answer：To find the cube root of 8, I need to determine which number, when multiplied by itself three times, equals 8.I start by recalling that the cube root of a number ( x ), denoted as ( sqrt[3]{x} ), is a number ( y ) such that ( y^3 = x ).Applying this to ( x = 8 ), I look for a number ( y ) where ( y^3 = 8 ).Testing the options provided:- For option A: ( 2^3 = 8 ), which satisfies the equation.- For option B: ( (-2)^3 = -8 ), which does not equal 8.- For option C: ( 4^3 = 64 ), which is much larger than 8.- For option D: ( (-4)^3 = -64 ), which also does not equal 8.Thus, the only number that meets the condition is 2.Therefore, the cube root of 8 is 2.

answer：First, I recognize that if two vectors are perpendicular, their dot product is zero. This means I can set up the equation (1 times x + 2 times 3 = 0).Solving for (x), I subtract 6 from both sides to get (x = -6).Next, I add the vectors (overrightarrow{p}) and (overrightarrow{q}) by adding their corresponding components: ((1 + (-6), 2 + 3)), which simplifies to ((-5, 5)).Finally, to find the magnitude of the resulting vector, I use the formula (sqrt{(-5)^2 + 5^2}), which calculates to (sqrt{25 + 25} = sqrt{50} = 5sqrt{2}).

answer：Okay, so I have this problem about an ellipse and a hyperbola sharing the same foci, and the ellipse has a specific eccentricity. Then there are two parts: part (I) is to find the standard equation of the ellipse, and part (II) has two subparts about points on the ellipse and their slopes, and then the dot product of vectors from the origin.Starting with part (I). I remember that for conic sections, both ellipses and hyperbolas have foci, and the distance between the foci is related to their parameters. The hyperbola given is ( y^2 - x^2 = 1 ). Let me recall the standard form of a hyperbola. It's usually ( frac{y^2}{a^2} - frac{x^2}{b^2} = 1 ) for a hyperbola opening upwards and downwards. Comparing this to the given equation, ( a^2 = 1 ) and ( b^2 = 1 ). So, for hyperbola, the distance between the foci is ( 2c ), where ( c^2 = a^2 + b^2 ). So here, ( c^2 = 1 + 1 = 2 ), so ( c = sqrt{2} ). Therefore, the foci are at ( (0, sqrt{2}) ) and ( (0, -sqrt{2}) ).Now, the ellipse shares these foci. So for the ellipse, the distance between the foci is also ( 2c ), which is ( 2sqrt{2} ). The eccentricity ( e ) of the ellipse is given as ( frac{sqrt{6}}{3} ). I remember that for an ellipse, ( e = frac{c}{a} ), where ( a ) is the semi-major axis. So, ( e = frac{sqrt{6}}{3} = frac{sqrt{2}}{a} ). Solving for ( a ), we get ( a = frac{sqrt{2} times 3}{sqrt{6}} ). Simplifying that, ( sqrt{6} ) is ( sqrt{2 times 3} ), so ( a = frac{3sqrt{2}}{sqrt{2}sqrt{3}} = frac{3}{sqrt{3}} = sqrt{3} ). So, ( a = sqrt{3} ).Now, for an ellipse, ( c^2 = a^2 - b^2 ). We know ( c = sqrt{2} ) and ( a = sqrt{3} ), so ( c^2 = 2 = 3 - b^2 ). Therefore, ( b^2 = 1 ), so ( b = 1 ).Since the foci are on the y-axis, the major axis of the ellipse is along the y-axis. So, the standard equation of the ellipse is ( frac{y^2}{a^2} + frac{x^2}{b^2} = 1 ), which becomes ( frac{y^2}{3} + x^2 = 1 ). So, that should be the answer for part (I).Moving on to part (II). Point ( A ) is the lower vertex of the ellipse. Since the ellipse is ( frac{y^2}{3} + x^2 = 1 ), the vertices along the y-axis are at ( (0, sqrt{3}) ) and ( (0, -sqrt{3}) ). So, the lower vertex ( A ) is at ( (0, -sqrt{3}) ).Points ( M ) and ( N ) are on the ellipse, different from ( A ). The product of the slopes of lines ( AM ) and ( AN ) is 1. So, if I let ( M = (x_1, y_1) ) and ( N = (x_2, y_2) ), then the slope of ( AM ) is ( frac{y_1 + sqrt{3}}{x_1 - 0} = frac{y_1 + sqrt{3}}{x_1} ), and similarly, the slope of ( AN ) is ( frac{y_2 + sqrt{3}}{x_2} ). The product of these slopes is 1, so:[left( frac{y_1 + sqrt{3}}{x_1} right) left( frac{y_2 + sqrt{3}}{x_2} right) = 1]Which simplifies to:[frac{(y_1 + sqrt{3})(y_2 + sqrt{3})}{x_1 x_2} = 1]So,[(y_1 + sqrt{3})(y_2 + sqrt{3}) = x_1 x_2]Expanding the left side:[y_1 y_2 + sqrt{3}(y_1 + y_2) + 3 = x_1 x_2]So, that's one equation. Now, I need to find something about line ( MN ). For part (i), I need to prove that line ( MN ) always passes through a fixed point and find that point.I think the approach here is to consider the equation of line ( MN ) and show that it passes through a specific point regardless of the positions of ( M ) and ( N ) on the ellipse, given the slope condition.Let me denote the equation of line ( MN ) as ( y = kx + t ), where ( k ) is the slope and ( t ) is the y-intercept. Since both ( M ) and ( N ) lie on this line and on the ellipse, substituting ( y = kx + t ) into the ellipse equation should give me a quadratic equation in ( x ).Substituting ( y = kx + t ) into ( frac{y^2}{3} + x^2 = 1 ):[frac{(kx + t)^2}{3} + x^2 = 1]Expanding:[frac{k^2 x^2 + 2ktx + t^2}{3} + x^2 = 1]Multiply through by 3 to eliminate the denominator:[k^2 x^2 + 2ktx + t^2 + 3x^2 = 3]Combine like terms:[(k^2 + 3)x^2 + 2ktx + (t^2 - 3) = 0]This is a quadratic in ( x ). Let me denote this as:[A x^2 + B x + C = 0]Where:- ( A = k^2 + 3 )- ( B = 2kt )- ( C = t^2 - 3 )So, the roots of this quadratic are ( x_1 ) and ( x_2 ), the x-coordinates of points ( M ) and ( N ). Therefore, from Vieta's formulas, we have:- ( x_1 + x_2 = -frac{B}{A} = -frac{2kt}{k^2 + 3} )- ( x_1 x_2 = frac{C}{A} = frac{t^2 - 3}{k^2 + 3} )Similarly, since ( y = kx + t ), the y-coordinates are:- ( y_1 = kx_1 + t )- ( y_2 = kx_2 + t )Therefore, ( y_1 + y_2 = k(x_1 + x_2) + 2t ). Substituting ( x_1 + x_2 ):[y_1 + y_2 = k left( -frac{2kt}{k^2 + 3} right) + 2t = -frac{2k^2 t}{k^2 + 3} + 2t = 2t - frac{2k^2 t}{k^2 + 3} = frac{6t}{k^2 + 3}]Similarly, ( y_1 y_2 = (k x_1 + t)(k x_2 + t) = k^2 x_1 x_2 + kt(x_1 + x_2) + t^2 ). Substituting the known values:[y_1 y_2 = k^2 left( frac{t^2 - 3}{k^2 + 3} right) + kt left( -frac{2kt}{k^2 + 3} right) + t^2][= frac{k^2(t^2 - 3)}{k^2 + 3} - frac{2k^2 t^2}{k^2 + 3} + t^2][= frac{k^2 t^2 - 3k^2 - 2k^2 t^2}{k^2 + 3} + t^2][= frac{-k^2 t^2 - 3k^2}{k^2 + 3} + t^2][= frac{-k^2(t^2 + 3)}{k^2 + 3} + t^2][= -k^2 + t^2]Wait, that seems a bit off. Let me check the calculation again.Starting from:[y_1 y_2 = k^2 x_1 x_2 + kt(x_1 + x_2) + t^2]Substitute ( x_1 x_2 = frac{t^2 - 3}{k^2 + 3} ) and ( x_1 + x_2 = -frac{2kt}{k^2 + 3} ):[y_1 y_2 = k^2 cdot frac{t^2 - 3}{k^2 + 3} + kt cdot left( -frac{2kt}{k^2 + 3} right) + t^2][= frac{k^2(t^2 - 3)}{k^2 + 3} - frac{2k^2 t^2}{k^2 + 3} + t^2]Combine the first two terms:[= frac{k^2 t^2 - 3k^2 - 2k^2 t^2}{k^2 + 3} + t^2][= frac{-k^2 t^2 - 3k^2}{k^2 + 3} + t^2]Factor out ( -k^2 ) from the numerator:[= frac{-k^2(t^2 + 3)}{k^2 + 3} + t^2][= -k^2 cdot frac{t^2 + 3}{k^2 + 3} + t^2]Hmm, that doesn't seem to simplify to ( -k^2 + t^2 ). Maybe I made a mistake in the algebra. Let me try another approach.Alternatively, perhaps I can express ( y_1 y_2 ) in terms of ( t ) and ( k ):Wait, maybe I should just proceed with the equation we had earlier:From the slope condition, we have:[x_1 x_2 = y_1 y_2 + sqrt{3}(y_1 + y_2) + 3]We have expressions for ( x_1 x_2 ), ( y_1 y_2 ), and ( y_1 + y_2 ) in terms of ( t ) and ( k ). Let me substitute them into this equation.So,[frac{t^2 - 3}{k^2 + 3} = left( frac{3t^2 - 3k^2}{k^2 + 3} right) + sqrt{3} cdot left( frac{6t}{k^2 + 3} right) + 3]Wait, hold on. Earlier, I tried to compute ( y_1 y_2 ) but maybe I made a mistake there. Let me recalculate ( y_1 y_2 ):Given ( y_1 = k x_1 + t ) and ( y_2 = k x_2 + t ), so:[y_1 y_2 = (k x_1 + t)(k x_2 + t) = k^2 x_1 x_2 + k t (x_1 + x_2) + t^2]We know ( x_1 x_2 = frac{t^2 - 3}{k^2 + 3} ) and ( x_1 + x_2 = -frac{2kt}{k^2 + 3} ). So,[y_1 y_2 = k^2 cdot frac{t^2 - 3}{k^2 + 3} + k t cdot left( -frac{2kt}{k^2 + 3} right) + t^2][= frac{k^2(t^2 - 3)}{k^2 + 3} - frac{2k^2 t^2}{k^2 + 3} + t^2]Combine the first two terms:[= frac{k^2 t^2 - 3k^2 - 2k^2 t^2}{k^2 + 3} + t^2][= frac{-k^2 t^2 - 3k^2}{k^2 + 3} + t^2]Factor out ( -k^2 ) from the numerator:[= frac{-k^2(t^2 + 3)}{k^2 + 3} + t^2][= -k^2 cdot frac{t^2 + 3}{k^2 + 3} + t^2]Hmm, this seems complicated. Maybe I can write it as:[y_1 y_2 = frac{-k^2(t^2 + 3) + t^2(k^2 + 3)}{k^2 + 3}][= frac{-k^2 t^2 - 3k^2 + k^2 t^2 + 3 t^2}{k^2 + 3}][= frac{-3k^2 + 3 t^2}{k^2 + 3}][= frac{3(t^2 - k^2)}{k^2 + 3}]Ah, that's better. So, ( y_1 y_2 = frac{3(t^2 - k^2)}{k^2 + 3} ).Now, going back to the slope condition equation:[x_1 x_2 = y_1 y_2 + sqrt{3}(y_1 + y_2) + 3]We have:- ( x_1 x_2 = frac{t^2 - 3}{k^2 + 3} )- ( y_1 y_2 = frac{3(t^2 - k^2)}{k^2 + 3} )- ( y_1 + y_2 = frac{6t}{k^2 + 3} )Substituting these into the equation:[frac{t^2 - 3}{k^2 + 3} = frac{3(t^2 - k^2)}{k^2 + 3} + sqrt{3} cdot frac{6t}{k^2 + 3} + 3]Multiply both sides by ( k^2 + 3 ) to eliminate denominators:[t^2 - 3 = 3(t^2 - k^2) + 6sqrt{3} t + 3(k^2 + 3)]Expand the right side:[t^2 - 3 = 3t^2 - 3k^2 + 6sqrt{3} t + 3k^2 + 9]Simplify terms:- The ( -3k^2 ) and ( +3k^2 ) cancel out.- So, we have:[t^2 - 3 = 3t^2 + 6sqrt{3} t + 9]Bring all terms to the left side:[t^2 - 3 - 3t^2 - 6sqrt{3} t - 9 = 0][-2t^2 - 6sqrt{3} t - 12 = 0]Multiply both sides by -1:[2t^2 + 6sqrt{3} t + 12 = 0]Divide both sides by 2:[t^2 + 3sqrt{3} t + 6 = 0]Now, solve for ( t ):Using quadratic formula:[t = frac{ -3sqrt{3} pm sqrt{(3sqrt{3})^2 - 4 cdot 1 cdot 6} }{2 cdot 1}][= frac{ -3sqrt{3} pm sqrt{27 - 24} }{2}][= frac{ -3sqrt{3} pm sqrt{3} }{2}]So, two solutions:1. ( t = frac{ -3sqrt{3} + sqrt{3} }{2} = frac{ -2sqrt{3} }{2} = -sqrt{3} )2. ( t = frac{ -3sqrt{3} - sqrt{3} }{2} = frac{ -4sqrt{3} }{2} = -2sqrt{3} )Now, we need to consider which solution is valid. If ( t = -sqrt{3} ), then the line ( MN ) would be ( y = kx - sqrt{3} ). But point ( A ) is at ( (0, -sqrt{3}) ). If line ( MN ) passes through ( A ), then points ( M ) and ( N ) would coincide with ( A ), but the problem states that ( M ) and ( N ) are different from ( A ). Therefore, ( t = -sqrt{3} ) is not acceptable because it would make the line pass through ( A ), which is excluded.Thus, the valid solution is ( t = -2sqrt{3} ). Therefore, the equation of line ( MN ) is ( y = kx - 2sqrt{3} ). This means that regardless of the value of ( k ), the line ( MN ) always has a y-intercept of ( -2sqrt{3} ). Therefore, it always passes through the point ( (0, -2sqrt{3}) ).So, part (i) is proved, and the fixed point is ( (0, -2sqrt{3}) ).Moving on to part (ii): If ( O ) is the origin, find the range of values for ( overrightarrow{OM} cdot overrightarrow{ON} ).First, ( overrightarrow{OM} cdot overrightarrow{ON} = x_1 x_2 + y_1 y_2 ).From earlier, we have expressions for ( x_1 x_2 ) and ( y_1 y_2 ):- ( x_1 x_2 = frac{t^2 - 3}{k^2 + 3} )- ( y_1 y_2 = frac{3(t^2 - k^2)}{k^2 + 3} )So,[overrightarrow{OM} cdot overrightarrow{ON} = frac{t^2 - 3}{k^2 + 3} + frac{3(t^2 - k^2)}{k^2 + 3}][= frac{t^2 - 3 + 3t^2 - 3k^2}{k^2 + 3}][= frac{4t^2 - 3 - 3k^2}{k^2 + 3}]But from part (i), we found that ( t = -2sqrt{3} ). So, substitute ( t = -2sqrt{3} ):[overrightarrow{OM} cdot overrightarrow{ON} = frac{4(-2sqrt{3})^2 - 3 - 3k^2}{k^2 + 3}][= frac{4(12) - 3 - 3k^2}{k^2 + 3}][= frac{48 - 3 - 3k^2}{k^2 + 3}][= frac{45 - 3k^2}{k^2 + 3}]Factor out 3 in the numerator:[= frac{3(15 - k^2)}{k^2 + 3}][= 3 cdot frac{15 - k^2}{k^2 + 3}]Let me denote ( m = k^2 ). Since ( k^2 ) is non-negative, ( m geq 0 ). So,[overrightarrow{OM} cdot overrightarrow{ON} = 3 cdot frac{15 - m}{m + 3}]Simplify:[= 3 cdot left( frac{15 - m}{m + 3} right )][= 3 cdot left( frac{15 + 3 - m - 3}{m + 3} right )][= 3 cdot left( frac{18 - (m + 3)}{m + 3} right )][= 3 cdot left( frac{18}{m + 3} - 1 right )][= 3 cdot left( frac{18}{m + 3} - 1 right )][= frac{54}{m + 3} - 3]So, ( overrightarrow{OM} cdot overrightarrow{ON} = frac{54}{m + 3} - 3 ), where ( m = k^2 ).Now, we need to find the range of this expression. To do that, we need to consider the possible values of ( m ). However, we also need to ensure that the quadratic equation for ( x ) has real solutions, meaning the discriminant must be positive.From the quadratic equation:[(k^2 + 3)x^2 + 2kt x + (t^2 - 3) = 0]The discriminant ( D ) is:[D = (2kt)^2 - 4(k^2 + 3)(t^2 - 3)][= 4k^2 t^2 - 4(k^2 + 3)(t^2 - 3)][= 4k^2 t^2 - 4[k^2 t^2 - 3k^2 + 3t^2 - 9]][= 4k^2 t^2 - 4k^2 t^2 + 12k^2 - 12t^2 + 36][= 12k^2 - 12t^2 + 36]Since ( t = -2sqrt{3} ), substitute:[D = 12k^2 - 12(12) + 36][= 12k^2 - 144 + 36][= 12k^2 - 108]For the quadratic to have real solutions, ( D > 0 ):[12k^2 - 108 > 0][12k^2 > 108][k^2 > 9]So, ( m = k^2 > 9 ).Therefore, ( m > 9 ). So, ( m + 3 > 12 ), meaning ( frac{54}{m + 3} < frac{54}{12} = 4.5 ). Also, as ( m ) increases, ( frac{54}{m + 3} ) approaches 0. So, ( frac{54}{m + 3} ) is between 0 and 4.5, but since ( m > 9 ), ( m + 3 > 12 ), so ( frac{54}{m + 3} < 4.5 ).Thus, ( overrightarrow{OM} cdot overrightarrow{ON} = frac{54}{m + 3} - 3 ). Let's analyze its range.When ( m ) approaches 9 from above, ( m + 3 ) approaches 12, so ( frac{54}{12} = 4.5 ), so ( overrightarrow{OM} cdot overrightarrow{ON} ) approaches ( 4.5 - 3 = 1.5 ).As ( m ) increases beyond 9, ( frac{54}{m + 3} ) decreases towards 0, so ( overrightarrow{OM} cdot overrightarrow{ON} ) approaches ( 0 - 3 = -3 ).Therefore, the range of ( overrightarrow{OM} cdot overrightarrow{ON} ) is ( (-3, 1.5) ).But let me express 1.5 as ( frac{3}{2} ) for exactness. So, the range is ( (-3, frac{3}{2}) ).So, summarizing:(I) The standard equation of the ellipse is ( frac{y^2}{3} + x^2 = 1 ).(II)(i) Line ( MN ) always passes through the fixed point ( (0, -2sqrt{3}) ).(II)(ii) The range of ( overrightarrow{OM} cdot overrightarrow{ON} ) is ( (-3, frac{3}{2}) ).**Final Answer**Ⅰ. The standard equation of the ellipse is boxed{dfrac{y^2}{3} + x^2 = 1}.Ⅱ. (i) The fixed point is boxed{(0, -2sqrt{3})}.(ii) The range of values for (overrightarrow{OM} cdot overrightarrow{ON}) is boxed{(-3, dfrac{3}{2})}.

answer：Okay, so I have this geometry problem here about triangle PQR. Let me try to visualize it first. There's a triangle PQR, and point S is the midpoint of side QR. That means QS equals SR. Then, point T is on PR such that PT to TR is 1:3. So, PT is one part, and TR is three parts. Then, point U is on PS such that PU to US is 2:1. So, PU is two parts, and US is one part. The area of triangle STU is given as 24, and I need to find the area of triangle PQR.Hmm, okay. Maybe I can use coordinate geometry to solve this. Let me assign coordinates to the points. Let's place point P at (0, 0) for simplicity. Let me assume point Q is at (2a, 0) so that the midpoint S will be at (a, 0). Then, point R can be at (0, 2b) so that the midpoint S is also at (a, 0). Wait, no, if Q is at (2a, 0) and R is at (0, 2b), then the midpoint S of QR would be at ((2a + 0)/2, (0 + 2b)/2) = (a, b). That makes more sense.So, let me correct that. Let me assign coordinates as follows:- P at (0, 0)- Q at (2a, 0)- R at (0, 2b)- Then, S, the midpoint of QR, is at ((2a + 0)/2, (0 + 2b)/2) = (a, b)Now, point T is on PR such that PT:TR = 1:3. Since PR goes from (0,0) to (0, 2b), it's a vertical line. So, T must be somewhere along this line. The ratio PT:TR is 1:3, so PT is one part and TR is three parts. That means T divides PR into four equal parts, with PT being one part. So, starting from P at (0,0), moving up 1/4 of the way to R at (0, 2b). Therefore, the coordinates of T would be (0, (1/4)*2b) = (0, b/2).Wait, is that correct? Let me think. If PT:TR = 1:3, then PT is 1 part and TR is 3 parts, so the total length PR is 4 parts. So, PT is 1/4 of PR. Since PR is from (0,0) to (0, 2b), the length is 2b. So, PT is (1/4)*2b = b/2. So, yes, T is at (0, b/2).Now, point U is on PS such that PU:US = 2:1. So, PU is two parts and US is one part. That means U divides PS into three equal parts, with PU being two of them. So, starting from P at (0,0) to S at (a, b), the coordinates of U can be found using the section formula. The section formula says that if a point divides a line segment joining (x1, y1) and (x2, y2) in the ratio m:n, then the coordinates are ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n)).Here, PU:US = 2:1, so m = 2, n = 1. So, U is ((2*a + 1*0)/(2 + 1), (2*b + 1*0)/(2 + 1)) = (2a/3, 2b/3).So, coordinates of U are (2a/3, 2b/3).Now, I need to find the area of triangle STU, which is given as 24. Let me find the coordinates of S, T, and U:- S is at (a, b)- T is at (0, b/2)- U is at (2a/3, 2b/3)To find the area of triangle STU, I can use the shoelace formula. The shoelace formula for three points (x1, y1), (x2, y2), (x3, y3) is:Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|Plugging in the coordinates:Area = |(a*(b/2 - 2b/3) + 0*(2b/3 - b) + (2a/3)*(b - b/2))/2|Let me compute each term step by step.First term: a*(b/2 - 2b/3) = a*( (3b - 4b)/6 ) = a*(-b/6) = -a*b/6Second term: 0*(2b/3 - b) = 0*( -b/3 ) = 0Third term: (2a/3)*(b - b/2) = (2a/3)*(b/2) = (2a/3)*(b/2) = (a*b)/3So, adding these up:- a*b/6 + 0 + a*b/3 = (-a*b/6 + 2a*b/6) = (a*b)/6Taking absolute value and dividing by 2:Area = |(a*b)/6| / 2 = |a*b| / 12But the area is given as 24, so:|a*b| / 12 = 24Therefore, |a*b| = 24 * 12 = 288So, a*b = 288 (since areas are positive, we can ignore the absolute value)Now, I need to find the area of triangle PQR. Let's recall that P is at (0,0), Q is at (2a, 0), and R is at (0, 2b). The area of triangle PQR can be found using the formula for the area of a triangle with base and height. The base can be PQ, which is 2a, and the height is the y-coordinate of R, which is 2b.So, area of PQR = (base * height)/2 = (2a * 2b)/2 = (4ab)/2 = 2abWe already found that ab = 288, so:Area of PQR = 2 * 288 = 576Wait, that's different from the initial thought process where the answer was 144. Did I make a mistake somewhere?Let me double-check my calculations.First, coordinates:- P(0,0), Q(2a,0), R(0,2b)- S is midpoint of QR: ((2a + 0)/2, (0 + 2b)/2) = (a, b)- T is on PR with PT:TR = 1:3. PR is from (0,0) to (0,2b). So, PT is 1/4 of PR, so T is at (0, b/2)- U is on PS with PU:US = 2:1. So, U divides PS into 3 parts, 2 from P. So, using section formula: (2a/3, 2b/3)Area of STU using shoelace formula:Coordinates:S(a, b), T(0, b/2), U(2a/3, 2b/3)Compute:Area = |(a*(b/2 - 2b/3) + 0*(2b/3 - b) + (2a/3)*(b - b/2))/2|Compute each term:a*(b/2 - 2b/3) = a*(3b/6 - 4b/6) = a*(-b/6) = -ab/60*(2b/3 - b) = 0(2a/3)*(b - b/2) = (2a/3)*(b/2) = (2a/3)*(b/2) = ab/3Sum: -ab/6 + 0 + ab/3 = (-ab/6 + 2ab/6) = ab/6Divide by 2: ab/12Set equal to 24: ab/12 = 24 => ab = 288Area of PQR is 2ab = 576Wait, but in the initial thought process, the answer was 144. So, which one is correct?Let me think again. Maybe I made a mistake in the area of triangle PQR. Let's calculate it differently.Using coordinates, the area of triangle PQR can be calculated using the shoelace formula as well.Coordinates of P(0,0), Q(2a,0), R(0,2b)Area = |(0*(0 - 2b) + 2a*(2b - 0) + 0*(0 - 0))/2|= |0 + 4ab + 0| / 2= |4ab| / 2= 2abSo, that's correct. So, if ab = 288, then area is 576.But the initial thought process got 144. Maybe the initial thought process was wrong.Wait, let me see the initial thought process:- First, noticed that |PSU| / |STU| = PU/US = 2, so |PSU| = 48- Then, |PSU| = |STU| + |PTU|, so |PTU| = 24- Then, since PT:TR = 1:3, |PTU| / |STU| = 1/3, so |STU| = 72- Then, |PTR| = 24 + 72 = 96- Then, since S is midpoint, |PSQ| = |PSR| = 1/2 |PQR|- |PSQ| = |PSU| + |PTU| = 48 + 24 = 72- So, |PQR| = 2 * 72 = 144Hmm, so in the initial thought process, they considered areas based on ratios, but perhaps they made a mistake in assuming certain areas.In my coordinate approach, I got 576, but the initial thought process got 144. Which one is correct?Wait, maybe I messed up the area of STU. Let me recalculate the area of STU.Using coordinates S(a, b), T(0, b/2), U(2a/3, 2b/3)Using shoelace formula:Arrange the points:S(a, b), T(0, b/2), U(2a/3, 2b/3), back to S(a, b)Compute the sum:a*(b/2 - 2b/3) + 0*(2b/3 - b) + (2a/3)*(b - b/2)Compute each term:First term: a*(b/2 - 2b/3) = a*(3b/6 - 4b/6) = a*(-b/6) = -ab/6Second term: 0*(2b/3 - b) = 0Third term: (2a/3)*(b - b/2) = (2a/3)*(b/2) = (2a/3)*(b/2) = ab/3Sum: -ab/6 + 0 + ab/3 = (-ab/6 + 2ab/6) = ab/6Divide by 2: ab/12So, area is ab/12 = 24 => ab = 288Thus, area of PQR is 2ab = 576Wait, but the initial thought process got 144. Maybe the initial thought process was wrong because they assumed certain areas without considering the correct ratios.Alternatively, perhaps I made a mistake in the coordinate system.Wait, another approach: mass point or area ratios.Let me try using area ratios.Let me denote the area of PQR as A.Since S is the midpoint of QR, the area of PQS and PRS are each A/2.Now, point T divides PR in the ratio 1:3, so PT:TR = 1:3. Therefore, the area of triangles PQT and RQT would be in the ratio 1:3 as well, because they share the same base QT and their heights from P and R are in the ratio 1:3.Wait, no, actually, triangles PQT and RQT share the same vertex Q, and their bases PT and TR are in ratio 1:3. So, their areas are in ratio 1:3.Therefore, area of PQT is (1/4)*A/2 = A/8, and area of RQT is (3/4)*A/2 = 3A/8.Wait, no, maybe not. Let me think again.Actually, since S is the midpoint of QR, area of PQS is A/2. Now, point T is on PR such that PT:TR = 1:3. So, the area of triangles PQT and RQT would be in the ratio of PT:TR, which is 1:3.Therefore, area of PQT = (1/4)*A/2 = A/8, and area of RQT = (3/4)*A/2 = 3A/8.Wait, no, perhaps not. Let me think differently.The area of triangle PQT: since T divides PR in 1:3, the height from Q to PR is the same for both triangles PQT and RQT. Therefore, their areas are proportional to PT and TR, which is 1:3.Therefore, area of PQT = (1/4)*area of PQR. Wait, no, because S is the midpoint, so area of PQS is A/2, and within that, PQT is a part.Wait, maybe it's better to consider the entire triangle.Alternatively, let's use vectors or barycentric coordinates.But perhaps my coordinate approach was correct, leading to area 576.But the initial thought process got 144, so maybe I need to see where the discrepancy is.Wait, in the initial thought process, they considered |PSU| = 48, |STU| =24, so |PTU| =24, then |PTR|=96, then |PSQ|=72, so |PQR|=144.But in my coordinate approach, |STU|=24 corresponds to ab=288, so |PQR|=576.So, which one is correct?Wait, perhaps the initial thought process made a mistake in assuming that |PSU| / |STU| = PU/US =2, but maybe that's not the case because the areas depend on both the base and the height.Wait, let's think about triangles PSU and STU. They share the same base SU, but their heights from P and T respectively to SU may not be in the ratio 2:1.Wait, no, actually, if PU:US=2:1, then the heights from P and T to SU would be in the ratio 2:1, because U divides PS in 2:1.Wait, but actually, triangles PSU and STU share the base SU, and their heights are from P and T respectively. The ratio of their areas would be equal to the ratio of their heights.But the heights are along PS, which is divided by U in the ratio 2:1. So, the height from P is PU = 2 parts, and the height from T is something else.Wait, no, T is not on PS. T is on PR. So, the heights from P and T to SU are not along the same line.Therefore, the ratio of areas of PSU and STU is not simply PU:US.So, the initial thought process might have made a mistake there.Therefore, perhaps the coordinate approach is more accurate, leading to area 576.But wait, let me think again.Alternatively, maybe using mass point geometry.Let me try mass point.Assign masses to the points.First, since S is the midpoint of QR, masses at Q and R are equal. Let's assign mass 1 to Q and R, so mass at S is 1+1=2.Point U is on PS such that PU:US=2:1. So, masses at P and S must be in the ratio 1:2. Since mass at S is 2, mass at P must be 1.Therefore, mass at U is 1+2=3.Point T is on PR such that PT:TR=1:3. So, masses at P and R are in the ratio 3:1. Since mass at P is 1, mass at R must be 3. But earlier, mass at R was 1. Contradiction.Wait, so maybe I need to adjust the masses.Let me try again.Since S is the midpoint of QR, masses at Q and R are both 1, so mass at S is 2.Point U is on PS with PU:US=2:1, so masses at P and S are in ratio 1:2. Since mass at S is 2, mass at P must be 1.Therefore, mass at U is 1+2=3.Now, point T is on PR with PT:TR=1:3. So, masses at P and R must be in ratio 3:1. But mass at P is 1, so mass at R must be 3. However, earlier, mass at R was 1 from the midpoint S. So, to reconcile, we need to scale the masses.Multiply masses at Q and R by 3 to make mass at R=3. So, masses at Q=3, R=3, so mass at S=6.Then, mass at P=1 (from PU:US=2:1), so mass at U=1+6=7.Now, on PR, PT:TR=1:3, so masses at P and R must be in ratio 3:1. Mass at P is 1, so mass at R must be 3, which is consistent with our earlier scaling.Therefore, masses are consistent.Now, using mass point, can we find the area ratios?Alternatively, maybe use the area ratios based on the masses.But I'm not sure. Maybe it's better to stick with the coordinate approach.Wait, in the coordinate approach, I got area of PQR as 576, but the initial thought process got 144. Maybe the initial thought process was wrong because they incorrectly assumed the ratio of areas based on segment ratios without considering the heights.Alternatively, perhaps I made a mistake in the coordinate approach.Wait, let me recalculate the area of STU.Coordinates:S(a, b), T(0, b/2), U(2a/3, 2b/3)Using shoelace formula:Area = |(a*(b/2 - 2b/3) + 0*(2b/3 - b) + (2a/3)*(b - b/2))/2|Compute each term:First term: a*(b/2 - 2b/3) = a*(3b/6 - 4b/6) = a*(-b/6) = -ab/6Second term: 0*(2b/3 - b) = 0Third term: (2a/3)*(b - b/2) = (2a/3)*(b/2) = (2a/3)*(b/2) = ab/3Sum: -ab/6 + 0 + ab/3 = (-ab/6 + 2ab/6) = ab/6Divide by 2: ab/12So, area of STU is ab/12 =24 => ab=288Therefore, area of PQR is 2ab=576So, I think my coordinate approach is correct, leading to 576.But the initial thought process got 144, so maybe they made a mistake.Alternatively, perhaps the initial thought process considered the area of STU as 24, but in reality, in the coordinate approach, it's ab/12=24, so ab=288, leading to PQR area=576.Therefore, I think the correct answer is 576.But wait, let me check with another method.Let me use vectors.Let me assign vectors to the points.Let me set P as the origin, so P=(0,0).Let vector PQ = 2a i, so Q=(2a, 0)Let vector PR = 2b j, so R=(0, 2b)Then, S is the midpoint of QR, so S=(a, b)Point T is on PR such that PT:TR=1:3. So, T divides PR in the ratio 1:3, so T=(0, b/2)Point U is on PS such that PU:US=2:1. So, U divides PS in the ratio 2:1, so U=(2a/3, 2b/3)Now, vectors for STU:S=(a, b), T=(0, b/2), U=(2a/3, 2b/3)Vectors ST = T - S = (-a, -b/2)Vectors SU = U - S = (-a/3, -b/3)The area of triangle STU is half the magnitude of the cross product of ST and SU.Compute the cross product:ST × SU = (-a)(-b/3) - (-b/2)(-a/3) = (ab/3) - (ab/6) = ab/6Therefore, area of STU is |ab/6| / 2 = ab/12Given that area is 24, so ab/12=24 => ab=288Thus, area of PQR is 2ab=576So, same result as coordinate approach.Therefore, I think the correct answer is 576.But the initial thought process got 144, so perhaps they made a mistake in their reasoning.Therefore, I think the area of triangle PQR is 576.

answer：First, I need to determine the original total cost of the gas. Let’s denote this total cost as ( x ).Initially, there are 5 friends sharing the cost equally. Therefore, each friend's share is ( frac{x}{5} ).When 3 more friends join, the total number of friends becomes 8. The new share for each friend is ( frac{x}{8} ).According to the problem, the share for each of the original 5 friends decreases by 15.00. This gives the equation:[frac{x}{5} - 15 = frac{x}{8}]To solve for ( x ), I'll first subtract ( frac{x}{8} ) from both sides:[frac{x}{5} - frac{x}{8} = 15]Next, I'll find a common denominator for the fractions, which is 40:[frac{8x - 5x}{40} = 15]Simplifying the numerator:[frac{3x}{40} = 15]Now, I'll multiply both sides by 40 to isolate ( 3x ):[3x = 600]Finally, I'll divide both sides by 3 to find ( x ):[x = 200]Therefore, the original total cost of the gas was 200.

answer：First, I need to identify all possible natural numbers less than 1000 that can be formed using the digits 1, 6, and 8 without repetition. This involves considering all permutations of the three digits.Next, I will list out all the possible permutations:- 168- 186- 618- 681- 816- 861After listing all permutations, I will arrange them in ascending order to determine the smallest and third smallest numbers:1. 168 (smallest)2. 1863. 618 (third smallest)4. 6815. 8166. 861Finally, I will add the smallest number (168) and the third smallest number (618) together to find the required sum:168 + 618 = 786