answer：Okay, so I have this problem where the terminal side of angle α passes through the point (3,4). I need to find sin α and cos α, and then calculate a trigonometric expression. Let me try to figure this out step by step.First, part (I): Find sin α and cos α.Hmm, I remember that in trigonometry, when you have a point (x, y) on the terminal side of an angle in standard position, you can find the sine and cosine using the coordinates and the distance from the origin. The distance r is calculated using the Pythagorean theorem, right? So, r is the square root of (x² + y²). Given the point (3,4), so x is 3 and y is 4. Let me compute r. r = √(3² + 4²) = √(9 + 16) = √25 = 5. Okay, so r is 5. That makes sense because 3-4-5 is a classic Pythagorean triple.Now, sin α is y/r, and cos α is x/r. So, sin α should be 4/5, and cos α should be 3/5. Let me write that down:sin α = 4/5 cos α = 3/5Wait, is that all? It seems straightforward. I think that's correct because the coordinates are positive, so the angle is in the first quadrant, and all trigonometric functions are positive there. Yeah, that makes sense.Moving on to part (II): Calculate the value of [2cos(π/2 - α) - cos(π + α)] / [2sin(π - α)].Hmm, this looks a bit more complicated. I need to simplify this expression. Maybe I can use some trigonometric identities to simplify each term.Let me recall some identities:1. cos(π/2 - α) is equal to sin α. That's one of the co-function identities.2. cos(π + α) is equal to -cos α. That's from the cosine addition formula or the unit circle; adding π to an angle reflects it to the opposite side, so cosine becomes negative.3. sin(π - α) is equal to sin α. That's another identity from the sine function's properties.Let me substitute these into the expression.First, the numerator: 2cos(π/2 - α) - cos(π + α) = 2 sin α - (-cos α) = 2 sin α + cos αAnd the denominator: 2 sin(π - α) = 2 sin αSo, putting it all together, the expression becomes:(2 sin α + cos α) / (2 sin α)Now, I can plug in the values of sin α and cos α that I found earlier. From part (I), sin α is 4/5 and cos α is 3/5.Substituting these in:Numerator: 2*(4/5) + (3/5) = (8/5) + (3/5) = 11/5 Denominator: 2*(4/5) = 8/5So, the expression becomes (11/5) divided by (8/5). Dividing fractions is the same as multiplying by the reciprocal, so:(11/5) * (5/8) = 11/8Wait, the 5s cancel out, so it's 11/8. That's a positive number, which makes sense because all the terms in the original expression should result in positive values given that α is in the first quadrant.Let me double-check my steps to make sure I didn't make a mistake.1. Calculated r correctly: √(9 + 16) = 5. Correct.2. sin α = 4/5, cos α = 3/5. Correct.3. Used identities: - cos(π/2 - α) = sin α. Correct. - cos(π + α) = -cos α. Correct. - sin(π - α) = sin α. Correct.4. Substituted into numerator: 2 sin α - (-cos α) = 2 sin α + cos α. Correct.5. Substituted into denominator: 2 sin α. Correct.6. Plugged in sin α = 4/5 and cos α = 3/5: - Numerator: 2*(4/5) + 3/5 = 8/5 + 3/5 = 11/5. Correct. - Denominator: 2*(4/5) = 8/5. Correct.7. Divided 11/5 by 8/5: (11/5)*(5/8) = 11/8. Correct.Everything seems to check out. I think I did this correctly. So, the final value is 11/8.**Final Answer**(I) (sin alpha = boxed{dfrac{4}{5}}) and (cos alpha = boxed{dfrac{3}{5}});(II) The value of the expression is (boxed{dfrac{11}{8}}).

answer：Okay, so I need to find the range of this function:[g(x) = left( arccos frac{x}{3} right)^2 + pi arcsin frac{x}{3} - left( arcsin frac{x}{3} right)^2 + frac{pi^2}{18} (x^2 + 9x + 27).]Hmm, that looks a bit complicated, but maybe I can simplify it. I remember that there's a relationship between arccos and arcsin. Specifically, for any x in the domain, arccos(x) + arcsin(x) = π/2. So, maybe I can use that identity here.Let me write that down:[arccos frac{x}{3} + arcsin frac{x}{3} = frac{pi}{2}.]That's true because the domain of x here is such that x/3 is between -1 and 1, so x is between -3 and 3. So, that identity holds for all x in [-3, 3].Looking back at the function g(x), I see that it has both arccos and arcsin terms. Maybe I can use the identity to substitute one in terms of the other. Let's see:Let me denote:[A = arccos frac{x}{3},][B = arcsin frac{x}{3}.]So, according to the identity, A + B = π/2. That might be useful.Now, let's rewrite g(x) in terms of A and B:[g(x) = A^2 + pi B - B^2 + frac{pi^2}{18}(x^2 + 9x + 27).]Hmm, so I have A^2 - B^2 + πB + the quadratic term.I notice that A^2 - B^2 is a difference of squares, so it can be factored as (A - B)(A + B). Let me try that:[A^2 - B^2 = (A - B)(A + B).]And since A + B = π/2, that becomes:[(A - B)left(frac{pi}{2}right).]So, substituting back into g(x):[g(x) = frac{pi}{2}(A - B) + pi B + frac{pi^2}{18}(x^2 + 9x + 27).]Let me simplify the terms involving A and B:First, expand the first term:[frac{pi}{2}A - frac{pi}{2}B + pi B.]Combine like terms:[frac{pi}{2}A + left(-frac{pi}{2}B + pi Bright).]Simplify the coefficients:[frac{pi}{2}A + left(frac{pi}{2}Bright).]So, that becomes:[frac{pi}{2}(A + B).]But wait, A + B is π/2, so:[frac{pi}{2} times frac{pi}{2} = frac{pi^2}{4}.]So, the entire expression simplifies to:[g(x) = frac{pi^2}{4} + frac{pi^2}{18}(x^2 + 9x + 27).]Wait, that seems a bit off. Let me double-check my steps.Starting from:[g(x) = frac{pi}{2}(A - B) + pi B + frac{pi^2}{18}(x^2 + 9x + 27).]Expanding:[frac{pi}{2}A - frac{pi}{2}B + pi B.]Combine the B terms:[-frac{pi}{2}B + pi B = frac{pi}{2}B.]So, total expression:[frac{pi}{2}A + frac{pi}{2}B = frac{pi}{2}(A + B).]Since A + B = π/2, this is:[frac{pi}{2} times frac{pi}{2} = frac{pi^2}{4}.]So, yes, that part is correct. So, g(x) simplifies to:[frac{pi^2}{4} + frac{pi^2}{18}(x^2 + 9x + 27).]Now, let's look at the quadratic term:[frac{pi^2}{18}(x^2 + 9x + 27).]I can factor out 9 from the quadratic:[x^2 + 9x + 27 = x^2 + 9x + frac{81}{4} + frac{81}{4} - 27.]Wait, that might not be the best approach. Alternatively, maybe complete the square.Let me try completing the square for x^2 + 9x + 27.First, take x^2 + 9x:x^2 + 9x = (x + 9/2)^2 - (81/4).So, adding 27:(x + 9/2)^2 - 81/4 + 27.Convert 27 to fourths: 27 = 108/4.So, (x + 9/2)^2 - 81/4 + 108/4 = (x + 9/2)^2 + 27/4.So, x^2 + 9x + 27 = (x + 9/2)^2 + 27/4.Therefore, the quadratic term becomes:[frac{pi^2}{18}left( (x + frac{9}{2})^2 + frac{27}{4} right).]But maybe that's not necessary. Alternatively, since the quadratic is in terms of x, and x is between -3 and 3, perhaps I can analyze the quadratic function over that interval.Wait, but let me first write the entire function:[g(x) = frac{pi^2}{4} + frac{pi^2}{18}(x^2 + 9x + 27).]Let me factor out π²/18:[g(x) = frac{pi^2}{18}(x^2 + 9x + 27) + frac{pi^2}{4}.]Alternatively, combine the constants:Let me compute the constants:First, write both terms with denominator 36:[frac{pi^2}{4} = frac{9pi^2}{36},][frac{pi^2}{18} = frac{2pi^2}{36}.]So, g(x) becomes:[frac{2pi^2}{36}(x^2 + 9x + 27) + frac{9pi^2}{36}.]Combine them:[frac{2pi^2(x^2 + 9x + 27) + 9pi^2}{36}.]Factor out π²:[frac{pi^2[2(x^2 + 9x + 27) + 9]}{36}.]Simplify inside the brackets:2x² + 18x + 54 + 9 = 2x² + 18x + 63.So,[g(x) = frac{pi^2(2x² + 18x + 63)}{36}.]Simplify numerator:Factor numerator:2x² + 18x + 63 = 2(x² + 9x) + 63.Wait, maybe factor out a 2:= 2(x² + 9x) + 63.But perhaps complete the square:x² + 9x = (x + 9/2)^2 - (81/4).So,2(x² + 9x) + 63 = 2[(x + 9/2)^2 - 81/4] + 63 = 2(x + 9/2)^2 - 81/2 + 63.Convert 63 to halves: 63 = 126/2.So,2(x + 9/2)^2 - 81/2 + 126/2 = 2(x + 9/2)^2 + 45/2.So,g(x) = π² [2(x + 9/2)^2 + 45/2] / 36.Simplify:= π² [2(x + 9/2)^2 + 45/2] / 36.Factor out 2:= π² [2{(x + 9/2)^2 + 45/4}] / 36.= π² [ (x + 9/2)^2 + 45/4 ] / 18.Hmm, not sure if that helps. Maybe instead, since x is in [-3, 3], I can analyze the quadratic function x² + 9x + 27 over this interval.Let me consider f(x) = x² + 9x + 27.It's a quadratic function opening upwards (since coefficient of x² is positive). Its vertex is at x = -b/(2a) = -9/(2*1) = -4.5.But our domain is x ∈ [-3, 3]. Since the vertex is at x = -4.5, which is outside the domain, the function is increasing on the interval [-3, 3] because the vertex is to the left of -3.Therefore, the minimum of f(x) on [-3, 3] is at x = -3, and the maximum is at x = 3.Compute f(-3):(-3)^2 + 9*(-3) + 27 = 9 - 27 + 27 = 9.Compute f(3):3² + 9*3 + 27 = 9 + 27 + 27 = 63.So, f(x) ranges from 9 to 63 as x goes from -3 to 3.Therefore, the quadratic term (π²/18)(x² + 9x + 27) ranges from (π²/18)*9 = π²/2 to (π²/18)*63 = (63/18)π² = (7/2)π².Wait, 63 divided by 18 is 3.5, which is 7/2.So, the quadratic term ranges from π²/2 to (7/2)π².Then, adding the constant term π²/4:So, g(x) = quadratic term + π²/4.Therefore, the minimum value of g(x) is π²/2 + π²/4 = (2π²/4 + π²/4) = 3π²/4.And the maximum value is (7/2)π² + π²/4 = (14π²/4 + π²/4) = 15π²/4.Wait, but earlier I thought the function simplified to π²/4 + quadratic term. But according to this, the range would be [3π²/4, 15π²/4].But wait, let me double-check my earlier steps because I might have made a mistake in simplifying.Wait, initially, I had:g(x) = π²/4 + (π²/18)(x² + 9x + 27).So, when x = -3, f(x) = 9, so:g(-3) = π²/4 + (π²/18)*9 = π²/4 + π²/2 = (π²/4 + 2π²/4) = 3π²/4.Similarly, when x = 3, f(x) = 63, so:g(3) = π²/4 + (π²/18)*63 = π²/4 + (7π²/2) = π²/4 + 14π²/4 = 15π²/4.So, that seems correct.But wait, in the initial simplification, I thought the function simplified to π²/6 + (π²/18)(x + 3)^2, but that might have been a mistake.Wait, let me go back to the initial steps.I had:g(x) = A² + πB - B² + (π²/18)(x² + 9x + 27).Then, I factored A² - B² as (A - B)(A + B) = (A - B)(π/2).Then, substituted back:g(x) = (π/2)(A - B) + πB + (π²/18)(x² + 9x + 27).Then, expanded:= (π/2)A - (π/2)B + πB + quadratic term.= (π/2)A + (π/2)B + quadratic term.= (π/2)(A + B) + quadratic term.Since A + B = π/2, this becomes:= (π/2)(π/2) + quadratic term.= π²/4 + quadratic term.So, that seems correct.Therefore, the function is indeed π²/4 + (π²/18)(x² + 9x + 27).Thus, as x varies from -3 to 3, the quadratic term varies from 9 to 63, so the function varies from π²/4 + π²/2 = 3π²/4 to π²/4 + 7π²/2 = 15π²/4.Wait, but in the initial problem, the user had a different simplification, arriving at π²/6 + (π²/18)(x + 3)^2, which led to a different range.I think I must have made a mistake in my initial steps. Let me check again.Wait, the user's initial steps were:They had g(x) = (arccos(x/3))² + π arcsin(x/3) - (arcsin(x/3))² + (π²/18)(x² + 9x + 27).They set A = arccos(x/3), B = arcsin(x/3), so A + B = π/2.Then, they rewrote g(x) as A² - B² + πB + quadratic term.Then, factored A² - B² as (A - B)(A + B) = (A - B)(π/2).So, g(x) = (π/2)(A - B) + πB + quadratic term.Then, expanded:= (π/2)A - (π/2)B + πB + quadratic term.= (π/2)A + (π/2)B + quadratic term.= (π/2)(A + B) + quadratic term.= (π/2)(π/2) + quadratic term.= π²/4 + quadratic term.But wait, the user then said that the quadratic term was (π²/18)(x + 3)^2, but I don't see how that comes from x² + 9x + 27.Wait, x² + 9x + 27 can be written as (x + 3)^2 + 18, because:(x + 3)^2 = x² + 6x + 9.So, x² + 9x + 27 = (x + 3)^2 + 3x + 18.Wait, that doesn't seem right. Let me compute (x + 3)^2:(x + 3)^2 = x² + 6x + 9.So, x² + 9x + 27 = (x + 3)^2 + 3x + 18.Hmm, that doesn't seem helpful. Alternatively, perhaps complete the square differently.Wait, x² + 9x + 27.Let me complete the square:x² + 9x = (x + 9/2)^2 - (81/4).So, x² + 9x + 27 = (x + 9/2)^2 - 81/4 + 27.Convert 27 to fourths: 27 = 108/4.So, (x + 9/2)^2 + (108/4 - 81/4) = (x + 9/2)^2 + 27/4.So, x² + 9x + 27 = (x + 9/2)^2 + 27/4.Therefore, the quadratic term is:(π²/18)[(x + 9/2)^2 + 27/4].But since x is in [-3, 3], the term (x + 9/2) varies from (-3 + 9/2) = 3/2 to (3 + 9/2) = 15/2.So, (x + 9/2)^2 varies from (3/2)^2 = 9/4 to (15/2)^2 = 225/4.Thus, the quadratic term becomes:(π²/18)[(x + 9/2)^2 + 27/4] = (π²/18)(x + 9/2)^2 + (π²/18)(27/4).Simplify the constant term:(π²/18)(27/4) = (3π²/4).So, g(x) = π²/4 + (π²/18)(x + 9/2)^2 + 3π²/4.Wait, that can't be right because earlier I had g(x) = π²/4 + quadratic term, but now it's π²/4 + something else.Wait, no, let me re-express:g(x) = π²/4 + (π²/18)[(x + 9/2)^2 + 27/4].= π²/4 + (π²/18)(x + 9/2)^2 + (π²/18)(27/4).= π²/4 + (π²/18)(x + 9/2)^2 + (3π²/4).So, combining constants:π²/4 + 3π²/4 = π².Thus, g(x) = π² + (π²/18)(x + 9/2)^2.Wait, that seems different from before. So, according to this, g(x) is π² plus a positive term, so the minimum value is π², and the maximum is when (x + 9/2)^2 is maximum.But x is in [-3, 3], so (x + 9/2) ranges from (-3 + 4.5) = 1.5 to (3 + 4.5) = 7.5.Thus, (x + 9/2)^2 ranges from (1.5)^2 = 2.25 to (7.5)^2 = 56.25.Therefore, the quadratic term (π²/18)(x + 9/2)^2 ranges from (π²/18)*2.25 = (π²/18)*(9/4) = π²/8 to (π²/18)*56.25 = (π²/18)*(225/4) = (25π²/8).So, g(x) ranges from π² + π²/8 = 9π²/8 to π² + 25π²/8 = 33π²/8.Wait, but earlier I had a different range. This is confusing.I think I made a mistake in the initial steps. Let me try a different approach.Let me go back to the original function:g(x) = (arccos(x/3))² + π arcsin(x/3) - (arcsin(x/3))² + (π²/18)(x² + 9x + 27).Let me denote B = arcsin(x/3). Then, arccos(x/3) = π/2 - B.So, substitute into g(x):g(x) = (π/2 - B)^2 + πB - B² + (π²/18)(x² + 9x + 27).Expand (π/2 - B)^2:= π²/4 - πB + B² + πB - B² + (π²/18)(x² + 9x + 27).Simplify term by term:π²/4 - πB + B² + πB - B².The -πB and +πB cancel, as do B² and -B².So, we're left with π²/4 + (π²/18)(x² + 9x + 27).So, that's the same as before: g(x) = π²/4 + (π²/18)(x² + 9x + 27).Therefore, the function is indeed π²/4 plus a quadratic function in x.Now, since x is in [-3, 3], let's analyze the quadratic term:f(x) = x² + 9x + 27.As before, it's a parabola opening upwards with vertex at x = -9/2 = -4.5, which is outside our domain.Therefore, on [-3, 3], the function f(x) is increasing because the vertex is to the left of -3.Thus, f(-3) = 9 - 27 + 27 = 9.f(3) = 9 + 27 + 27 = 63.So, f(x) ranges from 9 to 63.Therefore, the quadratic term (π²/18)f(x) ranges from (π²/18)*9 = π²/2 to (π²/18)*63 = 7π²/2.Adding the constant term π²/4, the total function g(x) ranges from:Minimum: π²/4 + π²/2 = 3π²/4.Maximum: π²/4 + 7π²/2 = 15π²/4.So, the range of g(x) is [3π²/4, 15π²/4].Wait, but the user's initial answer was [π²/6, 25π²/6], which is different.I think I must have made a mistake in my simplification. Let me check again.Wait, when I substituted A = π/2 - B, I had:g(x) = (π/2 - B)^2 + πB - B² + quadratic term.Expanding (π/2 - B)^2:= π²/4 - πB + B².So, adding the rest:π²/4 - πB + B² + πB - B² + quadratic term.Yes, the -πB and +πB cancel, as do B² and -B².So, g(x) = π²/4 + quadratic term.So, that part is correct.Therefore, the function is π²/4 + (π²/18)(x² + 9x + 27).Thus, the range is from 3π²/4 to 15π²/4.But wait, let me compute 3π²/4 and 15π²/4 in terms of π²:3π²/4 = 0.75π²,15π²/4 = 3.75π².But the user's answer was [π²/6, 25π²/6], which is approximately [0.166π², 4.166π²].So, my result is different. I must have made a mistake.Wait, perhaps I made a mistake in simplifying the quadratic term.Wait, let me compute the quadratic term again:(π²/18)(x² + 9x + 27).At x = -3:= (π²/18)(9 - 27 + 27) = (π²/18)(9) = π²/2.At x = 3:= (π²/18)(9 + 27 + 27) = (π²/18)(63) = 7π²/2.So, adding π²/4:At x = -3: π²/4 + π²/2 = 3π²/4.At x = 3: π²/4 + 7π²/2 = 15π²/4.So, that seems correct.But the user's answer was [π²/6, 25π²/6], which is different.Wait, perhaps I made a mistake in the initial substitution.Wait, let me check the initial function again:g(x) = (arccos(x/3))² + π arcsin(x/3) - (arcsin(x/3))² + (π²/18)(x² + 9x + 27).I set A = arccos(x/3), B = arcsin(x/3).So, A + B = π/2.Then, g(x) = A² + πB - B² + quadratic term.Then, A² - B² = (A - B)(A + B) = (A - B)(π/2).So, g(x) = (π/2)(A - B) + πB + quadratic term.= (π/2)A - (π/2)B + πB + quadratic term.= (π/2)A + (π/2)B + quadratic term.= (π/2)(A + B) + quadratic term.= (π/2)(π/2) + quadratic term.= π²/4 + quadratic term.Yes, that seems correct.So, the function is π²/4 + (π²/18)(x² + 9x + 27).Thus, the range is [3π²/4, 15π²/4].But the user's initial answer was [π²/6, 25π²/6].Wait, perhaps the user made a mistake in their simplification.Alternatively, maybe I made a mistake in interpreting the quadratic term.Wait, let me compute the quadratic term again:x² + 9x + 27.At x = -3: 9 - 27 + 27 = 9.At x = 3: 9 + 27 + 27 = 63.So, the quadratic term is between 9 and 63.Thus, (π²/18)*9 = π²/2,(π²/18)*63 = 7π²/2.Adding π²/4:Minimum: π²/2 + π²/4 = 3π²/4,Maximum: 7π²/2 + π²/4 = 15π²/4.So, the range is [3π²/4, 15π²/4].But the user's answer was [π²/6, 25π²/6].Wait, perhaps the user made a mistake in their simplification.Alternatively, maybe I made a mistake in the initial steps.Wait, let me check the initial function again:g(x) = (arccos(x/3))² + π arcsin(x/3) - (arcsin(x/3))² + (π²/18)(x² + 9x + 27).I set A = arccos(x/3), B = arcsin(x/3).So, A + B = π/2.Then, g(x) = A² + πB - B² + quadratic term.Then, A² - B² = (A - B)(A + B) = (A - B)(π/2).So, g(x) = (π/2)(A - B) + πB + quadratic term.= (π/2)A - (π/2)B + πB + quadratic term.= (π/2)A + (π/2)B + quadratic term.= (π/2)(A + B) + quadratic term.= (π/2)(π/2) + quadratic term.= π²/4 + quadratic term.Yes, that seems correct.Therefore, the function is π²/4 + (π²/18)(x² + 9x + 27).Thus, the range is [3π²/4, 15π²/4].But the user's initial answer was [π²/6, 25π²/6].Wait, perhaps the user made a mistake in their simplification.Alternatively, maybe I made a mistake in the initial steps.Wait, let me try to compute g(x) at x = -3 and x = 3 to see what values I get.At x = -3:arccos(-3/3) = arccos(-1) = π.arcsin(-3/3) = arcsin(-1) = -π/2.So, g(-3) = (π)^2 + π*(-π/2) - (-π/2)^2 + (π²/18)(9 - 27 + 27).Compute each term:(π)^2 = π²,π*(-π/2) = -π²/2,(-π/2)^2 = π²/4,Quadratic term: (π²/18)(9) = π²/2.So, g(-3) = π² - π²/2 - π²/4 + π²/2.Simplify:π² - π²/2 = π²/2,π²/2 - π²/4 = π²/4,π²/4 + π²/2 = 3π²/4.So, g(-3) = 3π²/4.Similarly, at x = 3:arccos(3/3) = arccos(1) = 0,arcsin(3/3) = arcsin(1) = π/2.So, g(3) = (0)^2 + π*(π/2) - (π/2)^2 + (π²/18)(9 + 27 + 27).Compute each term:0,π*(π/2) = π²/2,(π/2)^2 = π²/4,Quadratic term: (π²/18)(63) = 7π²/2.So, g(3) = 0 + π²/2 - π²/4 + 7π²/2.Simplify:π²/2 - π²/4 = π²/4,π²/4 + 7π²/2 = π²/4 + 14π²/4 = 15π²/4.So, g(3) = 15π²/4.Therefore, the function ranges from 3π²/4 to 15π²/4.Thus, the range is [3π²/4, 15π²/4].But the user's initial answer was [π²/6, 25π²/6], which is different.I think the user made a mistake in their simplification. They probably made an error when they said:g(x) = π²/6 + (π²/18)(x + 3)^2.Because when I compute g(x) at x = -3, I get 3π²/4, which is approximately 2.356, while π²/6 is approximately 1.645, which is lower.Similarly, at x = 3, I get 15π²/4 ≈ 11.781, while 25π²/6 ≈ 13.089, which is higher.So, the user's answer is incorrect.Therefore, the correct range is [3π²/4, 15π²/4].But wait, let me check if the quadratic term can be expressed differently.Wait, x² + 9x + 27 can be written as (x + 3)^2 + 18, because:(x + 3)^2 = x² + 6x + 9,So, x² + 9x + 27 = (x + 3)^2 + 3x + 18.Wait, that doesn't seem helpful.Alternatively, perhaps the user made a mistake in completing the square.Wait, let me try again:x² + 9x + 27.Complete the square:x² + 9x = (x + 9/2)^2 - (81/4).So, x² + 9x + 27 = (x + 9/2)^2 - 81/4 + 27.Convert 27 to fourths: 27 = 108/4.So, (x + 9/2)^2 + (108/4 - 81/4) = (x + 9/2)^2 + 27/4.Thus, x² + 9x + 27 = (x + 9/2)^2 + 27/4.So, the quadratic term is:(π²/18)[(x + 9/2)^2 + 27/4].= (π²/18)(x + 9/2)^2 + (π²/18)(27/4).= (π²/18)(x + 9/2)^2 + (3π²/4).So, g(x) = π²/4 + (π²/18)(x + 9/2)^2 + 3π²/4.= π²/4 + 3π²/4 + (π²/18)(x + 9/2)^2.= π² + (π²/18)(x + 9/2)^2.So, g(x) = π² + (π²/18)(x + 9/2)^2.Now, since x is in [-3, 3], (x + 9/2) ranges from (-3 + 4.5) = 1.5 to (3 + 4.5) = 7.5.Thus, (x + 9/2)^2 ranges from (1.5)^2 = 2.25 to (7.5)^2 = 56.25.Therefore, (π²/18)(x + 9/2)^2 ranges from (π²/18)*2.25 = π²/8 to (π²/18)*56.25 = 25π²/8.Thus, g(x) ranges from π² + π²/8 = 9π²/8 to π² + 25π²/8 = 33π²/8.Wait, that's different from before. But earlier, when I computed g(-3) and g(3), I got 3π²/4 and 15π²/4, which are 6π²/8 and 30π²/8, respectively.But according to this, g(x) ranges from 9π²/8 to 33π²/8, which is approximately 3.534 to 12.723.But when I computed g(-3) and g(3), I got 3π²/4 ≈ 2.356 and 15π²/4 ≈ 11.781.So, there's a discrepancy here.I think the mistake is in the way I expressed the quadratic term.Wait, when I wrote g(x) = π² + (π²/18)(x + 9/2)^2, that's correct.But when I computed g(-3), I got 3π²/4, but according to this expression, g(-3) should be π² + (π²/18)(1.5)^2 = π² + (π²/18)(2.25) = π² + π²/8 = 9π²/8 ≈ 3.534.But earlier, when I computed g(-3) directly, I got 3π²/4 ≈ 2.356.This inconsistency suggests that there's a mistake in the simplification.Wait, let me re-express g(x):g(x) = π²/4 + (π²/18)(x² + 9x + 27).But x² + 9x + 27 = (x + 9/2)^2 + 27/4.So, g(x) = π²/4 + (π²/18)( (x + 9/2)^2 + 27/4 ).= π²/4 + (π²/18)(x + 9/2)^2 + (π²/18)(27/4).= π²/4 + (π²/18)(x + 9/2)^2 + 3π²/4.= π²/4 + 3π²/4 + (π²/18)(x + 9/2)^2.= π² + (π²/18)(x + 9/2)^2.Yes, that's correct.But when I plug x = -3 into this expression:g(-3) = π² + (π²/18)(1.5)^2 = π² + (π²/18)(2.25) = π² + π²/8 = 9π²/8 ≈ 3.534.But earlier, when I computed g(-3) directly, I got 3π²/4 ≈ 2.356.This is a contradiction, which means I must have made a mistake in my simplification.Wait, let me compute g(-3) again directly:g(-3) = (arccos(-1))² + π arcsin(-1) - (arcsin(-1))² + (π²/18)(9 - 27 + 27).= (π)^2 + π*(-π/2) - (π/2)^2 + (π²/18)(9).= π² - π²/2 - π²/4 + π²/2.Simplify:π² - π²/2 = π²/2,π²/2 - π²/4 = π²/4,π²/4 + π²/2 = 3π²/4.So, g(-3) = 3π²/4.But according to the expression g(x) = π² + (π²/18)(x + 9/2)^2, g(-3) = π² + (π²/18)(1.5)^2 = π² + π²/8 = 9π²/8.This is a contradiction, meaning that my simplification is wrong.Therefore, I must have made a mistake in the simplification steps.Wait, let me go back to the initial substitution.I had:g(x) = A² + πB - B² + quadratic term.Then, A² - B² = (A - B)(A + B) = (A - B)(π/2).So, g(x) = (π/2)(A - B) + πB + quadratic term.= (π/2)A - (π/2)B + πB + quadratic term.= (π/2)A + (π/2)B + quadratic term.= (π/2)(A + B) + quadratic term.= (π/2)(π/2) + quadratic term.= π²/4 + quadratic term.But when I compute g(-3), I get 3π²/4, which is π²/4 + π²/2.So, the quadratic term at x = -3 is π²/2.But according to the expression, the quadratic term is (π²/18)(x² + 9x + 27).At x = -3, that's (π²/18)(9 - 27 + 27) = (π²/18)(9) = π²/2.So, g(-3) = π²/4 + π²/2 = 3π²/4.Similarly, at x = 3, quadratic term is (π²/18)(63) = 7π²/2.So, g(3) = π²/4 + 7π²/2 = 15π²/4.Thus, the function is correctly simplified as π²/4 + (π²/18)(x² + 9x + 27).Therefore, the range is [3π²/4, 15π²/4].But the user's initial answer was [π²/6, 25π²/6], which is different.I think the user made a mistake in their simplification, perhaps incorrectly completing the square or misapplying the identity.Therefore, the correct range is [3π²/4, 15π²/4].But to express this in terms of π², it's:3π²/4 = (3/4)π²,15π²/4 = (15/4)π².Alternatively, in terms of fractions with denominator 6:3π²/4 = 9π²/12 = 3π²/4,15π²/4 = 45π²/12 = 15π²/4.But perhaps it's better to leave it as [3π²/4, 15π²/4].Alternatively, factor out π²/4:= π²/4 [3, 15].But I think the simplest form is [3π²/4, 15π²/4].Therefore, the range of g(x) is [3π²/4, 15π²/4].

answer：First, I need to calculate the total cost for the tickets bought separately. This includes 20 students, 12 adults, 15 children, and 10 senior citizens. Each group has a specific ticket price: 6 for students, 8 for adults, 4 for children, and 7 for senior citizens.Next, I'll calculate the total cost for the group of 32 people who purchased tickets under the group discount. The group consists of 5 students, 8 adults, 10 children, and 9 senior citizens. I'll apply the same ticket prices as above to find the total cost before the discount.Since the group purchased more than 10 tickets, they are eligible for a 10% discount. I'll calculate the discount amount by taking 10% of the total cost before the discount.After applying the discount, I'll determine the total cost for the group purchase. Finally, I'll add the total costs from the separate purchases and the group purchase to find the overall total dollars' worth of tickets sold.

answer：Okay, so I have this problem where I need to find constants ( a ) and ( b ) such that when I multiply ( a ) by the vector ( begin{pmatrix} 1 4 end{pmatrix} ) and ( b ) by the vector ( begin{pmatrix} -1 6 end{pmatrix} ), and then add those two results together, I get the vector ( begin{pmatrix} 0 5 end{pmatrix} ). Hmm, let me write that out to make it clearer. The equation is:[ a begin{pmatrix} 1 4 end{pmatrix} + b begin{pmatrix} -1 6 end{pmatrix} = begin{pmatrix} 0 5 end{pmatrix} ]So, breaking this down into components, I can set up two separate equations, one for the first component (the top number) and one for the second component (the bottom number). For the first component, it's ( a times 1 + b times (-1) ), which should equal 0. So that gives me:[ a - b = 0 ]And for the second component, it's ( a times 4 + b times 6 ), which should equal 5. So that gives me:[ 4a + 6b = 5 ]Alright, so now I have a system of two equations:1. ( a - b = 0 )2. ( 4a + 6b = 5 )I need to solve this system to find the values of ( a ) and ( b ). Let me see. From the first equation, ( a - b = 0 ), I can solve for one variable in terms of the other. If I add ( b ) to both sides, I get:[ a = b ]So, ( a ) is equal to ( b ). That simplifies things because I can substitute ( a ) for ( b ) in the second equation. Let me do that.Substituting ( a = b ) into the second equation:[ 4a + 6a = 5 ]Wait, that's because ( b ) is ( a ), so ( 6b ) becomes ( 6a ). So, combining like terms:[ 4a + 6a = 10a ]So, ( 10a = 5 ). To solve for ( a ), I can divide both sides by 10:[ a = frac{5}{10} ][ a = 0.5 ]Since ( a = b ), that means ( b ) is also 0.5. So, both constants are 0.5. Let me check my work to make sure I didn't make a mistake. I'll plug ( a = 0.5 ) and ( b = 0.5 ) back into the original vector equation.Calculating the first component:[ 0.5 times 1 + 0.5 times (-1) = 0.5 - 0.5 = 0 ]That's correct because the first component of the resulting vector should be 0.Now, the second component:[ 0.5 times 4 + 0.5 times 6 = 2 + 3 = 5 ]That's also correct because the second component of the resulting vector should be 5.So, both components check out. Therefore, my solution seems to be correct.I think I've got it. The constants ( a ) and ( b ) are both 0.5.**Final Answer**The constants are ( a = boxed{0.5} ) and ( b = boxed{0.5} ).

answer：Okay, so I need to construct an isosceles triangle where the orthocenter bisects the altitude to the base. The base and the radius of the circle touching the extensions of the two legs are given. Hmm, let me try to break this down step by step.First, let me recall what an isosceles triangle is. It's a triangle with at least two sides equal. In this case, since it's an isosceles triangle with the base given, I can assume that the two legs (the equal sides) are the ones that are equal. So, if I denote the triangle as ABC with base BC, then AB = AC.Now, the orthocenter is the point where the three altitudes of the triangle intersect. In an isosceles triangle, the altitude from the apex (which is A in this case) to the base BC is also the median and the angle bisector. So, this altitude should be bisected by the orthocenter. That means the orthocenter lies exactly halfway along this altitude.Wait, so if the orthocenter bisects the altitude, that tells me something about the position of the orthocenter relative to the triangle. In an isosceles triangle, the orthocenter lies along the altitude from the apex. So, if it bisects this altitude, the orthocenter is at the midpoint of the altitude.Let me try to visualize this. If I draw triangle ABC with base BC and apex A, then the altitude from A to BC is AA1, where A1 is the midpoint of BC. The orthocenter H is the midpoint of AA1. So, H is halfway between A and A1.Now, the problem also mentions a circle that touches the extensions of the two legs. I think this is referring to an ex-circle, specifically the A-excircle, which touches the extensions of AB and AC and the base BC. The radius of this excircle is given.I remember that the radius of an excircle can be calculated using the formula:r_a = Δ / (s - a)where Δ is the area of the triangle, s is the semiperimeter, and a is the side opposite the vertex (in this case, BC).But wait, in this problem, the radius is given, so maybe I can use this formula to find some relationship between the sides of the triangle.Let me denote the length of the base BC as 'b' and the radius of the excircle as 'r'. Since the triangle is isosceles, the two equal sides AB and AC can be denoted as 'c'. The altitude AA1 can be calculated using the Pythagorean theorem:AA1 = sqrt(c² - (b/2)²)Since the orthocenter H is the midpoint of AA1, the length from A to H is half of AA1:AH = (1/2) * sqrt(c² - (b/2)²)But in an isosceles triangle, the orthocenter, centroid, and circumcenter all lie along the same line (the altitude AA1). However, their positions differ. The centroid divides the altitude in a 2:1 ratio, but the orthocenter here is bisecting the altitude, so it's different.Wait, maybe I can use coordinate geometry to model this. Let's place the triangle in a coordinate system to make it easier.Let me set point A at (0, h), and base BC on the x-axis with points B at (-b/2, 0) and C at (b/2, 0). The midpoint A1 is at (0, 0). The altitude AA1 is along the y-axis from (0, h) to (0, 0).The orthocenter H is the midpoint of AA1, so it's at (0, h/2).Now, the excircle opposite to A touches the extensions of AB and AC and the base BC. The center of this excircle, let's call it I_a, is located at a distance equal to the radius 'r' from each of these sides.In coordinate terms, the excenter opposite to A can be found using the formula:I_a = ( ( -a*A_x + b*B_x + c*C_x ) / ( -a + b + c ), ( -a*A_y + b*B_y + c*C_y ) / ( -a + b + c ) )But since this is an isosceles triangle, things might simplify.Wait, in our coordinate system, points are A(0, h), B(-b/2, 0), C(b/2, 0). The sides AB and AC have length c, and BC has length b.The exradius r_a is given by:r_a = Δ / (s - a)where Δ is the area, s is the semiperimeter, and a is BC.First, let's compute Δ, the area. Since it's an isosceles triangle, area is (base * height)/2 = (b * h)/2.The semiperimeter s is (AB + AC + BC)/2 = (c + c + b)/2 = (2c + b)/2 = c + b/2.So, s - a = (c + b/2) - b = c - b/2.Thus, r_a = ( (b * h)/2 ) / (c - b/2 ) = (b h) / (2(c - b/2)) = (b h) / (2c - b).But we are given r_a = r, so:r = (b h) / (2c - b)So, we have:r = (b h) / (2c - b)We can write this as:2c - b = (b h)/rSo,2c = (b h)/r + bc = (b h)/(2r) + b/2But we also know that in the triangle, the altitude h is related to c and b/2 by the Pythagorean theorem:h = sqrt(c² - (b/2)²)So, h = sqrt(c² - b²/4)Let me substitute h into the equation for c:c = (b * sqrt(c² - b²/4)) / (2r) + b/2This seems a bit complicated, but maybe I can solve for c in terms of b and r.Let me denote sqrt(c² - b²/4) as h, so:c = (b h)/(2r) + b/2But h = sqrt(c² - b²/4), so:c = (b sqrt(c² - b²/4))/(2r) + b/2Let me rearrange this equation:c - b/2 = (b sqrt(c² - b²/4))/(2r)Multiply both sides by 2r:2r(c - b/2) = b sqrt(c² - b²/4)Let me square both sides to eliminate the square root:[2r(c - b/2)]² = [b sqrt(c² - b²/4)]²Left side:4r²(c - b/2)² = 4r²(c² - b c + b²/4)Right side:b²(c² - b²/4)So, expanding left side:4r²c² - 4r²b c + r²b² = b²c² - (b^4)/4Bring all terms to one side:4r²c² - 4r²b c + r²b² - b²c² + (b^4)/4 = 0Combine like terms:(4r²c² - b²c²) + (-4r²b c) + (r²b² + (b^4)/4) = 0Factor c²:c²(4r² - b²) - 4r²b c + (r²b² + (b^4)/4) = 0This is a quadratic equation in terms of c:(4r² - b²)c² - 4r²b c + (r²b² + (b^4)/4) = 0Let me write it as:(4r² - b²)c² - 4r²b c + (r²b² + (b^4)/4) = 0Let me denote this as:A c² + B c + C = 0Where:A = 4r² - b²B = -4r²bC = r²b² + (b^4)/4We can solve for c using the quadratic formula:c = [4r²b ± sqrt( (4r²b)^2 - 4*(4r² - b²)*(r²b² + b^4/4) ) ] / [2*(4r² - b²)]Let me compute the discriminant D:D = (4r²b)^2 - 4*(4r² - b²)*(r²b² + b^4/4)First, compute (4r²b)^2:= 16r^4b²Now, compute 4*(4r² - b²)*(r²b² + b^4/4):First, compute (4r² - b²)*(r²b² + b^4/4):= 4r²*(r²b² + b^4/4) - b²*(r²b² + b^4/4)= 4r²*r²b² + 4r²*(b^4)/4 - b²*r²b² - b²*(b^4)/4Simplify each term:= 4r^4b² + r²b^4 - r²b^4 - (b^6)/4Simplify:= 4r^4b² - (b^6)/4So, 4*(4r² - b²)*(r²b² + b^4/4) = 4*(4r^4b² - (b^6)/4) = 16r^4b² - b^6Thus, discriminant D:= 16r^4b² - (16r^4b² - b^6) = 16r^4b² -16r^4b² + b^6 = b^6So, D = b^6Therefore, sqrt(D) = b^3So, plugging back into the quadratic formula:c = [4r²b ± b^3] / [2*(4r² - b²)]Factor numerator:= b[4r² ± b²] / [2*(4r² - b²)]So, two solutions:1. c = b[4r² + b²] / [2*(4r² - b²)]2. c = b[4r² - b²] / [2*(4r² - b²)] = b/2But c must be greater than b/2 because in a triangle, the sum of two sides must be greater than the third side. If c = b/2, then AB + AC = b, which is equal to BC, which would make it degenerate. So, we discard c = b/2.Thus, c = b[4r² + b²] / [2*(4r² - b²)]Simplify:c = [b(4r² + b²)] / [2(4r² - b²)]So, now we have c in terms of b and r.But we also have h = sqrt(c² - (b/2)²)So, let's compute h:h = sqrt(c² - b²/4)Plugging c:= sqrt( [ (b(4r² + b²))/(2(4r² - b²)) ) ]² - b²/4 )Let me compute this step by step.First, compute c²:c² = [b²(4r² + b²)^2] / [4(4r² - b²)^2]So,h² = c² - b²/4 = [b²(4r² + b²)^2] / [4(4r² - b²)^2] - b²/4Factor b²/4:= (b²/4)[ (4r² + b²)^2 / (4r² - b²)^2 - 1 ]Compute the term in brackets:= [ (4r² + b²)^2 - (4r² - b²)^2 ] / (4r² - b²)^2Expand numerator:= [ (16r^4 + 8r²b² + b^4) - (16r^4 - 8r²b² + b^4) ] / (4r² - b²)^2Simplify numerator:= (16r^4 + 8r²b² + b^4 -16r^4 +8r²b² -b^4) = 16r²b²Thus,h² = (b²/4) * [16r²b² / (4r² - b²)^2 ] = (b²/4) * (16r²b²) / (4r² - b²)^2Simplify:= (4r²b^4) / (4r² - b²)^2Thus,h = (2r b²) / (4r² - b²)So, h = (2r b²)/(4r² - b²)Now, recall that the orthocenter H is at (0, h/2) in our coordinate system.But in an isosceles triangle, the orthocenter, centroid, and circumcenter all lie along the altitude. However, their positions are different.Wait, in our earlier analysis, we found that the orthocenter is the midpoint of the altitude, so H is at (0, h/2). But in general, for an isosceles triangle, the centroid is at (0, h/3), and the circumcenter is at (0, h/2) if the triangle is acute, but actually, in an isosceles triangle, the circumradius can be found using the formula:R = (c²) / (2h)Wait, let me recall the formula for the circumradius R of a triangle:R = (a b c) / (4Δ)In our case, sides are c, c, b. So,R = (c * c * b) / (4 * (b h / 2)) = (c² b) / (2 b h) = c² / (2 h)So, R = c² / (2 h)But in our coordinate system, the circumcenter should lie along the altitude at (0, k). Let's find k.In an isosceles triangle, the circumcenter is located at a distance R from each vertex. So, the distance from (0, k) to A(0, h) is |h - k| = RSimilarly, the distance from (0, k) to B(-b/2, 0) is sqrt( (b/2)^2 + k² ) = RThus,sqrt( (b/2)^2 + k² ) = |h - k|Square both sides:(b²)/4 + k² = (h - k)² = h² - 2 h k + k²Simplify:(b²)/4 = h² - 2 h kThus,2 h k = h² - (b²)/4So,k = (h² - (b²)/4) / (2 h) = (h²)/(2 h) - (b²)/(8 h) = h/2 - b²/(8 h)But earlier, we found that h = (2 r b²)/(4 r² - b²)So, let's compute k:k = h/2 - b²/(8 h)= [ (2 r b²)/(4 r² - b²) ] / 2 - b² / [8 * (2 r b²)/(4 r² - b²) ]Simplify:= (r b²)/(4 r² - b²) - [ b² (4 r² - b²) ] / (16 r b² )Simplify second term:= (r b²)/(4 r² - b²) - (4 r² - b²)/(16 r )So, k = (r b²)/(4 r² - b²) - (4 r² - b²)/(16 r )Hmm, this seems a bit messy. Maybe I made a mistake earlier.Wait, let's recall that in our coordinate system, the circumradius R is equal to the distance from the circumcenter (0, k) to A(0, h), which is |h - k|, and also equal to the distance from (0, k) to B(-b/2, 0), which is sqrt( (b/2)^2 + k² ). So, setting these equal:sqrt( (b/2)^2 + k² ) = |h - k|We already did this and found k = h/2 - b²/(8 h)But we also have another expression for R, which is c²/(2 h). So,R = c²/(2 h) = |h - k|But R is also equal to sqrt( (b/2)^2 + k² )So, equating these two expressions:c²/(2 h) = sqrt( (b/2)^2 + k² )But we already used this to find k, so maybe this isn't helpful.Alternatively, since we have expressions for c and h in terms of b and r, maybe we can express k in terms of b and r.Given that h = (2 r b²)/(4 r² - b²), let's compute k:k = h/2 - b²/(8 h)= ( (2 r b²)/(4 r² - b²) ) / 2 - b² / [8 * (2 r b²)/(4 r² - b²) ]= (r b²)/(4 r² - b²) - [ b² (4 r² - b²) ] / (16 r b² )Simplify the second term:= (r b²)/(4 r² - b²) - (4 r² - b²)/(16 r )So,k = (r b²)/(4 r² - b²) - (4 r² - b²)/(16 r )Let me combine these terms:Find a common denominator, which is 16 r (4 r² - b²)So,k = [16 r * r b² - (4 r² - b²)^2 ] / [16 r (4 r² - b²) ]Compute numerator:= 16 r² b² - (16 r^4 - 8 r² b² + b^4 )= 16 r² b² -16 r^4 +8 r² b² - b^4= (16 r² b² +8 r² b²) -16 r^4 - b^4=24 r² b² -16 r^4 - b^4Factor:= -16 r^4 +24 r² b² - b^4Let me factor this expression:= -(16 r^4 -24 r² b² + b^4 )Notice that 16 r^4 -24 r² b² + b^4 = (4 r² - b²)^2 - 8 r² b²Wait, actually, let me check:(4 r² - b²)^2 = 16 r^4 -8 r² b² + b^4But we have 16 r^4 -24 r² b² + b^4, which is (4 r² - b²)^2 -16 r² b²Wait, maybe not helpful.Alternatively, let me factor it as:= -(16 r^4 -24 r² b² + b^4 )Let me see if this factors:Let me set x = r², then expression becomes:16 x² -24 x b² + b^4Let me compute discriminant:D = (24 b²)^2 -4*16* b^4 = 576 b^4 -64 b^4 = 512 b^4So, sqrt(D) = 16 sqrt(2) b²Thus, roots are:x = [24 b² ±16 sqrt(2) b² ] / (2*16) = [24 ±16 sqrt(2)] b² /32 = [3 ±2 sqrt(2)] b² /4Thus,16 x² -24 x b² + b^4 = 16(x - [3 +2 sqrt(2)] b² /4)(x - [3 -2 sqrt(2)] b² /4 )But this seems complicated, maybe not helpful.Alternatively, perhaps I made a mistake earlier in the calculation.Wait, let's recap:We have:k = h/2 - b²/(8 h)We have h = (2 r b²)/(4 r² - b²)So,h/2 = (r b²)/(4 r² - b²)And,b²/(8 h) = b² / [8*(2 r b²)/(4 r² - b²) ] = (4 r² - b²)/(16 r )Thus,k = (r b²)/(4 r² - b²) - (4 r² - b²)/(16 r )So, k = [16 r² b² - (4 r² - b²)^2 ] / [16 r (4 r² - b²) ]As before.So, numerator:16 r² b² - (16 r^4 -8 r² b² + b^4 ) =16 r² b² -16 r^4 +8 r² b² -b^4 =24 r² b² -16 r^4 -b^4So, numerator is -16 r^4 +24 r² b² -b^4Let me factor out a negative sign:= - (16 r^4 -24 r² b² + b^4 )I think this is as simplified as it gets.So, k = - (16 r^4 -24 r² b² + b^4 ) / [16 r (4 r² - b²) ]But this seems complicated. Maybe I can leave it as is.Alternatively, perhaps I can express k in terms of h and b.But maybe this is getting too algebraic. Perhaps I should think about the construction steps.Given that we have expressions for c and h in terms of b and r, maybe we can proceed to construct the triangle.Given base BC of length b, and radius r of the excircle.We can compute c and h using the formulas:c = [b(4r² + b²)] / [2(4r² - b²)]h = (2 r b²)/(4 r² - b²)So, with b and r given, we can compute c and h.Once we have c and h, we can construct the triangle.But let's think about the construction steps.1. Draw the base BC with length b.2. Find the midpoint A1 of BC.3. Draw the altitude AA1 perpendicular to BC. The length of AA1 is h, which we can compute.4. The orthocenter H is the midpoint of AA1, so it's at distance h/2 from A and A1.5. The excircle opposite to A has radius r and touches the extensions of AB and AC and the base BC.Wait, but in our earlier analysis, we found that the exradius r_a is related to b and h by r = (b h)/(2c - b)So, given b and r, we can compute c and h.Thus, the construction steps would involve:1. Draw base BC with length b.2. Compute c and h using the given r.3. Construct the altitude AA1 of length h.4. The orthocenter H is the midpoint of AA1.5. Then, construct the excircle with radius r.But perhaps a more geometric construction is needed without relying on algebraic formulas.Alternatively, maybe we can use the fact that the exradius is related to the sides and the semiperimeter.Given that, perhaps we can construct the excenter and then find the triangle accordingly.But I think the algebraic approach gives us the necessary lengths, so with that, we can proceed to construct the triangle.So, summarizing the steps:Given base BC = b and exradius r_a = r.1. Compute c = [b(4r² + b²)] / [2(4r² - b²)]2. Compute h = (2 r b²)/(4 r² - b²)3. Draw base BC with length b.4. Find midpoint A1 of BC.5. Draw perpendicular to BC at A1 with length h to locate point A.6. Connect A to B and C to form triangle ABC.7. Verify that the orthocenter H is the midpoint of AA1.Thus, the triangle is constructed.But let me check if this makes sense.Wait, if we compute c and h as above, then triangle ABC will have sides AB = AC = c, base BC = b, altitude h, and orthocenter at midpoint of AA1.Additionally, the exradius r_a = r is satisfied.So, yes, this should work.Therefore, the construction is possible by first computing c and h from the given b and r, then constructing the triangle accordingly.I think that's the way to go.

answer：Alright, so I have this problem where I need to divide the polynomial ( x^4 + 3x^2 - 4 ) by ( x^2 - 3 ) and find the remainder. Hmm, okay, let me think about how to approach this. I remember that when dividing polynomials, it's similar to long division with numbers, but with variables involved. So, I guess I need to set it up like a long division problem.First, I'll write the dividend, which is ( x^4 + 3x^2 - 4 ), and the divisor, which is ( x^2 - 3 ). I notice that the dividend has a degree of 4, and the divisor has a degree of 2. That means the quotient will have a degree of 2, right? Because when you subtract the degrees, 4 minus 2 is 2. So, the quotient will be a quadratic polynomial, and the remainder will have a degree less than 2, so it will be a linear term or a constant.Let me start by dividing the leading terms. The leading term of the dividend is ( x^4 ), and the leading term of the divisor is ( x^2 ). If I divide ( x^4 ) by ( x^2 ), I get ( x^2 ). So, the first term of the quotient is ( x^2 ).Next, I need to multiply the entire divisor by this term and subtract it from the dividend. Multiplying ( x^2 ) by ( x^2 - 3 ) gives me ( x^4 - 3x^2 ). Now, I subtract this from the original dividend:[(x^4 + 3x^2 - 4) - (x^4 - 3x^2) = x^4 - x^4 + 3x^2 + 3x^2 - 4 = 6x^2 - 4]Okay, so after the first step, I have a new polynomial: ( 6x^2 - 4 ). Now, I repeat the process. I divide the leading term of this new polynomial, which is ( 6x^2 ), by the leading term of the divisor, ( x^2 ), giving me 6. So, the next term in the quotient is 6.Now, I multiply the entire divisor by 6: ( 6 times (x^2 - 3) = 6x^2 - 18 ). I subtract this from the current polynomial:[(6x^2 - 4) - (6x^2 - 18) = 6x^2 - 6x^2 - 4 + 18 = 14]So, after this subtraction, I'm left with 14, which is a constant. Since the degree of the remainder (which is 0) is less than the degree of the divisor (which is 2), I can't continue the division any further.Therefore, the remainder when ( x^4 + 3x^2 - 4 ) is divided by ( x^2 - 3 ) is 14. Let me just double-check my steps to make sure I didn't make any mistakes. Divided ( x^4 ) by ( x^2 ) to get ( x^2 ), multiplied back, subtracted, got ( 6x^2 - 4 ). Then divided ( 6x^2 ) by ( x^2 ) to get 6, multiplied back, subtracted, and got 14. Yeah, that seems right.I think I'm confident with this answer. It's one of the options given, so I don't need to convert it or anything. Just 14 is the remainder.