💡Alright, so I have this problem where I need to find the equation of a circle M. The circle passes through two points, (3,0) and (0,1), and its center, point M, lies on the line 2x + y - 1 = 0. Hmm, okay, let me break this down step by step.First, I know that the general equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius. Since the center M lies on the line 2x + y - 1 = 0, I can express the coordinates of M in terms of one variable. Let's say M is (a, b). Then, according to the line equation, 2a + b - 1 = 0, which means b = 1 - 2a. So, the center M can be written as (a, 1 - 2a).Now, since both points (3,0) and (0,1) lie on the circle, the distance from M to each of these points should be equal to the radius. That gives me two equations:1. The distance from M(a, 1 - 2a) to (3,0): sqrt[(a - 3)^2 + (1 - 2a - 0)^2] = radius.2. The distance from M(a, 1 - 2a) to (0,1): sqrt[(a - 0)^2 + (1 - 2a - 1)^2] = radius.Since both distances are equal to the radius, I can set them equal to each other:sqrt[(a - 3)^2 + (1 - 2a)^2] = sqrt[a^2 + (-2a)^2].Let me square both sides to eliminate the square roots:(a - 3)^2 + (1 - 2a)^2 = a^2 + ( -2a)^2.Expanding both sides:Left side: (a^2 - 6a + 9) + (1 - 4a + 4a^2) = a^2 - 6a + 9 + 1 - 4a + 4a^2 = 5a^2 - 10a + 10.Right side: a^2 + 4a^2 = 5a^2.So, setting them equal:5a^2 - 10a + 10 = 5a^2.Subtracting 5a^2 from both sides:-10a + 10 = 0.Solving for a:-10a = -10 => a = 1.Okay, so a = 1. Now, substituting back into b = 1 - 2a:b = 1 - 2(1) = -1.So, the center M is at (1, -1).Now, let's find the radius. Using the distance from M(1, -1) to (3,0):sqrt[(1 - 3)^2 + (-1 - 0)^2] = sqrt[(-2)^2 + (-1)^2] = sqrt[4 + 1] = sqrt[5].So, the radius is sqrt(5).Therefore, the equation of the circle is:(x - 1)^2 + (y + 1)^2 = (sqrt(5))^2 => (x - 1)^2 + (y + 1)^2 = 5.Let me double-check to make sure both points lie on this circle.For (3,0):(3 - 1)^2 + (0 + 1)^2 = (2)^2 + (1)^2 = 4 + 1 = 5. That works.For (0,1):(0 - 1)^2 + (1 + 1)^2 = (-1)^2 + (2)^2 = 1 + 4 = 5. That also works.Looks good! So, the equation is correct.

💡Alright, so I have this math problem here: If x plus 1 over x equals some number, and then x squared plus 1 over x squared equals 2, what is that number? Hmm, okay, let's try to figure this out step by step.First, I'll denote the unknown number as N. So, according to the problem, x + (1/x) = N. And we also know that x² + (1/x²) = 2. I need to find the value of N.I remember that when dealing with expressions like x + 1/x, squaring them can sometimes help simplify things. So, maybe I can square both sides of the equation x + 1/x = N to see if that leads me somewhere.Let's try that. If I square both sides, I get:(x + 1/x)² = N²Expanding the left side, I recall that (a + b)² = a² + 2ab + b². Applying that here:x² + 2*(x)*(1/x) + (1/x)² = N²Simplifying the middle term, 2*(x)*(1/x) is just 2*(1) = 2. So now the equation becomes:x² + 2 + (1/x²) = N²But wait, the problem already tells me that x² + (1/x²) = 2. That's useful! I can substitute that into the equation:2 + 2 = N²So, 4 = N²To find N, I take the square root of both sides:N = √4Which gives me N = 2 or N = -2.But hold on a second. Can N be negative? Let's think about it. If x is a positive real number, then x + 1/x is always positive because both x and 1/x are positive. Similarly, if x is negative, x + 1/x would still be negative because both x and 1/x are negative, and their sum would be negative.However, the problem doesn't specify whether x is positive or negative. So, technically, N could be either 2 or -2. But let's check if both solutions make sense in the context of the problem.If N = 2, then x + 1/x = 2. Solving for x, we can multiply both sides by x to get x² + 1 = 2x, which simplifies to x² - 2x + 1 = 0. This factors to (x - 1)² = 0, so x = 1. Plugging back into x + 1/x, we get 1 + 1 = 2, which checks out.If N = -2, then x + 1/x = -2. Multiplying both sides by x gives x² + 1 = -2x, which simplifies to x² + 2x + 1 = 0. This factors to (x + 1)² = 0, so x = -1. Plugging back into x + 1/x, we get -1 + (-1) = -2, which also checks out.So, both N = 2 and N = -2 satisfy the equation x + 1/x = N and lead to x² + 1/x² = 2. Therefore, the possible values for N are 2 and -2.But the problem asks for "the number," implying a single answer. Maybe I need to consider the context or any restrictions on x. If x is a real number, both solutions are valid. However, if we're only considering positive real numbers, then N would be 2. If negative numbers are allowed, then both are possible.Since the problem doesn't specify, I think it's safe to consider both possibilities. But often in such problems, unless stated otherwise, they might expect the positive solution. So, I'll go with N = 2 as the primary answer, but I should note that N could also be -2.Let me double-check my work to make sure I didn't make any mistakes. Starting from x + 1/x = N, squaring both sides gives x² + 2 + 1/x² = N². Substituting x² + 1/x² = 2, we get 2 + 2 = N², so N² = 4, leading to N = ±2. That seems correct.Yes, I think I've covered all the bases here. Both 2 and -2 are valid solutions, but depending on the context, 2 might be the expected answer.

💡Okay, so I have this problem about calculating the total sample variance of high school students' heights based on stratified random sampling. The school took samples of 100 male students and 60 female students, and they have the mean and variance for both groups. I need to find the total sample variance for all the students combined.First, I remember that variance is a measure of how spread out the data is. Since the samples are stratified, meaning they're taken from different subgroups (males and females), I need to account for both the variance within each subgroup and the variance between the subgroups.Let me recall the formula for the total variance in stratified sampling. I think it involves the weighted average of the variances within each stratum plus the variance between the strata. So, the total variance ( s^2 ) can be calculated as:[s^2 = left( frac{n_1}{n} times s_1^2 right) + left( frac{n_2}{n} times s_2^2 right) + left( frac{n_1}{n} times (bar{x}_1 - bar{x})^2 right) + left( frac{n_2}{n} times (bar{x}_2 - bar{x})^2 right)]Where:- ( n_1 ) and ( n_2 ) are the sample sizes for males and females, respectively.- ( s_1^2 ) and ( s_2^2 ) are the variances for males and females.- ( bar{x}_1 ) and ( bar{x}_2 ) are the means for males and females.- ( bar{x} ) is the overall mean.- ( n ) is the total sample size.First, I need to calculate the overall mean ( bar{x} ). The overall mean is the weighted average of the two group means:[bar{x} = frac{n_1 times bar{x}_1 + n_2 times bar{x}_2}{n}]Plugging in the numbers:[bar{x} = frac{100 times 172 + 60 times 164}{100 + 60} = frac{17200 + 9840}{160} = frac{27040}{160} = 169]So, the overall mean height is 169 cm.Next, I need to calculate the total variance. Let's break it down step by step.First, calculate the weighted variances:[frac{n_1}{n} times s_1^2 = frac{100}{160} times 18 = frac{5}{8} times 18 = 11.25][frac{n_2}{n} times s_2^2 = frac{60}{160} times 30 = frac{3}{8} times 30 = 11.25]Now, calculate the weighted squared differences between each group mean and the overall mean:[frac{n_1}{n} times (bar{x}_1 - bar{x})^2 = frac{100}{160} times (172 - 169)^2 = frac{5}{8} times 9 = 5.625][frac{n_2}{n} times (bar{x}_2 - bar{x})^2 = frac{60}{160} times (164 - 169)^2 = frac{3}{8} times 25 = 9.375]Now, add all these components together to get the total variance:[s^2 = 11.25 + 11.25 + 5.625 + 9.375 = 37.5]So, the total sample variance of the high school students' heights is 37.5 cm².Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, the overall mean:[bar{x} = frac{100 times 172 + 60 times 164}{160} = frac{17200 + 9840}{160} = frac{27040}{160} = 169]That seems correct.Weighted variances:[frac{100}{160} times 18 = 11.25][frac{60}{160} times 30 = 11.25]That's correct.Squared differences:[frac{100}{160} times 9 = 5.625][frac{60}{160} times 25 = 9.375]Adding them up:11.25 + 11.25 = 22.55.625 + 9.375 = 1522.5 + 15 = 37.5Yes, that all checks out. So, the total sample variance is indeed 37.5 cm².

💡Alright, so I have this problem here about a function and a triangle. Let me try to figure it out step by step. First, the function is given as f(x) = sin(2x) - (cos²x - sin²x) - 1. Hmm, okay. I need to find the minimum value of this function and its smallest positive period. Then, in part (2), there's a triangle ABC with some conditions involving this function and vectors. Let me focus on part (1) first.Starting with f(x) = sin(2x) - (cos²x - sin²x) - 1. I remember that cos²x - sin²x is equal to cos(2x). So, maybe I can rewrite the function using that identity. Let me try that:f(x) = sin(2x) - cos(2x) - 1.Okay, that simplifies it a bit. Now, I have f(x) = sin(2x) - cos(2x) - 1. I need to find the minimum value and the period. I recall that expressions like a sinθ + b cosθ can be rewritten as a single sine or cosine function using the amplitude-phase form. Maybe I can apply that here. Let me see:Let me consider sin(2x) - cos(2x). This is of the form a sinθ + b cosθ where a = 1 and b = -1. The formula for combining these is R sin(θ + φ), where R = sqrt(a² + b²) and φ is the phase shift.Calculating R: R = sqrt(1² + (-1)²) = sqrt(1 + 1) = sqrt(2). Now, φ is given by arctan(b/a), but since b is negative, I need to make sure I get the correct quadrant. Alternatively, I can write it as R sin(θ - φ) because of the negative cosine term. Let me think. Alternatively, maybe it's easier to write it as R sin(2x - φ). Let me try that. So, sin(2x) - cos(2x) can be written as sqrt(2) sin(2x - π/4). Let me verify that:Using the identity sin(A - B) = sinA cosB - cosA sinB. So, sqrt(2) sin(2x - π/4) = sqrt(2)[sin(2x)cos(π/4) - cos(2x)sin(π/4)].Since cos(π/4) = sin(π/4) = sqrt(2)/2, this becomes sqrt(2)[sin(2x)(sqrt(2)/2) - cos(2x)(sqrt(2)/2)] = sqrt(2)*(sqrt(2)/2)(sin(2x) - cos(2x)) = (2/2)(sin(2x) - cos(2x)) = sin(2x) - cos(2x). Perfect, that works.So, f(x) = sqrt(2) sin(2x - π/4) - 1.Now, to find the minimum value of f(x). The sine function oscillates between -1 and 1, so sqrt(2) sin(2x - π/4) oscillates between -sqrt(2) and sqrt(2). Therefore, f(x) oscillates between -sqrt(2) - 1 and sqrt(2) - 1. So, the minimum value is -sqrt(2) - 1.Wait, but let me double-check. If sin(2x - π/4) = -1, then f(x) = sqrt(2)*(-1) - 1 = -sqrt(2) - 1, which is indeed the minimum. Okay, that seems right.Now, the period of f(x). The function sin(2x - π/4) has a period of π because the coefficient of x is 2, so the period is 2π / 2 = π. Therefore, the smallest positive period of f(x) is π.Alright, so for part (1), the minimum value is -sqrt(2) - 1 and the period is π.Moving on to part (2). We have triangle ABC with sides a, b, c opposite angles A, B, C respectively. Given that c = sqrt(7), and f(C) = 0. Also, vectors m = (1, sinA) and n = (3, sinB) are collinear. We need to find a and b.First, let's find angle C. Since f(C) = 0, let's use the function f(x) we have. From part (1), f(x) = sqrt(2) sin(2x - π/4) - 1. So, setting f(C) = 0:sqrt(2) sin(2C - π/4) - 1 = 0=> sin(2C - π/4) = 1/sqrt(2)So, 2C - π/4 = π/4 + 2πk or 2C - π/4 = 3π/4 + 2πk, where k is an integer.Solving for C:Case 1: 2C - π/4 = π/4 + 2πk=> 2C = π/2 + 2πk=> C = π/4 + πkSince C is an angle in a triangle, it must be between 0 and π. So, k = 0 gives C = π/4, which is valid. k = 1 would give C = 5π/4, which is more than π, so invalid.Case 2: 2C - π/4 = 3π/4 + 2πk=> 2C = π + 2πk=> C = π/2 + πkAgain, C must be between 0 and π. So, k = 0 gives C = π/2, which is valid. k = 1 gives C = 3π/2, which is invalid.So, possible values for C are π/4 and π/2.But let's check if both are possible. If C = π/2, then the triangle is right-angled at C. Then, using the Pythagorean theorem, a² + b² = c² = 7. But we also have vectors m and n collinear. Let's see if that's possible.Alternatively, if C = π/4, then we have an acute angle. Let's see which one fits with the other conditions.But let's first note that both C = π/4 and C = π/2 are possible, so we need to check both cases.Now, vectors m = (1, sinA) and n = (3, sinB) are collinear. Collinear vectors mean one is a scalar multiple of the other. So, there exists a scalar λ such that (1, sinA) = λ(3, sinB). Therefore:1 = 3λsinA = λ sinBFrom the first equation, λ = 1/3. Then, sinA = (1/3) sinB.So, sinA = (1/3) sinB.In triangle ABC, angles A + B + C = π. So, A + B = π - C.Depending on C, we have different expressions.Case 1: C = π/4Then, A + B = π - π/4 = 3π/4.We have sinA = (1/3) sinB.Let me write A = 3π/4 - B.So, sin(3π/4 - B) = (1/3) sinB.Using the sine subtraction formula:sin(3π/4) cosB - cos(3π/4) sinB = (1/3) sinB.We know sin(3π/4) = sqrt(2)/2 and cos(3π/4) = -sqrt(2)/2.So, sqrt(2)/2 cosB - (-sqrt(2)/2) sinB = (1/3) sinB=> sqrt(2)/2 cosB + sqrt(2)/2 sinB = (1/3) sinBLet me factor out sqrt(2)/2:sqrt(2)/2 (cosB + sinB) = (1/3) sinBMultiply both sides by 2/sqrt(2):cosB + sinB = (2/sqrt(2))*(1/3) sinB=> cosB + sinB = (sqrt(2)/3) sinBBring all terms to one side:cosB + sinB - (sqrt(2)/3) sinB = 0=> cosB + sinB (1 - sqrt(2)/3) = 0Let me write this as:cosB = -sinB (1 - sqrt(2)/3)Divide both sides by cosB (assuming cosB ≠ 0):1 = -tanB (1 - sqrt(2)/3)So,tanB = -1 / (1 - sqrt(2)/3)Let me rationalize the denominator:Multiply numerator and denominator by (1 + sqrt(2)/3):tanB = - (1 + sqrt(2)/3) / [ (1 - sqrt(2)/3)(1 + sqrt(2)/3) ]= - (1 + sqrt(2)/3) / [1 - (2)/9]= - (1 + sqrt(2)/3) / (7/9)= - (9/7)(1 + sqrt(2)/3)= - (9/7 + 3 sqrt(2)/7)= - (9 + 3 sqrt(2))/7So, tanB = - (9 + 3 sqrt(2))/7But angle B is between 0 and π, so tanB negative implies B is in the second quadrant. So, B = π - arctan( (9 + 3 sqrt(2))/7 )But this seems complicated. Maybe I made a mistake earlier. Let me check.Wait, when I had:sqrt(2)/2 (cosB + sinB) = (1/3) sinBLet me instead write it as:sqrt(2)/2 cosB + sqrt(2)/2 sinB = (1/3) sinBBring all terms to left:sqrt(2)/2 cosB + (sqrt(2)/2 - 1/3) sinB = 0Let me factor out sinB:sqrt(2)/2 cosB = (1/3 - sqrt(2)/2) sinBDivide both sides by cosB:sqrt(2)/2 = (1/3 - sqrt(2)/2) tanBSo,tanB = (sqrt(2)/2) / (1/3 - sqrt(2)/2)Let me compute the denominator:1/3 - sqrt(2)/2 = (2 - 3 sqrt(2))/6So,tanB = (sqrt(2)/2) / ( (2 - 3 sqrt(2))/6 ) = (sqrt(2)/2) * (6 / (2 - 3 sqrt(2))) = (3 sqrt(2)) / (2 - 3 sqrt(2))Multiply numerator and denominator by (2 + 3 sqrt(2)):tanB = [3 sqrt(2)(2 + 3 sqrt(2))] / [ (2 - 3 sqrt(2))(2 + 3 sqrt(2)) ]= [6 sqrt(2) + 9*2 ] / [4 - (9*2)]= [6 sqrt(2) + 18] / [4 - 18]= [6 sqrt(2) + 18] / (-14)= - (6 sqrt(2) + 18)/14= - (3 sqrt(2) + 9)/7So, tanB = - (3 sqrt(2) + 9)/7Which is the same as before. So, B is in the second quadrant, so B = π - arctan( (3 sqrt(2) + 9)/7 )This seems messy, but maybe we can find sinB and cosB from this.Alternatively, perhaps it's better to use the Law of Sines.In triangle ABC, a/sinA = b/sinB = c/sinC.Given c = sqrt(7), and C is either π/4 or π/2.Let me first assume C = π/4.Then, a/sinA = b/sinB = sqrt(7)/sin(π/4) = sqrt(7)/(sqrt(2)/2) = 2 sqrt(7)/sqrt(2) = sqrt(14).So, a = sqrt(14) sinA, b = sqrt(14) sinB.From earlier, sinA = (1/3) sinB.So, a = sqrt(14) * (1/3) sinB = (sqrt(14)/3) sinB.But b = sqrt(14) sinB, so a = (1/3) b.So, a = b/3.Now, using the Law of Cosines for angle C:c² = a² + b² - 2ab cosCGiven c = sqrt(7), C = π/4, so cosC = sqrt(2)/2.So,7 = a² + b² - 2ab*(sqrt(2)/2)= a² + b² - ab sqrt(2)But since a = b/3, substitute:7 = (b²/9) + b² - (b/3)*b*sqrt(2)= (b²/9) + b² - (b² sqrt(2))/3Combine terms:= (1/9 + 1 - sqrt(2)/3) b²= (10/9 - sqrt(2)/3) b²So,b² = 7 / (10/9 - sqrt(2)/3) = 7 / [ (10 - 3 sqrt(2))/9 ] = 7 * 9 / (10 - 3 sqrt(2)) = 63 / (10 - 3 sqrt(2))Rationalize the denominator:Multiply numerator and denominator by (10 + 3 sqrt(2)):b² = 63*(10 + 3 sqrt(2)) / [ (10 - 3 sqrt(2))(10 + 3 sqrt(2)) ]= 63*(10 + 3 sqrt(2)) / (100 - 18)= 63*(10 + 3 sqrt(2)) / 82So,b = sqrt(63*(10 + 3 sqrt(2))/82 )This seems complicated. Maybe I made a mistake earlier.Wait, perhaps C = π/2 is the correct case. Let me check that.Case 2: C = π/2Then, A + B = π - π/2 = π/2.From vectors m and n collinear, we have sinA = (1/3) sinB.Also, A + B = π/2, so B = π/2 - A.So, sinB = sin(π/2 - A) = cosA.Thus, sinA = (1/3) cosASo,tanA = 1/3So, A = arctan(1/3), and B = π/2 - arctan(1/3)Now, using the Law of Sines:a/sinA = b/sinB = c/sinCc = sqrt(7), sinC = sin(π/2) = 1. So,a = sqrt(7) sinAb = sqrt(7) sinBBut sinB = cosA, and sinA = 1/3 cosA, so:From tanA = 1/3, we have sinA = 1/√(10) and cosA = 3/√(10).Thus,a = sqrt(7) * (1/√(10)) = sqrt(7/10)b = sqrt(7) * (3/√(10)) = 3 sqrt(7/10)So, a = sqrt(70)/10 * 10? Wait, sqrt(7/10) can be written as sqrt(70)/10? Wait, no:sqrt(7/10) = sqrt(7)/sqrt(10) = (sqrt(7)*sqrt(10))/10 = sqrt(70)/10.Similarly, 3 sqrt(7/10) = 3 sqrt(70)/10.So, a = sqrt(70)/10 and b = 3 sqrt(70)/10.But let me check if this satisfies the Law of Cosines for C = π/2:c² = a² + b²c² = 7a² + b² = (7/10) + (9*7/10) = (7 + 63)/10 = 70/10 = 7. Perfect, it works.So, in this case, a = sqrt(70)/10 and b = 3 sqrt(70)/10.But wait, earlier when I assumed C = π/4, I got a complicated expression, but when I assumed C = π/2, it worked out neatly. So, perhaps C = π/2 is the correct case.But why did f(C) = 0 give both C = π/4 and C = π/2? Maybe both are possible, but in the context of the triangle, only one makes sense.Wait, if C = π/4, then the Law of Cosines would give a different result, but when I tried that, it led to a complicated expression. Whereas with C = π/2, everything worked out nicely.So, perhaps the correct value is C = π/2, leading to a = sqrt(70)/10 and b = 3 sqrt(70)/10.But let me double-check if C = π/4 is possible.If C = π/4, then using the Law of Sines:a = sqrt(14) sinAb = sqrt(14) sinBWith sinA = (1/3) sinB, so a = (1/3) b.Then, Law of Cosines:7 = a² + b² - ab sqrt(2)Substituting a = b/3:7 = (b²/9) + b² - (b² sqrt(2))/3= (10b²/9) - (b² sqrt(2))/3Factor out b²:b² (10/9 - sqrt(2)/3) = 7So,b² = 7 / (10/9 - sqrt(2)/3) = 7 / [ (10 - 3 sqrt(2))/9 ] = 63 / (10 - 3 sqrt(2))Rationalizing:b² = 63*(10 + 3 sqrt(2)) / (100 - 18) = 63*(10 + 3 sqrt(2))/82So,b = sqrt(63*(10 + 3 sqrt(2))/82 )This is a valid positive real number, so C = π/4 is possible. But then, why does the case with C = π/2 also work?Wait, perhaps both solutions are valid, but in the context of the problem, we need to see which one fits with the collinearity condition.Wait, in the case of C = π/4, we have a = b/3, and from the Law of Cosines, we get a specific value for b. But in the case of C = π/2, we have a = sqrt(70)/10 and b = 3 sqrt(70)/10, which are specific values.But the problem doesn't specify whether the triangle is acute or right-angled, so both could be possible. However, the function f(C) = 0 gives two possible solutions for C, but in the context of the triangle, we need to see which one leads to a valid triangle with the given conditions.Wait, but when C = π/2, the vectors m and n being collinear leads to a neat solution, whereas for C = π/4, it leads to a more complicated expression. Maybe both are valid, but the problem expects us to find both solutions.Alternatively, perhaps only one solution is valid because the other leads to an impossible triangle.Wait, let me check for C = π/4:From the Law of Sines, a = sqrt(14) sinA, b = sqrt(14) sinB, and sinA = (1/3) sinB.Also, A + B = 3π/4.So, A = 3π/4 - B.Thus, sinA = sin(3π/4 - B) = sin(3π/4) cosB - cos(3π/4) sinB = sqrt(2)/2 cosB + sqrt(2)/2 sinB.But sinA = (1/3) sinB, so:sqrt(2)/2 cosB + sqrt(2)/2 sinB = (1/3) sinBWhich simplifies to:sqrt(2)/2 cosB + (sqrt(2)/2 - 1/3) sinB = 0This is the same equation as before, leading to tanB = - (3 sqrt(2) + 9)/7, which is negative, implying B is in the second quadrant. But in a triangle, angles are between 0 and π, so B being in the second quadrant is possible, but then A would be in the first quadrant.But let's see if this leads to a valid triangle.From tanB = - (3 sqrt(2) + 9)/7, which is negative, so B = π - arctan( (3 sqrt(2) + 9)/7 )Let me compute arctan( (3 sqrt(2) + 9)/7 ). Let's approximate:3 sqrt(2) ≈ 4.2426, so 4.2426 + 9 = 13.2426. Divided by 7: ≈ 1.8918.So, arctan(1.8918) ≈ 62 degrees (since tan(60) ≈ 1.732, tan(62) ≈ 1.88). So, arctan(1.8918) ≈ 62 degrees.Thus, B ≈ π - 62 degrees ≈ 118 degrees.Then, A = 3π/4 - B ≈ 135 degrees - 118 degrees ≈ 17 degrees.So, angles would be approximately 17°, 118°, and 45°, which sum to 180°, so that's valid.Thus, both cases are possible. So, we have two possible solutions for a and b.But the problem doesn't specify whether C is acute or right, so perhaps both solutions are valid. However, the problem asks for "the values of a and b", implying there might be a unique solution. Maybe I missed something.Wait, in the case of C = π/2, the vectors m and n being collinear leads to a neat solution with a = sqrt(70)/10 and b = 3 sqrt(70)/10. Whereas for C = π/4, the solution is more complicated, but still valid.But perhaps the problem expects us to consider both cases. Alternatively, maybe only one case is possible due to the collinearity condition.Wait, in the case of C = π/4, we have a = b/3, and from the Law of Cosines, we get a specific value for b, which is sqrt(63*(10 + 3 sqrt(2))/82 ). Let me compute this numerically to see if it's a valid positive number.Compute denominator: 10 - 3 sqrt(2) ≈ 10 - 4.2426 ≈ 5.7574So, b² = 63 / 5.7574 ≈ 10.94Thus, b ≈ sqrt(10.94) ≈ 3.308Then, a = b/3 ≈ 1.103So, a ≈ 1.103, b ≈ 3.308, c = sqrt(7) ≈ 2.6458Checking the triangle inequality: a + b ≈ 4.411 > c ≈ 2.6458; a + c ≈ 3.749 > b ≈ 3.308; b + c ≈ 5.953 > a ≈ 1.103. So, valid.In the case of C = π/2, a ≈ sqrt(70)/10 ≈ 0.8367, b ≈ 3 sqrt(70)/10 ≈ 2.510, c ≈ 2.6458Checking triangle inequality: a + b ≈ 3.3467 > c ≈ 2.6458; a + c ≈ 3.4825 > b ≈ 2.510; b + c ≈ 5.1558 > a ≈ 0.8367. Valid.So, both cases are valid. Therefore, there are two possible solutions for a and b.But the problem doesn't specify whether C is acute or right, so perhaps both solutions are acceptable. However, the problem might expect us to provide both solutions.Alternatively, perhaps I made a mistake in assuming both cases are possible. Let me think again.Wait, when C = π/4, the function f(C) = 0, and when C = π/2, f(C) = 0 as well. So, both are valid. Therefore, the triangle could be either acute or right-angled at C, leading to two different sets of a and b.But the problem asks to "find the values of a and b", which might imply both solutions. However, in the initial problem statement, part (2) is a single question, so perhaps only one solution is expected. Maybe I need to consider which case is more likely.Alternatively, perhaps the function f(x) = 0 has solutions at C = π/4 and C = π/2, but in the context of the triangle, only one of them leads to a valid triangle with the given collinearity condition.Wait, in the case of C = π/2, the vectors m and n being collinear leads to a neat solution with a = sqrt(70)/10 and b = 3 sqrt(70)/10. Whereas for C = π/4, it leads to a more complicated expression, but still valid.But perhaps the problem expects us to consider both cases. However, since the problem didn't specify, maybe it's safer to provide both solutions.But looking back at the initial problem, part (2) says "find the values of a and b". It doesn't specify whether there are multiple solutions, so perhaps both are acceptable.Alternatively, maybe only one solution is valid because the other leads to an impossible triangle. But as I checked earlier, both cases are valid.Wait, perhaps I made a mistake in the calculation for C = π/4. Let me re-examine.From f(C) = 0, we have C = π/4 or C = π/2.In the case of C = π/4, we have:From vectors m and n collinear, sinA = (1/3) sinB.From the Law of Sines, a = sqrt(14) sinA, b = sqrt(14) sinB.Thus, a = (1/3) b.From the Law of Cosines:c² = a² + b² - 2ab cosCWith c = sqrt(7), C = π/4, cosC = sqrt(2)/2.So,7 = a² + b² - ab sqrt(2)But a = b/3, so:7 = (b²/9) + b² - (b² sqrt(2))/3= (10b²/9) - (b² sqrt(2))/3Factor out b²:b² (10/9 - sqrt(2)/3) = 7Thus,b² = 7 / (10/9 - sqrt(2)/3) = 7 / [ (10 - 3 sqrt(2))/9 ] = 63 / (10 - 3 sqrt(2))Rationalizing:b² = 63*(10 + 3 sqrt(2)) / (100 - 18) = 63*(10 + 3 sqrt(2))/82So,b = sqrt(63*(10 + 3 sqrt(2))/82 )This is a valid positive real number, so C = π/4 is possible.Therefore, both cases are valid, leading to two possible sets of a and b.But the problem didn't specify whether C is acute or right, so perhaps both solutions are acceptable.However, in the initial problem, part (2) is a single question, so maybe only one solution is expected. Alternatively, perhaps I need to consider that C = π/2 is the only valid solution because when C = π/4, the vectors m and n being collinear might not hold due to the negative tangent.Wait, in the case of C = π/4, we found that tanB is negative, implying B is in the second quadrant. But in a triangle, angles are between 0 and π, so B being in the second quadrant is possible, but then A would be in the first quadrant. So, it's still a valid triangle.Therefore, both solutions are valid, and the problem might expect both sets of a and b.But since the problem asks for "the values of a and b", perhaps both solutions are acceptable. However, in the initial problem, part (2) is a single question, so maybe only one solution is expected. Alternatively, perhaps I need to consider that C = π/2 is the only valid solution because when C = π/4, the vectors m and n being collinear might not hold due to the negative tangent.Wait, no, the collinearity condition is satisfied in both cases, as we derived the necessary relations for both C = π/4 and C = π/2.Therefore, perhaps the problem expects both solutions. However, since the problem didn't specify, maybe I should provide both.But looking back at the initial problem, part (2) is a single question, so perhaps only one solution is expected. Alternatively, perhaps I need to consider that C = π/2 is the only valid solution because when C = π/4, the vectors m and n being collinear might not hold due to the negative tangent.Wait, no, the collinearity condition is satisfied in both cases, as we derived the necessary relations for both C = π/4 and C = π/2.Therefore, perhaps the problem expects both solutions. However, since the problem didn't specify, maybe I should provide both.But given the complexity of the expressions, perhaps the intended solution is C = π/2, leading to a = sqrt(70)/10 and b = 3 sqrt(70)/10.Alternatively, maybe I made a mistake in assuming both cases are possible. Let me think again.Wait, when C = π/4, the function f(C) = 0, and when C = π/2, f(C) = 0 as well. So, both are valid. Therefore, the triangle could be either acute or right-angled at C, leading to two different sets of a and b.But the problem asks to "find the values of a and b", which might imply both solutions. However, in the initial problem statement, part (2) is a single question, so perhaps only one solution is expected. Maybe the problem expects the right-angled case because it leads to a simpler solution.Alternatively, perhaps the problem expects both solutions, but the user might have intended only one. Given the complexity, I think the intended solution is C = π/2, leading to a = sqrt(70)/10 and b = 3 sqrt(70)/10.But to be thorough, I should present both solutions.So, in conclusion:For part (1), the minimum value is -sqrt(2) - 1 and the period is π.For part (2), there are two possible solutions:Case 1: C = π/4a = sqrt(63*(10 + 3 sqrt(2))/82 ) ≈ 1.103b = 3 * sqrt(63*(10 + 3 sqrt(2))/82 ) ≈ 3.308Case 2: C = π/2a = sqrt(70)/10 ≈ 0.8367b = 3 sqrt(70)/10 ≈ 2.510But since the problem didn't specify, perhaps both are acceptable. However, the problem might expect the right-angled case, so I'll go with that.

💡Okay, so I have this quadrilateral with a dashed diagonal, and I need to figure out how many different whole numbers could be the length of that diagonal. The quadrilateral has sides labeled 8, 10, 16, and 12. Hmm, let me try to visualize it. There's an Asymptote figure provided, but I think I can just imagine it as a four-sided shape with those side lengths.First, I remember that in a quadrilateral, the length of a diagonal has to satisfy the triangle inequality in both triangles that the diagonal creates. So, if I draw the diagonal, it will split the quadrilateral into two triangles. Each of these triangles must satisfy the triangle inequality, which says that the sum of any two sides must be greater than the third side.Let me label the quadrilateral's vertices as A, B, C, D in order. The diagonal is from A to C, so it splits the quadrilateral into triangle ABC and triangle ADC. In triangle ABC, the sides are AB = 8, BC = 10, and AC = x (the diagonal we're trying to find). Applying the triangle inequality here:1. AB + BC > AC: 8 + 10 > x => 18 > x2. AB + AC > BC: 8 + x > 10 => x > 23. BC + AC > AB: 10 + x > 8 => x > -2 (but since lengths can't be negative, this doesn't add any new information)So from triangle ABC, x must be greater than 2 and less than 18.Now, looking at triangle ADC, the sides are AD = 12, DC = 16, and AC = x. Applying the triangle inequality here:1. AD + DC > AC: 12 + 16 > x => 28 > x2. AD + AC > DC: 12 + x > 16 => x > 43. DC + AC > AD: 16 + x > 12 => x > -4 (again, not useful since x is positive)So from triangle ADC, x must be greater than 4 and less than 28.Now, combining the inequalities from both triangles:From ABC: 2 < x < 18From ADC: 4 < x < 28The overlapping range where both conditions are satisfied is 4 < x < 18. So x must be greater than 4 and less than 18.But the question asks for how many different whole numbers x could be. So x can be 5, 6, 7, ..., up to 17.To count how many whole numbers that is, subtract the lower bound from the upper bound and add 1 (since both endpoints are inclusive):17 - 5 + 1 = 13So there are 13 different whole numbers that could be the length of the diagonal.Wait, let me double-check. If x is greater than 4, the smallest whole number is 5. If x is less than 18, the largest whole number is 17. So yes, from 5 to 17 inclusive, that's 13 numbers.I think that makes sense. The diagonal has to be long enough to satisfy both triangles, so it can't be too short or too long. The overlapping range gives the possible lengths, and counting them gives 13 whole numbers.**Final Answer**The number of different whole numbers that could be the length of the diagonal is boxed{13}.

💡Okay, so I need to figure out the volume of this cylindrical tank. It's 25 feet in diameter and 12 feet deep. Hmm, let's see. I remember that the formula for the volume of a cylinder is V = πr²h, where r is the radius and h is the height or depth. First, I need to find the radius. The diameter is 25 feet, so the radius would be half of that, right? So, 25 divided by 2 is 12.5 feet. Got that down. Now, plugging the numbers into the formula: V = π*(12.5)²*12. Let me calculate (12.5) squared first. 12.5 times 12.5 is... let's see, 12 times 12 is 144, and 0.5 times 0.5 is 0.25, but wait, that's not quite right. Actually, 12.5 squared is 156.25. Yeah, because 12*12 is 144, 12*0.5 is 6, 0.5*12 is another 6, and 0.5*0.5 is 0.25. So adding those up: 144 + 6 + 6 + 0.25 equals 156.25. Okay, so that part checks out.Next, multiply that by the height, which is 12 feet. So, 156.25 times 12. Let me do that step by step. 156 times 12 is... 156*10 is 1560, and 156*2 is 312, so adding those together gives 1872. Then, 0.25 times 12 is 3. So, adding that to 1872 gives 1875. So, the volume in cubic feet is 1875π.But the question asks for the volume in cubic yards. I remember that 1 yard is 3 feet, so 1 cubic yard is 3 feet by 3 feet by 3 feet, which is 27 cubic feet. Therefore, to convert cubic feet to cubic yards, I need to divide by 27. So, 1875 divided by 27. Let me do that division. 27 times 69 is 1863, because 27*70 is 1890, which is too much. So, 27*69 is 1863. Subtracting that from 1875 gives 12. So, 1875 divided by 27 is 69 with a remainder of 12. To express this as a fraction, 12/27 simplifies to 4/9. So, the volume is 69 and 4/9 π cubic yards.Wait, but the problem says to express the answer in terms of π and convert it into cubic yards. So, I think I can write it as 69.444...π cubic yards, but since 4/9 is approximately 0.444, that makes sense. Alternatively, I could leave it as a fraction, 69 4/9 π cubic yards.Let me double-check my calculations to make sure I didn't make any mistakes. Starting from the beginning: diameter 25 feet, radius 12.5 feet. Volume formula: πr²h. So, π*(12.5)²*12. 12.5 squared is 156.25, times 12 is 1875. So, 1875π cubic feet. Convert to cubic yards by dividing by 27: 1875/27. 27*69 is 1863, remainder 12, which is 4/9. So, 69 4/9 π cubic yards. Yep, that seems right.I think I've got it. The volume of the tank is 69 and 4/9 π cubic yards.