrow{OX}=(4,2).(2) When overrightarrow{OX}=(4,2), i.e., y=2, we have overrightarrow{XA}=(-3,5) and overrightarrow{XB}=(1,-1).Thus, |overrightarrow{XA}|=sqrt{34} and |overrightarrow{XB}|=sqrt{2}.Hence, cos angle AXB = frac{overrightarrow{XA} cdot overrightarrow{XB}}{|overrightarrow{XA}| |overrightarrow{XB}|} = -frac{4sqrt{17}}{17}.The final answers are:(1) boxed{overrightarrow{OX}=(4,2)}(2) boxed{cos angle AXB = -frac{4sqrt{17}}{17}}

🧠:[<think>Okay, so I need to figure out which of the three given functions are "limited growth functions" as defined. The definition says that a function f(x) is a limited growth function if there exist positive constants a and b such that for all real numbers x, the inequality f(x + a) ≤ f(x) + b holds. The options are ① x² + x + 1, ② sqrt(|x|), and ③ sin(x²). The answer choices are A: all three, B: ② and ③, C: ① and ③, D: only ③. I need to check each function one by one.Starting with function ①: f(x) = x² + x + 1. Let's see if we can find positive constants a and b such that f(x + a) ≤ f(x) + b for all x. Let's compute f(x + a):f(x + a) = (x + a)² + (x + a) + 1 = x² + 2a x + a² + x + a + 1 = x² + (2a + 1)x + (a² + a + 1).Now, compare this to f(x) + b, which is x² + x + 1 + b. So, subtracting f(x) from both sides of the inequality f(x + a) ≤ f(x) + b gives:[x² + (2a + 1)x + (a² + a + 1)] - [x² + x + 1] ≤ bSimplify the left-hand side:(2a + 1)x - x + (a² + a + 1 - 1) = 2a x + a² + a.So the inequality becomes 2a x + a² + a ≤ b for all x ∈ ℝ.Wait, hold on. If this inequality has to hold for all real numbers x, then the left-hand side is linear in x, specifically 2a x plus some constants. However, unless the coefficient of x is zero, this linear term will dominate as x becomes large in absolute value. Therefore, for the inequality 2a x + a² + a ≤ b to hold for all x, the coefficient of x must be zero. So 2a must equal zero. But a is a positive constant, so 2a = 0 implies a = 0, which contradicts the requirement that a is positive. Therefore, there's no positive a for which the coefficient of x disappears, which means that the left-hand side will grow without bound as x increases or decreases, depending on the sign. Therefore, for function ①, it's impossible to find such a and b, so ① is not a limited growth function. Wait, but hold on, maybe I made a mistake here. Let me double-check.Wait, when I subtract f(x), I should have:f(x + a) - f(x) = [ (x + a)^2 + (x + a) + 1 ] - [ x^2 + x + 1 ]= (x² + 2a x + a² + x + a + 1) - x² - x - 1= 2a x + a² + a.So, yes, that's correct. Therefore, f(x + a) - f(x) = 2a x + a² + a. To have this ≤ b for all x, we need 2a x + (a² + a) ≤ b for all x. However, since a is positive, 2a is positive, so as x → ∞, 2a x will go to infinity, which would exceed any finite b. Similarly, as x → -∞, 2a x will go to negative infinity, but the inequality is that this entire expression is ≤ b. Wait, but if x is negative, then 2a x is negative, so 2a x + a² + a could be less than b if a² + a ≤ b. But the problem is when x is positive. For x positive, 2a x increases without bound, so the left-hand side can be made larger than any b by choosing a large enough x. Therefore, there's no way to choose a and b such that this holds for all x. Therefore, function ① is not a limited growth function.Wait, but maybe there's a different approach? Maybe using some inequality? For example, can we bound f(x + a) - f(x) somehow? Let me see.Alternatively, perhaps choosing specific a and b. Let's suppose that a is 1. Then f(x + 1) - f(x) would be 2x + 1 + 1 + 1 = 2x + 3. Wait, no, earlier computation shows that when a = 1, f(x + 1) - f(x) = 2*1*x + 1² + 1 = 2x + 2. Wait, yes, if a =1, then 2a = 2, and a² + a = 1 +1 = 2. So f(x +1) - f(x) = 2x + 2. So if we set this ≤ b, then for all x, 2x + 2 ≤ b. But as x increases, this becomes impossible. So even for a =1, there is no such b. Similarly, for any a >0, we have f(x + a) - f(x) = 2a x + a² + a. The linear term in x is unavoidable here. Therefore, ① is not a limited growth function. So ① is out. So answer choices A and C can be eliminated. So possible answers are B or D.Now moving to function ②: f(x) = sqrt(|x|). Let's check if there exist positive constants a and b such that sqrt(|x + a|) ≤ sqrt(|x|) + b for all x.First, note that |x + a| can be written as |x| + a if x ≥ 0, but if x is negative, say x = -c where c >0, then |x + a| = | -c + a |. Hmm. So the behavior depends on the value of x.But perhaps we can use the inequality sqrt(|x + a|) ≤ sqrt(|x|) + b. Let me try to manipulate this.Square both sides (since both sides are non-negative):|x + a| ≤ (sqrt(|x|) + b)^2 = |x| + 2b sqrt(|x|) + b².So we need |x + a| ≤ |x| + 2b sqrt(|x|) + b² for all x ∈ ℝ.Let's analyze different cases based on the value of x.Case 1: x ≥ 0. Then |x + a| = x + a, since x + a ≥ 0. The right-hand side is x + 2b sqrt(x) + b². So the inequality becomes:x + a ≤ x + 2b sqrt(x) + b² => a ≤ 2b sqrt(x) + b².But this must hold for all x ≥0. However, as x approaches 0, sqrt(x) approaches 0, so the right-hand side approaches b². Therefore, we need a ≤ b². On the other hand, when x is large, the term 2b sqrt(x) dominates. So as x increases, the right-hand side grows without bound, so the inequality a ≤ 2b sqrt(x) + b² will hold for large x as long as a is a constant. Wait, but for x approaching 0, we need a ≤ b². So if we set a ≤ b², then for x approaching 0, the inequality holds. For x large, since the right-hand side is 2b sqrt(x) + b², which is greater than a because it's growing. But does this hold for all x ≥0?Wait, the problem is when x is between 0 and some value where 2b sqrt(x) + b² is minimized. But since sqrt(x) is increasing, the minimum of 2b sqrt(x) + b² occurs at x=0, which is b². Therefore, as long as a ≤ b², then for x ≥0, the inequality a ≤ 2b sqrt(x) + b² holds.Case 2: x < 0. Let x = -y where y >0. Then |x + a| = |-y + a| = |a - y|. The right-hand side is sqrt(|x|) + b = sqrt(y) + b, then squared is y + 2b sqrt(y) + b². So the inequality becomes |a - y| ≤ y + 2b sqrt(y) + b².Let's split into two subcases:Subcase 2a: y ≤ a. Then |a - y| = a - y. So the inequality is a - y ≤ y + 2b sqrt(y) + b² => a ≤ 2y + 2b sqrt(y) + b².Since y >0 and y ≤ a, the right-hand side is 2y + 2b sqrt(y) + b². Let's find the minimum of this expression for y ∈ (0, a]. Let me denote z = sqrt(y), so z ∈ (0, sqrt(a)]. Then the expression becomes 2z² + 2b z + b². To find its minimum, take derivative with respect to z: 4z + 2b. Setting derivative to zero: 4z + 2b = 0 => z = -b/2. But z >0, so minimum occurs at z=0 (but z approaches 0). So the expression approaches 0 + 0 + b² = b² as y approaches 0. Therefore, the minimal value of the right-hand side is b² (approached as y approaches 0), so we need a ≤ b². Since we already have a ≤ b² from case 1, this is the same condition.Subcase 2b: y > a. Then |a - y| = y - a. So the inequality is y - a ≤ y + 2b sqrt(y) + b² => -a ≤ 2b sqrt(y) + b². Since the right-hand side is always positive (b and y are positive), this inequality is automatically true. Therefore, in this subcase, the inequality holds for all y > a.Therefore, combining all cases, as long as a ≤ b², the inequality holds for all x ∈ ℝ. But wait, in the problem statement, we need to find positive constants a and b such that the inequality holds. So if we set, for example, b = sqrt(a), then a ≤ (sqrt(a))² = a. So equality holds. Therefore, if we pick a = b², then the condition is satisfied. Let me check with specific numbers.Suppose we choose a = 1, then we need b² ≥1, so b ≥1. Then for all x, sqrt(|x +1|) ≤ sqrt(|x|) + b. Let's take x=0: sqrt(1) ≤ sqrt(0) + b => 1 ≤ 0 + b => b ≥1. Which holds if b=1. Then check x approaching infinity. For x large, sqrt(x +1) ≈ sqrt(x) + 1/(2 sqrt(x)) (using the expansion sqrt(x +1) ≈ sqrt(x) + 1/(2 sqrt(x)) - ...). So sqrt(x +1) - sqrt(x) ≈ 1/(2 sqrt(x)) →0 as x→infty. So the difference between sqrt(x +1) and sqrt(x) becomes negligible, so sqrt(x +1) ≤ sqrt(x) +1 would hold for x large. Similarly, for x negative, if x = -2 (y =2), then |x +1|=1, so sqrt(1)=1, sqrt(|x|)=sqrt(2), so 1 ≤ sqrt(2) +1 ≈1.414 +1=2.414, which holds. For x = -1 + ε, approaching -1 from the right, x = -1 + ε, then |x +1|=ε, so sqrt(ε) ≤ sqrt(|x|) +1. sqrt(|x|)=sqrt(|-1 + ε|)=sqrt(1 - ε)≈1 - ε/2. So sqrt(ε) ≤ 1 - ε/2 +1=2 - ε/2. Which is true for small ε>0. Therefore, it seems that with a=1 and b=1, the inequality holds for all x. Wait, but earlier we had that in order to have a ≤ b², if a=1, then b ≥1, so with b=1, it's okay. Let me check x=0: sqrt(1) =1 ≤ sqrt(0) +1=0 +1=1, equality holds. For x approaching negative infinity, say x = -N where N is large. Then |x +a| = |-N +1|=N -1, sqrt(N -1). The right-hand side is sqrt(N) +1. So sqrt(N -1) ≤ sqrt(N) +1. Let's square both sides: N -1 ≤ N + 2 sqrt(N) +1 => -1 ≤ 2 sqrt(N) +1 => which is always true. So yes, for x negative and large in magnitude, the inequality holds. Therefore, choosing a=1 and b=1 works. Therefore, function ② is a limited growth function.Wait, but the problem states "there exist positive constants a and b", so as long as we can find some a and b. So yes, ② is a limited growth function.Now moving to function ③: f(x) = sin(x²). We need to check if there exist positive constants a and b such that sin((x + a)^2) ≤ sin(x²) + b for all x ∈ ℝ.First, note that the sine function is bounded between -1 and 1, so sin(x²) ∈ [-1,1], so sin(x²) + b ∈ [-1 + b, 1 + b]. Similarly, sin((x + a)^2) ∈ [-1,1]. So the maximum possible value of sin((x + a)^2) is 1, and the minimum of sin(x²) + b is -1 + b. So for the inequality sin((x + a)^2) ≤ sin(x²) + b to hold for all x, we need that even in the worst case, 1 ≤ sin(x²) + b. Since sin(x²) can be as low as -1, so 1 ≤ -1 + b => b ≥ 2. Similarly, the lower bound of sin((x + a)^2) is -1, so we need -1 ≤ sin(x²) + b. Since sin(x²) can be as high as 1, this becomes -1 ≤ 1 + b, which is always true if b is positive. So the main constraint here is b ≥2. However, even if we set b=2, does the inequality hold?But wait, this is only considering the bounds. However, the problem is that when sin(x²) is at its minimum (-1), we have sin((x + a)^2) ≤ -1 + b. Since sin((x + a)^2) ≥ -1, so if b ≥2, then -1 ≤ -1 + b, which is true. Similarly, when sin(x²) is at its maximum (1), sin((x + a)^2) ≤1 + b. Since sin((x + a)^2) ≤1, this is also true. However, these are just bounds. But the actual inequality may or may not hold for specific x.But perhaps there's a more clever way. Let's think about the difference sin((x + a)^2) - sin(x²). We can use the trigonometric identity:sin(A) - sin(B) = 2 cos((A + B)/2) sin((A - B)/2).So here, A = (x + a)^2, B = x². Then:sin((x + a)^2) - sin(x²) = 2 cos( [(x + a)^2 + x²]/2 ) sin( [(x + a)^2 - x²]/2 )Simplify:[(x + a)^2 + x²]/2 = (x² + 2a x + a² + x²)/2 = (2x² + 2a x + a²)/2 = x² + a x + a²/2[(x + a)^2 - x²]/2 = (2a x + a²)/2 = a x + a²/2Therefore, sin((x + a)^2) - sin(x²) = 2 cos(x² + a x + a²/2) sin(a x + a²/2)Therefore, the difference is 2 cos(...) sin(...). The maximum possible value of this expression is 2 *1 *1=2, and the minimum is -2. So sin((x + a)^2) - sin(x²) ∈ [-2, 2]. Therefore, sin((x + a)^2) ≤ sin(x²) + 2. Therefore, if we choose b=2, then for all x, sin((x + a)^2) ≤ sin(x²) + 2. But the question is whether there exists a positive constant a (so a fixed a) such that for all x, sin((x + a)^2) ≤ sin(x²) + b, where b is a constant. Wait, but according to the difference, the maximum possible difference is 2, regardless of a. Therefore, if we set b=2, then sin((x + a)^2) ≤ sin(x²) + 2 holds for any a. Wait, is that true?Wait, the difference sin((x + a)^2) - sin(x²) can be as large as 2, but if a is arbitrary, but we need to fix a and b such that for this a, the difference is always ≤b. However, the maximum possible difference is 2, which occurs when cos(...) =1 and sin(...)=1. But does this maximum actually occur for some x, given a fixed a?Suppose a is fixed. Let's see if there exists x such that sin((x + a)^2) - sin(x²)=2. For that to happen, we need both cos(x² + a x + a²/2)=1 and sin(a x + a²/2)=1.So sin(a x + a²/2)=1 implies that a x + a²/2 = π/2 + 2π k for some integer k. Then x = (π/2 + 2π k - a²/2)/a.At such x, we need cos(x² + a x + a²/2)=1. Let's substitute x from above:x = (π/2 + 2π k - a²/2)/aSo x² + a x + a²/2:First, compute x + a = (π/2 + 2π k - a²/2)/a + a = (π/2 + 2π k - a²/2 + a²)/a = (π/2 + 2π k + a²/2)/aTherefore, (x + a)^2 = [ (π/2 + 2π k + a²/2)/a ]²But x² + a x + a²/2:Compute x²:x² = [ (π/2 + 2π k - a²/2)/a ]²a x = a * [ (π/2 + 2π k - a²/2)/a ] = π/2 + 2π k - a²/2So x² + a x + a²/2 = x² + (π/2 + 2π k - a²/2) + a²/2 = x² + π/2 + 2π k.But x² is [ (π/2 + 2π k - a²/2)/a ]². Therefore, x² + π/2 + 2π k is equal to [ (π/2 + 2π k - a²/2)/a ]² + π/2 + 2π k. This seems complicated, but the key point is whether this expression is such that the cosine of it is 1. However, cosine of any real number is 1 only when its argument is 2π n for some integer n. Therefore, we need x² + a x + a²/2 = 2π n.But given the previous substitution, this expression equals x² + π/2 + 2π k. So unless [x² + π/2 + 2π k] = 2π n, which would require x² = 2π n - π/2 - 2π k. However, x is defined as (π/2 + 2π k - a²/2)/a. Therefore, unless this specific combination results in x² being equal to 2π(n - k) - π/2, which would be a very special case, this equality may not hold in general.Therefore, it's possible that for some a, there exists x such that sin((x + a)^2) - sin(x²)=2, but perhaps for other a, this maximum difference isn't achieved. However, regardless of a, the difference can approach 2. For example, take a very small a. Then, (x + a)^2 ≈ x² + 2a x. Then, sin(x² + 2a x) - sin(x²) ≈ 2a x cos(x²) (using the derivative approximation). But this is not precise. However, regardless of a, if we set x such that (x + a)^2 = π/2 + 2π k and x² = -π/2 + 2π k, then sin((x + a)^2)=1 and sin(x²)=-1, leading to a difference of 2. But such x may not exist because x² cannot be negative. Wait, x² is always non-negative, so sin(x²) ∈ [-1,1], but x² is ≥0. So to have sin(x²) = -1, we need x² = 3π/2 + 2π n for some integer n. Similarly, to have sin((x + a)^2)=1, we need (x + a)^2 = π/2 + 2π m.Therefore, suppose we set x such that x² = 3π/2 + 2π n and (x + a)^2 = π/2 + 2π m. Then sin(x²) = -1 and sin((x + a)^2)=1, so the difference is 2. Therefore, if such x exists for some n and m, then the difference can reach 2. Let's see if such x exists.Suppose x² = 3π/2 + 2π n => x = sqrt(3π/2 + 2π n). Then (x + a)^2 = 3π/2 + 2π n + 2a x + a². We need this to be equal to π/2 + 2π m. Therefore,3π/2 + 2π n + 2a x + a² = π/2 + 2π m => 2a x + a² = -π + 2π (m - n).So 2a x + a² = π(2(m - n) -1). Since x = sqrt(3π/2 + 2π n), this is possible if a is chosen appropriately. For example, if we fix a, then for some n, there exists an integer m such that π(2(m - n) -1) is approximately equal to 2a x + a². Since x can be made large by choosing large n, and m can also be chosen as needed, it's possible to find such n and m for any fixed a, making the difference reach 2. Therefore, for any fixed a, there exists x such that sin((x + a)^2) - sin(x²) =2, which would require b ≥2. However, even if we set b=2, the question is whether sin((x + a)^2) ≤ sin(x²) +2 for all x. Since the maximum difference is 2, then yes, sin((x + a)^2) ≤ sin(x²) +2 holds for all x, regardless of a. Wait, but does that hold?Wait, sin((x + a)^2) - sin(x²) ≤2 always, regardless of a. So sin((x + a)^2) ≤ sin(x²) +2. So regardless of the choice of a, if we set b=2, the inequality holds. But wait, the problem states that there must exist positive constants a and b. So if we can choose any a>0 and set b=2, then it works. Therefore, function ③ is a limited growth function with a=1 (for example) and b=2.But wait, earlier analysis suggested that for any a, the maximum difference is 2, so b=2 works for any a. Therefore, as long as we choose b=2, regardless of a, the inequality holds. Therefore, function ③ is a limited growth function. Therefore, ② and ③ are both limited growth functions. Hence, the answer should be B: ②③.But wait, the options are A: ①②③, B:②③, C:①③, D:③. Earlier, I concluded that ① is not a limited growth function, ② is, and ③ is. So that would be B. But wait, the problem statement says "there exist positive constants a and b", so for each function, we need to have some a and b. For function ③, as per the above, if we set b=2 and any a>0, then sin((x + a)^2) ≤ sin(x²) +2 holds for all x. Therefore, ③ is a limited growth function. For function ②, we saw that choosing a=1 and b=1 works. Therefore, both ② and ③ are limited growth functions, hence the answer is B. But the options given are B:②③, so that's the answer.But let me just verify function ③ again. Suppose someone argues that even though the maximum difference is 2, but for the inequality f(x + a) ≤ f(x) + b, we might need a specific a. However, according to the problem statement, we need there to exist some a and b. So even if for a=1, b=2 works, then that's okay. So regardless of a, as long as b=2, then sin((x + a)^2) ≤ sin(x²) +2. So for any a>0, setting b=2 suffices. Therefore, yes, ③ is a limited growth function. Therefore, the answer is B:②③.But wait, wait. Let's check with a concrete example. Take a=1 and b=2. Then, is sin((x +1)^2) ≤ sin(x²) +2 always true? Since sin((x +1)^2) ≤1 and sin(x²) ≥-1, so 1 ≤ -1 +2=1. So when sin(x²)= -1, sin((x +1)^2) ≤1, which is equal to -1 +2=1. Therefore, equality holds. But does there exist an x where sin((x +1)^2)=1 and sin(x²)=-1 simultaneously?As above, if x²=3π/2 +2π n and (x +1)^2=π/2 +2π m for some integers n and m. Then x= sqrt(3π/2 +2π n), and (sqrt(3π/2 +2π n)+1)^2=π/2 +2π m. Whether such n and m exist is a question of number theory. However, since the squares of real numbers are dense modulo 2π, it's possible that for some x, both x² and (x +1)^2 attain the required values to make sin(x²)=-1 and sin((x +1)^2)=1. Therefore, in such cases, the difference is exactly 2, so the inequality becomes equality. Therefore, for such x, sin((x +1)^2)=1 and sin(x²)+2=1, which is 1 ≤1, which holds. Wait, sin(x²)+2= -1 +2=1. Yes, so equality holds. Therefore, in these cases, the inequality is tight, but it still holds. Therefore, choosing b=2 is sufficient for any a. Hence, ③ is indeed a limited growth function. Therefore, the answer should be B:②③.Wait, but in the answer options given, B is ②③, which would be my conclusion. However, the original analysis for function ③ suggests that regardless of a, b=2 works, so yes, such constants exist. Therefore, the correct answer is B.Final Answerboxed{B}</think>To determine which functions are "limited growth functions," we need to check if there exist positive constants (a) and (b) such that (f(x + a) leq f(x) + b) for all (x in mathbb{R}).1. Function ①: (f(x) = x^2 + x + 1) - Compute (f(x + a) - f(x)): [ f(x + a) - f(x) = (x + a)^2 + (x + a) + 1 - (x^2 + x + 1) = 2ax + a^2 + a ] - This expression is linear in (x) with a positive coefficient (2a). As (x to infty), (2ax) will exceed any constant (b). Therefore, no such (a) and (b) exist. Hence, ① is not a limited growth function.2. Function ②: (f(x) = sqrt{|x|}) - We need (sqrt{|x + a|} leq sqrt{|x|} + b). Squaring both sides: [ |x + a| leq |x| + 2bsqrt{|x|} + b^2 ] - For (x geq 0), this simplifies to (x + a leq x + 2bsqrt{x} + b^2), which holds if (a leq 2bsqrt{x} + b^2). For (x < 0), let (x = -y), then (|a - y| leq y + 2bsqrt{y} + b^2). By choosing (a = 1) and (b = 1), the inequality holds for all (x). Hence, ② is a limited growth function.3. Function ③: (f(x) = sin(x^2)) - Using trigonometric identities, the difference (sin((x + a)^2) - sin(x^2)) is bounded by ([-2, 2]). Therefore, (sin((x + a)^2) leq sin(x^2) + 2) for any (a > 0). Choosing (b = 2) satisfies the inequality for all (x). Hence, ③ is a limited growth function.Thus, the limited growth functions are ② and ③, leading to the answer:[boxed{B}],<think>Okay, so I need to figure out which of the three given functions are "limited growth functions" as defined. The definition says that a function f(x) is a limited growth function if there exist positive constants a and b such that for all real numbers x, the inequality f(x + a) ≤ f(x) + b holds. Let me parse this step by step.First, let me understand the condition. For a function to be a limited growth function, adding some fixed value a to the input x should not increase the output by more than another fixed value b. So, no matter how big x gets, if I shift x by a, the function's value doesn't spike up by more than b. This seems like it's controlling the growth rate of the function, ensuring it doesn't grow too quickly. The key here is that a and b are fixed constants, so they can't depend on x. They have to work for every x in the real numbers.Now, let's look at each function one by one.Function ①: f(x) = x² + x + 1This is a quadratic function. Quadratic functions grow like x², which is faster than linear. Let's test if this satisfies the limited growth condition.We need to find a and b such that f(x + a) ≤ f(x) + b for all x.Compute f(x + a):f(x + a) = (x + a)² + (x + a) + 1 = x² + 2a x + a² + x + a + 1 = x² + (2a + 1)x + (a² + a + 1)Now, subtract f(x):f(x + a) - f(x) = [x² + (2a + 1)x + (a² + a + 1)] - [x² + x + 1] = (2a + 1)x - x + (a² + a + 1 - 1) = 2a x + a² + aSo, f(x + a) - f(x) = 2a x + a² + aWe need this difference to be ≤ b for all x. But 2a x + a² + a is a linear function in x. As x becomes large (either positive or negative), this expression will go to infinity or negative infinity. However, since a is a positive constant, 2a is positive, so as x approaches positive infinity, the expression 2a x + a² + a will go to positive infinity, which can't be bounded by a constant b. Similarly, if x goes to negative infinity, the expression would go to negative infinity, but since we have an inequality ≤ b, even if the left side goes to negative infinity, the inequality would hold, but the problem is when x is positive and large. So, the difference f(x + a) - f(x) becomes unbounded as x increases, meaning there's no constant b that can satisfy the inequality for all x. Therefore, function ① is not a limited growth function.Wait, but hold on. The problem states that a and b must be positive constants. So even though for negative x, the difference might be negative, but for positive x, it's going to infinity. Thus, regardless of how we choose a and b, as x grows, this difference will exceed any b. Therefore, function ① doesn't satisfy the condition. So ① is out.Function ②: f(x) = √|x|This is a square root function. Let's check the same condition.Compute f(x + a):f(x + a) = √|x + a|We need to check if there exist a, b > 0 such that √|x + a| ≤ √|x| + b for all x ∈ ℝ.Let me consider different cases for x.Case 1: x is non-negative. Then |x| = x, so f(x) = √x. Then f(x + a) = √(x + a). So the inequality becomes √(x + a) ≤ √x + b. Let's square both sides (since both sides are non-negative):x + a ≤ (√x + b)² = x + 2b√x + b²Subtract x:a ≤ 2b√x + b²So, we need 2b√x + b² ≥ a for all x ≥ 0.But as x increases, √x increases without bound. However, the left side 2b√x + b² will also increase without bound, which is greater than any fixed a. So, for x large enough, this inequality holds. But we need this inequality to hold for all x, including x near 0. Wait, actually, when x approaches 0 from the positive side, the left side becomes 2b*0 + b² = b². So, we need b² ≥ a. But for x approaching 0, we must have a ≤ b². For x large, since the left side is growing, but the inequality is that the left side is greater than or equal to a, which is okay as long as a is fixed.But wait, actually, the original inequality we have is √(x + a) ≤ √x + b. When x is large, √(x + a) ≈ √x + a/(2√x) (using the expansion for small a compared to x). So the difference is roughly a/(2√x), which goes to zero as x increases. Therefore, even without squaring, the left side is approximately √x + something small, and the right side is √x + b, so for large x, we need that something small (a/(2√x)) ≤ b. But since a/(2√x) approaches zero as x increases, this will eventually hold for any positive b. However, the problem is when x is near zero.For example, take x = 0. Then the inequality becomes √a ≤ 0 + b => √a ≤ b. So, we need b ≥ √a. But earlier, from x approaching 0, we had a ≤ b². Therefore, combining these two conditions: √a ≤ b and a ≤ b². These are actually the same condition. Because if a ≤ b², then √a ≤ b (since b is positive). So, as long as we choose a and b such that a ≤ b², then both conditions are satisfied.But we need the inequality √(x + a) ≤ √x + b to hold for all x. Let's check x = -a/2, which is a negative x. Wait, but x can be any real number. Let's test when x is negative. Suppose x is negative, such that x + a is still negative. Then |x + a| = |a + x|. If x is negative enough so that x + a is negative, then |x + a| = -x -a, so f(x + a) = √(-x -a). However, f(x) = √|x| = √(-x) when x is negative. So, let's suppose x is negative and x + a is negative. Let me denote x = -y where y > 0. Then x + a = -y + a. If a < y, then x + a is negative, otherwise, it's non-negative.So, case 2: x = -y, where y > a. Then x + a = -y + a < 0. Thus, f(x) = √y, and f(x + a) = √(y - a). So the inequality becomes √(y - a) ≤ √y + b.But y - a must be non-negative, so y ≥ a. So, in this case, y ≥ a. Then √(y - a) ≤ √y + b. Let's square both sides:y - a ≤ y + 2b√y + b²Subtract y:- a ≤ 2b√y + b²Which is equivalent to 2b√y + b² + a ≥ 0. Since all terms are positive, this is always true. So, for x negative with |x| ≥ a, the inequality holds.But if x + a is non-negative, even if x is negative. For example, x = -a/2. Then x + a = a/2 ≥ 0. So, f(x + a) = √(a/2), and f(x) = √(a/2). Wait, x = -a/2, so |x| = a/2. Thus, f(x) = √(a/2). Then the inequality is √(a/2) ≤ √(a/2) + b, which is 0 ≤ b, which is true since b is positive. So, in these cases, the inequality holds.Another case: x is negative, x + a is positive. So x = -c, where c < a. Then x + a = a - c. So, f(x + a) = √(a - c), and f(x) = √c. The inequality becomes √(a - c) ≤ √c + b.Since c is between 0 and a, we need to ensure that for all c in (0, a), √(a - c) - √c ≤ b.What's the maximum value of √(a - c) - √c for c in (0, a)? Let's find the maximum of this function. Let’s set derivative with respect to c to zero.Let g(c) = √(a - c) - √cg’(c) = (-1)/(2√(a - c)) - 1/(2√c)Set to zero:-1/(2√(a - c)) - 1/(2√c) = 0But this equation has no solution because both terms are negative for c in (0, a). Therefore, the function g(c) is decreasing on (0, a). Therefore, the maximum occurs at c approaching 0. When c approaches 0, g(c) approaches √a - 0 = √a. The minimum occurs at c approaching a, where g(c) approaches 0 - √a = -√a. So the maximum difference is √a. Therefore, to have √(a - c) ≤ √c + b for all c in (0, a), we need that √a ≤ b. Because when c approaches 0, the left side approaches √a, and the right side approaches 0 + b. Therefore, √a ≤ b.But previously, from the x = 0 case, we had √a ≤ b, and from x positive, we needed a ≤ b². So, combining these two conditions: a ≤ b² and √a ≤ b. But √a ≤ b implies a ≤ b². So actually, the only condition needed is a ≤ b².Therefore, if we choose a and b such that a ≤ b², then all the inequalities hold. For example, choose b = 1, then a can be up to 1. So, such a and b exist. For example, take a = 1 and b = 1. Then check:For x positive: √(x + 1) ≤ √x + 1. This is known to be true because √(x + 1) - √x = 1/(√(x + 1) + √x) ≤ 1, so adding 1 to √x gives something larger.For x negative with |x| ≥ 1: Already checked, it's okay.For x negative between -1 and 0: √(a - c) ≤ √c + 1, with a = 1. So √(1 - c) ≤ √c + 1. When c approaches 0, √1 = 1 ≤ 0 + 1, which is equality. When c approaches 1, √0 = 0 ≤ √1 + 1 = 1 + 1 = 2, which is true. And in between, since the function is decreasing, it's always less than or equal to 1. So this works.Therefore, function ② is a limited growth function. So ② is okay.Function ③: f(x) = sin(x²)This is a sine function with argument x squared. Let's check the condition.We need to find a, b > 0 such that sin((x + a)²) ≤ sin(x²) + b for all x ∈ ℝ.First, note that the sine function is bounded between -1 and 1. Therefore, sin((x + a)²) ≤ 1, and sin(x²) ≥ -1. Therefore, sin((x + a)²) - sin(x²) ≤ 1 - (-1) = 2. So the maximum possible difference is 2. But can we have a tighter bound?But wait, the problem is that even though the sine function is bounded, the difference sin((x + a)^2) - sin(x^2) can be up to 2 in theory, but maybe depending on a, can we make this difference bounded by a fixed b?However, the problem states that we need sin((x + a)^2) ≤ sin(x^2) + b for all x. Since sin is bounded, the left side is at most 1, and the right side is at least -1 + b. So if we choose b ≥ 2, then -1 + b ≥ 1, so 1 ≤ -1 + b => b ≥ 2. Therefore, if we take b = 2, then sin((x + a)^2) ≤ 1 ≤ sin(x²) + 2, since sin(x²) ≥ -1. Therefore, sin(x²) + 2 ≥ 1, and since left side is ≤1, the inequality holds. Wait, but this seems like using the boundedness of sine to make the inequality hold.But actually, let's check in detail. Let's take b = 2. Then for any x, sin((x + a)^2) ≤ 1, and sin(x²) + 2 ≥ -1 + 2 = 1. Therefore, since 1 ≤ sin(x²) + 2 and sin((x + a)^2) ≤ 1, then sin((x + a)^2) ≤ sin(x²) + 2. Therefore, with b = 2 and any a > 0, the inequality holds. So, for example, a = 1 and b = 2.Wait, that seems too easy. But is this correct? Let me test with some specific values.Take x such that sin(x²) = -1. Then sin(x²) + 2 = 1, and sin((x + a)^2) ≤ 1, so the inequality holds. If sin(x²) is something else, say sin(x²) = 0, then sin((x + a)^2) ≤ 0 + 2, which is always true because sin is ≤1. Similarly, if sin(x²) = 1, then sin((x + a)^2) ≤ 1 + 2 = 3, which is always true. So regardless of a, as long as b is at least 2, the inequality holds. Therefore, for any a > 0, choosing b = 2 would satisfy the condition.But wait, does this really work? Let me check an example where (x + a)^2 is a different point. Suppose x = 0, then sin(0) = 0, and sin(a²) ≤ 0 + b. So sin(a²) ≤ b. Since sin(a²) ≤ 1, then as long as b ≥1, this holds. But if we set b = 2, as before, that's okay.But actually, the problem states that there exist positive constants a and b. So, if we can find such constants, even if b is 2, it's acceptable. Therefore, since the sine function is bounded, the difference between sin((x + a)^2) and sin(x²) is at most 2 (since each can be at most 1 and at least -1). Therefore, choosing b = 2, regardless of a, works. But we can choose a to be any positive constant. For example, a = 1, b = 2. Therefore, function ③ satisfies the condition.Wait, but let me check if there are any cases where even with b = 2, the inequality might not hold. Suppose there's a point x where sin((x + a)^2) - sin(x²) > b. But since the maximum value of sin((x + a)^2) is 1, and the minimum value of sin(x²) is -1, then the difference is at most 1 - (-1) = 2. Therefore, sin((x + a)^2) - sin(x²) ≤ 2. Hence, setting b = 2 ensures that the inequality holds for all x. Therefore, function ③ is a limited growth function.Wait, but is there a trick here? Because x is over all real numbers. For example, if a is fixed, then (x + a)^2 can be very different from x², but the sine of both will still be bounded. The key is that the difference between sin((x + a)^2) and sin(x²) is bounded by 2, regardless of a. So even if we shift the argument by any a, the difference in their sine values can't exceed 2. Therefore, choosing b = 2 works for any a. So, for example, choosing a = 1 and b = 2, the inequality holds for all x. Therefore, function ③ is a limited growth function.But wait, actually, the problem requires that there exist a and b. So as long as there exists at least one pair (a, b). Since for any a, we can take b = 2, then such a pair exists. For example, a = 1 and b = 2. Therefore, ③ is a limited growth function.But let me verify with a specific example. Let’s take a = π/2. Then, for some x, x² could be, say, 0, so (x + a)^2 = (π/2)^2. Then sin((π/2)^2) ≈ sin(2.467) ≈ 0.664. Then sin(0) + b = 0 + b. So we need 0.664 ≤ b. If b = 2, that's okay. Another example: x such that x² = π/2, so sin(x²) = 1. Then (x + a)^2 = (sqrt(pi/2) + a)^2. The sine of that could be anything between -1 and 1. So sin((sqrt(pi/2) + a)^2) ≤ 1 ≤ 1 + b, which holds since b ≥0. So yes, even in these cases, it holds.Therefore, function ③ is indeed a limited growth function.Wait, but hold on a second. Suppose we pick a different a. Let's say a = π. Then, for x = 0, sin((0 + π)^2) = sin(π²) ≈ sin(9.8696) ≈ -0.4303. Then sin(0) + b = 0 + b. So we need -0.4303 ≤ b. Since b is positive, this holds. For another x, say x = sqrt(pi/2). Then x² = pi/2, so sin(x²) = 1. Then (x + a)^2 = (sqrt(pi/2) + pi)^2. The sine of that will be between -1 and 1, so sin((sqrt(pi/2) + pi)^2) ≤ 1 ≤ 1 + b. Which holds for any b >=0.But the key point is that no matter how we choose a, the maximum difference between sin((x + a)^2) and sin(x^2) is 2, so as long as we take b = 2, it works. Therefore, function ③ is okay.Therefore, among the three functions:①: Not a limited growth function.②: Limited growth function.③: Limited growth function.So the answer should be B: ②③.But let me cross-verify once again.For function ②, we saw that choosing a = 1 and b = 1 works (since a = 1 ≤ b² = 1, so equality holds). Then for all x, √|x + 1| ≤ √|x| + 1. This is true as per our earlier analysis.For function ③, choosing a = 1 and b = 2 works, as the maximum difference is 2.Function ①, as a quadratic, can't satisfy the condition because the difference f(x + a) - f(x) is linear in x, which becomes unbounded.Therefore, the correct answer is B: ②③.Final Answerboxed{B}</think>To determine which of the given functions are "limited growth functions," we need to check if there exist positive constants (a) and (b) such that (f(x + a) leq f(x) + b) for all (x in mathbb{R}).Function ①: (f(x) = x^2 + x + 1)This is a quadratic function. The difference (f(x + a) - f(x)) results in a linear term in (x), which grows without bound as (x) increases. Therefore, no constants (a) and (b) can satisfy the inequality for all (x). Hence, function ① is not a limited growth function.Function ②: (f(x) = sqrt{|x|})For this function, we need to check if (sqrt{|x + a|} leq sqrt{|x|} + b). By analyzing different cases for (x) (positive, negative, and near zero), we find that choosing (a) and (b) such that (a leq b^2) ensures the inequality holds. For example, (a = 1) and (b = 1) work. Therefore, function ② is a limited growth function.Function ③: (f(x) = sin(x^2))The sine function is bounded between -1 and 1. The difference (sin((x + a)^2) - sin(x^2)) is at most 2. By choosing (b = 2) (or any (b geq 2)), the inequality (sin((x + a)^2) leq sin(x^2) + b) holds for all (x). Therefore, function ③ is a limited growth function.Thus, the limited growth functions are ② and ③.[boxed{B}]]

🔑:For (①), transforming (f(x+a) leqslant f(x)+b) yields: ((x+a)^{2}+(x+a)+1 leqslant x^{2}+x+1+b), which simplifies to (2ax leqslant -a^{2}-a+b), and further to (x leqslant dfrac {-a^{2}-a+b}{2a}). This must hold for all (x in mathbb{R}), but given the domain of the function is (mathbb{R}), there do not exist positive constants (a) and (b) that satisfy the condition, hence (f(x)=x^{2}+x+1) is not a "limited growth function"; For (②), if (f(x)= sqrt {|x|}) is a "limited growth function", then (f(x+a) leqslant f(x)+b) can be transformed into: ( sqrt {|x+a|} leqslant sqrt {|x|}+b), (therefore |x+a| leqslant |x|+b^{2}+2b sqrt {|x|}) always holds, and since (|x+a| leqslant |x|+a), (therefore |x|+a leqslant |x|+b^{2}+2b sqrt {|x|}), (therefore sqrt {|x|} geqslant dfrac {a-b^{2}}{2b}), Clearly, when (a < b^{2}), the condition always holds, (therefore f(x)= sqrt {|x|}) is a "limited growth function"; For (③), since (-1 leqslant f(x)=sin (x^{2}) leqslant 1), (therefore f(x+a)-f(x) leqslant 2), (therefore) when (b geqslant 2), with (a) being any positive number, (f(x+a) leqslant f(x)+b) always holds, hence (f(x)=sin (x^{2})) is a "limited growth function". Therefore, the correct choice is B. Assuming each function is a "limited growth function", deducing the conditions under which (f(x+a) leqslant f(x)+b) always holds, and examining the existence of (a) and (b) leads to the answer. This question tests the understanding of new definitions, the existence of functions, and the study of conditions that always hold, and is considered a medium-level problem. boxed{text{B}}

🧠:[<think>Okay, let's try to tackle this problem step by step. So we have triangle ABC with medians AM and BP. The problem states that angle APB is equal to angle BMA. Also, we know that the cosine of angle ACB is 0.8, and BP is 1. We need to find the area of triangle ABC.First, let me recall some properties of medians in a triangle. A median connects a vertex to the midpoint of the opposite side. So, AM is the median from A to the midpoint M of BC, and BP is the median from B to the midpoint P of AC. Given that angles APB and BMA are equal. Let me try to visualize the triangle and the medians. Let's denote the midpoints: M is the midpoint of BC, so BM = MC. Similarly, P is the midpoint of AC, so AP = PC.Since BP is a median, its length is given as 1. The goal is to find the area of triangle ABC. We also know cos(angle ACB) = 0.8. So angle at C has cosine 0.8, which might be useful when applying the Law of Cosines later.Now, the key condition here is angle APB = angle BMA. Let's denote these angles as θ for simplicity. So, angle APB = θ and angle BMA = θ.Let me try to draw this triangle. Let me label the triangle ABC with points A, B, C. M is the midpoint of BC, and P is the midpoint of AC. Then BP and AM are medians intersecting at some point, maybe the centroid? Wait, in a triangle, the medians intersect at the centroid, which divides each median into a ratio 2:1. However, in this problem, maybe we need to use the given angle equality to find some relations between the sides.Alternatively, perhaps considering triangles APB and BMA. Hmm, angle APB is in triangle APB, and angle BMA is in triangle BMA. Maybe these triangles are similar? If two angles are equal, and maybe there's some proportionality in the sides? Let me check.But first, let's note the coordinates. Maybe coordinate geometry can help here. Assign coordinates to the triangle's vertices and use the given conditions to set up equations.Let me place point C at the origin (0,0) to simplify calculations. Let me denote coordinates as follows:Let’s set point C at (0, 0).Let’s denote point A as (2a, 0), so that midpoint P of AC is at (a, 0). Similarly, let's denote point B as (2b, 2c), so that midpoint M of BC is at (b, c).Wait, but if C is at (0,0), then midpoint of BC would be at ((2b + 0)/2, (2c + 0)/2) = (b, c). Similarly, midpoint P of AC is at ((2a + 0)/2, (0 + 0)/2) = (a, 0).So coordinates:A: (2a, 0)B: (2b, 2c)C: (0, 0)Midpoints:M: (b, c)P: (a, 0)Now, the median AM connects A(2a,0) to M(b,c). The median BP connects B(2b,2c) to P(a,0).We need to find the angles at APB and BMA.First, let's compute angle APB. Point P is at (a,0). Points A(2a,0), P(a,0), B(2b,2c). So angle at P between points A and B.Similarly, angle BMA is the angle at M between points B and A. Points B(2b,2c), M(b,c), A(2a,0). So angle at M between points B and A.Given that these two angles are equal. Let's compute the vectors involved in these angles to find the cosine of the angles, perhaps.For angle APB: At point P(a,0), the vectors PA and PB.PA is from P to A: (2a - a, 0 - 0) = (a, 0)PB is from P to B: (2b - a, 2c - 0) = (2b - a, 2c)The cosine of angle APB can be found using the dot product:cosθ = (PA · PB) / (|PA| |PB|)PA · PB = (a)(2b - a) + (0)(2c) = a(2b - a)|PA| = sqrt(a^2 + 0^2) = |a||PB| = sqrt((2b - a)^2 + (2c)^2)So cosθ = [a(2b - a)] / [ |a| sqrt((2b - a)^2 + (2c)^2) ]Similarly, for angle BMA at point M(b,c), the vectors MB and MA.MB is from M to B: (2b - b, 2c - c) = (b, c)MA is from M to A: (2a - b, 0 - c) = (2a - b, -c)The cosine of angle BMA is:cosθ = (MB · MA) / (|MB| |MA|)MB · MA = (b)(2a - b) + (c)(-c) = 2ab - b^2 - c^2|MB| = sqrt(b^2 + c^2)|MA| = sqrt((2a - b)^2 + (-c)^2) = sqrt((2a - b)^2 + c^2)So cosθ = (2ab - b^2 - c^2) / [ sqrt(b^2 + c^2) sqrt((2a - b)^2 + c^2) ]Given that angle APB = angle BMA, so their cosines must be equal. Therefore, we can set the two expressions equal:[ a(2b - a) ] / [ |a| sqrt((2b - a)^2 + (2c)^2) ] = (2ab - b^2 - c^2) / [ sqrt(b^2 + c^2) sqrt((2a - b)^2 + c^2) ]Hmm, this looks complicated. Maybe we can simplify this equation. Let's note that |a| is either a or -a, but since these are lengths in a triangle, coordinates are probably chosen such that a, b, c are positive. So |a| = a.Therefore, simplifying the left-hand side (LHS):[ a(2b - a) ] / [ a sqrt((2b - a)^2 + (2c)^2) ] = (2b - a) / sqrt((2b - a)^2 + (2c)^2 )So LHS = (2b - a)/sqrt((2b - a)^2 + 4c^2 )RHS = (2ab - b^2 - c^2)/[ sqrt(b^2 + c^2) sqrt((2a - b)^2 + c^2) ]Set LHS = RHS:(2b - a)/sqrt((2b - a)^2 + 4c^2 ) = (2ab - b^2 - c^2)/[ sqrt(b^2 + c^2) sqrt((2a - b)^2 + c^2) ]This equation seems quite involved. Maybe there's a better approach.Alternatively, maybe using vector analysis or coordinate geometry. Let me think. Alternatively, perhaps using the properties of medians and centroids.In a triangle, the centroid divides each median into a ratio 2:1. So if AM and BP are medians, their intersection is the centroid G. Then AG:GM = 2:1 and BG:GP = 2:1.But how does this help with the angle equality?Alternatively, maybe using the Law of Cosines on the triangles involved.Given that angle APB = angle BMA. Let's denote this common angle as θ.First, in triangle APB:AP = PC = half of AC. Since P is the midpoint, AP = AC/2. Similarly, BM = MC = BC/2.But maybe we need to express the sides in terms of the original triangle's sides.Let’s denote the sides of triangle ABC as follows:AB = c, BC = a, AC = b.Wait, but we need to be careful with notation. Alternatively, let's use standard notation: let’s denote BC = a, AC = b, AB = c.But given that angle at C is known, with cos(angle ACB) = 0.8, so angle C is known.Also, BP is a median with length 1. So the median from B to AC has length 1.Wait, the formula for the length of a median: the length of the median from B to AC is given by:m_b = (1/2)sqrt(2a^2 + 2c^2 - b^2)Wait, let me confirm. The formula for the length of a median from vertex B is:m_b = (1/2)sqrt(2a^2 + 2c^2 - b^2)Where a, b, c are the lengths of the sides opposite to angles A, B, C respectively.But in standard notation, usually:a = BC, b = AC, c = AB.Wait, in standard triangle notation, a is the side opposite angle A, but in this problem, angle C is given. Maybe we need to adjust accordingly.Alternatively, perhaps to avoid confusion, let's assign variables differently. Let me define:Let’s let sides of the triangle ABC be as follows:Let’s denote AC = x, BC = y, AB = z.Then, the median from B to AC (which is BP) has length 1. The formula for the median from B to AC is:BP^2 = (2AB^2 + 2BC^2 - AC^2)/4Wait, the formula for the median from vertex B to side AC (which is of length x) is:BP^2 = (2AB^2 + 2BC^2 - AC^2)/4So BP = (1/2)sqrt(2AB^2 + 2BC^2 - AC^2) = 1Therefore:sqrt(2z^2 + 2y^2 - x^2)/2 = 1=> sqrt(2z^2 + 2y^2 - x^2) = 2Squaring both sides:2z^2 + 2y^2 - x^2 = 4 ...(1)Also, we know that cos(angle ACB) = 0.8. Angle at C is between sides AC and BC, so by the Law of Cosines:cos(angle C) = (AC^2 + BC^2 - AB^2)/(2*AC*BC)So:0.8 = (x^2 + y^2 - z^2)/(2xy)=> 0.8*(2xy) = x^2 + y^2 - z^2=> 1.6xy = x^2 + y^2 - z^2 ...(2)So equations (1) and (2) relate x, y, z.But we need another equation from the given angle equality. The angles APB and BMA are equal.This seems complicated. Let me think if there's a way to relate these angles through the sides.Alternatively, maybe express the angles in terms of the sides using the Law of Cosines in triangles APB and BMA.First, triangle APB:AP is half of AC, so AP = x/2BP is given as 1AB is zWait, but triangle APB: sides AP = x/2, BP = 1, and AB is z. Wait, no, APB is a triangle with points A, P, B.Wait, AP is half of AC, so AP = x/2BP is the median, which is 1And what is the third side of triangle APB? It's the distance from A to B, which is AB = z.Wait, so in triangle APB, sides are:AP = x/2BP = 1AB = zSimilarly, angle at P is angle APB = θ.Similarly, in triangle BMA:BM is half of BC, since M is the midpoint. So BM = y/2AM is the median from A to BC. Let's compute its length.The median AM can be computed using the formula:AM^2 = (2AB^2 + 2AC^2 - BC^2)/4So AM = (1/2)sqrt(2z^2 + 2x^2 - y^2)But perhaps we can just consider triangle BMA.In triangle BMA, sides are:BM = y/2MA is the median from A, which we can denote as m_a. Let's keep it as m_a for now.AB = zWait, no. Wait, in triangle BMA, the sides are BM, MA, and AB? Wait, no, triangle BMA has vertices B, M, A.So BM = y/2MA is the median from A to BC, which is:MA = (1/2)sqrt(2AB^2 + 2AC^2 - BC^2) = (1/2)sqrt(2z^2 + 2x^2 - y^2)And BA = zWait, but in triangle BMA, the sides are BM = y/2, MA = (1/2)sqrt(2z^2 + 2x^2 - y^2), and BA = z? Wait, no, BA is not a side of triangle BMA. The sides of triangle BMA are BM, MA, and the third side is the distance from B to A, which is AB = z. Wait, no. Wait, triangle BMA: points B, M, A. So sides:BM: from B to M: length y/2MA: from M to A: that's the median AM, which we can denote as m_a = (1/2)sqrt(2z^2 + 2x^2 - y^2)AB: from A to B: but that's not part of triangle BMA. Wait, no, triangle BMA has sides BM, MA, and AB is not part of it. Wait, A to B is a different side. Wait, the third side of triangle BMA would be from B to A, but that's AB = z.Wait, but points B, M, A. So BM is from B to M, MA is from M to A, and AB is from A to B. But in triangle BMA, the sides are BM, MA, and AB. Wait, but AB is not between B and A through M. Wait, actually, triangle BMA is formed by connecting points B, M, and A. So the three sides are BM, MA, and AB.Wait, but AB is the side opposite angle M in triangle BMA. Wait, no. Let me check:In triangle BMA:- Vertex B is connected to M and A.Wait, no. Wait, triangle BMA: points B, M, A. So sides:BM: from B to M (length y/2)MA: from M to A (length m_a = (1/2)sqrt(2z^2 + 2x^2 - y^2))AB: from A to B (length z)Wait, but that can't be a triangle. Wait, actually, in triangle BMA, the sides are BM, MA, and AB. But AB is not part of the triangle; the sides should be BM, MA, and the segment connecting A to B, but AB is already a side of the original triangle. Wait, this is getting confusing. Let me clarify:No, triangle BMA is formed by points B, M, and A. Therefore, the sides are:1. BM: connects B to M (length y/2)2. MA: connects M to A (length m_a, the median from A)3. AB: connects A to B (length z)Wait, no. Wait, in triangle BMA, the three sides are BM, MA, and BA. Wait, BA is the same as AB. So triangle BMA has sides BM, MA, and AB. But AB is a different side. So in triangle BMA, sides BM = y/2, MA = m_a, AB = z.But in triangle BMA, angle at M is angle BMA = θ.Similarly, in triangle APB, angle at P is angle APB = θ.So both triangles APB and BMA have an angle θ. Maybe these triangles are similar? If so, their sides would be proportional.Alternatively, using the Law of Cosines on both angles θ in triangles APB and BMA.In triangle APB:AP = x/2BP = 1AB = zLaw of Cosines for angle at P (θ):AB² = AP² + BP² - 2*AP*BP*cosθSo:z² = (x/2)² + 1² - 2*(x/2)*1*cosθ=> z² = x²/4 + 1 - x*cosθ ...(3)In triangle BMA:BM = y/2MA = m_a = (1/2)sqrt(2z² + 2x² - y²)AB = zLaw of Cosines for angle at M (θ):AB² = BM² + MA² - 2*BM*MA*cosθSo:z² = (y/2)² + [ (1/2)sqrt(2z² + 2x² - y²) ]² - 2*(y/2)*(1/2)sqrt(2z² + 2x² - y²)*cosθSimplify:First, compute the square of MA:[ (1/2)sqrt(2z² + 2x² - y²) ]² = (1/4)(2z² + 2x² - y²) = (2z² + 2x² - y²)/4So:z² = (y²/4) + (2z² + 2x² - y²)/4 - 2*(y/2)*(1/2)sqrt(2z² + 2x² - y²)*cosθSimplify terms:First term: y²/4Second term: (2z² + 2x² - y²)/4Adding these two:[ y²/4 + (2z² + 2x² - y²)/4 ] = [ (y² + 2z² + 2x² - y²) ] /4 = (2z² + 2x²)/4 = (z² + x²)/2Third term: - 2*(y/2)*(1/2)sqrt(...)*cosθ = - (y/2)*sqrt(2z² + 2x² - y²)*cosθThus, equation becomes:z² = (z² + x²)/2 - (y/2)*sqrt(2z² + 2x² - y²)*cosθ ...(4)So now, from equation (3):z² = x²/4 + 1 - x*cosθAnd from equation (4):z² = (z² + x²)/2 - (y/2)*sqrt(2z² + 2x² - y²)*cosθThis seems very complicated. Maybe we can find a relation between x, y, z from equations (1) and (2), and then connect with equations (3) and (4).Recall equations (1) and (2):(1) 2z² + 2y² - x² = 4(2) 1.6xy = x² + y² - z²Let’s try to solve equations (1) and (2). Let’s express z² from equation (2):From (2):z² = x² + y² - 1.6xySubstitute this into equation (1):2(x² + y² - 1.6xy) + 2y² - x² = 4Expand:2x² + 2y² - 3.2xy + 2y² - x² = 4Combine like terms:(2x² - x²) + (2y² + 2y²) - 3.2xy = 4x² + 4y² - 3.2xy = 4Hmm, so:x² - 3.2xy + 4y² = 4 ...(5)This is a quadratic equation in terms of x and y. Maybe we can solve for x/y ratio.Let’s let k = x/y. Then x = ky.Substitute into equation (5):(k y)^2 - 3.2*(k y)*y + 4y^2 = 4k² y² - 3.2k y² + 4y² = 4Factor y²:y²(k² - 3.2k + 4) = 4So,y² = 4 / (k² - 3.2k + 4) ...(6)We can also express z² from equation (2):z² = x² + y² - 1.6xy = (k² y²) + y² - 1.6*(k y)*y = y²(k² + 1 - 1.6k)Thus,z² = y²(k² + 1 - 1.6k) ...(7)Now, recall from equation (3):z² = x²/4 + 1 - x*cosθBut we need to relate θ. However, θ is also present in equation (4). This seems too interconnected. Maybe we can find another relation.Alternatively, perhaps use equation (3) and (4) to find an expression for cosθ and set them equal.From equation (3):z² = x²/4 + 1 - x*cosθSo rearranged:x*cosθ = x²/4 + 1 - z²cosθ = (x²/4 + 1 - z²)/x ...(8)From equation (4):z² = (z² + x²)/2 - (y/2)*sqrt(2z² + 2x² - y²)*cosθLet’s rearrange:Multiply both sides by 2:2z² = z² + x² - y*sqrt(2z² + 2x² - y²)*cosθSubtract z² + x²:2z² - z² - x² = - y*sqrt(2z² + 2x² - y²)*cosθSo,z² - x² = - y*sqrt(2z² + 2x² - y²)*cosθMultiply both sides by -1:x² - z² = y*sqrt(2z² + 2x² - y²)*cosθSolve for cosθ:cosθ = (x² - z²)/( y*sqrt(2z² + 2x² - y²) ) ...(9)Now, set equation (8) equal to equation (9):(x²/4 + 1 - z²)/x = (x² - z²)/( y*sqrt(2z² + 2x² - y²) )This seems really complicated, but maybe substituting z² from equation (2):From equation (2): z² = x² + y² - 1.6xySo substitute z² into equations (8) and (9):First, substitute into equation (8):cosθ = (x²/4 + 1 - (x² + y² - 1.6xy))/x= (x²/4 + 1 - x² - y² + 1.6xy)/x= (-3x²/4 - y² + 1.6xy + 1)/x ...(10)Similarly, substitute z² into equation (9):cosθ = (x² - (x² + y² - 1.6xy)) / ( y*sqrt(2(x² + y² - 1.6xy) + 2x² - y²) )Simplify numerator:x² - x² - y² + 1.6xy = -y² + 1.6xyDenominator:sqrt(2x² + 2y² - 3.2xy + 2x² - y²) = sqrt(4x² + y² - 3.2xy)Thus,cosθ = (-y² + 1.6xy) / ( y*sqrt(4x² + y² - 3.2xy) )Factor numerator:= y(1.6x - y) / ( y*sqrt(4x² + y² - 3.2xy) )Cancel y:= (1.6x - y)/sqrt(4x² + y² - 3.2xy) ...(11)Now, set equation (10) equal to equation (11):(-3x²/4 - y² + 1.6xy + 1)/x = (1.6x - y)/sqrt(4x² + y² - 3.2xy)This is still very complex, but maybe we can substitute k = x/y as before. Let me try that.Let’s let k = x/y, so x = ky. Then:Left-hand side (LHS):[ -3(k y)^2/4 - y^2 + 1.6(k y)y + 1 ] / (k y)= [ (-3k² y²/4 - y² + 1.6k y² + 1) ] / (k y)Factor y² terms:= [ y²(-3k²/4 - 1 + 1.6k ) + 1 ] / (k y )= [ y²( -0.75k² -1 + 1.6k ) + 1 ] / (k y )Right-hand side (RHS):(1.6k y - y)/sqrt(4(k y)^2 + y² - 3.2k y² )= y(1.6k - 1)/sqrt(4k² y² + y² - 3.2k y² )= y(1.6k -1)/sqrt( y²(4k² +1 -3.2k ) )= y(1.6k -1)/( y sqrt(4k² +1 -3.2k ) )Cancel y:= (1.6k -1)/sqrt(4k² +1 -3.2k )Thus, equation becomes:[ y²( -0.75k² -1 + 1.6k ) + 1 ] / (k y ) = (1.6k -1)/sqrt(4k² +1 -3.2k )Multiply both sides by k y:y²( -0.75k² -1 + 1.6k ) + 1 = k y * (1.6k -1)/sqrt(4k² +1 -3.2k )This equation is still complicated, but recall from equation (6):y² = 4 / (k² - 3.2k +4 )So substitute y² = 4 / (k² - 3.2k +4 ) into the equation:[ 4/(k² -3.2k +4 )*( -0.75k² -1 +1.6k ) + 1 ] = k y * (1.6k -1)/sqrt(4k² +1 -3.2k )But we also need to express y in terms of k. From y² = 4 / (k² -3.2k +4 ), so y = 2 / sqrt(k² -3.2k +4 )Thus, substitute y = 2 / sqrt(k² -3.2k +4 ) into the equation:Left-hand side (LHS):4/(k² -3.2k +4 )*( -0.75k² -1 +1.6k ) +1= [4*(-0.75k² -1 +1.6k )]/(k² -3.2k +4 ) +1= [ -3k² -4 +6.4k ]/(k² -3.2k +4 ) +1= [ -3k² -4 +6.4k +k² -3.2k +4 ] / (k² -3.2k +4 )Combine like terms:( -3k² +k² ) + (6.4k -3.2k ) + (-4 +4 )= (-2k²) + (3.2k ) + 0Thus, LHS = (-2k² +3.2k ) / (k² -3.2k +4 )Right-hand side (RHS):k * [2 / sqrt(k² -3.2k +4 ) ] * (1.6k -1)/sqrt(4k² +1 -3.2k )= 2k (1.6k -1 ) / [ sqrt(k² -3.2k +4 ) * sqrt(4k² +1 -3.2k ) ]Let’s compute the denominator:sqrt(k² -3.2k +4 ) * sqrt(4k² +1 -3.2k )= sqrt( (k² -3.2k +4)(4k² +1 -3.2k ) )Let’s compute the product inside:Multiply (k² -3.2k +4)(4k² -3.2k +1 )Let’s denote A = k², B = -3.2k, C =4 and D=4k², E= -3.2k, F=1Wait, maybe expand term by term:(k²)(4k²) + k²(-3.2k) + k²(1) + (-3.2k)(4k²) + (-3.2k)(-3.2k) + (-3.2k)(1) +4(4k²) +4(-3.2k) +4(1)= 4k^4 -3.2k^3 +k^2 -12.8k^3 +10.24k^2 -3.2k +16k² -12.8k +4Combine like terms:k^4: 4k^4k^3: (-3.2 -12.8)k^3 = -16k^3k^2: (1 +10.24 +16)k^2 = 27.24k^2k terms: (-3.2k -12.8k ) = -16kConstants: 4So product is 4k^4 -16k^3 +27.24k^2 -16k +4Hmm, this seems messy. Maybe there's a better way.Alternatively, perhaps notice that 4k² +1 -3.2k = 4k² -3.2k +1, and k² -3.2k +4. Let's see if there's a relationship between them.Alternatively, maybe write both quadratics:First quadratic: k² -3.2k +4Second quadratic:4k² -3.2k +1Not obvious. Alternatively, perhaps compute 4*(k² -3.2k +4) =4k² -12.8k +16, which is different from 4k² -3.2k +1. Hmm.Alternatively, maybe note that 4k² -3.2k +1 = (2k)^2 - 3.2k +1, but not sure.Alternatively, compute the product:(k² -3.2k +4)(4k² -3.2k +1)Let’s compute:Multiply each term:k²*(4k²) =4k^4k²*(-3.2k) = -3.2k^3k²*(1) =k²-3.2k*(4k²)= -12.8k^3-3.2k*(-3.2k)=10.24k²-3.2k*(1)= -3.2k4*(4k²)=16k²4*(-3.2k)= -12.8k4*(1)=4Now sum all terms:4k^4-3.2k^3 -12.8k^3 = -16k^3k² +10.24k² +16k² = (1 +10.24 +16)k² =27.24k²-3.2k -12.8k =-16k+4So the product is 4k^4 -16k^3 +27.24k² -16k +4This seems complicated, but maybe factor it.Alternatively, perhaps approximate the values numerically. Let’s try plugging in values for k to see if we can find a solution.But this might take a while. Alternatively, notice that equation (5):x² - 3.2xy +4y² =4Expressed in terms of k=x/y:k² -3.2k +4 =4/y²Wait, from equation (6):y²=4/(k² -3.2k +4 )Which is always positive, so denominator must be positive. So k² -3.2k +4 >0. Since the discriminant is (3.2)^2 -16=10.24-16= -5.76<0, the quadratic is always positive. So y is real for any k.But this might not help.Alternatively, maybe there's a specific ratio k that simplifies the equation. Let's assume k is a rational number. Maybe k=2?Let’s try k=2:Then x=2y.From equation (5):(2y)^2 -3.2*(2y)*y +4y²=44y² -6.4y² +4y²= (4 -6.4 +4)y²=1.6y²=4 => y²=2.5 => y=√(2.5)=≈1.5811Then x=2y≈3.1623From equation (2):z² = x² + y² -1.6xy≈(10) +2.5 -1.6*3.1623*1.5811≈12.5 -1.6*5≈12.5 -8=4.5Thus z≈2.1213Now check equation (1):2z² +2y² -x²≈2*(4.5)+2*(2.5)-10≈9+5-10=4. Which matches. So k=2 is a valid solution.So with k=2, x=2y, and equations hold.Thus, maybe k=2 is the solution. Let’s check if this satisfies the angle equality.If k=2, then x=2y, z²=4.5, so z= sqrt(4.5)=3/sqrt(2)≈2.1213Now compute cosθ from equation (8):cosθ=(x²/4 +1 -z²)/xWith x=2y, and y²=2.5, so y=sqrt(2.5), x=2sqrt(2.5)x≈3.1623Then x²/4= (4*2.5)/4=2.5So cosθ=(2.5 +1 -4.5)/3.1623=( -1)/3.1623≈-0.316From equation (11):cosθ=(1.6x - y)/sqrt(4x² + y² -3.2xy )With x=2y:1.6x=3.2y, so numerator=3.2y - y=2.2yDenominator= sqrt(4*(4y²) + y² -3.2*(2y)*y )= sqrt(16y² + y² -6.4y² )= sqrt(10.6y² )=y*sqrt(10.6)Thus, cosθ=2.2y/(y*sqrt(10.6))=2.2/sqrt(10.6)≈2.2/3.255≈0.676But previously, from equation (8), cosθ≈-0.316. This doesn't match, so k=2 is not a valid solution for the angle equality. Hence, even though k=2 satisfies equations (1) and (2), it doesn't satisfy the angle condition. Therefore, my assumption is wrong.Hmm, so k=2 is not the correct ratio.This suggests that we need to solve the equation:(-2k² +3.2k ) / (k² -3.2k +4 ) = 2k (1.6k -1 ) / sqrt( (k² -3.2k +4)(4k² +1 -3.2k ) )This is equivalent to:(-2k² +3.2k ) / (k² -3.2k +4 ) = 2k (1.6k -1 ) / sqrt( (k² -3.2k +4)(4k² +1 -3.2k ) )Multiply both sides by sqrt( (k² -3.2k +4)(4k² +1 -3.2k ) ) / (k² -3.2k +4 )):Left side: (-2k² +3.2k ) / sqrt( (k² -3.2k +4)(4k² +1 -3.2k ) )Right side: 2k (1.6k -1 ) / (k² -3.2k +4 )But this seems too complicated. Maybe square both sides to eliminate the square root.Let’s denote LHS1 = (-2k² +3.2k ) / (k² -3.2k +4 )RHS1 = 2k (1.6k -1 ) / sqrt( (k² -3.2k +4)(4k² +1 -3.2k ) )So the equation is LHS1 = RHS1Square both sides:[ (-2k² +3.2k )^2 ] / (k² -3.2k +4 )^2 = [4k² (1.6k -1 )^2 ] / [ (k² -3.2k +4 )(4k² +1 -3.2k ) ]Multiply both sides by (k² -3.2k +4 )^2 (4k² +1 -3.2k ):Left side: (-2k² +3.2k )^2 (4k² +1 -3.2k )Right side:4k² (1.6k -1 )^2 (k² -3.2k +4 )Expand both sides:Left side:(-2k² +3.2k )^2 =4k^4 -12.8k³ +10.24k²Multiply by (4k² +1 -3.2k ):(4k^4 -12.8k³ +10.24k²)(4k² -3.2k +1 )This will be a lengthy expansion, but let's do term by term:4k^4*4k²=16k^64k^4*(-3.2k )= -12.8k^54k^4*1=4k^4-12.8k³*4k²= -51.2k^5-12.8k³*(-3.2k )=40.96k^4-12.8k³*1= -12.8k³10.24k²*4k²=40.96k^410.24k²*(-3.2k )= -32.768k³10.24k²*1=10.24k²Combine all terms:16k^6-12.8k^5 -51.2k^5= -64k^54k^4 +40.96k^4 +40.96k^4= (4 +40.96 +40.96)k^4=85.92k^4-12.8k³ -32.768k³= -45.568k³10.24k²So Left side expanded: 16k^6 -64k^5 +85.92k^4 -45.568k³ +10.24k²Right side:4k² (1.6k -1 )^2 (k² -3.2k +4 )First compute (1.6k -1 )^2 =2.56k² -3.2k +1Then multiply by (k² -3.2k +4 ):(2.56k² -3.2k +1)(k² -3.2k +4 )Expand term by term:2.56k²*k²=2.56k^42.56k²*(-3.2k )=-8.192k³2.56k²*4=10.24k²-3.2k*k²=-3.2k³-3.2k*(-3.2k )=10.24k²-3.2k*4= -12.8k1*k²=k²1*(-3.2k )=-3.2k1*4=4Combine terms:2.56k^4-8.192k³ -3.2k³= -11.392k³10.24k² +10.24k² +k²=21.48k²-12.8k -3.2k= -16k+4Then multiply by 4k²:4k²*(2.56k^4 -11.392k³ +21.48k² -16k +4 )=10.24k^6 -45.568k^5 +85.92k^4 -64k³ +16k²Thus, Right side expanded:10.24k^6 -45.568k^5 +85.92k^4 -64k³ +16k²Set Left side = Right side:16k^6 -64k^5 +85.92k^4 -45.568k³ +10.24k² =10.24k^6 -45.568k^5 +85.92k^4 -64k³ +16k²Subtract Right side from Left side:(16k^6 -10.24k^6 ) + (-64k^5 +45.568k^5 ) + (85.92k^4 -85.92k^4 ) + (-45.568k³ +64k³ ) + (10.24k² -16k² ) =0Compute each term:5.76k^6 -18.432k^5 +0 +18.432k³ -5.76k²=0Factor:5.76k^6 -18.432k^5 +18.432k³ -5.76k²=0Factor common terms:5.76k²(k^4 -3.2k³ +3.2k -1)=0Since k is a ratio of lengths, k ≠0, so:k^4 -3.2k³ +3.2k -1=0This quartic equation: k^4 -3.2k³ +3.2k -1=0Let’s try to factor this.Perhaps rational roots. Possible rational roots are ±1.Test k=1:1 -3.2 +3.2 -1=0. So k=1 is a root.Thus, factor (k -1). Use polynomial division.Divide k^4 -3.2k³ +0k² +3.2k -1 by (k -1):Using synthetic division:Coefficients:1 | -3.2 |0 |3.2 |-1Write root k=1:1 | -3.2 |0 |3.2 |-1Bring down 1Multiply by1:1Add to next term: -3.2 +1= -2.2Multiply by1: -2.2Add to next term:0 +(-2.2)= -2.2Multiply by1: -2.2Add to next term:3.2 +(-2.2)=1Multiply by1:1Add to last term: -1 +1=0Thus, the quotient polynomial is k³ -2.2k² -2.2k +1So, factor as (k -1)(k³ -2.2k² -2.2k +1 )=0Now, solve k³ -2.2k² -2.2k +1=0Try k=1 again:1 -2.2 -2.2 +1= -2.4 ≠0k=0.5:0.125 -0.55 -1.1 +1= -0.525 ≠0k=2:8 -8.8 -4.4 +1= -4.2 ≠0k= -1:-1 -2.2 +2.2 +1=0. So k=-1 is a root.Divide polynomial k³ -2.2k² -2.2k +1 by (k +1):Using synthetic division:Coefficients:1 | -2.2 | -2.2 |1Root k=-1:Bring down 1Multiply by -1: -1Add to next term: -2.2 + (-1)= -3.2Multiply by -1:3.2Add to next term: -2.2 +3.2=1Multiply by -1: -1Add to last term:1 + (-1)=0Thus quotient polynomial: k² -3.2k +1Thus, original polynomial factors as (k -1)(k +1)(k² -3.2k +1)=0Thus roots are k=1, k=-1, and roots of k² -3.2k +1=0.But since k=x/y and x,y are lengths, k>0. So possible roots are k=1 and roots of k² -3.2k +1=0.Solve k² -3.2k +1=0:k=(3.2±sqrt(10.24 -4))/2=(3.2±sqrt(6.24))/2≈(3.2±2.496)/2Thus,k≈(3.2+2.496)/2≈5.696/2≈2.848k≈(3.2-2.496)/2≈0.704/2≈0.352So possible k=1, 2.848, 0.352.Now, test k=1:If k=1, then x=y. Let's check if this satisfies the original equation.From equation (5):x² -3.2xy +4y²=4With x=y:y² -3.2y² +4y²= (1 -3.2 +4)y²=1.8y²=4 => y²=4/1.8≈2.222 => y≈1.491Then x=y≈1.491From equation (2):z² =x² + y² -1.6xy≈2.222 +2.222 -1.6*(1.491)^2≈4.444 -1.6*2.224≈4.444 -3.558≈0.886Thus z≈0.941Now check equation (1):2z² +2y² -x²≈2*0.886 +2*2.222 -2.222≈1.772 +4.444 -2.222≈4.0, which holds.Now check angle equality.From equation (8):cosθ=(x²/4 +1 -z²)/x≈(2.222/4 +1 -0.886)/1.491≈(0.5555 +1 -0.886)/1.491≈0.6695/1.491≈0.449From equation (11):cosθ=(1.6x - y)/sqrt(4x² + y² -3.2xy )With x=y=1.491:1.6x - y=1.6*1.491 -1.491≈2.3856 -1.491≈0.8946Denominator= sqrt(4*(2.224) +2.224 -3.2*(2.224))≈sqrt(8.896 +2.224 -7.117)≈sqrt(4.003)≈2.0Thus, cosθ≈0.8946/2.0≈0.4473Which is close to 0.449, considering rounding errors. Thus, k=1 is a valid solution.Thus, k=1, so x=y. Therefore, triangle ABC is isoceles with AC=BC=x=y. But wait, AC= x, BC=y, so if x=y, then AC=BC, so triangle ABC is isoceles with AB as base.Wait, but angle at C has cosine 0.8, so angle C is arccos(0.8)≈36.87 degrees. If AC=BC, then it's isoceles with apex at C, so angles at A and B are equal. But angle C is given. However, in this case, since AC=BC, angles at A and B are equal.But regardless, we can compute the area.Given that AC=BC=x=y≈1.491, AB=z≈0.941But wait, let's compute exactly.From k=1, x=y. From equation (6):y²=4/(k² -3.2k +4 )=4/(1 -3.2 +4 )=4/(1.8)=20/9≈2.222, so y=2/sqrt(1.8)=2*sqrt(10)/3≈2.108Wait, actually, exact calculation:k=1, so x=y. From equation (5):x² -3.2x² +4x²=4 => (1 -3.2 +4 )x²=1.8x²=4 =>x²=4/1.8=20/9 =>x=2*sqrt(5)/3≈2.108Similarly, z²=x² + y² -1.6xy= x² +x² -1.6x²= (2 -1.6)x²=0.4x²=0.4*(20/9)=8/9 =>z=2*sqrt(2)/3≈0.9428Thus, sides:AC=BC=2*sqrt(5)/3AB=2*sqrt(2)/3Angle at C: arccos(0.8)But wait, let's check using Law of Cosines:cos(angle C)= (AC² + BC² - AB²)/(2*AC*BC)AC=BC=2√5/3, AB=2√2/3AC²= (4*5)/9=20/9AB²=(4*2)/9=8/9Thus,cos(angle C)= (20/9 +20/9 -8/9)/(2*(20/9))= (32/9)/(40/9)=32/40=0.8, which matches.Therefore, this is a valid solution.Now, the area of triangle ABC can be computed using the formula:Area = (1/2)*AC*BC*sin(angle C)Since AC=BC=2√5/3, angle C has cos=0.8, so sin=sqrt(1 -0.64)=sqrt(0.36)=0.6Thus,Area=(1/2)*(2√5/3)*(2√5/3)*0.6= (1/2)*(4*5/9)*0.6= (1/2)*(20/9)*0.6= (10/9)*0.6= 6/9=2/3≈0.666...But wait, but we have another solution. We also have k≈2.848 and k≈0.352. Let's check if these also satisfy the angle equality.First, let's check k=2.848.k≈2.848, so x≈2.848y.From equation (6):y²=4/(k² -3.2k +4 )Calculate denominator:k² -3.2k +4≈(2.848)^2 -3.2*2.848 +4≈8.113 -9.114 +4≈2.999≈3Thus, y²≈4/3≈1.333, y≈1.154x≈2.848*1.154≈3.287From equation (2):z²=x² + y² -1.6xy≈(3.287)^2 + (1.154)^2 -1.6*3.287*1.154≈10.80 +1.33 -1.6*3.80≈12.13 -6.08≈6.05z≈2.46From equation (1):2z² +2y² -x²≈2*6.05 +2*1.33 -10.80≈12.10 +2.66 -10.80≈3.96≈4, which holds.Now, compute cosθ from equation (8):cosθ=(x²/4 +1 -z²)/x≈(10.80/4 +1 -6.05)/3.287≈(2.7 +1 -6.05)/3.287≈(-2.35)/3.287≈-0.715From equation (11):cosθ=(1.6x - y)/sqrt(4x² + y² -3.2xy )1.6x=1.6*3.287≈5.2595.259 -1.154≈4.105Denominator= sqrt(4*(3.287)^2 + (1.154)^2 -3.2*3.287*1.154 )≈sqrt(4*10.80 +1.33 -3.2*3.80 )≈sqrt(43.2 +1.33 -12.16 )≈sqrt(32.37 )≈5.69Thus, cosθ≈4.105/5.69≈0.721But from equation (8) it's ≈-0.715, which is different. So this solution doesn't satisfy the angle equality. Hence, discard.Similarly, check k≈0.352:x≈0.352yFrom equation (6):y²=4/(k² -3.2k +4 )=4/(0.352² -3.2*0.352 +4 )≈4/(0.1239 -1.126 +4 )≈4/3.0≈1.333, so y≈1.154x≈0.352*1.154≈0.407From equation (2):z²= x² + y² -1.6xy≈0.166 +1.333 -1.6*0.407*1.154≈1.499 -1.6*0.470≈1.499 -0.752≈0.747z≈0.865From equation (1):2z² +2y² -x²≈2*0.747 +2*1.333 -0.166≈1.494 +2.666 -0.166≈4.0, which holds.Compute cosθ from equation (8):cosθ=(x²/4 +1 -z²)/x≈(0.166/4 +1 -0.747)/0.407≈(0.0415 +1 -0.747)/0.407≈0.2945/0.407≈0.723From equation (11):cosθ=(1.6x - y)/sqrt(4x² + y² -3.2xy )1.6x=1.6*0.407≈0.6510.651 -1.154≈-0.503Denominator= sqrt(4x² + y² -3.2xy )≈sqrt(4*0.166 +1.333 -3.2*0.407*1.154 )≈sqrt(0.664 +1.333 -1.510 )≈sqrt(0.487 )≈0.698Thus, cosθ≈-0.503/0.698≈-0.721Which doesn't match equation (8). Thus, invalid.Thus, only valid solution is k=1, leading to isoceles triangle with AC=BC=2√5/3, AB=2√2/3, and area=2/3.But wait, according to our earlier calculation, the area was 2/3. But the problem states that BP=1. But in this solution, BP is the median from B to AC. Let's verify BP's length.BP is the median from B to AC. The formula for the median is:BP² = (2AB² + 2BC² - AC²)/4From our solution:AB=2√2/3BC=2√5/3AC=2√5/3Thus,BP²=(2*(8/9) +2*(20/9) -20/9)/4=(16/9 +40/9 -20/9)/4=(36/9)/4=4/4=1Thus, BP=1, which matches the problem's condition.Therefore, this is the correct solution.Thus, the area of triangle ABC is 2/3.But wait, wait. Let me confirm:Area=(1/2)*AC*BC*sin(angle C)AC=BC=2√5/3, sin(angle C)=0.6Thus,Area=0.5*(2√5/3)*(2√5/3)*0.6=0.5*(4*5/9)*0.6=0.5*(20/9)*0.6= (10/9)*0.6=6/9=2/3Yes, correct.Thus, the area is 2/3.Final Answerboxed{dfrac{2}{3}}</think>Given triangle (ABC) with medians (AM) and (BP), where (angle APB = angle BMA), (cos angle ACB = 0.8), and (BP = 1). We need to find the area of triangle (ABC).1. Key Properties and Setup: - Medians (AM) and (BP) intersect at the centroid (G), dividing each median into a 2:1 ratio. - Using coordinates and vector analysis, we set up the problem with (C) at the origin, and midpoints (M) and (P) on (BC) and (AC) respectively. - Using the Law of Cosines and properties of medians, we derive equations involving the sides of the triangle.2. Equations Derived: - Using the median length formula for (BP): [ BP^2 = frac{2AB^2 + 2BC^2 - AC^2}{4} = 1 ] - Using the Law of Cosines for (angle ACB): [ cos angle ACB = frac{AC^2 + BC^2 - AB^2}{2 cdot AC cdot BC} = 0.8 ]3. Solving for Sides: - By setting up ratios and solving quadratic equations, we find that (AC = BC = frac{2sqrt{5}}{3}) and (AB = frac{2sqrt{2}}{3}).4. Area Calculation: - Using the formula for the area of a triangle with two sides and the included angle: [ text{Area} = frac{1}{2} cdot AC cdot BC cdot sin angle ACB ] - Given (cos angle ACB = 0.8), (sin angle ACB = sqrt{1 - 0.8^2} = 0.6): [ text{Area} = frac{1}{2} cdot left(frac{2sqrt{5}}{3}right) cdot left(frac{2sqrt{5}}{3}right) cdot 0.6 = frac{1}{2} cdot frac{4 cdot 5}{9} cdot 0.6 = frac{1}{2} cdot frac{20}{9} cdot 0.6 = frac{10}{9} cdot 0.6 = frac{2}{3} ]Thus, the area of triangle (ABC) is (boxed{dfrac{2}{3}}).,<think>Alright, let's tackle this geometry problem step by step. I need to find the area of triangle ABC given some conditions about medians, angles, and the length of one median. Let me jot down what's given:1. In triangle ABC, medians AM and BP are drawn.2. Angle APB is equal to angle BMA.3. The cosine of angle ACB is 0.8, so cos(∠ACB) = 0.8.4. The length of median BP is 1.And the goal is to find the area of triangle ABC.First, let me recall some properties of medians in a triangle. A median connects a vertex to the midpoint of the opposite side. So, median AM connects vertex A to the midpoint M of side BC, and median BP connects vertex B to the midpoint P of side AC.Given that BP = 1, which is one of the medians. Also, the angles ∠APB and ∠BMA are equal. Hmm, that seems important. Let me visualize the triangle and the medians.Let me try to sketch this mentally. Triangle ABC, with medians AM and BP. Point M is the midpoint of BC, and point P is the midpoint of AC. So, AP = PC since P is the midpoint, and BM = MC since M is the midpoint.Now, angles ∠APB and ∠BMA are equal. Let me denote these angles as θ for simplicity. So, ∠APB = ∠BMA = θ.Given that cos(∠ACB) = 0.8, which is angle C. So, angle at C has cosine 0.8, meaning it's an acute angle (since cosine is positive) and we can find its sine using sin²θ + cos²θ = 1. So, sin(angle C) would be sqrt(1 - 0.64) = sqrt(0.36) = 0.6. Therefore, sin(∠ACB) = 0.6.Since we need to find the area of triangle ABC, and area can be calculated using (1/2)*ab*sinθ, where a and b are two sides and θ is the included angle. If we can find two sides and the included angle, we can compute the area. But here, we have information about medians and angles between them. Hmm. Maybe we need to relate the medians to the sides of the triangle.I remember that the length of a median can be calculated using the formula:m_a = (1/2) * sqrt(2b² + 2c² - a²)where m_a is the median from vertex A to side a (opposite to A). Similarly for other medians.But here, we have BP = 1. Let's denote sides as follows: Let's say in triangle ABC, side opposite to A is BC = a, side opposite to B is AC = b, and side opposite to C is AB = c.Therefore, the median from B to AC (which is midpoint P) would be:BP = (1/2) * sqrt(2a² + 2c² - b²) = 1.Similarly, the median from A to BC (midpoint M):AM = (1/2) * sqrt(2b² + 2c² - a²).But I don't know if we need the lengths of both medians. However, we also have the condition that ∠APB = ∠BMA.This seems like a key condition. Let me try to analyze triangles APB and BMA.Wait, but angles ∠APB and ∠BMA are equal. So, θ is present in both triangles APB and BMA. Maybe these triangles are similar? If two angles are equal, perhaps there's some similarity or congruence.Alternatively, perhaps we can apply the Law of Sines or Cosines in triangles APB and BMA.Let me consider triangle APB first. Points A, P, B. P is the midpoint of AC, so AP = PC = b/2. Similarly, in triangle BMA, points B, M, A. M is the midpoint of BC, so BM = MC = a/2.In triangle APB, sides are AP = b/2, BP = 1, and AB = c. The angle at P is θ.In triangle BMA, sides are BM = a/2, AM (unknown length), and AB = c. The angle at M is θ.Wait, but angles at P and M are equal. Hmm. Maybe we can set up equations using the Law of Sines in both triangles.In triangle APB:Applying the Law of Sines:AP / sin(∠ABP) = BP / sin(∠BAP) = AB / sin(θ)But I need to figure out the angles here. Maybe this is getting complicated. Let me label all the sides and angles properly.Alternatively, perhaps coordinate geometry can help here. Let me try to place the triangle in a coordinate system to simplify calculations.Let me place point C at the origin (0,0) for simplicity. Then, let me denote coordinates for points A, B, and C.Since cos(angle ACB) = 0.8, angle at C is between sides AC and BC. Let me denote coordinates such that point C is at (0,0), point A is along the x-axis at (b, 0), and point B is somewhere in the plane. Wait, but angle at C is between AC and BC. If I place C at (0,0), then AC is from C(0,0) to A(b,0), and BC is from C(0,0) to B(d,e). Then, the angle at C can be calculated using the dot product.But maybe this is a good approach. Let me proceed.Let’s set:- Point C at (0, 0).- Point A at (b, 0), since AC is a side. So AC has length b.- Point B at (d, e).Then, the angle at C (angle ACB) is the angle between vectors CA and CB.Vector CA is A - C = (b, 0) - (0,0) = (b, 0).Vector CB is B - C = (d, e) - (0,0) = (d, e).The cosine of the angle between them is (CA · CB) / (|CA| |CB|) = (b*d + 0*e) / (b * sqrt(d² + e²)) ) = (b d) / (b sqrt(d² + e²)) ) = d / sqrt(d² + e²).Given that cos(angle ACB) = 0.8, so:d / sqrt(d² + e²) = 0.8Squaring both sides:d² / (d² + e²) = 0.64=> d² = 0.64(d² + e²)=> d² = 0.64d² + 0.64e²=> d² - 0.64d² = 0.64e²=> 0.36d² = 0.64e²=> (d/e)² = 0.64 / 0.36 = 16/9=> d/e = 4/3 or -4/3Assuming the triangle is oriented such that d is positive (since angle at C is between AC and BC, and we placed A at (b,0)), so probably d is positive. So, d = (4/3)e.Therefore, coordinates of B can be written as ((4/3)e, e). Let me set e = 3k for some k, so d = 4k. Therefore, B is at (4k, 3k). Then, sqrt(d² + e²) = sqrt(16k² + 9k²) = 5k. So, length CB is 5k. Then, since cos(angle C) is 0.8, which checks out because d / 5k = 4k /5k = 0.8.So, coordinates:- C: (0,0)- A: (b, 0)- B: (4k, 3k)Now, medians AM and BP are drawn.Median AM connects A to midpoint M of BC.Coordinates of M: midpoint of BC: ((4k + 0)/2, (3k + 0)/2) = (2k, 1.5k)Median BP connects B to midpoint P of AC.Coordinates of P: midpoint of AC: ((b + 0)/2, (0 + 0)/2) = (b/2, 0)So, points:- M: (2k, 1.5k)- P: (b/2, 0)We need to find the angles ∠APB and ∠BMA.First, let's find ∠APB. This is the angle at point P between points A, P, B.Points:- A: (b,0)- P: (b/2,0)- B: (4k,3k)So, vectors PA and PB.Vector PA = A - P = (b - b/2, 0 - 0) = (b/2, 0)Vector PB = B - P = (4k - b/2, 3k - 0) = (4k - b/2, 3k)The angle at P between PA and PB is ∠APB.Similarly, ∠BMA is the angle at M between points B, M, A.Points:- B: (4k,3k)- M: (2k,1.5k)- A: (b,0)Vectors MB and MA.Vector MB = B - M = (4k - 2k, 3k - 1.5k) = (2k, 1.5k)Vector MA = A - M = (b - 2k, 0 - 1.5k) = (b - 2k, -1.5k)The angle at M between MB and MA is ∠BMA.Given that these two angles are equal, ∠APB = ∠BMA.So, the angle between vectors PA and PB at point P is equal to the angle between vectors MB and MA at point M.To find these angles, we can use the dot product formula. The cosine of the angle between two vectors u and v is (u · v) / (|u| |v|).Let's compute cos(∠APB):Vectors PA = (b/2, 0), PB = (4k - b/2, 3k)Dot product PA · PB = (b/2)(4k - b/2) + 0*3k = (b/2)(4k - b/2)|PA| = sqrt( (b/2)^2 + 0^2 ) = b/2|PB| = sqrt( (4k - b/2)^2 + (3k)^2 )Similarly, cos(∠BMA):Vectors MB = (2k, 1.5k), MA = (b - 2k, -1.5k)Dot product MB · MA = (2k)(b - 2k) + (1.5k)(-1.5k) = 2k(b - 2k) - 2.25k²|MB| = sqrt( (2k)^2 + (1.5k)^2 ) = sqrt(4k² + 2.25k²) = sqrt(6.25k²) = 2.5k|MA| = sqrt( (b - 2k)^2 + (-1.5k)^2 ) = sqrt( (b - 2k)^2 + 2.25k² )Given that ∠APB = ∠BMA, so their cosines are equal:[ (b/2)(4k - b/2) ] / [ (b/2) * |PB| ] = [ 2k(b - 2k) - 2.25k² ] / [ 2.5k * |MA| ]Wait, let's write the equations step by step.First, compute cos(∠APB):cos(θ) = [PA · PB] / (|PA| |PB|) = [ (b/2)(4k - b/2) ] / [ (b/2) * sqrt( (4k - b/2)^2 + (3k)^2 ) ]Simplify numerator and denominator:Numerator: (b/2)(4k - b/2)Denominator: (b/2) * sqrt( (4k - b/2)^2 + 9k² )Cancel out (b/2):cos(θ) = (4k - b/2) / sqrt( (4k - b/2)^2 + 9k² )Similarly, compute cos(∠BMA):cos(θ) = [MB · MA] / (|MB| |MA| )= [2k(b - 2k) - 2.25k²] / [2.5k * sqrt( (b - 2k)^2 + 2.25k² ) ]So, set these two expressions equal:(4k - b/2) / sqrt( (4k - b/2)^2 + 9k² ) = [2k(b - 2k) - 2.25k²] / [2.5k * sqrt( (b - 2k)^2 + 2.25k² ) ]This looks complicated, but maybe cross-multiplying and simplifying can help.First, let's compute the numerator of the second fraction:2k(b - 2k) - 2.25k² = 2kb - 4k² - 2.25k² = 2kb - 6.25k²Denominator of the second fraction: 2.5k * sqrt( (b - 2k)^2 + 2.25k² )So, putting it all together:(4k - b/2) / sqrt( (4k - b/2)^2 + 9k² ) = (2kb - 6.25k²) / [2.5k * sqrt( (b - 2k)^2 + 2.25k² ) ]Multiply both sides by sqrt( (4k - b/2)^2 + 9k² ) and by 2.5k * sqrt( (b - 2k)^2 + 2.25k² ) to eliminate denominators:(4k - b/2) * 2.5k * sqrt( (b - 2k)^2 + 2.25k² ) = (2kb - 6.25k²) * sqrt( (4k - b/2)^2 + 9k² )This is a very complicated equation. Maybe there's a smarter way or some substitution to simplify.Alternatively, since we have BP = 1. Let's compute BP in terms of coordinates.Point B is at (4k, 3k), point P is at (b/2, 0). So, BP is the distance between these two points:BP = sqrt( (4k - b/2)^2 + (3k - 0)^2 ) = sqrt( (4k - b/2)^2 + 9k² ) = 1Therefore, we have:sqrt( (4k - b/2)^2 + 9k² ) = 1Squaring both sides:(4k - b/2)^2 + 9k² = 1Let me note this as equation (1):(4k - b/2)^2 + 9k² = 1.So that's one equation relating k and b.Now, maybe we can find another equation from the equality of the angles. Let's see.Earlier, we had:cos(θ) from ∠APB:(4k - b/2) / sqrt( (4k - b/2)^2 + 9k² ) = (4k - b/2) / 1, since the denominator is sqrt(...) = 1 (from BP = 1). Wait, that's right! Because we just found that sqrt( (4k - b/2)^2 + 9k² ) = 1. Therefore, the denominator in cos(θ) for ∠APB is 1. So, cos(θ) = 4k - b/2.Wait, that's a key insight. Let me re-express cos(∠APB):From above:cos(θ) = (4k - b/2) / 1 = 4k - b/2.Similarly, for ∠BMA, let's compute cos(θ):We have:cos(θ) = [2kb - 6.25k²] / [2.5k * sqrt( (b - 2k)^2 + 2.25k² ) ]But we need to compute sqrt( (b - 2k)^2 + 2.25k² ), which is |MA|. Let's see if we can express |MA| in terms of known variables.Wait, AM is the other median. If I can express |MA| in terms of the sides, maybe using the median formula. Wait, median AM in triangle ABC:AM = (1/2) * sqrt(2b² + 2c² - a²)But here, sides:In triangle ABC:- a = BC = 5k (since point B is at (4k,3k), point C at (0,0), so distance BC = sqrt( (4k)^2 + (3k)^2 ) = 5k)- b = AC = distance from A(b,0) to C(0,0) is b.- c = AB = distance from A(b,0) to B(4k,3k): sqrt( (4k - b)^2 + (3k)^2 )Therefore, median AM is (1/2)*sqrt(2b² + 2c² - a²) = (1/2)*sqrt(2b² + 2[(4k - b)^2 + 9k²] - 25k² )Simplify inside the sqrt:2b² + 2[(4k - b)^2 + 9k²] - 25k²= 2b² + 2(16k² - 8kb + b² + 9k²) -25k²= 2b² + 2(25k² -8kb + b²) -25k²= 2b² + 50k² -16kb + 2b² -25k²= 4b² + 25k² -16kbTherefore, AM = (1/2)*sqrt(4b² +25k² -16kb) = (1/2)*sqrt(4b² -16kb +25k² )But we can also compute |MA| from coordinates. Since MA is from M(2k,1.5k) to A(b,0). So:|MA| = sqrt( (b - 2k)^2 + (0 - 1.5k)^2 ) = sqrt( (b - 2k)^2 + 2.25k² )So, equate the two expressions for |MA|:sqrt( (b - 2k)^2 + 2.25k² ) = (1/2)*sqrt(4b² -16kb +25k² )Square both sides:( (b - 2k)^2 + 2.25k² ) = (1/4)(4b² -16kb +25k² )Left side:b² -4kb +4k² +2.25k² = b² -4kb +6.25k²Right side:(1/4)(4b² -16kb +25k² ) = b² -4kb +6.25k²So, both sides are equal. Therefore, this doesn't give new information. So, perhaps we need another equation from the angle equality.Earlier, we had:cos(θ) from ∠APB is 4k - b/2.And cos(θ) from ∠BMA is [2kb - 6.25k²] / [2.5k * |MA| ]But since |MA| = sqrt( (b - 2k)^2 + 2.25k² ), and we already know that cos(theta) is equal for both angles.Thus:4k - b/2 = [2kb - 6.25k²] / [2.5k * sqrt( (b - 2k)^2 + 2.25k² ) ]But this still looks complicated. Let me see if we can express sqrt( (b - 2k)^2 + 2.25k² ) in terms of known quantities or relate it via another equation.Wait, from BP = 1, we have equation (1):(4k - b/2)^2 + 9k² = 1.Let me expand this:(16k² -4kb + b²/4) +9k² =1Combine like terms:16k² +9k² -4kb + b²/4 =125k² -4kb + b²/4 =1Multiply both sides by 4 to eliminate the fraction:100k² -16kb + b² =4So, we have:b² -16kb +100k² =4 ... (1a)That's equation (1a). Now, let's look back at the other equation from the angle equality.We have:4k - (b/2) = [2kb -6.25k²] / [2.5k * sqrt( (b -2k)^2 +2.25k² ) ]Let me compute the denominator first:2.5k * sqrt( (b -2k)^2 +2.25k² ) = 2.5k * |MA|But we can write |MA| as sqrt( (b -2k)^2 + 2.25k² )Alternatively, from the median formula earlier, |MA| is (1/2)*sqrt(4b² -16kb +25k² )But let me compute the denominator:2.5k * |MA| = 2.5k * sqrt( (b -2k)^2 +2.25k² )= 2.5k * sqrt( b² -4kb +4k² +2.25k² )= 2.5k * sqrt( b² -4kb +6.25k² )Which is similar to equation (1a). From equation (1a):b² -16kb +100k² =4But here we have b² -4kb +6.25k². Let's see if we can relate this.Wait, 6.25k² is (2.5k)^2. Hmm, not sure.Alternatively, let's denote D = sqrt(b² -4kb +6.25k²). Then the denominator is 2.5k * D.So, equation from angle equality:4k - (b/2) = [2kb -6.25k²] / (2.5k * D )Multiply numerator and denominator:Left side: 4k - b/2Right side: (2kb -6.25k²)/(2.5k D )Let me compute 2kb -6.25k² = k(2b -6.25k)Denominator: 2.5k DSo, right side becomes [k(2b -6.25k)] / [2.5k D] = (2b -6.25k)/(2.5 D )Therefore, equation becomes:4k - (b/2) = (2b -6.25k)/(2.5 D )Multiply both sides by 2.5 D:(4k - b/2)*2.5 D = 2b -6.25kLet me compute 4k - b/2:Let me denote x = 4k - b/2. Then, from equation (1a):From equation (1a):b² -16kb +100k² =4But I don't know if this helps. Alternatively, notice that D = sqrt(b² -4kb +6.25k²). Let's compute D²:D² = b² -4kb +6.25k²But from equation (1a):b² -16kb +100k² =4So, perhaps express b² from equation (1a):b² =4 +16kb -100k²Plug into D²:D² = (4 +16kb -100k²) -4kb +6.25k²=4 +12kb -93.75k²Therefore, D = sqrt(4 +12kb -93.75k²)Hmm, not sure if helpful.Alternatively, perhaps express variables in terms of k or b. Let's let’s say we set variables such that k = 1 or something. Wait, but we need to solve for k and b.Alternatively, maybe introduce variables substitution. Let me set t = b/k. Then, b = t k. Let me substitute this into the equations.Set b = t k. Then, equation (1a):b² -16kb +100k² =4=> (t k)^2 -16k*(t k) +100k² =4=> t² k² -16 t k² +100k² =4Factor out k²:k²(t² -16 t +100) =4Similarly, the equation from angle equality:Let me rewrite 4k - b/2 = [2kb -6.25k²]/(2.5k D )Substitute b = t k:Left side:4k - (t k)/2 = k(4 - t/2)Right side numerator:2k*(t k) -6.25k² = 2t k² -6.25k² =k²(2t -6.25)Denominator:2.5k * D, where D = sqrt( (b -2k)^2 +2.25k² ) = sqrt( (t k -2k)^2 +2.25k² ) =k sqrt( (t -2)^2 +2.25 )Therefore, denominator:2.5k * k sqrt( (t -2)^2 +2.25 ) =2.5k² sqrt( (t -2)^2 +2.25 )Thus, right side becomes: [k²(2t -6.25)] / [2.5k² sqrt( (t -2)^2 +2.25 ) ] = (2t -6.25)/(2.5 sqrt( (t -2)^2 +2.25 ) )So, equation becomes:k(4 - t/2) = (2t -6.25)/(2.5 sqrt( (t -2)^2 +2.25 ) )But from equation (1a):k²(t² -16 t +100) =4 => k² =4 / (t² -16 t +100)Thus, k = 2 / sqrt(t² -16t +100)Substitute into left side:k(4 - t/2) = [2 / sqrt(t² -16t +100)] * (4 - t/2 )So, left side: [2(4 - t/2)] / sqrt(t² -16t +100) = (8 - t) / sqrt(t² -16t +100)Right side: (2t -6.25)/(2.5 sqrt( (t -2)^2 +2.25 ) )Simplify right side denominator:sqrt( (t -2)^2 +2.25 ) = sqrt(t² -4t +4 + 2.25) = sqrt(t² -4t +6.25) = sqrt( (t -2)^2 + (sqrt(2.25))^2 )Wait, sqrt(6.25) = 2.5, so sqrt(t² -4t +6.25) is sqrt( (t - 2)^2 + (1.5)^2 ). Not sure if helpful.But let's compute denominator sqrt(t² -4t +6.25) = sqrt( (t - 2)^2 + 2.25 ). Anyway, proceed.So, right side is (2t -6.25)/(2.5 * sqrt(t² -4t +6.25) )Note that 2t -6.25 = 2(t - 3.125). Hmm, but maybe write 6.25 as 25/4, so 2t -25/4.Alternatively, let's factor numerator and denominator:Numerator: 2t -6.25 = 2t -25/4 = (8t -25)/4Denominator:2.5 * sqrt(t² -4t +6.25) = (5/2) * sqrt(t² -4t +25/4 )Notice that t² -4t +25/4 = (t - 2)^2 + (sqrt(25/4 - 4)) but wait, 25/4 -4 = 25/4 -16/4 =9/4. So, t² -4t +25/4 = (t -2)^2 + (3/2)^2. Not sure if helpful.But the denominator becomes (5/2)*sqrt( (t -2)^2 + (3/2)^2 )Thus, right side becomes ( (8t -25)/4 ) / ( (5/2) * sqrt( (t -2)^2 + (3/2)^2 ) ) = (8t -25)/4 * 2/(5 sqrt(...)) ) = (8t -25)/(10 sqrt(...)).Therefore, equation is:(8 - t)/sqrt(t² -16t +100) = (8t -25)/(10 sqrt(t² -4t +6.25))Cross-multiplying:(8 - t) * 10 sqrt(t² -4t +6.25) = (8t -25) * sqrt(t² -16t +100)Square both sides to eliminate the square roots:[ (8 - t)^2 * 100 (t² -4t +6.25) ] = [ (8t -25)^2 (t² -16t +100) ]This seems very complicated, but let's proceed step by step.Left side:(8 - t)^2 * 100 * (t² -4t +6.25)Right side:(8t -25)^2 * (t² -16t +100)First, compute (8 - t)^2:=64 -16t + t²Multiply by 100:100*(64 -16t + t²) = 6400 -1600t +100t²Multiply by (t² -4t +6.25):So left side:(6400 -1600t +100t²)(t² -4t +6.25)Similarly, right side:(8t -25)^2 = 64t² -400t +625Multiply by (t² -16t +100):(64t² -400t +625)(t² -16t +100)This expansion will be tedious. Let me see if there is a possible value of t that satisfies the equation. Maybe t is a rational number. Let's try to test some integer values.Let me first check if t is an integer. Let's try t =5.Check equation (1a):t² -16t +100 =25 -80 +100=45. Then k²=4/45, so k=2/3√5≈0.298.But substituting t=5 into the angle equality equation.Left side: (8 -5)/sqrt(25 -80 +100)=3/sqrt(45)=3/(3√5)=1/√5≈0.447.Right side: (8*5 -25)/(10*sqrt(25 -20 +6.25))= (40-25)/(10*sqrt(11.25))=15/(10*3.354)=≈15/33.54≈0.447. So, approximately equal. Therefore, t=5 is a solution.Wow, that works. So, t=5.Therefore, with t=5:From equation (1a):k²=4/(t² -16t +100)=4/(25 -80 +100)=4/45. Thus, k=2/(3√5)=2√5/15.Therefore, k=2√5/15.Then, b =t k=5*(2√5/15)= (10√5)/15= (2√5)/3.So, b=2√5/3.Now, we can find the coordinates:Point A: (b,0)=(2√5/3,0)Point B: (4k,3k)=4*(2√5/15), 3*(2√5/15)= (8√5/15, 6√5/15)= (8√5/15, 2√5/5)Point C: (0,0)Now, we need to find the area of triangle ABC.Since we have coordinates for A, B, and C, we can use the shoelace formula.Coordinates:A: (2√5/3, 0)B: (8√5/15, 2√5/5)C: (0,0)The area is (1/2)| (Ax(By - Cy) + Bx(Cy - Ay) + Cx(Ay - By) ) |Plugging in:Ax = 2√5/3, Ay=0Bx=8√5/15, By=2√5/5Cx=0, Cy=0Compute:Area = (1/2)| (2√5/3)(2√5/5 - 0) + (8√5/15)(0 - 0) + 0(0 - 2√5/5) |Simplify term by term:First term: (2√5/3)(2√5/5) = (4*5)/15 = 20/15 = 4/3Second term: (8√5/15)(0 -0)=0Third term:0So, total area = (1/2)|4/3 +0 +0| = (1/2)(4/3) = 2/3.But wait, that can't be right. Wait, but the problem states that BP=1. Let me verify the coordinates:BP is the distance from B(8√5/15, 2√5/5) to P, the midpoint of AC.Midpoint P of AC is ((2√5/3 +0)/2, (0 +0)/2)= (√5/3,0)So, BP distance is sqrt( (8√5/15 -√5/3)^2 + (2√5/5 -0)^2 )Compute the differences:x-coordinate:8√5/15 -√5/3 =8√5/15 -5√5/15=3√5/15=√5/5y-coordinate:2√5/5 -0=2√5/5Thus, BP distance is sqrt( (√5/5)^2 + (2√5/5)^2 )=sqrt(5/25 + 20/25)=sqrt(25/25)=1. Okay, good. So BP=1 as required.But the area computed via shoelace is 2/3. But let's check another way. The area can also be computed using sides and angles.We know angle at C has cosine 0.8 and sine 0.6. So, area is (1/2)*AC*BC*sin(angle C)AC = b =2√5/3BC =5k=5*(2√5/15)=10√5/15=2√5/3Thus, AC=2√5/3, BC=2√5/3Wait, both AC and BC are equal? So triangle ABC is isoceles with AC=BC=2√5/3, and angle at C with sin(theta)=0.6.Thus, area= (1/2)*(2√5/3)*(2√5/3)*0.6Compute:(1/2)*(4*5/9)*0.6 = (1/2)*(20/9)*0.6 = (10/9)*0.6 = 6/9=2/3.Yes, same result. So the area is 2/3. Therefore, the answer is 2/3.But the problem might expect the answer in a different form or maybe I made a mistake. Wait, 2/3 is 0.666..., but in the problem, they might expect a fractional form or something else. Wait, 2/3 is already a fraction.Alternatively, perhaps they want it as (2/3). Let me check calculations again.But wait, all steps led to area=2/3. Let me check via another method. Let's compute vectors AB and AC and take cross product.Vector AB = B - A = (8√5/15 -2√5/3, 2√5/5 -0)= (8√5/15 -10√5/15, 2√5/5)= (-2√5/15, 2√5/5)Vector AC = C - A = (-2√5/3, 0)Area is (1/2)|AB × AC|Compute cross product:(-2√5/15)(0) - (2√5/5)(-2√5/3) =0 + (4*5)/15=20/15=4/3Thus, area=(1/2)*(4/3)=2/3. Same result.Thus, the area of triangle ABC is 2/3.Final AnswerThe area of triangle ( ABC ) is (boxed{dfrac{2}{3}}).</think>Given triangle (ABC) with medians (AM) and (BP), where (angle APB = angle BMA), (cos angle ACB = 0.8), and (BP = 1), we need to find the area of triangle (ABC).1. Key Properties and Setup: - Medians (AM) and (BP) intersect at their midpoints. - (angle APB = angle BMA = theta). - (cos angle ACB = 0.8) implies (sin angle ACB = 0.6).2. Coordinate System: - Place (C) at ((0, 0)), (A) at ((b, 0)), and (B) at ((4k, 3k)). - Midpoints (M) (midpoint of (BC)) at ((2k, 1.5k)) and (P) (midpoint of (AC)) at ((b/2, 0)).3. Using Distance Formula: - Given (BP = 1), the distance from (B) to (P) is calculated and squared: [ sqrt{left(4k - frac{b}{2}right)^2 + (3k)^2} = 1 ] Squaring both sides: [ left(4k - frac{b}{2}right)^2 + 9k^2 = 1 ]4. Solving for (b) and (k): - Introduce (t = frac{b}{k}) and solve equations derived from the angle equality and median properties. - Find (t = 5) and (k = frac{2sqrt{5}}{15}), leading to (b = frac{2sqrt{5}}{3}).5. Area Calculation: - Coordinates of (A), (B), and (C) are used to compute the area via the shoelace formula: [ text{Area} = frac{1}{2} left| frac{2sqrt{5}}{3} cdot frac{2sqrt{5}}{5} right| = frac{1}{2} left| frac{4 cdot 5}{15} right| = frac{1}{2} cdot frac{4}{3} = frac{2}{3} ]Thus, the area of triangle (ABC) is (boxed{dfrac{2}{3}}).]

🔑:Points A, B, M and P lie on the same circle, so C M cdot C B = C P cdot C A, or 2 C M^2 = 2 C P^2. Therefore, C M = C P, C A = C B, which means triangle A B C is isosceles.Let M C = x. By the cosine rule, 1 = B P^2 = x^2 + 4 x^2 - 2 x cdot 2 x cdot 4 / 5, so x^2 = 5 / 9. Thus, S_{A B C} = 1 / 2 A C cdot B C sin angle C = 1 / 2 (2 x)^2 cdot 3 / 5 = 6 / 5 x^2 = 2 / 3.## Answer2 / 3.

🧠:[<think>Okay, so I have this equation to solve: (n + 2)! - (n + 1)! - n! = n² + n⁴. And I need to find all positive integer solutions. The answer is given as n = 3, but I need to figure out how to get there. Let me start by understanding what factorials are and how they work here.First, factorials grow really fast. So, n! means 1×2×3×…×n, right? So, (n + 2)! is (n + 2)(n + 1)n!, and similarly, (n + 1)! is (n + 1)n!. Maybe I can factor out n! from the left side to simplify the equation. Let me try that.Starting with the left-hand side (LHS): (n + 2)! - (n + 1)! - n!. Let me express each term in terms of n!.(n + 2)! = (n + 2)(n + 1)n!(n + 1)! = (n + 1)n!n! = n!So substituting back into the LHS:(n + 2)(n + 1)n! - (n + 1)n! - n!Now, factor out n! from all terms:n![(n + 2)(n + 1) - (n + 1) - 1]Let me compute the expression inside the brackets:First, expand (n + 2)(n + 1): that's n² + 3n + 2.So, substituting:n² + 3n + 2 - (n + 1) - 1Simplify term by term:n² + 3n + 2 - n - 1 - 1Combine like terms:n² + (3n - n) + (2 - 1 - 1) = n² + 2n + 0So the LHS simplifies to n! × (n² + 2n)Therefore, the original equation becomes:n! × (n² + 2n) = n² + n⁴Hmm, so maybe factor out n² on the right-hand side (RHS):n²(1 + n²)So equation is:n! × (n² + 2n) = n²(1 + n²)Alternatively, divide both sides by n² (assuming n ≠ 0, which is fine since n is positive integer):(n! / n²) × (n + 2) = 1 + n²Wait, let me check that step. Wait, if I factor n² + 2n = n(n + 2), so LHS is n! × n(n + 2) = n! × n(n + 2). So LHS is n! × n(n + 2) = (n+1)! × (n + 2). Wait, no. Wait, n! × n = (n+1)!.Wait, n! × n = (n+1)!.Wait, so n! × n(n + 2) = (n+1)! × (n + 2) = (n + 2)(n + 1)!.But perhaps I made a miscalculation here. Let me re-examine.Original LHS after factoring:n! × (n² + 2n) = n! × n(n + 2)Yes, because n² + 2n factors to n(n + 2). So, LHS is n! × n(n + 2) = n × n! × (n + 2). But n × n! is (n + 1)!.Wait, n × n! = (n + 1)! because (n + 1)! = (n + 1) × n! So, actually, n × n! is not (n + 1)! unless n × n! = (n + 1)! But that's only if (n + 1)! = (n + 1) × n! So, actually, n × n! is not equal to (n + 1)! unless n = n + 1, which is not possible. So, my mistake here.Wait, let's correct that. Let's see: n! × n(n + 2) = n! × n × (n + 2) = n × (n + 2) × n!But n × n! = (n + 1)! - n! ?Wait, maybe that's a different approach. Let's step back.Original equation after simplifying LHS:n! × (n² + 2n) = n² + n⁴So, n! × n(n + 2) = n²(1 + n²)Divide both sides by n²:n! × (n + 2) / n = 1 + n²Simplify n! / n: since n! = n × (n - 1)! So n! / n = (n - 1)!.Therefore, LHS after dividing by n² becomes (n - 1)! × (n + 2) = 1 + n²So the equation is now:(n - 1)! × (n + 2) = n² + 1That seems simpler. So, (n - 1)! × (n + 2) = n² + 1Now, we can check for positive integers n >= 1 (since factorial is defined for non-negative integers, but n positive integer here). Let's test small values of n.n = 1:Left side: (1 - 1)! × (1 + 2) = 0! × 3 = 1 × 3 = 3Right side: 1² + 1 = 23 ≠ 2, so no.n = 2:Left side: (2 - 1)! × (2 + 2) = 1! × 4 = 1 × 4 = 4Right side: 2² + 1 = 54 ≠ 5, nope.n = 3:Left side: (3 - 1)! × (3 + 2) = 2! × 5 = 2 × 5 = 10Right side: 3² + 1 = 1010 = 10. Bingo! So n = 3 is a solution.Check n = 4:Left side: (4 - 1)! × (4 + 2) = 3! × 6 = 6 × 6 = 36Right side: 4² + 1 = 1736 ≠ 17, nope.n = 5:Left side: 4! × 7 = 24 × 7 = 168Right side: 26. Not equal.n = 6:5! × 8 = 120 × 8 = 960Right side: 37. Not equal.Clearly, as n increases, left side grows factorially, which is way faster than the right side, which is polynomial. So after n = 3, left side becomes much larger. So maybe n = 3 is the only solution. But let's check n = 0, even though the problem says positive integers. If n = 0:Left side: (-1)! is undefined, so nope.So the only solution is n = 3.But wait, let me verify again the steps to make sure I didn't make a mistake in algebra.Original equation:(n + 2)! - (n + 1)! - n! = n² + n⁴Expressed in terms of n!:(n + 2)(n + 1)n! - (n + 1)n! - n! = n² + n⁴Factor out n!:n! [ (n + 2)(n + 1) - (n + 1) - 1 ] = n² + n⁴Compute inside the brackets:(n + 2)(n + 1) = n² + 3n + 2Then subtract (n + 1) and 1:n² + 3n + 2 - n - 1 - 1 = n² + 2nSo left side is n! (n² + 2n) = n² + n⁴Then factor right side: n²(1 + n²)So, n! (n² + 2n) = n²(1 + n²)Divide both sides by n² (n ≠ 0):(n! / n²)(n² + 2n) = 1 + n²Wait, actually, wait, perhaps a different approach here. Let's see:Wait, if we have n! (n² + 2n) = n² + n⁴Factor left side as n! * n(n + 2) = n²(1 + n²)So, divide both sides by n²:n! * (n + 2) / n = 1 + n²But n! / n = (n - 1)! as I had earlier, so:(n - 1)! * (n + 2) = 1 + n²Which is the same as before. So this seems correct. Then testing n=3 gives 2! *5=2*5=10 and 3² +1=10. Correct.So steps are correct, so answer is n=3.But perhaps to be thorough, check n=4 again:Left: 3! * 6=6*6=36Right:4² +1=17. Not equal.n=5: left 4! *7=24*7=168 vs 25 +1=26. Not equal.So yes, only n=3 works.Alternatively, can we argue that for n >=4, left side is divisible by higher primes, while the right side may not be? For example, for n >=4, (n -1)! is divisible by 4! =24 when n-1=4, i.e., n=5, so (5-1)! =24, so left side 24*(5+2)=24*7=168. Right side is 25 +1=26. 168 is divisible by 8, but 26 is only divisible by 2 and 13, so no.Similarly, n=4: left is 6*6=36, right is 16 +1=17, which is prime. So 36 vs17.n=3: 2!*5=10 vs 9 +1=10. Equal.Alternatively, maybe think in terms of inequalities. Let me see for n >=3, how does LHS compare to RHS.For n=3: equal.For n=4: LHS=36, RHS=17. So LHS > RHS.For n=5: LHS=168, RHS=26. LHS increases much faster.But let's check if for n >=4, LHS >= RHS. Let's see:We have (n -1)!*(n +2) vs n² +1.For n >=4, (n -1)! is at least 6 (for n=4, 3!=6). Then, (n +2) is at least 6 (for n=4, 6). So LHS >=6*6=36. RHS for n=4 is 17, for n=5 is 26, etc. But as n increases, LHS is factorial, which grows much faster than the quadratic RHS. So after n=3, LHS becomes much larger than RHS. So only possible solution is n=3.Alternatively, to formalize this, maybe use induction to show that for n >=4, LHS > RHS.Base case, n=4: 36 >17, true.Assume for some k >=4, (k -1)!*(k +2) > k² +1.Then for n =k +1:(k)!*(k +3) = k*(k -1)!*(k +3) = k*(k +3)/(k +2) * [(k -1)!*(k +2)].But since [(k -1)!*(k +2)] >k² +1 by hypothesis, then:If k*(k +3)/(k +2) * [something >k² +1] > ?But this might complicate. Alternatively, note that for n >=4, (n -1)! >=2^(n -1). Not sure. Alternatively, note that (n -1)! grows faster than exponential, while RHS is polynomial.But perhaps even simpler: for n >=4, (n -1)! >=2^(n -1), which for n=4: 6 >=8? No, that's not. Hmm, maybe Stirling's approximation.But maybe not necessary. Since for n=4, LHS=36, RHS=17. For n=5, 24*7=168 vs 26. So each time, LHS is multiplied by (n) and (n +2)/(n -1 +2)= (n +2)/(n +1). Wait, no. When n increases by 1, LHS becomes (n)!*(n +3) compared to previous (n -1)!*(n +2). So ratio is [n!*(n +3)] / [(n -1)!*(n +2)] = n*(n +3)/(n +2). For n >=4, n*(n +3)/(n +2) > n, since (n +3)/(n +2) >1. So LHS grows faster than n per step, while RHS grows as n². But factorial growth is faster than any polynomial, so eventually LHS will outpace. But even from n=4 onwards, LHS is already larger. So, yes, only n=3 is solution.Therefore, the answer is n=3.Final AnswerThe positive integer solution is boxed{3}.</think>To find all positive integer solutions of the equation ((n + 2)! - (n + 1)! - n! = n^2 + n^4), we start by simplifying the left-hand side (LHS):1. Express each term in terms of (n!): [ (n + 2)! = (n + 2)(n + 1)n! ] [ (n + 1)! = (n + 1)n! ] [ n! = n! ]2. Substitute these into the LHS: [ (n + 2)(n + 1)n! - (n + 1)n! - n! ]3. Factor out (n!): [ n! left[(n + 2)(n + 1) - (n + 1) - 1right] ]4. Simplify the expression inside the brackets: [ (n + 2)(n + 1) - (n + 1) - 1 = n^2 + 3n + 2 - n - 1 - 1 = n^2 + 2n ]5. Thus, the LHS becomes: [ n! cdot (n^2 + 2n) ]6. The original equation simplifies to: [ n! cdot (n^2 + 2n) = n^2 + n^4 ]7. Factor out (n^2) on the right-hand side (RHS): [ n^2(1 + n^2) ]8. Divide both sides by (n^2) (assuming (n neq 0)): [ frac{n! cdot (n + 2)}{n} = 1 + n^2 ]9. Simplify (frac{n!}{n}) to ((n - 1)!): [ (n - 1)! cdot (n + 2) = 1 + n^2 ]Testing small values of (n):- For (n = 1): [ 0! cdot 3 = 3 quad text{vs} quad 1 + 1 = 2 quad (text{not equal}) ]- For (n = 2): [ 1! cdot 4 = 4 quad text{vs} quad 4 + 1 = 5 quad (text{not equal}) ]- For (n = 3): [ 2! cdot 5 = 10 quad text{vs} quad 9 + 1 = 10 quad (text{equal}) ]- For (n geq 4), the left-hand side grows factorially, which is much faster than the polynomial growth of the right-hand side. Testing (n = 4), (n = 5), etc., shows the left-hand side is much larger.Thus, the only positive integer solution is (boxed{3}).,<think>Okay, so I have this equation to solve: (n+2)! - (n+1)! - n! = n² + n⁴. I need to find all positive integer solutions. The answer is supposed to be n=3, but I need to figure out how to get there. Let me start by understanding factorials and how they grow. Factorials get really big as n increases, so maybe this equation only works for small n? Let's try plugging in some small positive integers and see.First, n=1. Let's check the left-hand side (LHS) and right-hand side (RHS).LHS: (1+2)! - (1+1)! - 1! = 3! - 2! - 1! = 6 - 2 - 1 = 3.RHS: 1² + 1⁴ = 1 + 1 = 2.So 3 ≠ 2. Not equal. So n=1 is not a solution.Next, n=2.LHS: (2+2)! - (2+1)! - 2! = 4! - 3! - 2! = 24 - 6 - 2 = 16.RHS: 2² + 2⁴ = 4 + 16 = 20.16 ≠ 20. Not equal. So n=2 isn't a solution either.Let's try n=3.LHS: (3+2)! - (3+1)! - 3! = 5! - 4! - 6 = 120 - 24 - 6 = 90.RHS: 3² + 3⁴ = 9 + 81 = 90.90 = 90. That works! So n=3 is a solution.But the problem says "find all positive integer solutions," so I need to check if there are any others. Let's check n=4.LHS: 6! - 5! - 4! = 720 - 120 - 24 = 720 - 144 = 576.RHS: 4² + 4⁴ = 16 + 256 = 272.576 ≠ 272. Not equal. So n=4 isn't a solution.How about n=5?LHS: 7! - 6! - 5! = 5040 - 720 - 120 = 5040 - 840 = 4200.RHS: 5² + 5⁴ = 25 + 625 = 650.4200 vs 650. Way bigger. The LHS is growing much faster than the RHS. Let's see if this trend continues.But maybe I should check n=0, but the problem says positive integers, so n=0 is not considered. What about n=6?LHS: 8! -7! -6! = 40320 - 5040 - 720 = 40320 - 5760 = 34560.RHS: 6² + 6⁴ = 36 + 1296 = 1332. Still way smaller. So the LHS is growing factorially, while the RHS is growing polynomially (quartic). So as n increases, LHS will dominate. So maybe after n=3, there's no solution. But just to confirm, let's see n=3 is the only one.Alternatively, maybe I can manipulate the equation algebraically instead of testing values. Let's try to factor the left-hand side.(n+2)! - (n+1)! - n!.First, note that (n+2)! = (n+2)(n+1)n!.Similarly, (n+1)! = (n+1)n!.So substituting back:(n+2)(n+1)n! - (n+1)n! - n! = n² + n⁴.Factor out n! from all terms on the left:n! [ (n+2)(n+1) - (n+1) - 1 ] = n² + n⁴.Let me compute the expression inside the brackets:(n+2)(n+1) - (n+1) - 1.First, expand (n+2)(n+1):= n² + 3n + 2.Then subtract (n+1):= n² + 3n + 2 - n -1 = n² + 2n +1.Subtract 1:= n² + 2n +1 -1 = n² + 2n.Therefore, the left-hand side becomes:n! (n² + 2n) = n² + n⁴.So:n! (n² + 2n) = n² + n⁴.We can factor the right-hand side as n²(1 + n²).Similarly, on the left-hand side, factor n from n² + 2n:n(n + 2). So:n! * n(n + 2) = n²(1 + n²).Simplify left-hand side:n! * n(n + 2) = n²(1 + n²).But n! * n = (n+1)! So:(n+1)! (n + 2) = n²(1 + n²).So, (n+1)! (n + 2) = n²(n² + 1).Hmm, this seems a bit complicated. Maybe divide both sides by n², assuming n ≠ 0.So:(n+1)! (n + 2) / n² = n² + 1.Let me compute (n+1)! (n + 2) / n².First, (n+1)! = (n+1) n! So:(n+1) n! (n + 2) / n².Which is (n+1)(n+2) n! / n².But n! / n² = (n-1)! / n.Wait, n! = n*(n-1)! So:(n+1)(n+2) * n*(n-1)! / n² = (n+1)(n+2)(n-1)! / n.Therefore, the equation becomes:(n+1)(n+2)(n-1)! / n = n² + 1.Hmm. Not sure if this helps. Maybe another approach.Let me write the original equation again:(n+2)! - (n+1)! - n! = n^4 + n².Earlier, I saw that when n=3, it works. Let's see for n=3:Left: 5! -4! -3! = 120 -24 -6=90.Right: 81 +9=90. Okay.So maybe this is the only solution. How do I show that?Since factorials grow faster than polynomials. The equation ( cos theta = frac{a + 1 cos theta = frac{a + 1 cos theta = frac{a + 1 cos theta: frac{a + 1 cos theta: frac{a +71. Wait, for p=2, 4+71=75=3*5². So two distinct primes, but with exponents. So the number of divisors is (1+1)(2+1)=2*3=6. For p=2:75=3*5². So number of divisors is (1+1)(2+1)=6. Correct.For p=3:80=2^4*5. So exponents are 4 and 1. So divisors count (4+1)*(1+1)=5*2=10. Correct.Any other primes where p² +71 has ≤10 divisors?Let's see. For p=5:96=2^5*3. Number of divisors is (5 points C and D.Let me proceed step by step.Position centers O1 and O2 above AB.But first, we need to find coordinates of O1 and O2.For Circle 1, θ1 = 117.3833 degrees. θ1/2 = 58.69165 ≈ 58.69165 degreesR1 = a / sin(θ1/2) ≈ a / sin(58.69165 degrees)Similarly, θ2 = 42.6167 degrees, θ2/2 = 21.30835 degreesR2 = a / sin(21.30835 degrees)Coordinates of O1: since O1 is on the perpendicular bisector of AB at (a,0), above AB.Coordinates O1: (a, h1), where h1 = R1*cos(θ1/2)Similarly, coordinates O2: (a, -h2), where h2 = R2*cos(θ2/2)Compute h1 and h2.h1 = R1*cos(θ1/2) = (a / sin(θ1/2)) * cos(θ1/2) = a * cot(θ1/2)Similarly, h2 = a * cot(θ2/2)Therefore, O1 is at (a, a*cot(θ1/2)) and O2 is at (a, -a*cot(θ2/2))Coordinates of point C: on Circle 1, midpoint of arc ACB.Since arc AC is θ1/2, the central angle from A to C is θ1/2. So, angle from positive x-axis to point C is θ1/2 measured from point A. Wait, point A is at (0,0), but O1 is at (a, h1). Therefore, to find coordinates of point C, which is located at an angle θ1/2 from the line AO1.Wait, perhaps better to parametrize.Coordinates of point C:Since O1 is at (a, h1). The angle AO1C is θ1/2. The coordinates of C can be found by rotating point A around O1 by θ1/2.Wait, point A is at (0,0). The vector from O1 to A is (-a, -h1). Rotating this vector by θ1/2 degrees to get to point C.Rotation matrix:x' = x cos θ - y sin θy' = x sin θ + y cos θSo, vector OA is (-a, -h1). Rotate by θ1/2 (58.69165 degrees):x' = (-a)cosθ - (-h1)sinθ = -a cosθ + h1 sinθy' = (-a)sinθ + (-h1)cosθ= -a sinθ - h1 cosθTherefore, coordinates of point C relative to O1 are:(x', y') = (-a cosθ + h1 sinθ, -a sinθ - h1 cosθ)But we need to add O1's coordinates (a, h1) to get the global coordinates.Wait, no. The coordinates of C are calculated as:Original vector from O1 to A: ( -a, -h1 )Rotated by θ1/2 degrees, so the new vector from O1 to C is ( -a cosθ + h1 sinθ, -a sinθ - h1 cosθ )But then the global coordinates of C are O1 plus this vector:C_x = a + (-a cosθ + h1 sinθ )C_y = h1 + (-a sinθ - h1 cosθ )= h1 - a sinθ - h1 cosθ= h1(1 - cosθ ) - a sinθSimilarly, C_x = a(1 - cosθ ) + h1 sinθBut h1 = a*cot(θ1/2 )So, substituting h1 = a cot(theta1/2 )Then,C_x = a(1 - cos(theta1/2 )) + a cot(theta1/2 ) sin(theta1/2 )Simplify:cot(theta1/2 ) sin(theta1/2 ) = cos(theta1/2 )Therefore,C_x = a(1 - cos(theta1/2 ) + cos(theta1/2 )) = a(1 ) = aWait, that's interesting. So, C_x = aSimilarly, C_y = h1(1 - cos(theta1/2 )) - a sin(theta1/2 )But h1 = a cot(theta1/2 ) = a ( cos(theta1/2 ) / sin(theta1/2 ) )Thus,C_y = a ( cos(theta1/2 ) / sin(theta1/2 ) ) (1 - cos(theta1/2 )) - a sin(theta1/2 )= a [ ( cos(theta1/2 )(1 - cos(theta1/2 )) ) / sin(theta1/2 ) - sin(theta1/2 ) ]= a [ ( cos(theta1/2 ) - cos^2(theta1/2 ) - sin^2(theta1/2 ) ) / sin(theta1/2 ) ]Since sin^2(theta1/2 ) + cos^2(theta1/2 ) =1, so:= a [ ( cos(theta1/2 ) - 1 ) / sin(theta1/2 ) ]= a [ ( - (1 - cos(theta1/2 )) ) / sin(theta1/2 ) ]= - a (1 - cos(theta1/2 )) / sin(theta1/2 )But (1 - cos(theta )) / sin(theta ) = tan(theta /2 )So, substituting theta = theta1/2:(1 - cos(theta1/2 )) / sin(theta1/2 ) = tan(theta1/4 )But theta1 is 117.3833 degrees, theta1/2 is 58.69165 degrees, theta1/4 is ~14.6729 degrees.Therefore,C_y = -a tan(theta1/4 )So, coordinates of point C: (a, -a tan(theta1/4 ) )Similarly, coordinates of point D: midpoint of arc ADB on Circle 2.Following similar steps, but on the lower circle.Point D will have coordinates (a, a tan(theta2/4 ) )Wait, let's check:For Circle 2, the central angle AO2B is theta2 = 42.6167 degrees.Similarly, the midpoint D of arc ADB is theta2/2 = 21.30835 degrees.Following similar steps as above:Coordinates of D:D_x = aD_y = a tan(theta2/4 )Wait, but let's verify:For Circle 2, the center O2 is at (a, -h2 ), where h2 = a cot(theta2/2 )Following the rotation, but in the opposite direction.The vector from O2 to A is (-a, h2 )Rotating by theta2/2 =21.30835 degrees.But since the arc ADB is on the lower circle, rotating in the opposite direction? Hmm, maybe the same calculation applies, but since O2 is below AB, the coordinates will result in positive y.Wait, let's do it step by step.Vector from O2 to A: (-a, h2 )Rotating this vector by theta2/2 =21.30835 degrees. Using the rotation matrix:x' = (-a) cos(theta2/2 ) - h2 sin(theta2/2 )y' = (-a) sin(theta2/2 ) + h2 cos(theta2/2 )Then, coordinates of D relative to O2 are (x', y') = [ -a cos(theta2/2 ) - h2 sin(theta2/2 ), -a sin(theta2/2 ) + h2 cos(theta2/2 ) ]Adding O2's coordinates (a, -h2 ):D_x = a + x' = a - a cos(theta2/2 ) - h2 sin(theta2/2 )D_y = -h2 + y' = -h2 - a sin(theta2/2 ) + h2 cos(theta2/2 )= -h2 (1 - cos(theta2/2 )) - a sin(theta2/2 )But h2 = a cot(theta2/2 )So,D_x = a(1 - cos(theta2/2 )) - a cot(theta2/2 ) sin(theta2/2 )= a(1 - cos(theta2/2 )) - a cos(theta2/2 )= a(1 - 2 cos(theta2/2 ) )Wait, that's different from point C's x-coordinate. Wait, but earlier for point C, we found C_x = a. But here, D_x is a(1 - 2 cos(theta2/2 ) )This contradicts, so I must have made a mistake.Wait, let's redo the calculation for point D.Coordinates of O2: (a, -h2 )Vector from O2 to A: (0 - a, 0 - (-h2 )) = (-a, h2 )Rotating this vector by theta2/2 degrees. The rotation is counterclockwise, assuming standard rotation. But since we are dealing with the lower circle, the direction might be clockwise. Wait, but the angle theta2 is the central angle for arc ADB, which is 42.6167 degrees. So, the rotation is theta2/2 =21.30835 degrees.But if we rotate the vector (-a, h2 ) by theta2/2 degrees counterclockwise, we get:x' = (-a) cos(theta2/2 ) - h2 sin(theta2/2 )y' = (-a) sin(theta2/2 ) + h2 cos(theta2/2 )Then, adding O2's coordinates (a, -h2 ):D_x = a + x' = a - a cos(theta2/2 ) - h2 sin(theta2/2 )D_y = -h2 + y' = -h2 - a sin(theta2/2 ) + h2 cos(theta2/2 )= -h2 (1 - cos(theta2/2 )) - a sin(theta2/2 )But h2 = a cot(theta2/2 )So, substituting:D_x = a - a cos(theta2/2 ) - a cot(theta2/2 ) sin(theta2/2 )= a - a cos(theta2/2 ) - a cos(theta2/2 )= a - 2a cos(theta2/2 )Similarly,D_y = -a cot(theta2/2 )(1 - cos(theta2/2 )) - a sin(theta2/2 )= -a [ cot(theta2/2 )(1 - cos(theta2/2 )) + sin(theta2/2 ) ]But cot(theta2/2 ) = cos(theta2/2 ) / sin(theta2/2 )Thus,D_y = -a [ (cos(theta2/2 ) / sin(theta2/2 )) (1 - cos(theta2/2 )) + sin(theta2/2 ) ]= -a [ (cos(theta2/2 )(1 - cos(theta2/2 )) + sin^2(theta2/2 )) / sin(theta2/2 ) ]= -a [ (cos(theta2/2 ) - cos^2(theta2/2 ) + sin^2(theta2/2 )) / sin(theta2/2 ) ]But sin^2(theta2/2 ) =1 - cos^2(theta2/2 )So,= -a [ (cos(theta2/2 ) - cos^2(theta2/2 ) +1 - cos^2(theta2/2 )) / sin(theta2/2 ) ]= -a [ (1 + cos(theta2/2 ) - 2 cos^2(theta2/2 )) / sin(theta2/2 ) ]Hmm, this seems complex, maybe there's a simplification.Alternatively, maybe using the same approach as for point C.But for point C, we arrived at coordinates (a, -a tan(theta1/4 ) )Similarly, for point D, maybe (a, a tan(theta2/4 ) )But let's see.Wait, for point C, we had C_y = -a tan(theta1/4 )Similarly, for point D, we might get D_y = a tan(theta2/4 )If that's the case, then coordinates of C: (a, -a tan(theta1/4 ) )Coordinates of D: (a, a tan(theta2/4 ) )Then, vectors AC and AD.Point A is at (0,0 ), so vector AC is (a, -a tan(theta1/4 ) )Vector AD is (a, a tan(theta2/4 ) )Then, angle between vectors AC and AD is angle CAD.Which can be found via the dot product formula:cos(angle CAD ) = (AC . AD) / (|AC| |AD| )Compute AC . AD = a*a + (-a tan(theta1/4 ))*(a tan(theta2/4 )) = a² - a² tan(theta1/4 ) tan(theta2/3 )|AC| = sqrt(a² + (a tan(theta1/4 ))^2 ) = a sqrt(1 + tan^2(theta1/4 )) = a sec(theta1/4 )Similarly, |AD| = a sec(theta2/4 )Therefore,cos(angle CAD ) = [1 - tan(theta1/4 ) tan(theta2/4 ) ] / [ sec(theta1/4 ) sec(theta2/4 ) ]= [1 - tan(theta1/4 ) tan(theta2/4 ) ] / [1 / (cos(theta1/4 ) cos(theta2/4 )) ]= [1 - tan(theta1/4 ) tan(theta2/4 ) ] cos(theta1/4 ) cos(theta2/4 )= cos(theta1/4 ) cos(theta2/4 ) - sin(theta1/4 ) sin(theta2/4 )= cos(theta1/4 + theta2/4 )Therefore,angle CAD = theta1/4 + theta2/4= (theta1 + theta2 ) /4So, angle CAD is (theta1 + theta2 ) /4Where theta1 =117.3833 degrees, theta2=42.6167 degrees.So theta1 + theta2 =160 degrees. Therefore, angle CAD =160 /4 =40 degrees.Wait, really? That's elegant. So, angle CAD is just (theta1 + theta2 ) /4.But let me check the calculation again. Because according to this, angle CAD = ( theta1 + theta2 ) /4. Which is (117°23' +42°37')/4. Let's check the sum:117°23' +42°37' =159°60' =160°.Therefore, 160° /4=40°.So angle CAD is 40 degrees.Wow, that's a neat result. So, regardless of the actual lengths of AB, the angle CAD is just (sum of the two arcs)/4.But why is that?Because when we calculated the coordinates, we found that points C and D are both at x=a, but different y-coordinates. So, vectors AC and AD are (a, y1 ) and (a, y2 ), so they form angles with the x-axis, and the angle between them is the difference between their angles. But according to the calculation, it turned out that the angle between them is (theta1 + theta2)/4.But let me verify with another approach.Given that angle CAD is formed by the lines AC and AD, where C and D are midpoints of arcs ACB and ADB with central angles theta1 and theta2 respectively.Since points C and D are midpoints of their respective arcs, the central angles for arcs AC and AD are theta1/2 and theta2/2.But in the circles, the position of points C and D can be represented in terms of angles relative to AB.But angle CAD is essentially the angle between two lines from A to points on different circles. Since AB is a common chord, and the arcs are on different circles, the angle at A depends on the positions of C and D.But according to the coordinate approach, angle CAD is 40 degrees.Alternatively, there's a formula in geometry for such an angle. If two arcs are drawn on different circles through two points A and B, with midpoints C and D, then the angle at A between the lines AC and AD is equal to half the sum of half the measures of the two arcs. Wait, perhaps not. But according to our coordinate calculation, angle CAD is (theta1 + theta2)/4.Given that theta1 and theta2 sum to 160 degrees, so angle CAD is 40 degrees.Alternatively, since the central angles theta1 and theta2 are 117°23' and42°37', their sum is 160°, which is 160/360 = 4/9 of a circle. But angle CAD is (theta1 + theta2 ) /4=40°, which is the answer.Thus, the answer is 40 degrees.But let me check with actual values.Compute theta1/4 and theta2/4:theta1=117.3833°, theta1/4=29.3458°theta2=42.6167°, theta2/4=10.6542°Sum=29.3458+10.6542≈40.0°Yes, so angle CAD is indeed 40 degrees.Therefore, the answer is 40 degrees, which is 40°0'0".So, boxed, that is boxed{40^circ}Final Answerboxed{40^circ}</think>To find the angle ( angle CAD ), we start by noting that points ( C ) and ( D ) are the midpoints of the arcs ( ACB ) and ( ADB ) respectively. The given measures of these arcs are ( 117^circ 23' ) and ( 42^circ 37' ).1. Convert the angles to decimal degrees: - ( 117^circ 23' = 117 + frac{23}{60} approx 117.3833^circ ) - ( 42^circ 37' = 42 + frac{37}{60} approx 42.6167^circ )2. Sum the angles: - ( 117.3833^circ + 42.6167^circ = 160^circ )3. Calculate the angle ( angle CAD ): - The angle ( angle CAD ) is one-fourth of the sum of the arcs because each arc is divided by two (midpoints) and the angle at ( A ) is half of the resulting angles. - ( angle CAD = frac{160^circ}{4} = 40^circ )Thus, the angle ( angle CAD ) is (boxed{40^circ}).]

🔑:Solution. Rewrite the equation as n!=left(n^{*}left(n^{2}+1right)right) /(n+2). Transforming the right side, we get n!=n^{2}-2 n+5-10:(n+2). The last fraction will be an integer for n=3 and n=8, but the latter number is not a solution (substitute and check!)Grading criteria. Acquiring extraneous solutions: minus 3 points. Guessed and verified answer: 1 point.

🧠:[<think>Okay, so I need to find the smallest natural number that when divided by 2 leaves a remainder of 1, divided by 3 leaves a remainder of 2, divided by 4 leaves a remainder of 3, and divided by 5 leaves a remainder of 4. Hmm, let me think about how to approach this. First, let me write down the conditions mathematically to make it clearer. If the number is N, then:- N ≡ 1 mod 2- N ≡ 2 mod 3- N ≡ 3 mod 4- N ≡ 4 mod 5So, each time when N is divided by these numbers, the remainder is one less than the divisor. That seems like a pattern. For example, 2-1=1, 3-2=1, 4-3=1, 5-4=1. So, each divisor minus the remainder is 1. Therefore, maybe N is one less than a common multiple of these divisors? Because if N +1 is divisible by each of these numbers, then N would leave a remainder of one less. Let me check that. Suppose N +1 is divisible by 2, 3, 4, and 5. Then, N +1 would be a common multiple of these numbers, and the smallest such N would be the least common multiple (LCM) of 2, 3, 4, 5 minus 1. So, if I can find the LCM of 2, 3, 4, and 5, then subtract 1, that should give me the answer. Let me compute the LCM. Breaking down the numbers into prime factors:- 2 is prime: 2- 3 is prime: 3- 4 is 2²- 5 is prime: 5The LCM is the product of the highest powers of all primes present. So that would be 2² * 3 * 5 = 4 * 3 * 5 = 60. Therefore, LCM(2,3,4,5) = 60. Therefore, N +1 = 60, so N = 60 -1 = 59. Wait, let me verify if 59 satisfies all the conditions.Divided by 2: 59 ÷ 2 = 29 with a remainder of 1. Correct.Divided by 3: 59 ÷ 3 = 19 with a remainder of 2. Correct.Divided by 4: 59 ÷ 4 = 14 with a remainder of 3. Correct.Divided by 5: 59 ÷ 5 = 11 with a remainder of 4. Correct.So 59 seems to work. But wait, just to be thorough, is there a smaller number that also satisfies these conditions? Let me check numbers less than 60 that might satisfy these remainders.Starting from the largest remainder condition, which is when divided by 5 gives a remainder of 4. So numbers that give remainder 4 when divided by 5 are 4, 9, 14, 19, 24, 29, 34, 39, 44, 49, 54, 59, etc. Let's check these numbers against the other conditions.Check 4: Divided by 4 gives 0 remainder. But we need remainder 3. So 4 is out.Check 9: 9 ÷ 4 = 2 with remainder 1. We need remainder 3. Not good.14 ÷ 4 = 3 remainder 2. Not 3. Next.19 ÷ 4 = 4 remainder 3. Okay, so 19 passes the divided by 4 test. Then check divided by 3: 19 ÷ 3 = 6 remainder 1. But we need remainder 2. So 19 fails.Next, 24 ÷ 4 = 6 remainder 0. Not 3. Next.29 ÷ 4 = 7 remainder 1. Not 3. Next.34 ÷ 4 = 8 remainder 2. Not 3. Next.39 ÷ 4 = 9 remainder 3. Okay, check 39. Then divided by 3: 39 ÷ 3 = 13 remainder 0. Needs to be 2. So fail.44 ÷ 4 = 11 remainder 0. No good.49 ÷ 4 = 12 remainder 1. No good.54 ÷ 4 = 13 remainder 2. No good.59 ÷ 4 = 14 remainder 3. We already checked 59 and it works. So in this list, the next number is 59, which works. So no smaller number than 59 in this list satisfies all the conditions.Alternatively, perhaps there's a smarter way to check. Let me see. If N ≡ -1 mod 2,3,4,5. So N ≡ -1 mod LCM(2,3,4,5)=60. So the solutions are N=60k -1, where k is a positive integer. Therefore, the smallest natural number is 60*1 -1=59. So that's the answer. But wait, just to make sure, let's check if there is any number smaller than 60 that satisfies all four conditions. Let's think step by step.Start from the first condition: N ≡1 mod 2. That means N is odd. So possible numbers: 1,3,5,7,9,... Second condition: N ≡2 mod 3. Let's write numbers that are 2 mod 3: 2,5,8,11,14,17,20,23,26,29,32,35,38,41,44,47,50,53,56,59,...Since N must be odd, from the first condition, the intersection of odd numbers and 2 mod 3 numbers is: 5, 11, 17, 23, 29, 35, 41, 47, 53, 59,...Third condition: N ≡3 mod 4. Let's see numbers that are 3 mod 4: 3,7,11,15,19,23,27,31,35,39,43,47,51,55,59,...Intersecting with previous list (5, 11, 17, 23, 29, 35, 41, 47, 53, 59,...) gives: 11,23,35,47,59,...Fourth condition: N ≡4 mod 5. Let's see numbers that are 4 mod 5: 4,9,14,19,24,29,34,39,44,49,54,59,...Intersecting with the previous list (11,23,35,47,59,...) gives 59. So the only number common to all conditions less than 60 is 59. Therefore, 59 is indeed the smallest such natural number.Alternatively, solving the system of congruences step by step.Start with N ≡1 mod 2. So N=2a+1 for some integer a≥0.Substitute into second condition: 2a+1 ≡2 mod 3 => 2a ≡1 mod 3 => Multiply both sides by 2 inverse mod3. Since 2*2=4≡1 mod3, so inverse is 2. Thus, a≡2*1≡2 mod3. So a=3b+2. Hence, N=2*(3b+2)+1=6b+5.So N≡5 mod6.Now, substitute into third condition: 6b+5 ≡3 mod4. So 6b ≡ -2 mod4 => 6b≡2 mod4. But 6≡2 mod4, so 2b≡2 mod4 => b≡1 mod2. So b=2c+1. Therefore, N=6*(2c+1)+5=12c+6+5=12c+11.Thus, N≡11 mod12.Now, substitute into the fourth condition: 12c+11 ≡4 mod5. 12c ≡ -7 mod5 => 12c ≡ -7+10=3 mod5. 12≡2 mod5, so 2c≡3 mod5. Multiply both sides by inverse of 2 mod5, which is 3, since 2*3=6≡1 mod5. Thus, c≡3*3=9≡4 mod5. So c=5d+4. Therefore, N=12*(5d+4)+11=60d+48+11=60d+59.Hence, the solutions are N≡59 mod60. The smallest natural number is 59 when d=0. Therefore, 59 is the answer. So both methods confirm that 59 is the smallest number meeting all four conditions. Therefore, the answer is 59.Final AnswerThe smallest natural number is boxed{59}.</think>To find the smallest natural number ( N ) that satisfies the given conditions:- ( N equiv 1 pmod{2} )- ( N equiv 2 pmod{3} )- ( N equiv 3 pmod{4} )- ( N equiv 4 pmod{5} )We observe that each condition can be rewritten as ( N equiv -1 pmod{k} ) for ( k = 2, 3, 4, 5 ). This implies that ( N + 1 ) must be a common multiple of 2, 3, 4, and 5. The least common multiple (LCM) of these numbers is calculated as follows:- Prime factorization: - ( 2 = 2 ) - ( 3 = 3 ) - ( 4 = 2^2 ) - ( 5 = 5 )- The LCM is the product of the highest powers of all primes: ( 2^2 times 3 times 5 = 60 ).Thus, ( N + 1 = 60 ), so ( N = 60 - 1 = 59 ).Verification:- ( 59 div 2 ) gives remainder 1- ( 59 div 3 ) gives remainder 2- ( 59 div 4 ) gives remainder 3- ( 59 div 5 ) gives remainder 4All conditions are satisfied, confirming that the smallest natural number is (boxed{59}).,<think>Okay, so I need to find the smallest natural number that when divided by 2 gives a remainder of 1, divided by 3 gives a remainder of 2, divided by 4 gives a remainder of 3, and divided by 5 gives a remainder of 4. Hmm, let's break this down step by step. First, let me restate the problem in my own words. We need a number that, when you divide it by 2, 3, 4, or 5, leaves a remainder that's one less than the divisor each time. For example, divided by 2 leaves 1, which is 2-1. Divided by 3 leaves 2, which is 3-1, and so on. So, each remainder is exactly one less than the divisor. That seems like a pattern. Maybe there's a way to use that pattern to find the number instead of solving each equation separately?Let me write down the congruences based on the problem:1. When divided by 2, remainder 1: So, the number ≡ 1 mod 2.2. When divided by 3, remainder 2: So, the number ≡ 2 mod 3.3. When divided by 4, remainder 3: So, the number ≡ 3 mod 4.4. When divided by 5, remainder 4: So, the number ≡ 4 mod 5.Looking at these congruences, each of them is of the form: the number ≡ (divisor - 1) mod divisor. So, if the number is N, then:N ≡ -1 mod 2 N ≡ -1 mod 3 N ≡ -1 mod 4 N ≡ -1 mod 5Ah! So, N + 1 is divisible by each of these divisors: 2, 3, 4, 5. Because if N ≡ -1 mod k, then N + 1 ≡ 0 mod k, which means k divides N + 1. So, N + 1 is a common multiple of 2, 3, 4, and 5. Therefore, the smallest such N would be the least common multiple (LCM) of these numbers minus 1.So, N = LCM(2, 3, 4, 5) - 1.Now, I need to compute LCM(2, 3, 4, 5). Let me recall how to calculate LCM. The LCM of several numbers is the smallest number that is a multiple of each of them. To compute it, we can break each number into prime factors:- 2 is prime: 2- 3 is prime: 3- 4 factors into 2²- 5 is prime: 5The LCM is the product of the highest power of each prime number present in the factors. So:- For prime 2: the highest power is 2² (from 4)- For prime 3: the highest power is 3¹- For prime 5: the highest power is 5¹Therefore, LCM = 2² × 3 × 5 = 4 × 3 × 5 = 60.So, LCM(2, 3, 4, 5) is 60. Then, N = 60 - 1 = 59.Wait, let me check if 59 satisfies all the original conditions:- Divided by 2: 59 ÷ 2 = 29 with remainder 1. Good.- Divided by 3: 59 ÷ 3 = 19 with remainder 2. Good.- Divided by 4: 59 ÷ 4 = 14 with remainder 3. Good.- Divided by 5: 59 ÷ 5 = 11 with remainder 4. Good.So, 59 works. But wait, is there a smaller number that also satisfies these conditions? The problem asks for the smallest natural number. Let's check numbers smaller than 60 - 1 = 59. Since N + 1 must be a multiple of 60, the next smaller multiple would be 60 - 60 = 0. But N + 1 = 0 would mean N = -1, which is not a natural number. Hence, 59 is indeed the smallest such number.But wait, let me make sure. Maybe there's a different approach where N + 1 is a common multiple of 2, 3, 4, 5, but maybe there's a smaller common multiple? But LCM of 2, 3, 4, 5 is 60, so that's the smallest common multiple. Therefore, N = 60 - 1 = 59 is the smallest natural number. So, that seems right.Alternatively, let's see if we can solve the system of congruences step by step, using the Chinese Remainder Theorem. Maybe that would confirm it.Starting with the first two congruences:N ≡ 1 mod 2 N ≡ 2 mod 3Let’s represent N as 2k + 1 for some integer k. Then substitute into the second equation:2k + 1 ≡ 2 mod 3 Subtract 1: 2k ≡ 1 mod 3 Multiply both sides by 2 inverse mod 3. Since 2×2=4≡1 mod3, inverse of 2 is 2. So, k ≡ 2×1 ≡ 2 mod3 Thus, k = 3m + 2 for some integer m. Then N = 2*(3m + 2) +1 = 6m +5. So, N ≡5 mod6.Now, next congruence: N≡3 mod4. So, 6m +5 ≡3 mod4 6m ≡ -2 mod4 But 6 mod4 is 2, so 2m ≡ -2 mod4 Which is equivalent to 2m ≡2 mod4 Divide both sides by 2: m ≡1 mod2 So, m = 2n +1 for some integer n. Then N=6*(2n +1)+5=12n +6 +5=12n +11 Thus, N≡11 mod12.Next congruence: N≡4 mod5. So, 12n +11 ≡4 mod5 12n ≡-7 mod5 12 mod5 is 2, and -7 mod5 is -7 +10=3 Thus, 2n ≡3 mod5 Multiply both sides by inverse of 2 mod5. The inverse of 2 mod5 is 3, since 2×3=6≡1 mod5. So, n ≡3×3=9≡4 mod5 Thus, n=5p +4 for some integer p. Then N=12*(5p +4)+11=60p +48 +11=60p +59 Therefore, the general solution is N≡59 mod60. Since we want the smallest natural number, p=0 gives N=59, which is the same as before. So, that confirms it. The answer is 59. But just to be thorough, let's check if maybe there's a number smaller than 59 that satisfies all these conditions. Let's list numbers that give remainder 4 when divided by 5 (since that's the modulus with the largest remainder) and see if they satisfy the other conditions.Numbers ≡4 mod5 are: 4,9,14,19,24,29,34,39,44,49,54,59,...Check 4: 4 divided by 2 gives remainder 0, need 1. Not good.9: 9/2=4 rem1 (good), 9/3=3 rem0 (need 2). Not good.14: 14/2=7 rem0 (nope)19: 19/2=9 rem1 (good), 19/3=6 rem1 (need 2). Not good.24: 24/2=12 rem0 (nope)29: 29/2=14 rem1 (good), 29/3=9 rem2 (good), 29/4=7 rem1 (need 3). Not good.34: 34/2=17 rem0 (nope)39: 39/2=19 rem1 (good), 39/3=13 rem0 (nope)44: 44/2=22 rem0 (nope)49: 49/2=24 rem1 (good), 49/3=16 rem1 (nope)54: 54/2=27 rem0 (nope)59: 59/2=29 rem1, 59/3=19 rem2, 59/4=14 rem3, 59/5=11 rem4. All good.So, yes, 59 is indeed the first number in the list that satisfies all conditions. Therefore, the answer is 59. Alternatively, if I didn't notice the pattern that N +1 is a multiple of all divisors, I could have solved the system step by step using the Chinese Remainder Theorem, which I did, and arrived at the same answer. So, both methods lead to 59, and checking with the list also confirms it. So, I can be confident that 59 is the correct answer.Final AnswerThe smallest natural number is boxed{59}.</think>We need to find the smallest natural number ( N ) that satisfies the following congruences:1. ( N equiv 1 pmod{2} )2. ( N equiv 2 pmod{3} )3. ( N equiv 3 pmod{4} )4. ( N equiv 4 pmod{5} )Observing that each remainder is one less than the divisor, we recognize that ( N + 1 ) must be a common multiple of 2, 3, 4, and 5. The least common multiple (LCM) of these numbers is calculated as follows:- Prime factorization: ( 2 = 2 ), ( 3 = 3 ), ( 4 = 2^2 ), ( 5 = 5 )- LCM is the product of the highest powers of all prime factors: ( 2^2 times 3 times 5 = 60 )Thus, ( N + 1 = 60 ), so ( N = 60 - 1 = 59 ).Verification:- ( 59 div 2 = 29 ) remainder 1- ( 59 div 3 = 19 ) remainder 2- ( 59 div 4 = 14 ) remainder 3- ( 59 div 5 = 11 ) remainder 4All conditions are satisfied, confirming that the smallest natural number is (boxed{59}).]

🔑:Solution. Let the required number be x; then we havex=2 x_{1}+1=3 x_{2}+2=4 x_{3}+3=5 x_{4}+4For the number x+1, we will havex+1=2left(x_{1}+1right)=3left(x_{2}+1right)=4left(x_{3}+1right)=5left(x_{4}+1right)which means that x+1 is divisible by 2,3,4 and 5. Therefore, the number x+1 is the least common multiple of the numbers 2,3,4 and 5, i.e., x+1=60, from which we get x=59.

🧠:[<think>Okay, so I need to figure out for which positive integers ( n ) the expression ( 3^{2n+1} - 2^{2n+1} - 6^{n} ) is composite. The answer given is all ( n neq 1 ), but I need to verify this step by step. Let me start by understanding the problem and then testing small values of ( n ).First, let's rewrite the expression to see if there's a way to factor or simplify it. The terms are ( 3^{2n+1} ), ( -2^{2n+1} ), and ( -6^{n} ). Hmm. Let me see if I can manipulate these terms.Note that ( 3^{2n+1} = 3 cdot 3^{2n} = 3 cdot 9^{n} ). Similarly, ( 2^{2n+1} = 2 cdot 4^{n} ). So, substituting these into the original expression:( 3 cdot 9^{n} - 2 cdot 4^{n} - 6^{n} ).Hmm, maybe I can factor something out here. Let's see. The terms involve powers of 9, 4, and 6. 9 and 4 are squares, and 6 is their product. Perhaps there's a common factor or a way to factor the entire expression.Alternatively, let's compute the value for small ( n ) and see if a pattern emerges.Starting with ( n = 1 ):( 3^{2(1)+1} - 2^{2(1)+1} - 6^{1} = 3^{3} - 2^{3} - 6 = 27 - 8 - 6 = 13 ).13 is a prime number. So, for ( n = 1 ), the result is prime. According to the answer, all ( n neq 1 ) should yield composite numbers, so let's check ( n = 2 ):( 3^{5} - 2^{5} - 6^{2} = 243 - 32 - 36 = 243 - 68 = 175 ).175 is 5 squared times 7, which is 25*7=175, so that's composite. Good.( n = 3 ):( 3^{7} - 2^{7} - 6^{3} = 2187 - 128 - 216 = 2187 - 344 = 1843 ).Wait, is 1843 composite? Let me check. Let's divide by small primes:1843 divided by 7: 7*263=1841, so remainder 2.Divide by 13: 13*141=1833, remainder 10.Divide by 17: 17*108=1836, remainder 7.Divide by 19: 19*97=1843. Yes! 19*97=1843. Both primes, so composite. So, n=3 is composite.n=4:3^{9} - 2^{9} -6^{4}= 19683 - 512 - 1296 = 19683 - 1808 = 17875.17875: ends with 5, so divisible by 5. 17875/5=3575. That's 5*3575, which is composite.n=5:3^{11} -2^{11} -6^5=177147 - 2048 - 7776=177147 - 9824=167323.Check if 167323 is composite. Let's divide by small primes:167323 ÷ 7: 7*23903=167321, remainder 2.Divide by 13: 13*12871=167323? Let's see, 13*10000=130000, 13*2871=37323, so 130000+37323=167323. Yes! So 13*12871. Now, check if 12871 is prime.12871 ÷ 7: 7*1838=12866, remainder 5.12871 ÷ 11: 11*1170=12870, remainder 1.12871 ÷ 13: 13*989=12857, remainder 14. Not divisible.Divide by 17: 17*757=12869, remainder 2. Not divisible.So 12871 might be prime? Wait, maybe not. Let me check with 3: 1+2+8+7+1=19, not divisible by 3. 5: Doesn't end with 0 or 5. 7: Tried. 11: Tried. 13: Tried. 17: Tried. Maybe 12871 is prime. Wait, but even if 12871 is prime, then 167323 would be 13*12871, which is composite. So, yes, 167323 is composite.So, up to n=5, it's composite except n=1.Wait, but perhaps n=0? Although n is positive integer, so n starts at 1. So, the answer is all n except n=1. But why is n=1 prime and others composite? Let's see if we can find a general formula or factorization.Let me try to factor the expression for general n. Let's write:3^{2n+1} - 2^{2n+1} -6^n.Note that 3^{2n+1} = 3*(3^n)^2, 2^{2n+1} = 2*(2^n)^2, and 6^n = (3*2)^n = 3^n*2^n.So, perhaps substituting variables. Let me set a = 3^n and b = 2^n. Then the expression becomes:3*a^2 - 2*b^2 - a*b.So, 3a² - ab - 2b². Hmm, maybe this quadratic in terms of a and b can be factored.Let me try factoring 3a² - ab - 2b².Looking for factors of the form (ma + pb)(na + qb). The product would be mna² + (mq + pn)ab + pqb². Comparing to 3a² - ab -2b².So, mn=3, pq=-2, and mq + pn = -1.Since 3 is prime, mn=3 can be (3,1) or (1,3). Similarly, pq=-2 can be (2,-1), (-2,1), (1,-2), (-1,2).Let me try m=3, n=1. Then, mq + pn = 3*q + p*1 = -1.pq=-2. So, p and q must satisfy p*q=-2.Let me see. Let's suppose that with m=3, n=1.Then 3q + p = -1.Also, pq=-2.Let me solve for p and q.From pq=-2, possible pairs (p,q): (2,-1), (-2,1), (1,-2), (-1,2).Let's plug into 3q + p = -1.For (p=2, q=-1): 3*(-1) + 2 = -3 +2 = -1. Bingo!So, p=2, q=-1.Therefore, the factors would be (3a + 2b)(1a -1b) = (3a + 2b)(a - b).Therefore, 3a² -ab -2b² factors as (3a + 2b)(a - b).So, substituting back a=3^n and b=2^n, the original expression is:(3*3^n + 2*2^n)(3^n - 2^n).Simplify the first term:3*3^n = 3^{n+1}, and 2*2^n = 2^{n+1}. So, the first factor is 3^{n+1} + 2^{n+1}, and the second factor is 3^n - 2^n.Therefore, ( 3^{2n+1} - 2^{2n+1} - 6^{n} = (3^{n+1} + 2^{n+1})(3^{n} - 2^{n}) ).Wow, so the expression factors into two terms: ( (3^{n+1} + 2^{n+1}) ) and ( (3^{n} - 2^{n}) ). Therefore, unless one of these factors is 1, the entire expression will be composite. Because if both factors are greater than 1, then their product is composite.So, we need to check if either of the factors is 1 for any positive integer n.First, check ( 3^{n} - 2^{n} = 1 ). When is this true?For n=1: 3 - 2 =1. So, n=1. Then, the second factor is 1. Then, the entire expression becomes ( (3^{2} + 2^{2}) * 1 = 9 + 4 = 13 ), which is prime. That's exactly the case when n=1.For n >1, ( 3^{n} - 2^{n} ) is greater than 1? Let's check n=2: 9 -4=5>1. n=3:27-8=19>1. So yes, for n≥2, ( 3^{n} -2^{n} ≥5 >1 ).What about the first factor ( 3^{n+1} + 2^{n+1} ). For n≥1, this is certainly greater than 1: 3^{2} + 2^{2}=9+4=13 for n=1, which is prime, but for n>1, even if one factor is prime, the other is greater than 1, so the product would be composite.Wait, but for n≥2, both factors ( 3^{n+1} + 2^{n+1} ) and ( 3^{n} -2^{n} ) are greater than 1, so their product is composite. For n=1, one of the factors is 1, so the expression is 13*1=13, which is prime. Therefore, the conclusion is that for all positive integers n except n=1, the expression is composite.Therefore, the answer is all positive integers n except n=1.But let me confirm for n=1:As before, 3^{3} -2^{3} -6^1=27-8-6=13, prime. For n=2, factors are 3^{3} +2^{3}=27+8=35 and 3^{2} -2^{2}=9-4=5, so 35*5=175, which is composite. Similarly, n=3: 3^4 +2^4=81+16=97 and 3^3 -2^3=27-8=19. 97*19=1843, which is composite. So yes, the factorization works.Therefore, except when n=1, where one factor is 1, the expression is composite. Hence, the answer is all positive integers n except n=1.Final AnswerThe positive integers ( n ) for which ( 3^{2n+1} - 2^{2n+1} - 6^{n} ) is composite are all ( n neq 1 ). Thus, the answer is boxed{text{all } n neq 1}.</think>To determine for which positive integers ( n ) the expression ( 3^{2n+1} - 2^{2n+1} - 6^n ) is composite, we start by examining small values of ( n ):- For ( n = 1 ): [ 3^{2(1)+1} - 2^{2(1)+1} - 6^1 = 3^3 - 2^3 - 6 = 27 - 8 - 6 = 13 ] which is prime.- For ( n = 2 ): [ 3^5 - 2^5 - 6^2 = 243 - 32 - 36 = 175 ] which is composite (5² × 7).- For ( n = 3 ): [ 3^7 - 2^7 - 6^3 = 2187 - 128 - 216 = 1843 ] which is composite (19 × 97).Next, we factor the expression for general ( n ). Let ( a = 3^n ) and ( b = 2^n ). The expression becomes:[3a^2 - ab - 2b^2]Factoring this quadratic expression, we get:[(3a + 2b)(a - b)]Substituting back ( a = 3^n ) and ( b = 2^n ), we have:[(3^{n+1} + 2^{n+1})(3^n - 2^n)]For ( n = 1 ), the second factor ( 3^1 - 2^1 = 1 ), resulting in the prime number 13. For ( n geq 2 ), both factors ( 3^{n+1} + 2^{n+1} ) and ( 3^n - 2^n ) are greater than 1, making their product composite.Thus, the expression ( 3^{2n+1} - 2^{2n+1} - 6^n ) is composite for all positive integers ( n ) except ( n = 1 ).[boxed{text{all } n neq 1}],<think>Okay, so I have this problem: For which positive integers (n) is (3^{2n+1} - 2^{2n+1} - 6^{n}) composite? The answer given is all (n neq 1), but I need to figure out why. Let me start by understanding what the expression looks like for small values of (n) and see if I can spot a pattern or maybe factor it somehow.First, let's try plugging in (n = 1). Then the expression becomes (3^{3} - 2^{3} - 6^{1}). Calculating each term:(3^3 = 27), (2^3 = 8), and (6^1 = 6). So, (27 - 8 - 6 = 13). Hmm, 13 is a prime number. So for (n = 1), the result is prime, which means the answer should exclude (n = 1). That's why the answer is all (n neq 1). But I need to verify this for other values.Let's try (n = 2). Then the exponents become (2*2 + 1 = 5), so (3^5 - 2^5 - 6^2). Calculating each term:(3^5 = 243), (2^5 = 32), (6^2 = 36). So, (243 - 32 - 36 = 243 - 68 = 175). 175 is 5*35, which is 5*5*7, so definitely composite. Okay, that's composite.Next, (n = 3): (3^{7} - 2^{7} - 6^{3}). Compute each term:(3^7 = 2187), (2^7 = 128), (6^3 = 216). So, (2187 - 128 - 216 = 2187 - 344 = 1843). Wait, is 1843 prime? Let me check.Divide by small primes: 1843 ÷ 7 = 263.285... Not divisible. 1843 ÷ 11 = 167.545... Not divisible. 13: 1843 ÷13 ≈141.769... Not. 17: 1843/17 ≈108.411... Hmm. 19: 1843 ÷19 = 97, exactly? 19*97 = 1843? Let me check: 20*97 = 1940, subtract 1*97, so 1940 - 97 = 1843. Yes! So 19*97 = 1843. Both primes? 97 is prime, and 19 is prime. So 1843 is composite. Therefore, for (n = 3), it's composite.How about (n = 4): (3^{9} - 2^{9} - 6^{4}). Let's compute:(3^9 = 19683), (2^9 = 512), (6^4 = 1296). So, (19683 - 512 - 1296 = 19683 - 1808 = 17875). 17875 ends with 5, so divisible by 5. 17875 ÷5 = 3575. Which is also divisible by 5: 3575 ÷5 = 715. Again ÷5: 715 ÷5 = 143. 143 is 11*13. So, 17875 = 5^3 *11*13. Composite. So that's composite.n=5: (3^{11} - 2^{11} - 6^5). Compute each term:3^11 = 177147, 2^11 = 2048, 6^5 = 7776. So 177147 - 2048 - 7776 = 177147 - 9824 = 167323. Let's check if this is composite. It ends with 3, so not even, not divisible by 5. Let's try dividing by 7: 167323 ÷7 ≈23903.285... Not exact. Divide by 11: 167323 ÷11 = 15211.181... Not. 13: 167323 ÷13 ≈12871.0... Let's do 13*12871 = 167,323? 13*12000=156000, 13*871=11323, 156000 +11323=167323. Yes! So 13*12871. Now, check if 12871 is prime. Let's see. 12871 ÷7= 1838.714… Not. 12871 ÷11=1170.09… Not. 13: 12871 ÷13=990.07… Not. Maybe 17: 12871 ÷17≈757.117… Not. 19: 12871 ÷19≈677.421… Hmm. 23: 12871 ÷23≈559.6… Not. 29: 12871 ÷29≈443.827… Not. 31: 12871 ÷31≈415.193… Not. Maybe it's prime? Wait, let's check 12871. 12871: sum of digits is 1+2+8+7+1=19, not divisible by 3. Let's try 7: 12871 ÷7=1838.714… Not. 11: same as before. 13: already checked. Maybe 37: 12871 ÷37≈347.86… Not. 43: 12871 ÷43≈299.325… Not. 47: 12871 ÷47≈273.85… Not. 53: 12871 ÷53≈242.849… Not. 59: 12871 ÷59≈218.15… Not. So maybe 12871 is prime? Wait, 12871 divided by 61: 12871 ÷61=211. So 61*211=12871? Let's check: 60*211=12660, 1*211=211, so total 12660 +211=12871. Yes! So 12871=61*211. Both primes? 61 is prime, 211 is prime. So 167323=13*61*211. Therefore composite.So up to n=5, except n=1, it's composite. So seems like the answer is correct.But perhaps there's a general approach here. Let's try to factor the expression (3^{2n+1} - 2^{2n+1} - 6^{n}).First, notice that (6^n = (3*2)^n = 3^n * 2^n). So, the expression is (3^{2n+1} - 2^{2n+1} - 3^n 2^n). Let me see if we can factor this somehow.Alternatively, maybe rewrite terms:(3^{2n+1} = 3 cdot 3^{2n} = 3 cdot 9^n)Similarly, (2^{2n+1} = 2 cdot 4^n)So the expression becomes (3 cdot 9^n - 2 cdot 4^n - 6^n). Hmm. Maybe express all terms with base 3 and 2?Alternatively, let me set (a = 3^n) and (b = 2^n). Then the expression is (3 a^2 - 2 b^2 - a b). Let's see:(3 a^2 - a b - 2 b^2). Maybe factor this quadratic in terms of a and b.Looking at (3 a^2 - a b - 2 b^2), let's try to factor it. Let's treat this as a quadratic in variable a:(3 a^2 - b a - 2 b^2). Let's factor this.Looking for two numbers m and n such that m * n = 3*(-2 b^2) = -6 b^2 and m + n = -b. Wait, but coefficients are in terms of b. Alternatively, factor as:(3 a + ...)(a + ...). Let me attempt to factor.Suppose it factors as (3a + k b)(a + m b). Then expanding:3a * a + 3a * m b + k b * a + k m b^2 = 3 a^2 + (3 m + k) a b + k m b^2.Compare with original: 3 a^2 - a b - 2 b^2.So, coefficients must satisfy:3 m + k = -1 (coefficient of a b term)k m = -2 (coefficient of b^2 term)So, solve for integers m and k:From first equation: k = -1 -3 m.Substitute into second equation: (-1 -3 m) * m = -2So, -m -3 m^2 = -2Multiply both sides by -1: m + 3 m^2 = 2Thus, 3 m^2 + m - 2 = 0Solve quadratic equation: m = [-1 ± sqrt(1 + 24)] /6 = [-1 ±5]/6Solutions: ( -1 +5 )/6 = 4/6 = 2/3, and (-1 -5)/6 = -6/6 = -1.So m = -1 or m = 2/3. Since m should be integer (if we want integer coefficients), m = -1.Then k = -1 -3*(-1) = -1 +3 = 2.So, the factorization is (3a + 2 b)(a - b).Check: (3a + 2b)(a - b) = 3a^2 -3a b +2a b -2b^2 = 3a^2 -a b -2b^2. Perfect, matches.Therefore, the original expression factors as:(3 a^2 - a b - 2 b^2 = (3 a + 2 b)(a - b)).Substituting back (a = 3^n) and (b = 2^n):((3 cdot 3^n + 2 cdot 2^n)(3^n - 2^n)).Simplify the first term: (3^{n+1} + 2^{n+1}), and the second term is (3^n - 2^n).So, the expression factors as:((3^{n+1} + 2^{n+1})(3^n - 2^n)).Therefore, (3^{2n+1} - 2^{2n+1} - 6^n = (3^{n+1} + 2^{n+1})(3^n - 2^n)).Now, for this product to be composite, both factors must be greater than 1. If either factor is 1, then the product could be prime. Let's check when each factor is 1.First factor: (3^{n+1} + 2^{n+1}). Since (3^{n+1}) and (2^{n+1}) are both positive and increasing, the smallest value occurs when n=1: (3^2 + 2^2 = 9 +4 =13). For n=1, the first factor is 13. The second factor when n=1 is (3^1 -2^1 =1). So, the product is 13*1=13, which is prime. Hence, for n=1, the expression is prime.For n>1, let's check both factors:Second factor: (3^n - 2^n). For n=2: 9-4=5>1; n=3:27-8=19>1; n=4:81-16=65>1, etc. So for n≥2, (3^n -2^n ≥5 >1).First factor: (3^{n+1} + 2^{n+1}). For n≥1, this is at least 13 (as above) for n=1, and increases from there. So, both factors are greater than 1 when n≥2, hence the product is composite.For n=1, the second factor is 1, so the product is 13*1=13, which is prime. Hence, only for n=1, the expression is prime. For all other positive integers n≥2, the expression factors into two integers greater than 1, making it composite.Therefore, the answer is all positive integers n except n=1.Final AnswerThe positive integers (n) for which (3^{2n+1} - 2^{2n+1} - 6^n) is composite are all (n neq 1). Thus, the answer is boxed{all n neq 1}.</think>To determine for which positive integers (n) the expression (3^{2n+1} - 2^{2n+1} - 6^n) is composite, we start by evaluating the expression for small values of (n):- For (n = 1), the expression is (3^3 - 2^3 - 6^1 = 27 - 8 - 6 = 13), which is prime.- For (n = 2), the expression is (3^5 - 2^5 - 6^2 = 243 - 32 - 36 = 175), which is composite.- For (n = 3), the expression is (3^7 - 2^7 - 6^3 = 2187 - 128 - 216 = 1843), which is composite (19 times 97).- For (n = 4), the expression is (3^9 - 2^9 - 6^4 = 19683 - 512 - 1296 = 17875), which is composite (5^3 times 11 times 13).- For (n = 5), the expression is (3^{11} - 2^{11} - 6^5 = 177147 - 2048 - 7776 = 167323), which is composite (13 times 61 times 211).Next, we factor the expression (3^{2n+1} - 2^{2n+1} - 6^n). Let (a = 3^n) and (b = 2^n). The expression becomes (3a^2 - ab - 2b^2), which factors as ((3a + 2b)(a - b)). Substituting back (a = 3^n) and (b = 2^n), we get:[(3^{n+1} + 2^{n+1})(3^n - 2^n)]For (n geq 2), both factors (3^{n+1} + 2^{n+1}) and (3^n - 2^n) are greater than 1, making the product composite. For (n = 1), the factors are (13) and (1), resulting in the prime number 13.Thus, the expression is composite for all positive integers (n) except (n = 1).[boxed{all n neq 1}]]

🔑:section*{Solution}(3^{2 n+1}-2^{2 n+1}-6^{n}=left(3^{n}-2^{n}right)left(3^{n+1}+2^{n+1}right)), so it is certainly composite for (n>1). For (n=1), it is (27-8-6=13), which is prime.