🧠:[<think>Okay, let's try to tackle this geometry problem. First, let me parse the problem statement again to make sure I understand all the components.We have triangle ABC with circumcircle Γ. D is the midpoint of the minor arc BC. Then, points E and F are on Γ such that DE is perpendicular to AC and DF is perpendicular to AB. Lines BE and DF intersect at G, and lines CF and DE intersect at H. We need to find the area of quadrilateral BCHG given that AB = 8, AC = 10, and angle BAC is 60 degrees.Alright, so first things first, maybe I can draw a rough sketch of the triangle and the circumcircle. Since angle BAC is 60 degrees, sides AB = 8, AC = 10, so BC can be found using the Law of Cosines. Let me compute BC first.Law of Cosines: BC² = AB² + AC² - 2*AB*AC*cos(angle BAC)Plugging in the numbers: BC² = 8² + 10² - 2*8*10*cos(60°)Calculates to: 64 + 100 - 160*0.5 = 164 - 80 = 84So BC = sqrt(84) = 2*sqrt(21). Let me note that down.Now, since D is the midpoint of the minor arc BC, that means D is equidistant from B and C on the circumcircle. So BD = DC. Also, since D is on the circumcircle, angles BAD and CAD might have some relation. Hmm. Maybe.Given DE perpendicular to AC and DF perpendicular to AB. So DE is an altitude from D to AC, and DF is an altitude from D to AB. But since E and F are on Γ, DE and DF are chords of Γ that are perpendicular to AC and AB respectively.So DE ⊥ AC implies that DE is the altitude from D to AC, but since D is on the circumcircle, E must be the foot of the perpendicular from D to AC extended onto the circumcircle? Wait, not necessarily. Hmm.Wait, maybe DE is a line such that DE is perpendicular to AC, and E is on Γ. So starting at D, draw a line perpendicular to AC; where this line intersects Γ again is E. Similarly for F. So DE is a diameter? Wait, not necessarily, unless AC is a diameter, which it isn't in a general triangle. So maybe DE is just a chord passing through D and perpendicular to AC.Similarly, DF is a chord passing through D and perpendicular to AB.Then lines BE and DF intersect at G, and lines CF and DE intersect at H. Then quadrilateral BCHG's area is to be found.Given the specific lengths and angle, maybe coordinate geometry can help here. Let me try that approach.First, let's set up a coordinate system. Let me place point A at the origin (0,0). Since angle BAC is 60 degrees, and AB = 8, AC = 10, we can place point B at (8,0). Then point C would be at (10*cos(60°), 10*sin(60°)) = (5, 5*sqrt(3)). Let me confirm that:Yes, since from A(0,0) to C is 10 units at 60 degrees, so coordinates (10*cos60°, 10*sin60°) = (5, 5√3). AB is along the x-axis from (0,0) to (8,0). Then BC can be calculated as between (8,0) and (5,5√3), which we already found as 2√21.Now, the circumcircle Γ of triangle ABC. Let me find the coordinates of the circumcircle. Since we have coordinates for A, B, and C, we can compute the circumradius and the center.The circumradius R of triangle ABC can be found using the formula:R = (a*b*c)/(4*Δ)where a, b, c are the sides, and Δ is the area.We have sides:AB = 8, AC = 10, BC = 2√21.Compute area Δ using the formula (1/2)*AB*AC*sin(angle BAC) = 0.5*8*10*sin60° = 40*(√3/2) = 20√3.Thus, R = (8*10*2√21)/(4*20√3) = (160√21)/(80√3) = 2√21/√3 = 2√7.Wait, let me check:Wait, sides a, b, c in the formula correspond to BC, AC, AB. Wait, actually, the formula is R = (a*b*c)/(4*Δ), where a, b, c are the lengths of the sides. So here, sides are:a = BC = 2√21b = AC = 10c = AB = 8Thus R = (2√21 *10 *8)/(4*20√3) = (160√21)/(80√3) = 2√21 / √3 = 2*(√21/√3) = 2*√7 ≈ 2*2.6458 ≈ 5.2915.Alternatively, we can compute the circumradius using the formula:R = a/(2*sin A)Here, angle A is 60°, and side a is BC = 2√21.Thus R = (2√21)/(2*sin60°) = √21/(√3/2) ) = √21 * 2/√3 = 2√7. Which matches the previous result.So the circumradius is 2√7. Now, to find the coordinates of the circumcircle's center. Since we have coordinates for A, B, C, we can compute the circumcenter.The circumcenter is the intersection of the perpendicular bisectors of AB and AC.Coordinates of A: (0,0), B: (8,0), C: (5,5√3).Midpoint of AB: (4,0). The perpendicular bisector of AB is the line perpendicular to AB (which is along the x-axis) through (4,0), so that's the line x = 4.Midpoint of AC: (2.5, 2.5√3). The slope of AC is (5√3 - 0)/(5 - 0) = √3, so the perpendicular bisector will have slope -1/√3.Thus, the equation of the perpendicular bisector of AC is:y - 2.5√3 = (-1/√3)(x - 2.5)We can find the intersection of this line with x=4.Substitute x=4 into the equation:y - 2.5√3 = (-1/√3)(4 - 2.5) = (-1/√3)(1.5) = -1.5/√3 = -√3/2Thus, y = 2.5√3 - √3/2 = (5/2 - 1/2)√3 = 2√3Therefore, the circumcenter is at (4, 2√3). Let me verify that.Wait, midpoint of AC is (2.5, 2.5√3), slope of AC is √3, so perpendicular bisector slope is -1/√3. Equation is y - 2.5√3 = (-1/√3)(x - 2.5). Then at x=4, y = 2.5√3 + (-1/√3)(4 - 2.5) = 2.5√3 - 1.5/√3.Convert 1.5/√3 to (3/2)/√3 = (3)/(2√3) = √3/2. So y = 2.5√3 - √3/2 = (5/2 - 1/2)√3 = 2√3. Yes, correct. So circumcenter at (4, 2√3).Therefore, Γ is the circle with center (4, 2√3) and radius 2√7.Now, we need to find point D, the midpoint of minor arc BC. Since D is the midpoint of arc BC, it is equidistant from B and C on the circumcircle. Also, since the center is at (4, 2√3), the arc midpoint D can be found by rotating point B around the center by half the arc length to point C. Alternatively, the midpoint of arc BC not containing A is the point such that BD = DC, and lies on the circumcircle.Alternatively, since arc BC's midpoint is the point where the angle bisector of angle BOC meets the circumcircle, where O is the center. Wait, but since O is the circumcenter, angle BOC is twice angle BAC. Since angle BAC is 60°, angle BOC is 120°, so the midpoint D of arc BC would split angle BOC into two 60° angles.Alternatively, perhaps we can parameterize the circumcircle and find point D.Given that center is at (4, 2√3), radius 2√7. Let me write the equation of Γ:(x - 4)^2 + (y - 2√3)^2 = (2√7)^2 = 4*7 = 28.Points B and C are on this circle.Coordinates of B: (8, 0). Check distance from center: (8-4)^2 + (0 - 2√3)^2 = 16 + 12 = 28. Correct.Coordinates of C: (5, 5√3). Distance from center: (5 -4)^2 + (5√3 - 2√3)^2 = 1 + (3√3)^2 = 1 + 27 = 28. Correct.So now, to find D, the midpoint of minor arc BC. The midpoint of arc BC can be found by finding the point on the circumcircle such that it's equidistant from B and C, and lying on the arc BC that doesn't contain A. Alternatively, since the center is at (4, 2√3), and points B(8,0) and C(5,5√3), the arc midpoint can be found by rotating the center towards the direction perpendicular to BC.Alternatively, since the midpoint of arc BC corresponds to the excenter opposite A in some contexts, but maybe not exactly here. Alternatively, parametrize points B and C on the circle and find the midpoint.Alternatively, compute the angle for points B and C with respect to the center O(4, 2√3), and find the midpoint.Let me compute the angles for points B and C.Vector from O to B: (8 - 4, 0 - 2√3) = (4, -2√3)Vector from O to C: (5 - 4, 5√3 - 2√3) = (1, 3√3)We can find the angle of these vectors with respect to the positive x-axis.For point B: the angle θ_B is arctangent of (-2√3)/4 = -√3/2. So θ_B = arctan(-√3/2). But since it's in the fourth quadrant relative to O, which is at (4, 2√3). Wait, maybe better to compute the angle from the positive x-axis.Wait, coordinates of O are (4, 2√3). The vector from O to B is (4, -2√3). So the angle for OB is arctan((-2√3)/4) = arctan(-√3/2). Similarly, the vector from O to C is (1, 3√3), so angle θ_C = arctan(3√3/1) = arctan(3√3).But to find the midpoint of arc BC, we can average the angles θ_B and θ_C. However, since they are in different quadrants, this might be tricky. Alternatively, we can parametrize the arc.Alternatively, we can use complex numbers. Let me consider the center O at (4, 2√3). Let me translate the coordinate system so that O is at the origin. Then points B and C become:B': (8 - 4, 0 - 2√3) = (4, -2√3)C': (5 - 4, 5√3 - 2√3) = (1, 3√3)Now, in this translated system, the midpoint of arc BC is the complex number on the circle with radius 2√7, which when translated back will be point D.The arc midpoint can be found by rotating the vector from O to B towards C by half the angle between them. Alternatively, since in complex plane, if we represent B' and C' as complex numbers, the arc midpoint would be the normalized sum of B' and C' multiplied by the radius.Wait, the midpoint of the arc in complex numbers can be found by normalizing the sum of the vectors B' and C', as the angle between them would be bisected by that direction.So, compute B' + C' = (4 + 1, -2√3 + 3√3) = (5, √3). Then the direction of the arc midpoint is (5, √3). The length of this vector is sqrt(5² + (√3)^2) = sqrt(25 + 3) = sqrt(28) = 2√7, which is exactly the radius. So, this vector (5, √3) is already a point on the circle. Therefore, in the translated system, the midpoint D' is (5, √3). Translating back, D is (5 + 4, √3 + 2√3) = (9, 3√3). Wait, is that correct?Wait, hold on. If in the translated system (with O at origin), B' is (4, -2√3), C' is (1, 3√3). Then adding B' + C' gives (5, √3). The length of that vector is sqrt(25 + 3) = sqrt(28) = 2√7, which is equal to the radius, so scaling is not needed. Therefore, the point D' in the translated system is (5, √3). Translating back to original coordinates: add (4, 2√3), so D is (5 + 4, √3 + 2√3) = (9, 3√3).Wait, but let's check if this point is on the original circle. (x - 4)^2 + (y - 2√3)^2 = (9 - 4)^2 + (3√3 - 2√3)^2 = 25 + (√3)^2 = 25 + 3 = 28. Yes, that's correct. So D is at (9, 3√3). Let me verify the arc midpoint.Alternatively, since angle BOC is 120 degrees, as angle BAC is 60 degrees, so angle BOC = 2*angle BAC = 120 degrees. Therefore, the arc BC is 120 degrees, so the midpoint D divides it into two 60 degrees arcs. Therefore, in the translated system, rotating point B' by 60 degrees around the center (which is the origin in translated coordinates) should give point D'.Let me check: Rotating B'(4, -2√3) by 60 degrees counterclockwise. The rotation matrix is [cos60 -sin60; sin60 cos60]. So:x' = 4*cos60 - (-2√3)*sin60 = 4*(0.5) + 2√3*(√3/2) = 2 + 3 = 5y' = 4*sin60 + (-2√3)*cos60 = 4*(√3/2) - 2√3*(0.5) = 2√3 - √3 = √3So after rotation, we get (5, √3), which matches the previous result. Therefore, D is indeed at (9, 3√3) in original coordinates.Great, so D is (9, 3√3).Now, next step: find points E and F on Γ such that DE ⊥ AC and DF ⊥ AB.First, let's find DE perpendicular to AC. AC has coordinates from A(0,0) to C(5,5√3). The slope of AC is (5√3 - 0)/(5 - 0) = √3. Therefore, DE is perpendicular to AC, so the slope of DE is -1/√3.But DE is a line passing through D(9, 3√3) with slope -1/√3. We need to find the other intersection point E of this line with Γ (since D is already on Γ). So parametrize line DE and find its other intersection with Γ.Similarly, DF is perpendicular to AB. AB is along the x-axis from (0,0) to (8,0), so slope is 0. Therefore, DF is perpendicular to AB, so DF is vertical (slope undefined). Wait, AB is horizontal, so perpendicular is vertical. Therefore, DF is a vertical line passing through D(9, 3√3). But DF is supposed to be a line from D perpendicular to AB, so since AB is horizontal, DF is vertical. Therefore, DF is the line x = 9. We need to find the other intersection point F of this line with Γ.Wait, but D is at (9, 3√3), and DF is vertical, so x=9. Let's check if x=9 intersects Γ again.Γ equation: (x -4)^2 + (y - 2√3)^2 = 28. Substitute x=9:(9 -4)^2 + (y - 2√3)^2 = 25 + (y - 2√3)^2 = 28 → (y - 2√3)^2 = 3 → y - 2√3 = ±√3 → y = 2√3 ± √3 → y = 3√3 or y = √3.Since D is at (9, 3√3), the other intersection is (9, √3). Therefore, point F is (9, √3).Similarly, for DE: line DE is perpendicular to AC, which has slope √3, so DE has slope -1/√3. Equation of DE: y - 3√3 = (-1/√3)(x - 9).We need to find the other intersection point E of this line with Γ.So let's solve the system:1. (x -4)^2 + (y - 2√3)^2 = 282. y = (-1/√3)(x - 9) + 3√3Substitute equation 2 into equation 1.First, simplify equation 2:y = (-1/√3)x + 9/√3 + 3√3Convert 9/√3 to 3√3:y = (-1/√3)x + 3√3 + 3√3 = (-1/√3)x + 6√3Now, substitute into equation 1:(x -4)^2 + [(-1/√3 x + 6√3 - 2√3)]^2 = 28Simplify the y-component:(-1/√3 x + 6√3 - 2√3) = (-1/√3 x + 4√3)So, equation becomes:(x -4)^2 + [(-1/√3 x + 4√3)]^2 = 28Let me compute each term:First term: (x -4)^2 = x² -8x +16Second term: [(-1/√3 x + 4√3)]^2Let me compute this:= [(-x/√3 + 4√3)]²= ( (-x/√3) + 4√3 )²= ( (-x/√3) )² + 2*(-x/√3)*(4√3) + (4√3)²= (x² / 3) + 2*(-x/√3)*(4√3) + 16*3Simplify term by term:First term: x² /3Second term: 2*(-x/√3)*(4√3) = 2*(-x)*(4√3)/√3 = 2*(-x)*4 = -8xThird term: 48So altogether, second term is x²/3 -8x +48Therefore, the equation becomes:(x² -8x +16) + (x²/3 -8x +48) = 28Combine like terms:x² + x²/3 -8x -8x +16 +48 = 28Compute coefficients:(1 + 1/3)x² + (-16x) + 64 = 28Convert to thirds:(4/3)x² -16x +64 =28Subtract 28:4/3 x² -16x +36 =0Multiply both sides by 3 to eliminate fractions:4x² -48x +108 =0Divide by 4:x² -12x +27=0Solve quadratic equation:x = [12 ± sqrt(144 - 108)] /2 = [12 ± sqrt(36)] /2 = [12 ±6]/2Thus, x = (12 +6)/2=9 or x=(12-6)/2=3We already know that D is at (9, 3√3), so the other intersection point E is at x=3. Plugging back into equation 2:y = (-1/√3)(3 -9) + 3√3 = (-1/√3)(-6) + 3√3 = 6/√3 + 3√3 = 2√3 + 3√3 =5√3Therefore, point E is (3,5√3).Wait, let's check this point on Γ:(x-4)^2 + (y -2√3)^2 = (3 -4)^2 + (5√3 -2√3)^2 = (-1)^2 + (3√3)^2 =1 +27=28. Correct.So E is (3,5√3). Good.Similarly, we found F is (9, √3). Let me confirm F is on Γ:(9 -4)^2 + (√3 -2√3)^2 =25 + (-√3)^2=25 +3=28. Correct.Okay, so E(3,5√3) and F(9,√3).Next step: lines BE and DF intersect at G, and lines CF and DE intersect at H.First, let's find equations for lines BE and DF.Point B is at (8,0), point E is at (3,5√3). So line BE:Slope of BE: (5√3 -0)/(3 -8) = 5√3 / (-5) = -√3Equation: y -0 = -√3(x -8)So y = -√3 x +8√3Line DF: points D(9,3√3) and F(9,√3). Since both have x=9, this is a vertical line at x=9.Therefore, the intersection G of BE and DF is at x=9. Substitute x=9 into BE's equation:y = -√3*9 +8√3 = (-9 +8)√3 = -√3Therefore, G is at (9, -√3). Wait, but DF is from D(9,3√3) to F(9,√3), which is the line x=9 from y=√3 to y=3√3. However, the line BE intersects DF at (9, -√3), which is below F. Is this possible?Wait, but point F is at (9,√3), and DF is the vertical line from D(9,3√3) to F(9,√3). However, line BE is y = -√3 x +8√3. When x=9, y= -9√3 +8√3= -√3, which is indeed below F. But since DF is the line segment from D to F, or is it the entire line? The problem says lines BE and DF meet at G. So DF is the entire line, not just the segment. So even though G is below F on the line DF, it's still a valid intersection.Similarly, line DF is x=9, which is a vertical line, so G is at (9, -√3). Okay.Now, lines CF and DE intersect at H. Let's find equations for CF and DE.First, CF: points C(5,5√3) and F(9,√3). Let's find the equation.Slope of CF: (√3 -5√3)/(9 -5) = (-4√3)/4 = -√3Equation: y -5√3 = -√3(x -5)Simplify: y = -√3 x +5√3 +5√3 = -√3 x +10√3Next, DE: points D(9,3√3) and E(3,5√3). Let's find the equation.Slope of DE: (5√3 -3√3)/(3 -9) = (2√3)/(-6) = -√3/3Equation: y -3√3 = (-√3/3)(x -9)Simplify:y = (-√3/3)x + (9√3)/3 +3√3 = (-√3/3)x +3√3 +3√3 = (-√3/3)x +6√3So DE: y = (-√3/3)x +6√3Now, intersection H of CF and DE:CF: y = -√3 x +10√3DE: y = (-√3/3)x +6√3Set equal:-√3 x +10√3 = (-√3/3)x +6√3Multiply both sides by 3 to eliminate denominators:-3√3 x +30√3 = -√3 x +18√3Bring all terms to left:-3√3 x +30√3 +√3 x -18√3 =0(-2√3 x) +12√3=0-2√3 x = -12√3x = (-12√3)/(-2√3) =6So x=6. Substitute back into one of the equations, say CF:y = -√3*6 +10√3 = (-6 +10)√3 =4√3Thus, H is at (6,4√3)Now, we have points:B(8,0), C(5,5√3), H(6,4√3), G(9,-√3)We need to find the area of quadrilateral BCHG.Quadrilateral BCHG has vertices at B(8,0), C(5,5√3), H(6,4√3), G(9,-√3). Let me list them in order:B(8,0), C(5,5√3), H(6,4√3), G(9,-√3). To compute the area, we can use the shoelace formula.First, list the coordinates in order:B: (8,0)C: (5,5√3)H: (6,4√3)G: (9,-√3)Then back to B: (8,0)Shoelace formula:Area = 1/2 |sum_{i=1 to n} (x_i y_{i+1} - x_{i+1} y_i)|Compute each term:First term: x_B y_C - x_C y_B =8*5√3 -5*0=40√3 -0=40√3Second term: x_C y_H - x_H y_C =5*4√3 -6*5√3=20√3 -30√3= -10√3Third term: x_H y_G - x_G y_H =6*(-√3) -9*4√3= -6√3 -36√3= -42√3Fourth term: x_G y_B - x_B y_G =9*0 -8*(-√3)=0 +8√3=8√3Sum these terms:40√3 -10√3 -42√3 +8√3 = (40 -10 -42 +8)√3 = (-4)√3Take absolute value and multiply by 1/2:Area = (1/2)*| -4√3 | = (1/2)*4√3=2√3Wait, that gives area 2√3. But is that correct? Let me verify the calculations step by step.Coordinates:B(8,0); C(5,5√3); H(6,4√3); G(9,-√3)Compute shoelace sum:Term1: x_B y_C =8*5√3=40√3; x_C y_B=5*0=0; term1 - term2=40√3 -0=40√3Term2: x_C y_H=5*4√3=20√3; x_H y_C=6*5√3=30√3; term2 - term3=20√3 -30√3= -10√3Term3: x_H y_G=6*(-√3)= -6√3; x_G y_H=9*4√3=36√3; term3 - term4= -6√3 -36√3= -42√3Term4: x_G y_B=9*0=0; x_B y_G=8*(-√3)= -8√3; term4 - term1=0 - (-8√3)=8√3Wait, no, actually, the shoelace formula is:Sum over (x_i * y_{i+1} - x_{i+1} * y_i)So for the four points:1. (8,0) to (5,5√3): term is 8*5√3 -5*0 =40√32. (5,5√3) to (6,4√3): term is5*4√3 -6*5√3=20√3 -30√3=-10√33. (6,4√3) to (9,-√3): term is6*(-√3) -9*4√3= -6√3 -36√3= -42√34. (9,-√3) to (8,0): term is9*0 -8*(-√3)=0 +8√3=8√3Total sum:40√3 -10√3 -42√3 +8√3= (40-10-42+8)√3= (-4)√3Absolute value:4√3. Multiply by 1/2:2√3. So the area is 2√3. Hmm. That seems small given the triangle's area was 20√3. But maybe it's correct.Wait, let me check if the order of the points is correct. The problem says quadrilateral BCHG. So the order should be B -> C -> H -> G. Let me confirm the coordinates:B(8,0), C(5,5√3), H(6,4√3), G(9,-√3). Connecting these in order, the shape might be a quadrilateral that folds over? Wait, perhaps I made a mistake in the order. Maybe the correct order is B -> C -> H -> G, but when I connected them, some lines cross?Alternatively, maybe I should plot the points to check.Point B is at (8,0), which is on the x-axis. Point C is at (5,5√3), which is up and left. Point H is at (6,4√3), which is between C and G? Point G is at (9,-√3), which is below B. Connecting B(8,0) to C(5,5√3) to H(6,4√3) to G(9,-√3) to B(8,0). The quadrilateral seems to have a "twist". Alternatively, maybe the order is different? The problem says "area of BCHG", so the quadrilateral is B connected to C connected to H connected to G connected to B.Alternatively, maybe the shoelace formula requires the points to be ordered either clockwise or counter-clockwise without crossing. Let's check the order.List the coordinates:B(8,0), C(5,5√3), H(6,4√3), G(9,-√3)Plotting these:- B is at (8,0)- C is at (5,5√3) ≈ (5,8.66)- H is at (6,4√3) ≈ (6,6.928)- G is at (9,-1.732)Connecting B(8,0) to C(5,8.66): this is a line going up and left.C(5,8.66) to H(6,6.928): down and right.H(6,6.928) to G(9,-1.732): steeply down and right.G(9,-1.732) to B(8,0): up and left.This creates a quadrilateral that is not self-intersecting, but the area calculated via shoelace is 2√3. Given that the entire triangle's area is 20√3, 2√3 seems possible if BCHG is a small quadrilateral inside.Alternatively, maybe I made a mistake in the coordinates of points G and H.Let me double-check the coordinates of G and H.For G: intersection of BE and DF.BE is from B(8,0) to E(3,5√3). We found the equation of BE as y = -√3 x +8√3. DF is x=9. So when x=9, y= -√3*9 +8√3= -9√3 +8√3= -√3. So G is (9, -√3). Correct.For H: intersection of CF and DE.CF is from C(5,5√3) to F(9,√3). Equation: y = -√3 x +10√3. DE is from D(9,3√3) to E(3,5√3). Equation: y = (-√3/3)x +6√3. Solving these gives x=6, y=4√3. So H is (6,4√3). Correct.Coordinates are correct. Then the shoelace formula must be right. Let me try another approach to compute the area.Alternatively, divide the quadrilateral into triangles. For example, triangle BCH and triangle BCG, but wait, not sure. Alternatively, split into BCH and BHG.Wait, quadrilateral BCHG can be split into triangle BCH and triangle BGH. Let's compute their areas.First, coordinates:B(8,0), C(5,5√3), H(6,4√3), G(9,-√3)Compute area of triangle BCH.Using coordinates:B(8,0); C(5,5√3); H(6,4√3)Use shoelace formula:Area = 1/2 |(8*5√3 +5*4√3 +6*0) - (0*5 +5√3*6 +4√3*8)|Compute term1:8*5√3=40√3;5*4√3=20√3;6*0=0. Sum:60√3term2:0*5=0;5√3*6=30√3;4√3*8=32√3. Sum:62√3Difference:60√3 -62√3= -2√3. Absolute value:2√3. Area=1/2*2√3=√3Area of triangle BCH is √3.Now area of triangle BGH.Points B(8,0); G(9,-√3); H(6,4√3)Shoelace formula:Area=1/2 |(8*(-√3) +9*4√3 +6*0) - (0*9 + (-√3)*6 +4√3*8)|Compute term1:8*(-√3)= -8√3;9*4√3=36√3;6*0=0. Sum:28√3term2:0*9=0; (-√3)*6= -6√3;4√3*8=32√3. Sum:26√3Difference:28√3 -26√3=2√3. Absolute value:2√3. Area=1/2*2√3=√3Thus total area of quadrilateral BCHG is √3 +√3=2√3. So matches the previous result.Hence, the area is 2√3.But wait, the problem states "find the area of BCHG," and according to both methods, it's 2√3. Given that the original triangle has area 20√3, 2√3 is plausible. However, given the construction, maybe there's a synthetic approach that could confirm this.Alternatively, perhaps I made a mistake in the coordinates.Wait, let's check if point H(6,4√3) is indeed inside the triangle.Original triangle ABC has points A(0,0), B(8,0), C(5,5√3). Point H is at (6,4√3). Since 5 <6 <8 in x-coordinate, and 4√3≈6.928 <5√3≈8.66. So H is inside the triangle? Wait, but the problem involves intersections outside as well.Similarly, point G is at (9,-√3), which is outside the original triangle.But regardless, coordinates are correct as per construction. So the area via shoelace formula is 2√3.But let me confirm once again with another approach.Alternatively, use vectors.Compute vectors for the sides of the quadrilateral and apply the shoelace formula again.Alternatively, coordinate geometry.Alternatively, check if quadrilateral BCHG is a trapezoid or something else.But given that the coordinates are correctly calculated, and the shoelace formula is reliable here, probably 2√3 is the answer.Therefore, the area of BCHG is 2√3, which in boxed form is boxed{2sqrt{3}}.Final Answerboxed{2sqrt{3}}</think>Given triangle (ABC) with (AB = 8), (AC = 10), and (angle BAC = 60^circ), we need to find the area of quadrilateral (BCHG). 1. Calculate (BC) using the Law of Cosines: [ BC^2 = AB^2 + AC^2 - 2 cdot AB cdot AC cdot cos(60^circ) = 8^2 + 10^2 - 2 cdot 8 cdot 10 cdot 0.5 = 84 implies BC = 2sqrt{21} ]2. Coordinates of points: - (A(0, 0)), (B(8, 0)), (C(5, 5sqrt{3})) - Circumcenter (O(4, 2sqrt{3})) with radius (2sqrt{7}) - Midpoint (D) of minor arc (BC) is ((9, 3sqrt{3}))3. Find points (E) and (F): - (E) on (Gamma) such that (DE perp AC): (E(3, 5sqrt{3})) - (F) on (Gamma) such that (DF perp AB): (F(9, sqrt{3}))4. Equations of lines: - (BE: y = -sqrt{3}x + 8sqrt{3}) - (DF: x = 9) - (CF: y = -sqrt{3}x + 10sqrt{3}) - (DE: y = -frac{sqrt{3}}{3}x + 6sqrt{3})5. Intersection points: - (G) is intersection of (BE) and (DF): (G(9, -sqrt{3})) - (H) is intersection of (CF) and (DE): (H(6, 4sqrt{3}))6. Area of quadrilateral (BCHG) using shoelace formula: - Coordinates: (B(8, 0)), (C(5, 5sqrt{3})), (H(6, 4sqrt{3})), (G(9, -sqrt{3})) - Shoelace formula calculation: [ text{Area} = frac{1}{2} left| 8 cdot 5sqrt{3} + 5 cdot 4sqrt{3} + 6 cdot (-sqrt{3}) + 9 cdot 0 - (0 cdot 5 + 5sqrt{3} cdot 6 + 4sqrt{3} cdot 9 + (-sqrt{3}) cdot 8) right| ] Simplifying, the area is (boxed{2sqrt{3}}).,<think>Okay, let's see. The problem is about triangle ABC with circumcircle Γ. D is the midpoint of the minor arc BC. Then there are points E and F on Γ such that DE is perpendicular to AC and DF is perpendicular to AB. Then lines BE and DF meet at G, and lines CF and DE meet at H. We need to find the area of quadrilateral BCHG given that AB = 8, AC = 10, and angle BAC = 60 degrees.First, I need to visualize this. Let me try to sketch the triangle ABC. Given angle BAC is 60 degrees, AB is 8, and AC is 10. So vertex A is between B and C, with AB = 8, AC = 10, angle at A is 60 degrees. The circumcircle Γ passes through all three vertices. D is the midpoint of minor arc BC, so it's the point opposite to A in some sense? Wait, no. The midpoint of the minor arc BC would be the point such that BD = DC in terms of arc length.Then E and F are points on Γ such that DE is perpendicular to AC and DF is perpendicular to AB. So DE ⊥ AC and DF ⊥ AB. Then BE and DF intersect at G, and CF and DE intersect at H. Then need to find area of BCHG.Hmm. Maybe coordinate geometry would help here? Given the specific lengths and angle, perhaps we can assign coordinates to points A, B, C, and then find coordinates for D, E, F, G, H?Let me start by setting up coordinate system. Let me place point A at the origin (0,0). Since angle BAC is 60 degrees, and AB is 8, AC is 10, we can place point B at (8, 0). Then point C would be at (10 cos 60°, 10 sin 60°). Cos 60° is 0.5, so 10*0.5 = 5, and sin 60° is (√3)/2, so 10*(√3)/2 = 5√3. Therefore, coordinates are:A: (0, 0)B: (8, 0)C: (5, 5√3)Now, need to find the circumcircle Γ of triangle ABC. The circumcircle can be found by finding the perpendicular bisectors of two sides and finding their intersection.First, find midpoint of AB. AB is from (0,0) to (8,0), so midpoint is (4,0). The slope of AB is 0, so the perpendicular bisector is vertical line x = 4.Next, find midpoint of AC. AC is from (0,0) to (5,5√3). Midpoint is (2.5, (5√3)/2). The slope of AC is (5√3 - 0)/(5 - 0) = √3. Therefore, the perpendicular bisector will have slope -1/√3. Equation: y - (5√3)/2 = (-1/√3)(x - 2.5)Find intersection of x=4 and this line.Substitute x=4 into the equation:y - (5√3)/2 = (-1/√3)(4 - 2.5) = (-1/√3)(1.5) = -3/(2√3) = -√3/2Therefore, y = (5√3)/2 - √3/2 = (4√3)/2 = 2√3Thus, circumcircle center is at (4, 2√3). Let's confirm radius. Distance from center (4, 2√3) to point A (0,0):√[(4)^2 + (2√3)^2] = √[16 + 12] = √28 = 2√7. So radius is 2√7.Wait, check distance to B (8,0):√[(8 - 4)^2 + (0 - 2√3)^2] = √[16 + 12] = √28. Correct. Distance to C (5,5√3):√[(5 -4)^2 + (5√3 -2√3)^2] = √[1 + (3√3)^2] = √[1 + 27] = √28. Correct. So Γ has center (4, 2√3) and radius 2√7.Now, point D is the midpoint of minor arc BC. Since D is the midpoint of arc BC not containing A. To find coordinates of D, perhaps we can parametrize the arc BC?Alternatively, since D is the midpoint of arc BC, the center of the circle lies on the angle bisector of angle BAC? Wait, no. The midpoint of arc BC is equidistant from B and C, so it lies on the perpendicular bisector of BC. Wait, but in the circumcircle, the midpoint of arc BC is the point such that BD = DC in arc length, which is equivalent to being the intersection of the angle bisector of angle BAC with the circumcircle? Hmm, maybe not.Alternatively, in triangle ABC, the midpoint of arc BC is the circumcircle point such that angles BDC and BAC are equal? Wait, perhaps there's a better way. Since we have coordinates for B, C, and the center of the circle, maybe we can parametrize points B and C on the circle and find the midpoint.But parametrizing might be complicated. Alternatively, since D is the midpoint of arc BC, the line AD is the angle bisector of angle BAC? Wait, in a triangle, the midpoint of arc BC does lie on the angle bisector. So if angle BAC is 60 degrees, then AD bisects angle BAC into 30 degrees each. Wait, but actually, the midpoint of arc BC (not containing A) is the ex-bisector? Wait, no. Let me recall that in a triangle, the midpoint of arc BC not containing A is the center of the circle tangent to BC and the circumcircle? Hmm, maybe not. Wait, actually, the midpoint of arc BC not containing A is equidistant from B and C, and lies on the circumcircle. Moreover, the angle bisector of angle BAC does pass through the midpoint of arc BC containing A, while the ex-bisector passes through the midpoint of arc BC not containing A. Wait, in our case, D is the midpoint of the minor arc BC. If the triangle is such that arc BC is less than 180 degrees, then D would be the midpoint of that arc. But given that angle BAC is 60 degrees, the arc BC is 2*angle BAC = 120 degrees? Wait, no. The central angle corresponding to arc BC is equal to 2*angle BAC. Wait, no, the central angle is twice the inscribed angle. So if angle BAC is 60 degrees, then the central angle over arc BC is 120 degrees. Therefore, minor arc BC is 120 degrees. So D is the midpoint, so 60 degrees from both B and C along the arc.Given the center of the circle is (4, 2√3). Let's find coordinates of B and C. B is (8,0), C is (5,5√3). So points B and C on the circle. Let me compute the angle from the center to B and to C.First, vector from center (4, 2√3) to B (8,0): (8-4, 0 - 2√3) = (4, -2√3)Vector from center to C (5,5√3): (5-4, 5√3 - 2√3) = (1, 3√3)To find the angle between these two vectors. The angle between vectors (4, -2√3) and (1, 3√3).Dot product: 4*1 + (-2√3)(3√3) = 4 - 6*3 = 4 - 18 = -14The magnitude of first vector: √(16 + 12) = √28 = 2√7Second vector: √(1 + 27) = √28 = 2√7So cosθ = (-14)/(2√7 * 2√7) = (-14)/(4*7) = (-14)/28 = -0.5Thus θ = 120 degrees. So the central angle for arc BC is 120 degrees. Therefore, the midpoint D divides arc BC into two arcs of 60 degrees each. So to find point D, we can rotate vector from center to B by 60 degrees towards C.But maybe it's easier to parameterize the arc. Let's parameterize point D.Since the center is at (4, 2√3), radius 2√7. Let's find the coordinates of D.From center O (4, 2√3), the midpoint of arc BC can be found by rotating point B around the center by 60 degrees towards C. Since the central angle is 120 degrees, so moving 60 degrees from B towards C would reach D.To compute the rotation, let's use rotation matrix. The vector from O to B is (4, -2√3). Rotating this vector by 60 degrees clockwise (since arc BC is 120 degrees, and D is the midpoint, so direction from B to C is clockwise? Let's check.Looking at the coordinates, point B is at (8,0), center at (4, 2√3), point C is at (5,5√3). The angle from OB to OC is 120 degrees. If we go from OB to OD, where D is the midpoint, we need to rotate OB by 60 degrees towards OC. Since the central angle is 120 degrees, rotating OB by 60 degrees clockwise would reach D.Alternatively, let's compute the rotation. Rotating the vector (4, -2√3) by 60 degrees clockwise. The rotation matrix for clockwise rotation by θ is [cosθ, sinθ; -sinθ, cosθ].So θ = 60 degrees. Cosθ = 0.5, sinθ = (√3)/2.So new vector:x' = 4*0.5 + (-2√3)*(√3/2) = 2 + (-2√3*(√3)/2) = 2 - (6/2) = 2 - 3 = -1y' = 4*(√3/2) + (-2√3)*0.5 = 2√3 - √3 = √3Therefore, the vector from O to D is (-1, √3). Therefore, coordinates of D are center (4,2√3) plus vector (-1, √3): (4 -1, 2√3 + √3) = (3, 3√3)Wait, let me check the calculation again:Original vector from O to B is (4, -2√3). Rotated 60 degrees clockwise.x component: 4*cos60 + (-2√3)*sin60 = 4*0.5 + (-2√3)*(√3/2) = 2 + (-2*3/2) = 2 - 3 = -1y component: -4*sin60 + (-2√3)*cos60 = -4*(√3/2) + (-2√3)*0.5 = -2√3 - √3 = -3√3Wait, wait, maybe I messed up the direction. Wait, rotation matrix for clockwise rotation is:[ cosθ, sinθ ][ -sinθ, cosθ ]So original vector (x, y) becomes (x cosθ + y sinθ, -x sinθ + y cosθ )So here, x = 4, y = -2√3. θ = 60 degrees.x’ = 4*0.5 + (-2√3)*(√3/2) = 2 - (2*3)/2 = 2 - 3 = -1y’ = -4*(√3/2) + (-2√3)*0.5 = -2√3 - √3 = -3√3Therefore, the new vector after rotation is (-1, -3√3). Therefore, the coordinates of D are (4 + (-1), 2√3 + (-3√3)) = (3, -√3)Wait, that can't be right because point D should be on the circumcircle above the triangle. Wait, point B is at (8,0), center at (4, 2√3), so if we rotate OB 60 degrees clockwise, we might end up below the center. But point C is at (5,5√3), which is above the center. Hmm. Maybe the rotation should be counterclockwise?Wait, maybe I got the direction wrong. If the central angle from B to C is 120 degrees, then rotating OB 60 degrees counterclockwise would reach D. Let's try that.Rotation counterclockwise by 60 degrees: rotation matrix is [cosθ, -sinθ; sinθ, cosθ]So θ = 60 degrees.x’ = 4*0.5 - (-2√3)*(√3/2) = 2 + (2√3 * √3)/2 = 2 + (6)/2 = 2 + 3 = 5y’ = 4*(√3/2) + (-2√3)*0.5 = 2√3 - √3 = √3Therefore, the vector from O to D would be (5, √3). Therefore, coordinates of D are center (4,2√3) + (5, √3) = (4 +5, 2√3 + √3) = (9, 3√3). Wait, but point (9,3√3) is outside the triangle. Let's check distance from center to D: √[(5)^2 + (√3)^2] = √(25 + 3) = √28, which matches the radius. So D is (9, 3√3). But does that make sense?Wait, original points: B is (8,0), C is (5,5√3). The arc BC is 120 degrees. If D is the midpoint of arc BC, then it's 60 degrees from both B and C along the arc. If we rotate OB 60 degrees counterclockwise, we get to D. Let's see. The original vector OB is (4, -2√3). Rotating 60 degrees counterclockwise gives us the new coordinates. Hmm, but earlier calculation gave D at (9, 3√3). Let's check if that's correct.Wait, the rotation is applied to the vector from the center, right? So if the original vector from O to B is (4, -2√3), rotating 60 degrees counterclockwise would result in a new vector. Let me compute that again.Rotation matrix for counterclockwise:x’ = x cosθ - y sinθy’ = x sinθ + y cosθSo x = 4, y = -2√3, θ = 60 degrees.x’ = 4*0.5 - (-2√3)*(√3/2) = 2 + (2√3 * √3)/2 = 2 + 6/2 = 2 + 3 = 5y’ = 4*(√3/2) + (-2√3)*0.5 = 2√3 - √3 = √3Therefore, vector from O is (5, √3). Therefore, D is at (4 + 5, 2√3 + √3) = (9, 3√3). Let's check if this point is on the circumcircle. Distance from O: √(5² + (√3)²) = √(25 + 3) = √28, correct. Now, does this make sense as the midpoint of arc BC?Point D is at (9,3√3). Let's see angle from B to D to C. Alternatively, compute the central angles. From O to B is (4, -2√3), to D is (5, √3), to C is (1, 3√3). Let's compute the angle between OD and OB, and OD and OC.Vector OD: (5, √3)Vector OB: (4, -2√3)Dot product: 5*4 + √3*(-2√3) = 20 - 2*3 = 20 - 6 = 14Magnitudes: |OD| = √(25 + 3) = √28, |OB| = √28. So cosθ = 14/(√28 * √28) = 14/28 = 0.5, so θ = 60 degrees. Similarly, between OD and OC:Vector OC is (1, 3√3)Dot product OD . OC = 5*1 + √3*3√3 = 5 + 9 = 14Same as above, cosθ = 14/(√28 * √28) = 0.5, θ = 60 degrees. So indeed, OD is 60 degrees from both OB and OC. Therefore, D is the midpoint of arc BC. So coordinates of D are (9, 3√3).Wait, but D is supposed to be the midpoint of the minor arc BC. If arc BC is 120 degrees, then D is 60 degrees from both B and C. So this seems correct.But point D is at (9, 3√3). Let me confirm that BD and DC are each 60 degrees. Since central angles from B to D and D to C are 60 degrees each, yes.Okay, moving on. Now we need to find E and F on Γ such that DE ⊥ AC and DF ⊥ AB.First, let's find DE perpendicular to AC. AC is from A (0,0) to C (5,5√3). The slope of AC is (5√3 - 0)/(5 - 0) = √3. Therefore, DE must be perpendicular to AC, so slope of DE is -1/√3.Similarly, DF is perpendicular to AB. AB is from A (0,0) to B (8,0), so slope is 0. Therefore, DF must be vertical (slope undefined) to be perpendicular. Wait, DF is perpendicular to AB. Since AB is horizontal, DF must be vertical. So DF is a vertical line passing through D (9, 3√3). So DF is the line x = 9. Since F is on Γ, the intersection of x=9 with Γ will give point F.Similarly, DE is a line with slope -1/√3 passing through D (9, 3√3). We need to find E as the other intersection point of this line with Γ.So let's find coordinates for E and F.First, let's find F. DF is vertical line x=9. The circumcircle Γ has equation (x - 4)^2 + (y - 2√3)^2 = (2√7)^2 = 28.Substitute x=9 into the circle equation:(9 - 4)^2 + (y - 2√3)^2 = 2825 + (y - 2√3)^2 = 28(y - 2√3)^2 = 3Therefore, y = 2√3 ± √3Thus, y = 3√3 or y = √3But D is (9, 3√3), so F is the other intersection point, which is (9, √3).Therefore, F is (9, √3).Now, find E. DE is the line through D (9, 3√3) with slope -1/√3.Equation of DE: y - 3√3 = (-1/√3)(x - 9)We need to find the other intersection point E of this line with Γ.Substitute y = (-1/√3)(x - 9) + 3√3 into the circle equation.First, simplify the equation of DE:y = (-1/√3)x + 9/√3 + 3√3Convert 9/√3 to 3√3:So y = (-1/√3)x + 3√3 + 3√3 = (-1/√3)x + 6√3Now, substitute into the circle equation:(x - 4)^2 + [(-1/√3 x + 6√3 - 2√3)]^2 = 28Simplify:(x - 4)^2 + [(-1/√3 x + 4√3)]^2 = 28Let's compute each term.First term: (x - 4)^2 = x² - 8x + 16Second term: [(-1/√3 x + 4√3)]^2Let me square that:(-1/√3 x + 4√3)^2 = ( (-1/√3 x)^2 + 2*(-1/√3 x)(4√3) + (4√3)^2 )= ( (1/3)x² + 2*(-1/√3)(4√3)x + 16*3 )Simplify each term:First term: (1/3)x²Second term: 2*(-1/√3)(4√3)x = 2*(-4*3/√3*√3)x? Wait,Wait, (-1/√3)(4√3) = (-4√3)/√3 = -4. Then multiplied by 2: -8xThird term: 16*3 = 48Therefore, the second term is (1/3)x² - 8x + 48Therefore, the circle equation becomes:(x² - 8x + 16) + (1/3 x² - 8x + 48) = 28Combine like terms:x² + (1/3)x² - 8x -8x + 16 + 48 = 28(4/3)x² - 16x + 64 = 28Multiply both sides by 3 to eliminate fraction:4x² - 48x + 192 = 84Subtract 84:4x² - 48x + 108 = 0Divide by 4:x² - 12x + 27 = 0Solve quadratic equation:x = [12 ± √(144 - 108)] / 2 = [12 ± √36]/2 = [12 ± 6]/2Thus, x = (12 + 6)/2 = 9 or x = (12 - 6)/2 = 3We already know that D is at x=9, so the other intersection point E is at x=3.Find y-coordinate when x=3:From DE's equation: y = (-1/√3)(3 - 9) + 3√3 = (-1/√3)(-6) + 3√3 = 6/√3 + 3√3 = 2√3 + 3√3 = 5√3Therefore, point E is (3,5√3).So E is (3,5√3), F is (9, √3)Now, need to find lines BE and DF intersect at G, and lines CF and DE intersect at H.First, find G as intersection of BE and DF.Line BE: connects B (8,0) and E (3,5√3). Let's find its equation.Slope of BE: (5√3 - 0)/(3 - 8) = 5√3 / (-5) = -√3Equation: y - 0 = -√3(x - 8) => y = -√3 x + 8√3Line DF: connects D (9,3√3) and F (9,√3). But DF is vertical line x=9.Intersection G is at x=9. Substitute x=9 into BE's equation: y = -√3*9 + 8√3 = (-9 + 8)√3 = -√3Therefore, G is at (9, -√3)Wait, but DF is from D (9,3√3) to F (9, √3). The line DF is vertical line x=9 from (9,3√3) to (9, √3). However, BE is the line from B (8,0) to E (3,5√3). When x=9, y is -√3, which is below F. So G is at (9, -√3), which is outside the segment DF, but since lines are extended, that's okay.Now, find H as intersection of CF and DE.Line CF: connects C (5,5√3) and F (9, √3). Let's find its equation.Slope of CF: (√3 - 5√3)/(9 - 5) = (-4√3)/4 = -√3Equation: y - 5√3 = -√3(x - 5)=> y = -√3 x + 5√3 + 5√3 = -√3 x + 10√3Line DE: connects D (9,3√3) and E (3,5√3). We already know DE's equation is y = (-1/√3)x + 6√3Find intersection H of CF and DE.Set equations equal:-√3 x + 10√3 = (-1/√3)x + 6√3Multiply both sides by √3 to eliminate denominator:-3x + 30 = -x + 18Bring variables to left and constants to right:-3x + x = 18 - 30-2x = -12x = 6Substitute x=6 into DE's equation: y = (-1/√3)(6) + 6√3 = (-6/√3) + 6√3 = (-2√3) + 6√3 = 4√3Thus, H is at (6,4√3)Now, we have points:B: (8,0)C: (5,5√3)H: (6,4√3)G: (9, -√3)Need to find the area of quadrilateral BCHG.Quadrilateral BCHG has vertices at B (8,0), C (5,5√3), H (6,4√3), G (9,-√3)To compute the area, can use the shoelace formula.Let me list the coordinates in order:B: (8, 0)C: (5, 5√3)H: (6, 4√3)G: (9, -√3)Back to B: (8, 0)Shoelace formula:Area = 1/2 |sum over i (x_i y_{i+1} - x_{i+1} y_i)|Compute each term:First pair: B to C:x_i = 8, y_i = 0x_{i+1} =5, y_{i+1} =5√3Term: 8*5√3 -5*0 = 40√3Second pair: C to H:x_i=5, y_i=5√3x_{i+1}=6, y_{i+1}=4√3Term:5*4√3 -6*5√3 =20√3 -30√3= -10√3Third pair: H to G:x_i=6, y_i=4√3x_{i+1}=9, y_{i+1}=-√3Term:6*(-√3) -9*4√3= -6√3 -36√3= -42√3Fourth pair: G to B:x_i=9, y_i=-√3x_{i+1}=8, y_{i+1}=0Term:9*0 -8*(-√3)=0 +8√3=8√3Sum all terms:40√3 -10√3 -42√3 +8√3 = (40 -10 -42 +8)√3 = (-4)√3Take absolute value and multiply by 1/2:Area= (1/2)*| -4√3 | = (1/2)*4√3=2√3Wait, but the area can't be negative, so absolute value. So 2√3. Hmm, but is this correct?Wait, let me check the calculation again step by step.First term: B to C: 8*5√3 -5*0 =40√3Second term: C to H:5*4√3 -6*5√3 =20√3 -30√3= -10√3Third term: H to G:6*(-√3) -9*4√3= -6√3 -36√3= -42√3Fourth term: G to B:9*0 -8*(-√3)=0 +8√3=8√3Total sum:40√3 -10√3 -42√3 +8√3 = (40 -10 -42 +8)√3 = (-4)√3So absolute value is 4√3, half of that is 2√3. Hmm. But let me check if the order of the points is correct.Quadrilateral BCHG: B (8,0), C (5,5√3), H (6,4√3), G (9,-√3). Let's plot these mentally.Point B is at (8,0), C is at (5,5√3) which is up and left. H is at (6,4√3), which is a bit right and down from C. G is at (9,-√3), which is right and down a lot. Connecting these points: B to C to H to G to B. The shoelace formula should work as long as the polygon is simple (non-intersecting). Let me check if the order is correct.Alternatively, maybe the order is not correct? Let me check the connectivity:B is connected to C, C to H, H to G, G to B. So the quadrilateral is B-C-H-G-B.But to apply shoelace, the order must be either clockwise or counter-clockwise without crossing.Alternatively, maybe I missed the order. Let's check coordinates:B: (8,0)C: (5,5√3) ≈ (5, 8.66)H: (6,4√3) ≈ (6, 6.928)G: (9, -√3) ≈ (9, -1.732)Plotting these points:B is on the right at (8,0), C is upper left, H is between C and B but higher than B, G is below B to the right.Wait, connecting B-C-H-G would create a quadrilateral that might have a crossing? Wait, no. From B to C (upper left), then to H (right a bit, down a bit), then to G (far right, down), then back to B. This should form a quadrilateral without crossing. But maybe the shoelace formula requires the points to be ordered either clockwise or counterclockwise. Let me check.Order B (8,0), C (5,5√3), H (6,4√3), G (9,-√3). Let's see:Going from B to C: moving left and up.From C to H: moving right and down.From H to G: moving right and down sharply.From G to B: moving left and up.This seems to form a quadrilateral that is a trapezoid or some kind of kite? Not sure. But according to shoelace formula, the area is 2√3.But given the original triangle has area, let's compute the area of triangle ABC to see if 2√3 is plausible.Area of triangle ABC with sides AB=8, AC=10, angle 60 degrees:Area = 1/2 * AB * AC * sin(60°) = 0.5 *8*10*(√3/2) = 20√3 ≈34.64The quadrilateral BCHG is part of this triangle? Wait, but according to coordinates, G is outside the triangle (at (9,-√3)), so the quadrilateral is partly inside and partly outside. But area 2√3 seems small. Maybe I made a mistake in the order of points.Alternatively, maybe I need to order the points correctly for shoelace. Let me try a different order. Maybe B, G, H, C.Let me list points as B (8,0), G (9,-√3), H (6,4√3), C (5,5√3). Let's apply shoelace:First term: B to G:8*(-√3) -9*0= -8√3Second term: G to H:9*4√3 -6*(-√3)=36√3 +6√3=42√3Third term: H to C:6*5√3 -5*4√3=30√3 -20√3=10√3Fourth term: C to B:5*0 -8*5√3=0 -40√3= -40√3Sum terms: -8√3 +42√3 +10√3 -40√3= ( -8 +42 +10 -40 )√3=4√3Take absolute value and multiply by 1/2: 1/2 *4√3=2√3. Same result. So regardless of the order, the area is 2√3. Hmm.But the problem states "find the area of BCHG". If the shoelace formula gives 2√3, but the triangle area is 20√3, and BCHG is a quadrilateral involving parts outside the triangle, maybe 2√3 is correct? Let me verify by alternative method.Alternatively, compute the area by breaking into triangles.For example, area of BCHG can be area of BCH plus area of BCG? Wait, not sure. Alternatively, split into triangles BCH and BCG? Wait, need to see the shape.Alternatively, use vectors or coordinate geometry.Alternatively, check coordinates:Point B: (8,0)Point C: (5,5√3)Point H: (6,4√3)Point G: (9,-√3)Compute vectors for sides and use cross products.Alternatively, use coordinates and compute the area step by step.Wait, coordinates:B (8,0)C (5,5√3)H (6,4√3)G (9,-√3)Let me plot these:- B is at (8,0), which is on the x-axis.- C is at (5,5√3) ≈ (5, 8.66)- H is at (6,4√3) ≈ (6, 6.928)- G is at (9, -√3) ≈ (9, -1.732)So quadrilateral B-C-H-G:From B (8,0) to C (5,5√3): up and left.From C to H (6,4√3): right and down.From H to G (9,-√3): right and sharply down.From G to B: left and up.This forms a quadrilateral that is somewhat "twisted". Wait, but according to coordinates, is it convex?Check the angles:From B to C to H: the turn is from a line going up-left to a line going right-down.From C to H to G: turn from right-down to right-sharply down.From H to G to B: turn from right-sharply down to left-up.From G to B to C: turn from left-up to up-left.It seems convex. So the shoelace formula should work.But according to calculations, area is 2√3. Let me confirm once more.Alternatively, use vectors:The area can be calculated using the shoelace formula as follows:List the coordinates in order (either clockwise or counter-clockwise):Let's take B (8,0), C (5,5√3), H (6,4√3), G (9,-√3)Compute shoelace sum:Sum1 = (8*5√3) + (5*4√3) + (6*(-√3)) + (9*0) = 40√3 +20√3 -6√3 +0 = 54√3Sum2 = (0*5) + (5√3*6) + (4√3*9) + (-√3*8) = 0 +30√3 +36√3 -8√3 = 58√3Area = 1/2 |Sum1 - Sum2| =1/2 |54√3 -58√3| =1/2 | -4√3 | =2√3. Same result.Therefore, area is 2√3. But given that the original triangle has area 20√3, is 2√3 plausible? It seems small, but considering that G and H are constructed points, maybe it's correct.Wait, but let's check with another approach. For instance, compute the coordinates again step by step to ensure no mistake.First, points A,B,C: confirmed.Circumcircle: center at (4,2√3), radius 2√7. Correct.Point D: (9,3√3). Correct.Point E: found by intersecting DE (slope -1/√3 through D) with Γ, other intersection at (3,5√3). Correct.Point F: vertical line DF at x=9, other intersection at (9,√3). Correct.Lines BE: from B (8,0) to E (3,5√3). Equation y = -√3 x +8√3. Correct.Intersection G with DF (x=9): y= -√3*9 +8√3= -√3. So G(9,-√3). Correct.Lines CF: from C (5,5√3) to F (9,√3). Slope -√3. Equation y = -√3 x +10√3. Correct.Lines DE: from D (9,3√3) to E (3,5√3). Equation y= -1/√3 x +6√3. Correct.Intersection H: solving -√3 x +10√3 = -1/√3 x +6√3. Got x=6, y=4√3. Correct.Thus coordinates all correct. Shoelace formula applied correctly gives area 2√3.But the problem states "find the area of BCHG". If all steps are correct, answer is 2√3.But wait, in the original problem, the answer might need to be 14√3 or something else. Let me check if I messed up in some calculation.Wait, when solving for H, when we set the equations equal:-√3 x +10√3 = (-1/√3)x +6√3Multiply both sides by √3:Left side: -3x + 30Right side: -x + 18Thus: -3x +30 = -x +18 => -2x = -12 => x=6. Correct.Then y= -√3*6 +10√3=4√3. Correct.Similarly, for G: x=9, y= -√3*9 +8√3= -9√3 +8√3= -√3. Correct.Coordinates all correct. Shoelace formula gives 2√3. Therefore, unless there is a miscalculation in the shoelace steps.Wait, when I computed Sum1 and Sum2:Sum1 = 8*5√3 +5*4√3 +6*(-√3) +9*0=40√3 +20√3 -6√3 +0=54√3Sum2=0*5 +5√3*6 +4√3*9 +(-√3)*8=0 +30√3 +36√3 -8√3=58√3Thus, 54√3 -58√3= -4√3, absolute value 4√3, half is 2√3. Correct.Alternatively, maybe the problem expects a different answer. Let me think.Alternatively, maybe the quadrilateral is not B-C-H-G, but another combination. Let me check the problem statement again."Lines BE and DF meet at G, and lines CF and DE meet at H. Given that AB = 8, AC = 10, and ∠BAC = 60°, find the area of BCHG."Yes, so BCHG is the quadrilateral with vertices B, C, H, G. The shoelace formula gives 2√3, but I feel maybe there's a miscalculation because 2√3 seems small.Wait, but let's check the positions of the points again.Point H is at (6,4√3). Point G is at (9,-√3). So quadrilateral BCHG is quite spread out.Alternatively, compute the area by breaking into triangles.Compute area of BCH and BCG.Wait, but need to see how the quadrilateral is formed. BCHG is a quadrilateral, not necessarily convex. Wait, but according to coordinates, it's convex.Alternatively, use vectors.Coordinates:B (8,0)C (5,5√3)H (6,4√3)G (9,-√3)Compute vectors for sides:From B to C: (-3,5√3)From C to H: (1,-√3)From H to G: (3,-5√3)From G to B: (-1,√3)Alternatively, use the formula for area of polygon with coordinates:Area = 1/2 |sum_{i=1}^n (x_i y_{i+1} - x_{i+1} y_i)|Which is exactly the shoelace formula. So if that gives 2√3, then unless there's an error in coordinates, which seem correct, the answer should be 2√3.But given the problem's complexity, maybe 2√3 is correct. Alternatively, the area is 14√3, but I need to check.Wait, maybe there is a miscalculation when I rotated the vector. Wait, point D was found by rotating OB 60 degrees counterclockwise, resulting in D at (9,3√3). But is this correct?Wait, the central angle between B and D is 60 degrees. Let's compute the angle between OB and OD.Vector OB: (4, -2√3)Vector OD: (5, √3)Dot product:4*5 + (-2√3)(√3)=20 -6=14Lengths of vectors: |OB|=|OD|=√(4² + (2√3)^2)=√(16+12)=√28Therefore, cosθ=14/(√28 * √28)=14/28=0.5, so θ=60 degrees. Correct.Therefore, D is indeed 60 degrees from B, so arc BD is 60 degrees, arc DC is 60 degrees. Correct.Therefore, coordinates of D are correct.Thus, unless there's an error in my calculation, which I can't find, the answer is 2√3.But to double-check, let me compute the area step by step using coordinates.Compute the area of quadrilateral BCHG.Coordinates:B (8,0)C (5,5√3)H (6,4√3)G (9,-√3)Divide the quadrilateral into two triangles: BCH and BCG.Wait, but which triangles?Alternatively, divide into triangle BCH and triangle BGH.Wait, not sure. Alternatively, use the shoelace formula with coordinates ordered properly.Alternatively, use vector cross products.Alternatively, compute the area of BCHG by using coordinates:First, find the vectors for the sides and apply the shoelace formula once again.Wait, let's use determinant method for quadrilaterals.The shoelace formula is reliable here. The only thing is the order of the points. Let me check once again.Ordering B (8,0), C (5,5√3), H (6,4√3), G (9,-√3).Plotting these points in order:Start at B (8,0). Move to C (5,5√3): up and left.Then to H (6,4√3): right and down.Then to G (9,-√3): right and down more.Then back to B (8,0): left and up.This forms a quadrilateral that is a trapezoid? No, since sides are not parallel. But shoelace formula still applies.Alternatively, check if the sides BH and CG intersect, but since it's a convex quadrilateral, they shouldn't.Given that the shoelace formula gives 2√3, and all steps are correct, I think the answer is 2√3.But wait, let me check with another method. For instance, compute the area using coordinates and vectors.Compute vectors BC, BH, BG, etc., but it might complicate.Alternatively, consider coordinates:B (8,0)C (5,5√3)H (6,4√3)G (9,-√3)Compute the area by using the formula:Area = 1/2 | (x_B(y_C - y_H) + x_C(y_H - y_G) + x_H(y_G - y_B) + x_G(y_B - y_C)) | ?Wait, not sure. Alternatively, translate the coordinates so that B is at the origin.But maybe not. Alternatively, use the surveyor's formula (shoelace). It should be the same.Alternatively, I might have a miscalculation in the shoelace terms.Let me recompute the shoelace formula step by step:Order: B, C, H, G, BCoordinates:B: (8,0)C: (5,5√3)H: (6,4√3)G: (9,-√3)B: (8,0)Compute terms:x_i * y_{i+1} - x_{i+1} * y_iFirst pair (B to C):8 * 5√3 -5 *0 =40√3 -0=40√3Second pair (C to H):5 *4√3 -6 *5√3=20√3 -30√3=-10√3Third pair (H to G):6*(-√3) -9*4√3=-6√3 -36√3=-42√3Fourth pair (G to B):9*0 -8*(-√3)=0 +8√3=8√3Sum all terms:40√3 -10√3 -42√3 +8√3=(40-10-42+8)√3=(-4)√3Absolute value:4√3Area=1/2*4√3=2√3Same result. Therefore, must be correct.But the problem says "find the area of BCHG". The answer is 2√3. Thus, I think the answer is 2√3.Final Answerboxed{2sqrt{3}}</think>Given triangle (ABC) with (AB = 8), (AC = 10), and (angle BAC = 60^circ), we need to find the area of quadrilateral (BCHG). 1. Coordinates of Points: - (A(0, 0)), (B(8, 0)), (C(5, 5sqrt{3})). - Circumcircle (Gamma) has center ((4, 2sqrt{3})) and radius (2sqrt{7}).2. Midpoint of Minor Arc (BC): - Point (D) is found by rotating vector (overrightarrow{OB}) (from center (O) to (B)) 60 degrees counterclockwise, resulting in (D(9, 3sqrt{3})).3. Finding Points (E) and (F): - (E) is the intersection of line (DE) (perpendicular to (AC)) with (Gamma). Coordinates of (E) are ((3, 5sqrt{3})). - (F) is the intersection of line (DF) (perpendicular to (AB)) with (Gamma). Coordinates of (F) are ((9, sqrt{3})).4. Finding Points (G) and (H): - (G) is the intersection of lines (BE) and (DF). Coordinates of (G) are ((9, -sqrt{3})). - (H) is the intersection of lines (CF) and (DE). Coordinates of (H) are ((6, 4sqrt{3})).5. Area Calculation using Shoelace Formula: - Coordinates of quadrilateral (BCHG): (B(8, 0)), (C(5, 5sqrt{3})), (H(6, 4sqrt{3})), (G(9, -sqrt{3})). - Applying the shoelace formula: [ text{Area} = frac{1}{2} left| 8 cdot 5sqrt{3} + 5 cdot 4sqrt{3} + 6 cdot (-sqrt{3}) + 9 cdot 0 - (0 cdot 5 + 5sqrt{3} cdot 6 + 4sqrt{3} cdot 9 + (-sqrt{3}) cdot 8) right| ] Simplifying: [ text{Area} = frac{1}{2} left| 40sqrt{3} - 10sqrt{3} - 42sqrt{3} + 8sqrt{3} right| = frac{1}{2} left| -4sqrt{3} right| = 2sqrt{3} ]Thus, the area of quadrilateral (BCHG) is (boxed{2sqrt{3}}).]

🔑:1. Identify Key Points and Properties: - Given triangle (ABC) with circumcircle (Gamma). - (D) is the midpoint of the minor arc (BC). - (E) and (F) are points on (Gamma) such that (DE perp AC) and (DF perp AB). - Lines (BE) and (DF) intersect at (G), and lines (CF) and (DE) intersect at (H). - Given (AB = 8), (AC = 10), and (angle BAC = 60^circ).2. Determine the Parallelogram Property: - Note that (angle BED = angle BAD = angle CAD = 90^circ - angle ADE), so (BE perp AD). - Similarly, (CF perp AD). - By applying Pascal's theorem on hexagon (DDEBCF), we conclude that (BCHG) is a parallelogram.3. Calculate the Area of Parallelogram (BCHG): - The area of a parallelogram is given by (BC cdot CH cdot sin angle BCH).4. Determine (CH) and (sin angle BCH): - (H) is the orthocenter of (Delta CAD), so (E) is the reflection of (H) over (AC). Thus, (CH = CE). - Since (BE perp AD) and (angle ABE = 60^circ), if (J = BE cap AC), then (Delta CJE) is equilateral, implying (CE = 2).5. Calculate (sin angle BCH): - Since (BE) and (CF) are perpendicular to (AD), (sin angle BCH = sin angle FAB). - By the extended Law of Sines, (sin angle FAB = frac{BF}{2R} = frac{CE}{2R}) because (BECF) is an isosceles trapezoid.6. Final Calculation: - The area of (BCHG) is (BC cdot CE cdot frac{CE}{2R}). - Given (CE = 2), we have: [ BC cdot CE cdot frac{CE}{2R} = 2^2 cdot sin angle BAC = 2^2 cdot frac{sqrt{3}}{2} = 4 cdot frac{sqrt{3}}{2} = 2sqrt{3}. ]The final answer is (boxed{2sqrt{3}}).

🧠:[<think>Okay, let's see. The problem is: √3 tan α + 1 = 0, with α between 0 degrees and 180 degrees. I need to find α. Hmm. Alright, so first, let's make sure I understand the equation correctly. It's the square root of 3 multiplied by tangent of alpha, plus 1 equals zero. So, maybe I can rearrange this equation to solve for tan α first. Let me try that.Starting with the equation:√3 tan α + 1 = 0Subtract 1 from both sides:√3 tan α = -1Then, divide both sides by √3:tan α = -1/√3Okay, so tan α equals negative 1 over √3. Now, I need to find the angle α between 0° and 180° where the tangent is -1/√3. Hmm. Let me recall the unit circle and the values of tangent. Tangent is sine over cosine, and it's positive in the first and third quadrants, and negative in the second and fourth. But since our angle is between 0° and 180°, which covers the first, second, and part of the third quadrants? Wait, no. 0° to 180° is from the positive x-axis, going counterclockwise to the negative x-axis. So that's the first and second quadrants. Third and fourth quadrants are from 180° to 360°, right?So, tangent is positive in the first and third quadrants, negative in the second and fourth. But since we're limited to 0° to 180°, which are quadrants I and II. So tangent is positive in QI and negative in QII. Therefore, the angle we are looking for is in QII because tangent is negative there.Now, what's the reference angle? Since tan α = -1/√3, the reference angle would be the angle whose tangent is 1/√3. Let me think. What angle has a tangent of 1/√3?I remember that tan 30° = 1/√3. Because tan 30° is sin 30° / cos 30° = (1/2) / (√3/2) = 1/√3. So the reference angle is 30 degrees. Since the angle is in the second quadrant, we subtract the reference angle from 180° to get the actual angle. Wait, no. Wait, in QII, the angles are 180° - θ, where θ is the reference angle. So if the reference angle is 30°, then the actual angle is 180° - 30° = 150°. Let me check that.So tan 150° = tan(180° - 30°) = -tan 30° = -1/√3. Yes, that's correct. So α is 150 degrees. Let me verify that this is the only solution between 0° and 180°. Since the tangent function has a period of 180°, so in the interval from 0° to 180°, there's only one solution where tan α is -1/√3, which is 150°. If we added another 180°, it would be outside the given interval, so 150° is the only solution.Therefore, the answer should be α = 150°. Let me just make sure there's no calculation mistake here. Starting from the original equation:√3 tan α + 1 = 0tan α = -1/√3Reference angle 30°, in QII, so 180 - 30 = 150°. Plugging back in:tan 150° = tan(180° - 30°) = -tan 30° = -1/√3. So yes, that works. So substituting back into the original equation:√3 * (-1/√3) + 1 = -1 + 1 = 0. Perfect, that checks out.So I think that's all. The solution is 150 degrees. No other angles in the interval would satisfy the equation because adding or subtracting 180° would take it out of the 0° to 180° range. So 150° is the only answer.Final AnswerThe value of α is boxed{150^circ}.</think>Given the equation (sqrt{3} tan alpha + 1 = 0) with (0^circ leqslant alpha leqslant 180^circ), we need to find the value of (alpha).Starting with the given equation:[sqrt{3} tan alpha + 1 = 0]Subtract 1 from both sides:[sqrt{3} tan alpha = -1]Divide both sides by (sqrt{3}):[tan alpha = -frac{1}{sqrt{3}}]We recognize that (tan alpha = -frac{1}{sqrt{3}}) implies a reference angle of (30^circ) because (tan 30^circ = frac{1}{sqrt{3}}). Since the tangent is negative, the angle must be in the second quadrant (where tangent is negative) within the given range (0^circ leqslant alpha leqslant 180^circ).The angle in the second quadrant is calculated as:[alpha = 180^circ - 30^circ = 150^circ]Verification by substituting (alpha = 150^circ) back into the original equation confirms the solution:[sqrt{3} tan 150^circ + 1 = sqrt{3} left(-frac{1}{sqrt{3}}right) + 1 = -1 + 1 = 0]Thus, the value of (alpha) is (boxed{150^circ}).,<think>Okay, so I have this equation to solve: √3 tan α + 1 = 0, and the domain is from 0 degrees to 180 degrees. Hmm, let's see. First, I need to recall what tan α means. Tangent is sine over cosine, right? So tan α = sin α / cos α. But maybe I don't need to get into that right away. Let's try to solve for tan α first.The equation is √3 tan α + 1 = 0. If I subtract 1 from both sides, I get √3 tan α = -1. Then, dividing both sides by √3 gives tan α = -1/√3. Hmm, okay. So I need to find all angles α between 0° and 180° where the tangent of α is -1/√3.Wait, tangent is positive in which quadrants? Let me remember. Since tangent is sine over cosine, it's positive in the first and third quadrants. But since our domain is 0° to 180°, the third quadrant isn't included here. So in 0° to 180°, tangent is positive in the first quadrant (0° to 90°) and negative in the second quadrant (90° to 180°), because in the second quadrant, sine is positive and cosine is negative, making tangent negative.So tan α = -1/√3 must be in the second quadrant. Got it. Now, what angle has a tangent of 1/√3? Let me think. I know that tan 30° = 1/√3. Because in a 30-60-90 triangle, the sides are 1, √3, and 2. So opposite to 30° is 1, adjacent is √3, so tan 30° = 1/√3. Therefore, the reference angle here is 30°, right?Since we're in the second quadrant where tangent is negative, the angle would be 180° - 30° = 150°. Let me verify that. If α is 150°, then tan 150° = tan (180° - 30°) = -tan 30° = -1/√3. Yes, that works. So that should be the solution.Wait, but let me check if there are any other angles in the domain 0° to 180° that satisfy tan α = -1/√3. Since tangent has a period of 180°, normally, but in the interval from 0° to 180°, the only solutions would be in the second quadrant, because in the first quadrant tangent is positive, and in the third quadrant, which is beyond 180°, so it's outside our domain. So only 150° is the solution here.Let me double-check. Let's plug α = 150° into the original equation. √3 tan 150° + 1 = √3 * (-1/√3) + 1 = -1 + 1 = 0. Yep, that works. What about 30°? Tan 30° is positive 1/√3, so plugging in would give √3*(1/√3) + 1 = 1 + 1 = 2 ≠ 0. So 30° isn't a solution. How about 180°? Tan 180° is 0, so √3*0 +1 =1≠0. Nope. 0°? Tan 0° is 0, same result. So only 150° works.Therefore, the solution is α = 150°. I think that's it. But let me make sure I didn't make a mistake in the reference angle or the quadrant. So the reference angle is 30°, correct. Since tangent is -1/√3, which is negative, and we're in the second quadrant, angle is 180° - 30° = 150°, yes. So that's the answer.Final AnswerThe solution is boxed{150^circ}.</think>Given the equation (sqrt{3} tan alpha + 1 = 0) with the domain (0^circ leqslant alpha leqslant 180^circ):1. Subtract 1 from both sides: (sqrt{3} tan alpha = -1).2. Divide both sides by (sqrt{3}): (tan alpha = -frac{1}{sqrt{3}}).We need to find the angle (alpha) between (0^circ) and (180^circ) where the tangent is (-frac{1}{sqrt{3}}). - The reference angle for (tan theta = frac{1}{sqrt{3}}) is (30^circ) because (tan 30^circ = frac{1}{sqrt{3}}).- Since (tan alpha) is negative, (alpha) must be in the second quadrant (where tangent is negative) because the domain is (0^circ) to (180^circ).The angle in the second quadrant is (180^circ - 30^circ = 150^circ).Verification:- (tan 150^circ = tan (180^circ - 30^circ) = -tan 30^circ = -frac{1}{sqrt{3}}).- Substituting (alpha = 150^circ) into the original equation: (sqrt{3} tan 150^circ + 1 = sqrt{3} left(-frac{1}{sqrt{3}}right) + 1 = -1 + 1 = 0).Thus, the solution is (boxed{150^circ}).]

🔑:6.150^{circ}

🧠:[<think>Okay, so I have this quadratic equation: ( x^2 - 2kx + k^2 - k - 1 = 0 ). It says there are two distinct real roots ( x_1 ) and ( x_2 ). The problem has two parts. Let me tackle them one by one.Part (1): If ( k = 5 ), find the value of ( x_1x_2^2 + x_1^2x_2 ).Alright, let's start by plugging ( k = 5 ) into the equation. That should make things concrete. Substituting 5 for k:( x^2 - 2*5*x + 5^2 - 5 - 1 = 0 )Calculating each term:- ( 2*5 = 10 ), so the second term is ( -10x )- ( 5^2 = 25 )- ( 25 - 5 - 1 = 19 )So the equation becomes:( x^2 - 10x + 19 = 0 )Now, this quadratic equation has roots ( x_1 ) and ( x_2 ). The problem asks for ( x_1x_2^2 + x_1^2x_2 ). Hmm, that expression looks a bit complicated. Let me see if I can factor it or relate it to Vieta's formulas.First, let's recall Vieta's formulas for a quadratic equation ( ax^2 + bx + c = 0 ):- Sum of roots: ( x_1 + x_2 = -b/a )- Product of roots: ( x_1x_2 = c/a )In our equation, ( a = 1 ), ( b = -10 ), ( c = 19 ). So:- Sum: ( x_1 + x_2 = 10 )- Product: ( x_1x_2 = 19 )Now, the expression ( x_1x_2^2 + x_1^2x_2 ). Let's factor that:( x_1x_2^2 + x_1^2x_2 = x_1x_2(x_2 + x_1) )Oh! That's nice. So it's equal to the product of the roots times the sum of the roots. Which we already know from Vieta's formulas. So substituting the values:( x_1x_2(x_1 + x_2) = 19 * 10 = 190 )So the answer should be 190. Wait, let me verify that.Alternatively, maybe I can compute the roots explicitly and then compute the expression. Let's check if that gives the same result.The quadratic equation is ( x^2 - 10x + 19 = 0 ). Using the quadratic formula:( x = [10 ± sqrt(100 - 76)] / 2 = [10 ± sqrt(24)] / 2 = [10 ± 2*sqrt(6)] / 2 = 5 ± sqrt(6) )So the roots are ( 5 + sqrt(6) ) and ( 5 - sqrt(6) ). Let's call them ( x_1 = 5 + sqrt(6) ), ( x_2 = 5 - sqrt(6) ).Now compute ( x_1x_2^2 + x_1^2x_2 ).First, compute ( x_1x_2 ). That's (5 + sqrt(6))(5 - sqrt(6)) = 25 - 6 = 19. Which matches Vieta's.Now compute ( x_1 + x_2 = 10 ), which also matches.But let's compute ( x_1x_2^2 + x_1^2x_2 ):Take x_1x_2^2 = x_1*(x_2)^2 and x_1^2x_2 = (x_1)^2*x_2. Let's compute each term.First, x_2^2 = (5 - sqrt(6))^2 = 25 - 10sqrt(6) + 6 = 31 - 10sqrt(6)Multiply by x_1: (5 + sqrt(6))(31 - 10sqrt(6)).Let me compute that:= 5*31 + 5*(-10sqrt(6)) + sqrt(6)*31 + sqrt(6)*(-10sqrt(6))= 155 - 50sqrt(6) + 31sqrt(6) - 10*6= 155 - 50sqrt(6) + 31sqrt(6) - 60= (155 - 60) + (-50sqrt(6) + 31sqrt(6))= 95 - 19sqrt(6)Similarly, x_1^2 = (5 + sqrt(6))^2 = 25 + 10sqrt(6) + 6 = 31 + 10sqrt(6)Multiply by x_2: (31 + 10sqrt(6))(5 - sqrt(6))Compute that:= 31*5 + 31*(-sqrt(6)) + 10sqrt(6)*5 + 10sqrt(6)*(-sqrt(6))= 155 - 31sqrt(6) + 50sqrt(6) - 10*6= 155 - 31sqrt(6) + 50sqrt(6) - 60= (155 - 60) + (-31sqrt(6) + 50sqrt(6))= 95 + 19sqrt(6)Now, add the two results together:(95 - 19sqrt(6)) + (95 + 19sqrt(6)) = 95 + 95 + (-19sqrt(6) + 19sqrt(6)) = 190 + 0 = 190So that confirms it. The answer is 190. So part (1) is 190.Part (2): If ( x_1 - 3x_2 = 2 ), find the value of the real number ( k ).Alright, now this is more complex. We need to find k such that one root minus three times the other root equals 2. Let's see.Given the quadratic equation: ( x^2 - 2kx + k^2 - k -1 = 0 ). Let's recall that for this equation, the sum and product of roots are:Sum: ( x_1 + x_2 = 2k )Product: ( x_1x_2 = k^2 - k -1 )We are given the equation ( x_1 - 3x_2 = 2 ). So, we have two equations:1. ( x_1 + x_2 = 2k )2. ( x_1 - 3x_2 = 2 )We can solve this system of equations to find expressions for ( x_1 ) and ( x_2 ) in terms of k, then use the product of the roots to find k.Let me subtract the second equation from the first, but perhaps better to use substitution.Let me write equations:From equation 2: ( x_1 = 2 + 3x_2 )Substitute this into equation 1:( (2 + 3x_2) + x_2 = 2k )Simplify:2 + 4x_2 = 2kDivide both sides by 2:1 + 2x_2 = kSo, ( k = 1 + 2x_2 )Now, we can express x_1 in terms of x_2: ( x_1 = 2 + 3x_2 )Now, we also know the product ( x_1x_2 = k^2 -k -1 )So, substitute x_1 and k in terms of x_2:First, let me write the product:( x_1x_2 = (2 + 3x_2)x_2 = 2x_2 + 3x_2^2 )Also, ( k = 1 + 2x_2 ), so k^2 -k -1 becomes:( (1 + 2x_2)^2 - (1 + 2x_2) -1 )Let me compute that:First, expand ( (1 + 2x_2)^2 = 1 + 4x_2 + 4x_2^2 )Then subtract (1 + 2x_2): ( 1 + 4x_2 + 4x_2^2 -1 -2x_2 = 2x_2 + 4x_2^2 )Subtract 1 more: ( 2x_2 + 4x_2^2 -1 )So, putting it all together:Product from Vieta: ( 2x_2 + 3x_2^2 = 4x_2^2 + 2x_2 -1 )Let me write that equation:( 2x_2 + 3x_2^2 = 4x_2^2 + 2x_2 -1 )Subtract ( 2x_2 + 3x_2^2 ) from both sides:0 = 4x_2^2 + 2x_2 -1 -2x_2 -3x_2^2Simplify:0 = (4x_2^2 -3x_2^2) + (2x_2 -2x_2) -1Which is:0 = x_2^2 -1Therefore:x_2^2 -1 = 0So, x_2 = ±1So, two possibilities: x_2 = 1 or x_2 = -1Let's consider each case:Case 1: x_2 = 1Then, from earlier: k = 1 + 2x_2 = 1 + 2*1 = 3Also, x_1 = 2 + 3x_2 = 2 + 3*1 = 5Check if these roots satisfy the original quadratic equation with k=3.Original equation: x^2 - 2*3x + 3^2 -3 -1 = x^2 -6x +9 -3 -1 = x^2 -6x +5 =0Factoring: x^2 -6x +5 = (x -1)(x -5). So roots 1 and 5. Which matches x_2=1, x_1=5. Good.Case 2: x_2 = -1Then, k = 1 + 2*(-1) =1 -2 = -1x_1 = 2 + 3*(-1) = 2 -3 = -1Check if these roots satisfy the original equation with k = -1.Original equation: x^2 -2*(-1)x + (-1)^2 - (-1) -1 = x^2 +2x +1 +1 -1 = x^2 +2x +1 =0Which factors as (x +1)^2 =0, so a double root at x=-1. However, the problem states that the equation has two distinct real roots. Therefore, this case is invalid because we have a repeated root. So k=-1 is not acceptable.Therefore, the only valid solution is k=3.Let me confirm once again.Given that when x_2 = -1, k=-1, but the quadratic equation would have a repeated root, which contradicts the problem statement that there are two distinct real roots. Hence, we must discard k=-1.Therefore, the answer is k=3.But just to be thorough, let's check the discriminant for k=3 and k=-1.Original quadratic equation discriminant:D = [(-2k)^2 -4*1*(k^2 -k -1)] = 4k^2 -4(k^2 -k -1) = 4k^2 -4k^2 +4k +4 =4k +4For the equation to have two distinct real roots, D>0:4k +4 >0 => 4(k +1) >0 => k +1 >0 =>k> -1So when k=3, D=4*3 +4=12+4=16>0, so two distinct roots. Good.When k=-1, D=4*(-1)+4=-4+4=0, which is a repeated root. So not allowed. Hence, only k=3 is valid.Therefore, part (2) answer is 3.Final Answer(1) boxed{190} (2) boxed{3}</think>Given the quadratic equation ( x^2 - 2kx + k^2 - k - 1 = 0 ) with two distinct real roots ( x_1 ) and ( x_2 ):1. For ( k = 5 ): - Substitute ( k = 5 ) into the equation: ( x^2 - 10x + 19 = 0 ). - Using Vieta's formulas, the sum of the roots ( x_1 + x_2 = 10 ) and the product ( x_1x_2 = 19 ). - The expression ( x_1x_2^2 + x_1^2x_2 ) can be factored as ( x_1x_2(x_1 + x_2) ). - Substituting the values, we get ( 19 times 10 = 190 ). [ boxed{190} ]2. Given ( x_1 - 3x_2 = 2 ): - Using Vieta's formulas, we know ( x_1 + x_2 = 2k ) and ( x_1x_2 = k^2 - k - 1 ). - From the given condition ( x_1 - 3x_2 = 2 ), we solve the system of equations: - ( x_1 = 2 + 3x_2 ) - Substitute ( x_1 ) into ( x_1 + x_2 = 2k ): ( 2 + 4x_2 = 2k Rightarrow k = 1 + 2x_2 ). - Substitute ( x_1 ) and ( k ) into the product equation: - ( (2 + 3x_2)x_2 = (1 + 2x_2)^2 - (1 + 2x_2) - 1 ) - Simplifying, we find ( x_2^2 = 1 Rightarrow x_2 = pm 1 ). - For ( x_2 = 1 ), ( k = 3 ) and ( x_1 = 5 ), which are valid as the quadratic equation has distinct roots. - For ( x_2 = -1 ), ( k = -1 ) leads to a repeated root, which is invalid. [ boxed{3} ],<think>Okay, let's tackle these two problems step by step. First, I need to understand what each part is asking for and how to approach them using the given quadratic equation. The equation is ( x^2 - 2kx + k^2 - k - 1 = 0 ), which has two distinct real roots ( x_1 ) and ( x_2 ).Starting with part (1): If ( k = 5 ), find the value of ( x_1x_2^2 + x_1^2x_2 ).Hmm, so they want a combination of the roots. I remember that for a quadratic equation ( ax^2 + bx + c = 0 ), the sum of the roots ( x_1 + x_2 = -b/a ) and the product ( x_1x_2 = c/a ). Maybe these relationships can help here.Let me first write down the equation when ( k = 5 ). Substituting 5 into the equation:( x^2 - 2*5*x + 5^2 -5 -1 = 0 )Simplify that:( x^2 -10x +25 -5 -1 = 0 )Which becomes:( x^2 -10x +19 = 0 )So, the quadratic equation is ( x^2 -10x +19 = 0 ).Now, the sum of the roots ( x_1 + x_2 = -b/a ). Here, a = 1, b = -10, so sum is 10. The product ( x_1x_2 = c/a = 19 ).The expression we need is ( x_1x_2^2 + x_1^2x_2 ). Let's factor that:( x_1x_2(x_2 + x_1) )Ah, so it's the product of the roots times the sum of the roots. That simplifies things. Since we already have both the sum and the product, we can just plug in the numbers.So, ( x_1x_2^2 + x_1^2x_2 = x_1x_2(x_1 + x_2) = 19 * 10 = 190 ).Wait, that seems straightforward. Let me double-check. If I expand ( x_1x_2^2 + x_1^2x_2 ), that's ( x_1x_2(x_2 + x_1) ). Yes, that's correct. So substituting 19 and 10 gives 190. Seems right. So part (1) is 190.Moving on to part (2): If ( x_1 - 3x_2 = 2 ), find the value of the real number ( k ).This is a bit more complex. We need to relate the roots with the coefficients of the equation. Let's recall the original equation: ( x^2 - 2kx + (k^2 - k -1) = 0 ). So, as before, sum of roots ( x_1 + x_2 = 2k ), product ( x_1x_2 = k^2 - k -1 ).But we have another equation given: ( x_1 - 3x_2 = 2 ). So now we have two equations involving ( x_1 ) and ( x_2 ):1. ( x_1 + x_2 = 2k )2. ( x_1 - 3x_2 = 2 )We can solve these two equations simultaneously to find expressions for ( x_1 ) and ( x_2 ) in terms of ( k ), and then use the product of the roots to form an equation in ( k ).Let me set up the system:Equation (1): ( x_1 + x_2 = 2k )Equation (2): ( x_1 - 3x_2 = 2 )Let's subtract equation (2) from equation (1):Wait, actually, maybe solving for one variable. Let's express ( x_1 ) from equation (2):From equation (2): ( x_1 = 2 + 3x_2 )Substitute this into equation (1):( (2 + 3x_2) + x_2 = 2k )Simplify:( 2 + 4x_2 = 2k )Divide both sides by 2:( 1 + 2x_2 = k )So, ( k = 1 + 2x_2 ). Now, we need to relate this to the product of the roots.The product ( x_1x_2 = k^2 - k -1 ). But ( x_1 = 2 + 3x_2 ), so substitute that into the product:( (2 + 3x_2)x_2 = k^2 - k -1 )So, ( 2x_2 + 3x_2^2 = k^2 - k -1 )But we also have ( k = 1 + 2x_2 ). Let's substitute ( k ) in terms of ( x_2 ) into this equation.First, compute ( k^2 ):( k = 1 + 2x_2 )So,( k^2 = (1 + 2x_2)^2 = 1 + 4x_2 + 4x_2^2 )Then, ( k^2 -k -1 = (1 + 4x_2 + 4x_2^2) - (1 + 2x_2) -1 )Simplify term by term:1. ( 1 + 4x_2 + 4x_2^2 )2. Subtract ( 1 + 2x_2 ): ( -1 -2x_2 )3. Subtract 1 more: ( -1 )Combine all together:( 1 + 4x_2 + 4x_2^2 -1 -2x_2 -1 )Simplify:( (1 -1 -1) + (4x_2 -2x_2) + 4x_2^2 )Which is:( (-1) + 2x_2 + 4x_2^2 )Therefore, ( k^2 -k -1 = 4x_2^2 + 2x_2 -1 )But from the product equation:( 2x_2 + 3x_2^2 = 4x_2^2 + 2x_2 -1 )Bring all terms to one side:( 2x_2 + 3x_2^2 -4x_2^2 -2x_2 +1 =0 )Simplify:( (3x_2^2 -4x_2^2) + (2x_2 -2x_2) +1 =0 )Which simplifies to:( -x_2^2 +0 +1 =0 )So,( -x_2^2 +1 =0 )Multiply both sides by -1:( x_2^2 -1 =0 )Factor:( (x_2 -1)(x_2 +1) =0 )So, ( x_2 =1 ) or ( x_2 = -1 )Now, let's find the corresponding k for each case.Case 1: ( x_2 =1 )From earlier, ( k =1 +2x_2 =1 +2(1)=3 )Case 2: ( x_2 = -1 )Then, ( k =1 +2*(-1)=1 -2= -1 )Now, we need to check whether these k values result in the original quadratic equation having two distinct real roots. The discriminant must be positive.The discriminant D of the equation ( x^2 -2kx +k^2 -k -1=0 ) is:( D = ( -2k )^2 -4*1*(k^2 -k -1 ) =4k^2 -4(k^2 -k -1)=4k^2 -4k^2 +4k +4=4k +4 )So, D=4(k +1). For distinct real roots, D>0:( 4(k +1) >0 implies k +1 >0 impliesk > -1 )So, check the two k values:Case 1: k=3. Since 3 > -1, it's okay. Valid.Case 2: k=-1. Then k +1 =0, so D=0. But the problem states that the equation has two distinct real roots. Therefore, k=-1 is invalid because discriminant is zero (roots are equal). So, we discard k=-1.Hence, the only valid solution is k=3.Wait, let me verify this again. If k=3, then discriminant D=4*(3 +1)=16>0, so yes, two distinct roots. For k=-1, D=4*(-1 +1)=0, so repeated roots, which contradicts the problem statement. Therefore, only k=3 is valid.Therefore, the answer to part (2) is k=3.But let me cross-validate. Let's take k=3 and check if x1 -3x2=2.First, with k=3, the quadratic equation is:( x^2 -2*3x +3^2 -3 -1= x^2 -6x +9 -3 -1= x^2 -6x +5=0 )So, equation is x² -6x +5=0. Factor: (x-5)(x-1)=0. Roots are x=5 and x=1.So, x1=5, x2=1. Then x1 -3x2=5 -3*1=5-3=2. Correct.Alternatively, if the roots were x1=1 and x2=5, then x1 -3x2=1 -15=-14≠2. So, order matters here. Wait, but in the quadratic equation, the roots are x1 and x2, but which one is which? Since quadratic equations have two roots, but unless specified, they can be in any order. However, in our case, when we solved the system:From equation (2): x1 =2 +3x2. If x2=1, then x1=2 +3*1=5. So, in this case, x1=5 and x2=1. So, the assignment is correct. Thus, it's correct that k=3 gives the roots 5 and 1, satisfying x1 -3x2=2.Therefore, k=3 is indeed the correct answer.So, summarizing:Part (1): 190Part (2): 3Final Answer(1) boxed{190} (2) boxed{3}</think>Given the quadratic equation ( x^2 - 2kx + k^2 - k - 1 = 0 ) with two distinct real roots ( x_1 ) and ( x_2 ):1. If ( k = 5 ), find the value of ( x_1x_2^2 + x_1^2x_2 ): Substitute ( k = 5 ) into the equation: [ x^2 - 10x + 19 = 0 ] The sum of the roots ( x_1 + x_2 = 10 ) and the product ( x_1x_2 = 19 ). The expression ( x_1x_2^2 + x_1^2x_2 ) can be factored as: [ x_1x_2(x_1 + x_2) = 19 times 10 = 190 ] Thus, the value is (boxed{190}).2. If ( x_1 - 3x_2 = 2 ), find the value of ( k ): The sum of the roots ( x_1 + x_2 = 2k ) and the product ( x_1x_2 = k^2 - k - 1 ). We have the system of equations: [ begin{cases} x_1 + x_2 = 2k x_1 - 3x_2 = 2 end{cases} ] Solving for ( x_1 ) from the second equation: [ x_1 = 2 + 3x_2 ] Substitute into the first equation: [ 2 + 3x_2 + x_2 = 2k implies 2 + 4x_2 = 2k implies k = 1 + 2x_2 ] Using the product of the roots: [ (2 + 3x_2)x_2 = k^2 - k - 1 ] Substitute ( k = 1 + 2x_2 ): [ 2x_2 + 3x_2^2 = (1 + 2x_2)^2 - (1 + 2x_2) - 1 ] Simplifying: [ 2x_2 + 3x_2^2 = 4x_2^2 + 2x_2 - 1 ] [ -x_2^2 + 1 = 0 implies x_2^2 = 1 implies x_2 = 1 text{ or } x_2 = -1 ] For ( x_2 = 1 ): [ k = 1 + 2(1) = 3 ] For ( x_2 = -1 ): [ k = 1 + 2(-1) = -1 ] Check the discriminant ( D = 4(k + 1) ). For distinct roots, ( D > 0 implies k > -1 ). Thus, ( k = -1 ) is invalid. Valid solution: ( k = 3 ). Verify with roots ( x_1 = 5 ) and ( x_2 = 1 ): [ x_1 - 3x_2 = 5 - 3 times 1 = 2 ] Thus, the value is (boxed{3}).]

🔑: Solution:# Part (1):Given the equation x^{2}-2kx+k^{2}-k-1=0.1. Sum and Product of Roots: - The sum of roots x_1 + x_2 = 2k. - The product of roots x_1 x_2 = k^2 - k - 1.2. Substituting k=5: - The sum of roots becomes x_1 + x_2 = 2 cdot 5 = 10. - The product of roots becomes x_1 x_2 = 5^2 - 5 - 1 = 19.3. Finding {x_1}x_2^2 + x_1^2{x_2}: - It equals {x_1}{x_2}({x_1} + {x_2}) = 19 cdot 10 = 190.So, {x_1}x_2^2+x_1^2{x_2} = boxed{190}.# Part (2):Given x_1 - 3x_2 = 2, we aim to find k.1. Given Relationships: - The sum of roots is x_1 + x_2 = 2k. - The product of roots is x_1 x_2 = k^2 - k - 1. - The discriminant Delta = 4k^{2} - 4(k^{2} - k - 1) = 4k + 4 > 0.2. Substituting the Given Equation: - Solving x_1 - 3x_2 = 2, we obtain x_2 = frac{k-1}{2} and x_1 = frac{3k+1}{2}.3. Finding k: - Substituting x_1 and x_2 into the product of roots, we get frac{(3k+1)}{2} cdot frac{(k-1)}{2} = k^2 - k - 1. - Simplifying, we discover k^2 - 2k - 3 = (k-3)(k+1) = 0, leading to k = 3 or k = -1.4. Considering Delta > 0: - Since Delta = 4k + 4 > 0, we conclude k > -1.Hence, the value of k is boxed{3}.

🧠:[<think>Okay, let's see. I need to find the real number m such that the angle between vectors a and b is 30 degrees. The vectors are given as a = (m, 3) and b = (√3, 1). Hmm, right. I remember there's a formula involving the dot product and the angle between two vectors. Let me recall that.The formula is: the dot product of vectors a and b equals the product of their magnitudes times the cosine of the angle between them. So, mathematically, that's:a · b = |a| |b| cosθYes, that sounds right. So I need to compute the dot product of a and b, find the magnitudes of both vectors, and set up the equation using the given angle of 30 degrees. Then solve for m. Let's break it down step by step.First, let's compute the dot product a · b. The dot product is calculated by multiplying the corresponding components and then adding them together. So for vectors a = (m, 3) and b = (√3, 1), the dot product is:a · b = m * √3 + 3 * 1 = m√3 + 3Okay, that's straightforward. Next, I need to find the magnitudes of vectors a and b.The magnitude of vector a, |a|, is sqrt(m² + 3²) = sqrt(m² + 9)Similarly, the magnitude of vector b, |b|, is sqrt((√3)² + 1²) = sqrt(3 + 1) = sqrt(4) = 2Got that. So |b| is 2. The magnitude of a is sqrt(m² + 9), which depends on m, the variable we need to find.Now, the angle between them is 30 degrees, so θ = 30°, and cosθ is cos(30°). I remember that cos(30°) is √3/2. So substituting all these into the formula:m√3 + 3 = |a| |b| cosθ = sqrt(m² + 9) * 2 * (√3 / 2)Wait, the 2 and the denominator 2 will cancel out. Let me check that again.Yes, the right-hand side becomes sqrt(m² + 9) * 2 * (√3 / 2) = sqrt(m² + 9) * √3So simplifying the equation:m√3 + 3 = √3 * sqrt(m² + 9)Okay, now I have an equation with m. Let's write that down clearly:m√3 + 3 = √3 * sqrt(m² + 9)Hmm, I need to solve for m here. Let me see. Let's subtract m√3 from both sides to isolate the square root term. Wait, but maybe squaring both sides would be better here. However, squaring might introduce extraneous solutions, so I need to check later. Let's try squaring both sides.But before that, maybe rearrange the equation:√3 * sqrt(m² + 9) = m√3 + 3Let me divide both sides by √3 to simplify:sqrt(m² + 9) = m + 3/√3Simplify 3/√3: 3 divided by √3 is √3, since 3/√3 = √3 * √3 / √3 = √3. So:sqrt(m² + 9) = m + √3Now, this seems manageable. Let's square both sides to eliminate the square root:(m² + 9) = (m + √3)²Expanding the right-hand side:(m + √3)² = m² + 2m√3 + (√3)² = m² + 2m√3 + 3So, substituting back into the equation:m² + 9 = m² + 2m√3 + 3Hmm, now subtract m² from both sides:9 = 2m√3 + 3Subtract 3 from both sides:6 = 2m√3Divide both sides by 2√3:6 / (2√3) = mSimplify numerator and denominator:3 / √3 = mAgain, 3 divided by √3 is √3, as before. So:m = √3Wait, so m is √3? Let me check this solution to make sure it's valid. Because sometimes when we square both sides, we might get extraneous roots.Let's substitute m = √3 back into the original equation to verify.First, compute the left-hand side (LHS): a · b = m√3 + 3 = √3 * √3 + 3 = 3 + 3 = 6Now compute the right-hand side (RHS): |a| |b| cosθ|a| = sqrt(m² + 9) = sqrt((√3)^2 + 9) = sqrt(3 + 9) = sqrt(12) = 2√3|b| = 2cosθ = √3/2So RHS = 2√3 * 2 * (√3/2) = 2√3 * 2 * √3 / 2Wait, let me compute that step by step.First, |a| |b| = 2√3 * 2 = 4√3Then multiply by cosθ (√3/2): 4√3 * √3 / 2 = (4 * 3) / 2 = 12 / 2 = 6So LHS is 6 and RHS is 6. That checks out. So m = √3 is a valid solution.But wait, let's also check if there's another possible solution. When we squared both sides, sometimes equations can have another solution. Let me check if there could be another m that satisfies the original equation.Suppose m + √3 is negative. However, since sqrt(m² + 9) is always non-negative, the right-hand side before squaring was sqrt(m² + 9) = m + √3. Since sqrt(m² + 9) is non-negative, m + √3 must be non-negative. Therefore, m + √3 ≥ 0 ⇒ m ≥ -√3.But when we squared, we might have another solution. Let me check.Suppose we had sqrt(m² + 9) = m + √3. If m + √3 is positive, then squaring gives m² +9 = m² +2m√3 +3. Then 9 =2m√3 +3, leading to m=√3.But if m + √3 were negative, then sqrt(m² +9) = negative number, which is impossible. So there are no other solutions.Therefore, the only solution is m=√3.Wait, but hold on. Let me check if m=√3 indeed makes the angle 30 degrees. Alternatively, maybe m could be another value. Let me think.Alternatively, could the angle between vectors be 30 degrees in another configuration? For example, if the angle is 30 degrees, but maybe the dot product is negative? Wait, but the angle between two vectors is always between 0 and 180 degrees, so cosine can be positive or negative. However, in the problem statement, they just mention the angle is 30 degrees. So if the cosine is positive, as 30 degrees is acute. So the dot product should be positive.But in our case, with m=√3, the dot product is 6, which is positive. So that's fine. If there was another solution where m is negative, but with m + √3 still positive, then maybe. Let me check.Suppose m is negative. Let's see. Suppose m is negative, but m + √3 is still positive. For example, if m is -1, then m + √3 ≈ -1 + 1.732 ≈ 0.732, which is positive. So sqrt(m² +9) = |m + √3|, but since m + √3 is positive, sqrt(m² +9) = m + √3. Then squaring would give same equation. But wait, when we squared, we might have another solution.Wait, let me take m as a variable. Let's consider the equation sqrt(m² +9) = m + √3.Suppose m + √3 is positive. Then the equation is as before, leading to m=√3.But suppose m + √3 is negative. Then sqrt(m² +9) = negative number, which is impossible. Hence, the equation sqrt(m² +9) = m + √3 has solutions only when m + √3 is non-negative, and the solution is unique: m=√3.Thus, m=√3 is the only solution.Wait, but let's verify with m=√3. The vectors would be a=(√3, 3) and b=(√3,1). Let's compute the angle between them.Compute the dot product: √3*√3 +3*1= 3 +3=6.Compute |a|=sqrt((√3)^2 +3^2)=sqrt(3+9)=sqrt(12)=2√3.Compute |b|=sqrt( (√3)^2 +1^2)=sqrt(3+1)=2.Then cosθ= (a·b)/(|a||b|)=6/(2√3 *2)=6/(4√3)= (3)/(2√3)= (√3)/2. Which is cos30°, so θ=30°, which matches. So it's correct.Alternatively, suppose there was another m. Let me suppose m=0. Then vectors a=(0,3) and b=(√3,1). The dot product is 0*√3 +3*1=3. |a|=3, |b|=2, so cosθ=3/(3*2)=0.5. So θ=60°, which is different. So m=0 gives 60°, not 30°, so that's not the answer.Alternatively, m= something else. Let's say m=1. Then a=(1,3), dot product=1*√3 +3*1=√3 +3. |a|=sqrt(1+9)=sqrt(10). |b|=2. So cosθ=(√3 +3)/(2*sqrt(10)). Let's compute that numerically. sqrt(3)≈1.732, so numerator≈1.732+3=4.732. Denominator≈2*3.162≈6.324. So cosθ≈4.732/6.324≈0.748. So angle≈acos(0.748)≈41.5 degrees. Not 30. So m=1 is not.Alternatively, m=2. Then a=(2,3). Dot product=2*√3 +3*1≈2*1.732 +3≈6.464. |a|=sqrt(4 +9)=sqrt(13)≈3.606. |b|=2. So cosθ≈6.464/(3.606*2)=6.464/7.212≈0.896. Angle≈26.6 degrees. Not 30.Hmm. What if m is negative? Let's try m=-√3. Then a=(-√3,3). Compute the dot product with b=(√3,1): (-√3)(√3) +3*1= -3 +3=0. So the dot product is zero, which means the angle is 90°, not 30°. So that's not it.Wait, so m=-√3 gives orthogonal vectors. So that's a different case. So m=√3 is the only solution. Therefore, the answer is m=√3.But let me make sure. Let me check once again.The equation after squaring both sides was:m² +9 = m² +2m√3 +3Then subtract m²: 9=2m√3 +3Subtract 3: 6=2m√3Divide by 2√3: m=6/(2√3)=3/√3=√3Yes, that's correct. So m=√3. So the answer should be √3. So boxed, it's boxed{sqrt{3}}.Final Answerboxed{sqrt{3}}</think>Given vectors (overrightarrow{a} = (m, 3)) and (overrightarrow{b} = (sqrt{3}, 1)), we need to find the real number (m) such that the angle between them is (30^circ).First, we use the dot product formula involving the angle between two vectors:[overrightarrow{a} cdot overrightarrow{b} = |overrightarrow{a}| |overrightarrow{b}| cos 30^circ]Calculating the dot product (overrightarrow{a} cdot overrightarrow{b}):[overrightarrow{a} cdot overrightarrow{b} = m cdot sqrt{3} + 3 cdot 1 = msqrt{3} + 3]Finding the magnitudes of the vectors:[|overrightarrow{a}| = sqrt{m^2 + 9}][|overrightarrow{b}| = sqrt{(sqrt{3})^2 + 1^2} = sqrt{3 + 1} = 2]Using (cos 30^circ = frac{sqrt{3}}{2}), we substitute into the formula:[msqrt{3} + 3 = sqrt{m^2 + 9} cdot 2 cdot frac{sqrt{3}}{2}]Simplifying the right-hand side:[msqrt{3} + 3 = sqrt{m^2 + 9} cdot sqrt{3}]Dividing both sides by (sqrt{3}):[m + frac{3}{sqrt{3}} = sqrt{m^2 + 9}]Simplifying (frac{3}{sqrt{3}} = sqrt{3}):[m + sqrt{3} = sqrt{m^2 + 9}]Squaring both sides to eliminate the square root:[(m + sqrt{3})^2 = m^2 + 9]Expanding the left-hand side:[m^2 + 2msqrt{3} + 3 = m^2 + 9]Subtracting (m^2) from both sides:[2msqrt{3} + 3 = 9]Solving for (m):[2msqrt{3} = 6][msqrt{3} = 3][m = frac{3}{sqrt{3}} = sqrt{3}]Verification confirms that (m = sqrt{3}) satisfies the original equation. Thus, the real number (m) is (boxed{sqrt{3}}).,<think>Okay, so I have this problem here: Given vectors a = (m, 3) and b = (√3, 1), and the angle between them is 30 degrees. I need to find the real number m. Hmm, let's see. I remember that the angle between two vectors can be found using the dot product formula. Right, the formula is:cosθ = (a · b) / (|a| |b|)Where θ is the angle between them, a · b is the dot product, and |a| and |b| are the magnitudes of the vectors. So since the angle is 30 degrees, I can set up this equation and solve for m. Let me write that down step by step.First, let's compute the dot product of vectors a and b. The dot product is just the sum of the products of the corresponding components. So for vectors a = (m, 3) and b = (√3, 1), the dot product is:a · b = m * √3 + 3 * 1 = m√3 + 3Okay, got that part. Now, I need the magnitudes of both vectors. The magnitude of vector a, |a|, is sqrt(m² + 3²) = sqrt(m² + 9). The magnitude of vector b, |b|, is sqrt((√3)² + 1²) = sqrt(3 + 1) = sqrt(4) = 2. That's straightforward.So putting this into the formula for the cosine of the angle:cos(30°) = (m√3 + 3) / (sqrt(m² + 9) * 2)I know that cos(30°) is √3 / 2. So substituting that in:√3 / 2 = (m√3 + 3) / (2 * sqrt(m² + 9))Hmm, okay, let me simplify this equation. Both sides have a denominator of 2, so I can multiply both sides by 2 to eliminate the denominators:√3 = (m√3 + 3) / sqrt(m² + 9)Now, let's multiply both sides by sqrt(m² + 9) to get rid of the denominator:√3 * sqrt(m² + 9) = m√3 + 3Hmm, this looks a bit complicated, but maybe we can square both sides to eliminate the square roots. Let's try that.Squaring the left side: (√3 * sqrt(m² + 9))² = (√3)² * (sqrt(m² + 9))² = 3 * (m² + 9)Squaring the right side: (m√3 + 3)² = (m√3)² + 2 * m√3 * 3 + 3² = 3m² + 6√3 m + 9So now the equation becomes:3(m² + 9) = 3m² + 6√3 m + 9Let's expand the left side:3m² + 27 = 3m² + 6√3 m + 9Hmm, subtract 3m² from both sides:27 = 6√3 m + 9Then subtract 9 from both sides:18 = 6√3 mDivide both sides by 6√3:18 / (6√3) = mSimplify numerator and denominator:3 / √3 = mBut 3 divided by √3 is the same as √3, because √3 times √3 is 3. So 3 / √3 = √3. Wait, so m = √3?Wait a second, let me check that again. 3 divided by √3. Multiply numerator and denominator by √3 to rationalize the denominator:(3 * √3) / (√3 * √3) = (3√3) / 3 = √3. Yup, that's correct. So m = √3.But wait, let me make sure that this is the only solution. When I squared both sides of the equation, sometimes extraneous solutions can pop up, right? So I should check if m = √3 satisfies the original equation.Let's substitute m = √3 back into the original equation before squaring:Left side: √3 * sqrt((√3)² + 9) = √3 * sqrt(3 + 9) = √3 * sqrt(12) = √3 * 2√3 = 2 * 3 = 6Right side: m√3 + 3 = √3 * √3 + 3 = 3 + 3 = 6Okay, so both sides equal 6. That works. So m = √3 is a valid solution.Wait, but is there another solution? When we square equations, sometimes there can be another solution. Let me see. Let's go back to the equation before squaring:√3 * sqrt(m² + 9) = m√3 + 3Suppose I write this as:sqrt(m² + 9) = (m√3 + 3)/√3 = m + 3/√3 = m + √3Wait, 3/√3 is √3, yes. So sqrt(m² + 9) = m + √3Now, normally when we square both sides, we might get another solution, but in this case, let's see:sqrt(m² + 9) = m + √3If m is a real number, the right side must be positive. So m + √3 > 0. Therefore, m > -√3.But when we squared the equation, we got 3(m² + 9) = 3m² + 6√3 m + 9Which simplified to 27 = 6√3 m + 9, leading to m = 18/(6√3) = √3. So only one solution here. However, sometimes squaring can lead to another solution, but in this case, we need to check if m + √3 is positive. If m = -√3, then m + √3 = 0, but sqrt(m² + 9) would be sqrt(3 + 9) = sqrt(12) ≈ 3.464, which is not zero. So no other solution here. Therefore, m = √3 is the only solution.But let me check if m can take another value. Suppose m is negative. Let's try m = -√3. Then let's compute the angle.If m = -√3, then vector a is (-√3, 3). Let's compute the dot product:a · b = (-√3)(√3) + 3*1 = (-3) + 3 = 0Wait, so the dot product is zero, meaning the angle is 90 degrees, not 30 degrees. So that can't be. Therefore, m = -√3 is invalid, which makes sense because when we squared the equation, we only got m = √3 as the solution. So that's the only one.Alternatively, maybe if I made a mistake in the squaring step. Let me retrace.Starting from:√3 * sqrt(m² + 9) = m√3 + 3Square both sides:Left: 3(m² + 9) = 3m² + 27Right: (m√3 + 3)^2 = 3m² + 6√3 m + 9Set equal:3m² + 27 = 3m² + 6√3 m + 9Subtract 3m²:27 = 6√3 m + 9Subtract 9:18 = 6√3 mDivide by 6√3:m = 18 / (6√3) = 3 / √3 = √3Yes, that's correct. So no mistake here.Alternatively, maybe using another method. For instance, using the formula for the tangent of the angle between two vectors? But I think the dot product formula is the standard approach here.Alternatively, maybe express the vectors in terms of components and use trigonometry. Let's see. The angle between them is 30 degrees. Vector a is (m, 3), vector b is (√3, 1). So maybe we can express vector a in polar coordinates and relate the angles? But that might complicate things.Alternatively, since vector b has components (√3, 1), maybe compute its angle with the x-axis first. Let's see. The angle θ_b for vector b can be found by tanθ_b = 1/√3, so θ_b = 30 degrees. Because tan(30°) = 1/√3. So vector b is at 30 degrees from the x-axis.Similarly, vector a is (m, 3). Let's say the angle of vector a with the x-axis is θ_a. Then the angle between vectors a and b is θ_a - θ_b = 30 degrees. So θ_a = θ_b + 30° = 30° + 30° = 60°, or θ_a = θ_b - 30° = 0°, but since the angle between them is 30°, it could be either way.Wait, but maybe not. Because the angle between two vectors is the smallest angle between them, so it's 30 degrees. So if vector b is at 30°, then vector a could be at 30° + 30° = 60° or 30° - 30° = 0°. Let's check both possibilities.If θ_a = 60°, then the components of vector a would be |a|(cos60°, sin60°) = |a|(0.5, (√3)/2). But vector a is given as (m, 3). So equate components:m = |a| * 0.53 = |a| * (√3)/2From the second equation:|a| = 3 / (√3/2) = 3 * 2 / √3 = 6 / √3 = 2√3Then m = |a| * 0.5 = 2√3 * 0.5 = √3Which is the solution we found earlier. Alternatively, if θ_a = 0°, then vector a would be along the x-axis: components (|a|, 0). But vector a is (m, 3), so unless 3 = 0, which is impossible, this is not possible. Therefore, only θ_a = 60° is valid, leading to m = √3. So that's another way to confirm it.Therefore, the answer is m = √3. I think that's correct. Let me just check with another approach.Alternatively, using the sine of the angle. Since the angle is 30°, we can use the formula involving the cross product magnitude. The formula is:sinθ = |a × b| / (|a| |b|)But since we are in 2D, the magnitude of the cross product is |a_x b_y - a_y b_x| = |m*1 - 3*√3| = |m - 3√3|So sin(30°) = |m - 3√3| / (sqrt(m² + 9) * 2)But sin(30°) is 1/2, so:1/2 = |m - 3√3| / (2 sqrt(m² + 9))Multiply both sides by 2 sqrt(m² + 9):sqrt(m² + 9) = |m - 3√3|Hmm, so this gives sqrt(m² + 9) = |m - 3√3|Now, square both sides:m² + 9 = (m - 3√3)^2Expand the right side:m² - 6√3 m + 27Set equal to left side:m² + 9 = m² - 6√3 m + 27Subtract m² from both sides:9 = -6√3 m + 27Subtract 27:-18 = -6√3 mDivide by -6√3:m = (-18)/(-6√3) = 3/√3 = √3Same result! So using the sine formula also leads to m = √3. Therefore, this further confirms the answer.Additionally, in this case, the absolute value was |m - 3√3|. If we consider the case where m - 3√3 is positive or negative.Case 1: m - 3√3 ≥ 0 → m ≥ 3√3Then sqrt(m² + 9) = m - 3√3But the left side sqrt(m² + 9) is always positive, and the right side m - 3√3 would be non-negative. However, let's see if there's a solution here.Square both sides:m² + 9 = m² -6√3 m + 27Again, same as before:9 = -6√3 m +27Leads to m = (27 -9)/6√3 = 18/6√3 = 3/√3 = √3But √3 ≈ 1.732, and 3√3 ≈ 5.196, so m = √3 is less than 3√3, so this contradicts the case assumption that m ≥ 3√3. Therefore, this case doesn't hold.Case 2: m - 3√3 < 0 → m < 3√3Then |m - 3√3| = -(m - 3√3) = 3√3 - mSo sqrt(m² +9) = 3√3 - mNow, since sqrt(m² +9) is positive, 3√3 - m must be positive. So 3√3 - m >0 → m < 3√3, which is consistent with the case.Square both sides:m² +9 = (3√3 -m)^2 = 27 -6√3 m + m²Subtract m²:9 = 27 -6√3 mSubtract 27:-18 = -6√3 mDivide:m = (-18)/(-6√3) = 3/√3 = √3Which is valid because √3 < 3√3. So this is the valid case. Therefore, m = √3 is the solution.Therefore, all methods lead to m = √3. So I think that's the correct answer. I can't find any mistakes in the reasoning, and both the cosine and sine approaches give the same result. So m must be √3.Final AnswerThe real number ( m ) is (boxed{sqrt{3}}).</think>Given vectors (overrightarrow{a} = (m, 3)) and (overrightarrow{b} = (sqrt{3}, 1)), and the angle between them is (30^circ), we need to find the real number (m).First, we use the dot product formula to relate the angle between the vectors:[cos(30^circ) = frac{overrightarrow{a} cdot overrightarrow{b}}{|overrightarrow{a}| |overrightarrow{b}|}]Calculating the dot product:[overrightarrow{a} cdot overrightarrow{b} = m cdot sqrt{3} + 3 cdot 1 = msqrt{3} + 3]Finding the magnitudes of the vectors:[|overrightarrow{a}| = sqrt{m^2 + 9}, quad |overrightarrow{b}| = sqrt{(sqrt{3})^2 + 1^2} = sqrt{4} = 2]Substituting these into the cosine formula:[frac{sqrt{3}}{2} = frac{msqrt{3} + 3}{2 sqrt{m^2 + 9}}]Simplifying:[sqrt{3} = frac{msqrt{3} + 3}{sqrt{m^2 + 9}}]Multiplying both sides by (sqrt{m^2 + 9}):[sqrt{3} sqrt{m^2 + 9} = msqrt{3} + 3]Squaring both sides:[3(m^2 + 9) = (msqrt{3} + 3)^2]Expanding both sides:[3m^2 + 27 = 3m^2 + 6sqrt{3}m + 9]Simplifying:[27 = 6sqrt{3}m + 9]Solving for (m):[18 = 6sqrt{3}m implies m = frac{18}{6sqrt{3}} = frac{3}{sqrt{3}} = sqrt{3}]Verification shows that (m = sqrt{3}) satisfies the original equation. Thus, the real number (m) is (boxed{sqrt{3}}).]

🔑:Since overrightarrow{a}=(m,3) and overrightarrow{b}=(sqrt{3},1), and the angle between vectors overrightarrow{a} and overrightarrow{b} is 30^{circ},therefore, overrightarrow{a}cdot overrightarrow{b}= sqrt{3}m+3= sqrt{m^{2}+9}cdot 2cdot cos 30^{circ}. Solving this, we get m= sqrt{3}.Hence, the answer is: boxed{sqrt{3}}.By using the definition of the dot product of two vectors and the formula for the dot product, we find the value of m.This problem mainly tests the definition of the dot product of two vectors and the application of the formula for the dot product, and it is considered a basic question.

🧠:[<think>Okay, let's see here. The problem is about a bridge renovation. The bridge is 36 meters long. The construction team increased their daily work efficiency by 50%, which allowed them to finish 2 days ahead of schedule. We need to find the original planned daily renovation distance.Hmm. Let's break this down. First, let me parse the information given. The total length of the bridge is 36 meters. The team was supposed to renovate this bridge at a certain daily rate, but they increased their efficiency by 50%, meaning they did more each day than originally planned. Because of this increase, they finished two days earlier than planned. So, we need to find out what their original daily renovation distance was.Let me define some variables to make this clearer. Let's say the original planned daily renovation distance is x meters per day. Then, under the original plan, the number of days it would take to complete the 36 meters would be 36 divided by x, right? So, original time = 36 / x days.But then they increased their efficiency by 50%. Increasing efficiency by 50% means they are now working at 1.5 times their original rate. So, their new daily renovation distance is x + 0.5x = 1.5x meters per day.With this increased rate, the time taken to complete the 36 meters would be 36 divided by 1.5x. So, new time = 36 / (1.5x).According to the problem, the new time is 2 days less than the original time. So, original time minus new time equals 2 days. Let me write that equation out:Original time - New time = 2 days(36 / x) - (36 / (1.5x)) = 2Now, let's solve this equation for x.First, simplify the equation. Let's compute 36 / x - 36 / (1.5x).Factor out 36/x from both terms:36/x * [1 - 1/1.5] = 2Wait, maybe I can compute 1 - 1/1.5 first. Let's see. 1/1.5 is equal to 2/3, right? Because 1.5 is 3/2, so 1 divided by 3/2 is 2/3. So, 1 - 2/3 = 1/3. Therefore, the equation becomes:36/x * (1/3) = 2Simplify that:(36/3x) = 212/x = 2Then, solving for x:12 = 2xx = 12 / 2x = 6Wait, so the original planned daily renovation distance was 6 meters per day?Let me check that. If originally they planned to do 6 meters per day, the total time would be 36 / 6 = 6 days. But then they increased their efficiency by 50%, so 6 + 3 = 9 meters per day. Then, the new time is 36 / 9 = 4 days. 6 days minus 4 days is 2 days, which matches the problem statement. So that works out.Wait, but let me double-check my math to be sure. Let me go through the equations again.Original time: 36 / xNew rate: 1.5xNew time: 36 / (1.5x)Difference in time: (36/x) - (36/(1.5x)) = 2Compute 36/x - 36/(1.5x). Let's combine the fractions.First, 36 divided by x minus 36 divided by 1.5x. Let's factor out 36/x:36/x [1 - 1/(1.5)] = 2As before, 1/(1.5) is 2/3, so 1 - 2/3 = 1/3. So, 36/x * 1/3 = 2.Then, 12/x = 2. So x = 12 / 2 = 6. Correct.Another way to approach it: Let's set up the equations step by step.Let’s denote:Original daily rate: x meters/dayOriginal time: t = 36 / x daysNew daily rate: 1.5x meters/dayNew time: t' = 36 / (1.5x) daysGiven that t - t' = 2 days.So:36/x - 36/(1.5x) = 2Multiply both sides by 1.5x to eliminate denominators:36 * 1.5 - 36 = 2 * 1.5xCompute 36*1.5: 36*1 + 36*0.5 = 36 + 18 = 54So 54 - 36 = 18 = 3xTherefore, 18 = 3x => x = 6.Wait, that's another way to solve it. Wait, but here we get 18 = 3x? Wait, let's check.Wait, 36/x - 36/(1.5x) = 2. Multiply both sides by 1.5x:36/x * 1.5x - 36/(1.5x) * 1.5x = 2 * 1.5xSimplify:36*1.5 - 36 = 3xWhich is 54 - 36 = 3x => 18 = 3x => x = 6. Correct. So same answer.Therefore, the original planned daily renovation distance was 6 meters per day.Let me verify again with actual numbers. Original plan: 6 meters/day. Total days: 36/6=6 days.Increased efficiency: 50% more, so 6*1.5=9 meters/day. Time taken: 36/9=4 days. 6 - 4 = 2 days ahead. Perfect.So the answer is 6 meters per day.Final AnswerThe original planned daily renovation distance was boxed{6} meters.</think>Let ( x ) be the original planned daily renovation distance in meters. The original time to complete the 36-meter bridge renovation would be ( frac{36}{x} ) days. When the construction team increased their daily work efficiency by 50%, their new daily renovation distance became ( 1.5x ) meters. The new time to complete the renovation is ( frac{36}{1.5x} ) days. According to the problem, the new time is 2 days less than the original time. Therefore, we set up the equation:[frac{36}{x} - frac{36}{1.5x} = 2]Simplifying the equation:[frac{36}{x} - frac{36}{1.5x} = 2]Factor out ( frac{36}{x} ):[frac{36}{x} left(1 - frac{1}{1.5}right) = 2]Since ( frac{1}{1.5} = frac{2}{3} ):[frac{36}{x} left(1 - frac{2}{3}right) = 2][frac{36}{x} cdot frac{1}{3} = 2][frac{12}{x} = 2]Solving for ( x ):[12 = 2x][x = 6]Verification: Original time is ( frac{36}{6} = 6 ) days. New rate is ( 1.5 times 6 = 9 ) meters per day, leading to a new time of ( frac{36}{9} = 4 ) days. The difference is ( 6 - 4 = 2 ) days, which matches the problem statement.Thus, the original planned daily renovation distance was (boxed{6}) meters.,<think>Okay, let me try to work through this problem step by step. So, the problem is about a bridge renovation. The bridge is 36 meters long. The construction team increased their daily work efficiency by 50%, which allowed them to finish 2 days ahead of schedule. We need to find out what the original planned daily renovation distance was.Hmm, let's parse this again. Originally, they had a plan to renovate the bridge at a certain daily rate. But then they increased that rate by 50%, which made them finish 2 days earlier. The total length of the bridge is 36 meters, so that's the total work done.I think I need to set up some variables here. Let me denote the original planned daily renovation distance as ( x ) meters per day. Then, the original time they planned to take would be the total distance divided by the daily rate, which is ( frac{36}{x} ) days.But when they increased their efficiency by 50%, their new daily rate became ( x + 0.5x = 1.5x ) meters per day. With this increased rate, the time taken would be ( frac{36}{1.5x} ) days. According to the problem, this new time is 2 days less than the original planned time. So, the original time minus the new time equals 2 days. That gives us the equation:Original time - New time = 2 daysWhich translates to:( frac{36}{x} - frac{36}{1.5x} = 2 )Okay, that seems like a solid equation. Let me check if this makes sense. If they work faster, the time should be less, so the difference in times is positive, which is 2 days. That lines up with the problem statement.Now, let's solve this equation for ( x ). Let's first simplify the terms.First, compute ( frac{36}{1.5x} ). Since 1.5 is the same as ( frac{3}{2} ), so ( 1.5x = frac{3}{2}x ). Therefore, ( frac{36}{1.5x} = frac{36}{frac{3}{2}x} = frac{36 times 2}{3x} = frac{72}{3x} = frac{24}{x} ).So substituting back into the equation:( frac{36}{x} - frac{24}{x} = 2 )Combine the fractions:( frac{36 - 24}{x} = 2 )( frac{12}{x} = 2 )Now, solve for ( x ):Multiply both sides by ( x ):( 12 = 2x )Divide both sides by 2:( x = 6 )Wait, so the original planned daily renovation distance was 6 meters per day? Let me check this answer to make sure it makes sense.Original plan: 6 meters per day. So time planned was ( 36 / 6 = 6 ) days.Increased efficiency: 50% more, so 6 + 3 = 9 meters per day. Time taken: ( 36 / 9 = 4 ) days.Difference in time: 6 - 4 = 2 days. Which matches the problem statement. So that checks out.Hmm, seems correct. Let me just go through the steps again to be thorough.Original daily rate: ( x ), original time: ( 36/x ).Increased rate: ( 1.5x ), new time: ( 36/(1.5x) ).Difference in time: ( 36/x - 36/(1.5x) = 2 ).Simplify ( 36/(1.5x) ): 1.5 is 3/2, so dividing by 3/2 is multiplying by 2/3. So 36*(2/3)/x = 24/x.Therefore, ( 36/x - 24/x = 12/x = 2 ). So ( x = 6 ). Yep, same result.Another way to think about it: If increasing the rate by 50% reduces the time by 2 days, then the time ratio is original time to new time. Let me see.Let the original time be T, new time is T - 2.Since the rate is increased by 50%, the new rate is 1.5 times original rate. Since rate and time are inversely proportional for the same work, then:Original rate / New rate = New time / Original timeBut original rate is x, new rate is 1.5x.So x / 1.5x = (T - 2)/TSimplify: 1 / 1.5 = (T - 2)/TWhich is 2/3 = (T - 2)/TCross multiply: 2T = 3(T - 2)2T = 3T - 62T - 3T = -6-T = -6T = 6 days.So original time was 6 days, original rate was 36 / 6 = 6 meters per day. Same answer. This confirms it again.Therefore, the original planned daily renovation distance was 6 meters.Final AnswerThe original planned daily renovation distance was boxed{6} meters.</think>Let ( x ) be the original planned daily renovation distance in meters. The original time planned to complete the renovation is ( frac{36}{x} ) days. With a 50% increase in efficiency, the new daily renovation distance becomes ( 1.5x ) meters, and the new time taken is ( frac{36}{1.5x} ) days. The problem states that the new time is 2 days less than the original time. We set up the equation:[frac{36}{x} - frac{36}{1.5x} = 2]Simplifying ( frac{36}{1.5x} ):[frac{36}{1.5x} = frac{36}{frac{3}{2}x} = frac{36 times 2}{3x} = frac{24}{x}]Substituting back into the equation:[frac{36}{x} - frac{24}{x} = 2]Combining the fractions:[frac{12}{x} = 2]Solving for ( x ):[12 = 2x implies x = 6]Checking the answer:- Original time: ( frac{36}{6} = 6 ) days.- New rate: ( 1.5 times 6 = 9 ) meters per day.- New time: ( frac{36}{9} = 4 ) days.- Difference in time: ( 6 - 4 = 2 ) days, which matches the problem statement.Thus, the original planned daily renovation distance was boxed{6} meters.]

🔑:To solve the problem, let's denote the original planned daily renovation distance for the construction team as x meters per day. The total length of the bridge to be renovated is 36 meters.Given that the construction team increased their daily work efficiency by 50%, their new daily renovation distance becomes (1 + 50%)x = 1.5x meters per day.According to the problem, the time it takes to complete the renovation at the original pace is frac{36}{x} days, and the time it takes to complete the renovation at the increased pace is frac{36}{1.5x} days. The construction team finished the task 2 days ahead of schedule, which gives us the equation:[frac{36}{x} = frac{36}{1.5x} + 2]Let's solve this equation step by step:[frac{36}{x} = frac{36}{1.5x} + 2]Multiplying every term by 1.5x to eliminate the denominators, we get:[1.5 cdot 36 = 36 + 2 cdot 1.5x]Simplifying the equation:[54 = 36 + 3x]Subtracting 36 from both sides:[18 = 3x]Dividing both sides by 3:[x = 6]Upon verification, substituting x = 6 into the original equation:[frac{36}{6} = frac{36}{1.5 cdot 6} + 2]Simplifies to:[6 = 4 + 2]Which holds true. Therefore, the original planned daily renovation distance for the construction team was boxed{6 text{ meters per day}}.

🧠:[<think>Alright, let me try to figure out how to solve this problem. So, we have the equation:( x^3 + (1 - 3a)x^2 + 2a^2x - 2ax + x + a^2 - a = 0 )And we need to find the range of values for ( a ) such that this equation has exactly one root. Hmm, okay. Let's start by simplifying the equation as much as possible. Maybe combining like terms will make it easier to analyze.Looking at the equation term by term:1. The cubic term: ( x^3 )2. Quadratic term: ( (1 - 3a)x^2 )3. Linear terms: ( 2a^2x ), ( -2ax ), and ( x )4. Constant term: ( a^2 - a )Let me combine the linear terms first. So, ( 2a^2x - 2ax + x ). Let's factor out an x to see if that helps:( x(2a^2 - 2a + 1) )So combining those, the linear term becomes ( (2a^2 - 2a + 1)x )Now, let me rewrite the entire equation with the simplified linear term:( x^3 + (1 - 3a)x^2 + (2a^2 - 2a + 1)x + (a^2 - a) = 0 )Okay, so now the equation is a cubic polynomial set to zero. The problem states that it has exactly one root. Now, I remember that a cubic polynomial can have one real root and two complex roots, or three real roots. So, for it to have exactly one real root, the other two roots must be complex conjugates. That happens when the discriminant of the cubic is negative. Alternatively, maybe there's a multiple root and another distinct root, but in that case, the total number of real roots would still be two (counting multiplicity). Wait, but the problem says "exactly one root," which I think means exactly one real root, counting multiplicity? Or exactly one distinct real root? Hmm. The wording is a bit ambiguous. But in math problems, when they say a polynomial equation has exactly one root, sometimes they mean exactly one real root, with multiplicity considered. But sometimes they might mean exactly one distinct real root. I need to clarify.But since it's a cubic, which has degree three, over the real numbers, it must have at least one real root, and up to three. So, if it's supposed to have exactly one real root, then the other two must be complex. So, that would be the case when the cubic has one real root and two complex conjugate roots. So, the discriminant of the cubic would be negative. Alternatively, if the cubic has a multiple root, then the discriminant is zero. But in that case, depending on the multiplicity, the number of real roots could be two or three. For example, a cubic with a double root and a simple root would have two distinct real roots (one with multiplicity two, another with multiplicity one). So, if the problem is asking for exactly one real root, counting multiplicity, that's impossible because the total multiplicity would still be three. But if it's asking for exactly one distinct real root, then even if there's a multiple root, as long as there's only one distinct root, which would require a triple root. So, a triple root would have multiplicity three, but that's still one distinct real root.So, maybe the problem is considering "exactly one root" as exactly one distinct real root, regardless of multiplicity. In that case, the equation would have a triple root, which is the case when the cubic can be written as ( (x - r)^3 = 0 ). Alternatively, there could be a double root and a simple root, but that would give two distinct real roots, which is more than one. So, to have exactly one distinct real root, the cubic must have a triple root. Alternatively, maybe a double root and a complex root? Wait, no. If there's a double root, it's still a real root. So, if there's a double root and another simple root, that's two distinct real roots. So, the only way for a cubic to have exactly one distinct real root is if it's a triple root. Therefore, the cubic must factor into ( (x - r)^3 ).Alternatively, maybe the problem is considering complex roots as well, but when it says "exactly one root", it's referring to real roots. Because over the complex numbers, a cubic always has three roots (counting multiplicity). So, probably, the problem is asking for exactly one real root, with the other two being complex. So, in that case, the discriminant of the cubic is negative.Alternatively, maybe the equation reduces to a lower degree polynomial, such as quadratic or linear, but given that the leading term is ( x^3 ), it's a cubic unless the coefficients of ( x^3 ) cancel out, but here the coefficient is 1, so it's definitely a cubic.Wait, let me check again if there's any cancellation. The original equation:( x^3 + (1 - 3a)x^2 + 2a^2x - 2ax + x + a^2 - a = 0 )So, the cubic term is ( x^3 ), quadratic term is ( (1 - 3a)x^2 ), linear terms are ( 2a^2x - 2ax + x ), and constants ( a^2 - a ). So, the linear terms can be combined as:( (2a^2 - 2a + 1)x )So, the equation is:( x^3 + (1 - 3a)x^2 + (2a^2 - 2a + 1)x + (a^2 - a) = 0 )So, it's definitely a cubic equation. Therefore, over the real numbers, it can have one or three real roots. So, if it has exactly one real root, then the discriminant of the cubic must be negative. Alternatively, if the discriminant is zero, it has a multiple root and all roots are real. So, in order to have exactly one real root, the discriminant must be negative.So, perhaps we need to compute the discriminant of this cubic and find for which values of ( a ) it is negative. But computing the discriminant of a cubic is a bit involved.Alternatively, maybe we can factor the cubic equation. Let's try to factor it. Let's see if we can factor by grouping or find a rational root.First, let's attempt to find a rational root using the Rational Root Theorem. The possible rational roots would be the divisors of the constant term divided by the divisors of the leading coefficient. The constant term here is ( a^2 - a ), and the leading coefficient is 1. Therefore, the possible rational roots are the divisors of ( a^2 - a ), which are ( pm1, pm a, pm (a - 1), pm (a^2 - a) ). But since ( a ) is a parameter here, not a constant, this might not be helpful. Alternatively, perhaps we can factor the cubic in terms of ( x ) and ( a ).Alternatively, maybe substitute a value for ( x ) that makes the equation true for all ( a ), but that might not work.Alternatively, let's try to factor the cubic. Let me write the equation again:( x^3 + (1 - 3a)x^2 + (2a^2 - 2a + 1)x + (a^2 - a) = 0 )Suppose we try to factor this as ( (x + k)(x^2 + mx + n) ), where ( k ), ( m ), ( n ) are expressions in terms of ( a ). Let's attempt to factor it.Multiplying out ( (x + k)(x^2 + mx + n) ):( x^3 + (m + k)x^2 + (n + mk)x + nk )Comparing coefficients with the original cubic:1. Coefficient of ( x^3 ): 1, which matches.2. Coefficient of ( x^2 ): ( m + k = 1 - 3a )3. Coefficient of ( x ): ( n + mk = 2a^2 - 2a + 1 )4. Constant term: ( nk = a^2 - a )So, we have the system:1. ( m + k = 1 - 3a )2. ( n + mk = 2a^2 - 2a + 1 )3. ( nk = a^2 - a )We need to solve for ( m ), ( n ), and ( k ) in terms of ( a ). Let's see if we can find such ( k ), ( m ), ( n ).From equation 3: ( nk = a^2 - a ). Let's factor ( a^2 - a = a(a - 1) ). So, possible pairs for ( n ) and ( k ) could be ( (a, a - 1) ) or ( ( -a, - (a - 1) ) ), etc. Let's try ( n = a ), ( k = a - 1 ). Then, check equation 3: ( a(a - 1) = a^2 - a ), which is correct.Now, from equation 1: ( m + k = 1 - 3a ). Since ( k = a - 1 ), then ( m + (a - 1) = 1 - 3a ), so ( m = 1 - 3a - (a - 1) = 1 - 3a - a + 1 = 2 - 4a ).Now, check equation 2: ( n + mk = 2a^2 - 2a + 1 ). Substitute ( n = a ), ( m = 2 - 4a ), ( k = a - 1 ):Left-hand side: ( a + (2 - 4a)(a - 1) )Compute ( (2 - 4a)(a - 1) ):( 2(a - 1) - 4a(a - 1) = 2a - 2 - 4a^2 + 4a = (-4a^2) + (2a + 4a) + (-2) = -4a^2 + 6a - 2 )Adding ( a ):Total left-hand side: ( -4a^2 + 6a - 2 + a = -4a^2 + 7a - 2 )Compare to right-hand side: ( 2a^2 - 2a + 1 )These are not equal unless:( -4a^2 + 7a - 2 = 2a^2 - 2a + 1 )Bring all terms to left-hand side:( -4a^2 + 7a - 2 - 2a^2 + 2a - 1 = -6a^2 + 9a - 3 = 0 )Multiply both sides by -1:( 6a^2 - 9a + 3 = 0 )Divide by 3:( 2a^2 - 3a + 1 = 0 )Solving:( a = [3 ± sqrt(9 - 8)] / 4 = [3 ± 1]/4 ), so ( a = 1 ) or ( a = 1/2 ).Hmm, interesting. So, only when ( a = 1 ) or ( a = 1/2 ), the factorization works? Wait, that seems like the values where the cubic can be factored into a linear term and a quadratic. But maybe for other values of ( a ), the cubic is irreducible. But wait, maybe if we try different factors for ( n ) and ( k ).Alternatively, maybe ( n = -a ), ( k = -(a - 1) = -a + 1 ). Then, check equation 3: ( (-a)(-a + 1) = a(a - 1) = a^2 - a ), which still works. Then, proceed similarly.But perhaps this approach is not the best. Alternatively, since we found that for ( a = 1 ) or ( a = 1/2 ), the cubic factors into ( (x + k)(x^2 + mx + n) ), maybe these are special cases. Let's check ( a = 1 ):If ( a = 1 ), then the original equation becomes:( x^3 + (1 - 3(1))x^2 + (2(1)^2 - 2(1) + 1)x + (1^2 - 1) = x^3 - 2x^2 + (2 - 2 + 1)x + (1 - 1) = x^3 - 2x^2 + x )Factor this: ( x(x^2 - 2x + 1) = x(x - 1)^2 ). So, the roots are ( x = 0 ) and ( x = 1 ) (double root). So, in this case, there are two distinct real roots. But the problem states "exactly one root", so perhaps ( a = 1 ) is not acceptable. Similarly, check ( a = 1/2 ):Original equation with ( a = 1/2 ):( x^3 + (1 - 3*(1/2))x^2 + (2*(1/2)^2 - 2*(1/2) + 1)x + ((1/2)^2 - 1/2) )Compute each term:1. ( 1 - 3*(1/2) = 1 - 3/2 = -1/2 )2. ( 2*(1/2)^2 - 2*(1/2) + 1 = 2*(1/4) - 1 + 1 = 1/2 - 1 + 1 = 1/2 )3. ( (1/2)^2 - 1/2 = 1/4 - 1/2 = -1/4 )So, the equation becomes:( x^3 - (1/2)x^2 + (1/2)x - 1/4 = 0 )Let me check if this factors. Suppose we factor as ( (x + k)(x^2 + mx + n) ). If ( a = 1/2 ), then earlier factorization would give ( k = a - 1 = -1/2 ), ( m = 2 - 4a = 2 - 2 = 0 ), ( n = a = 1/2 ). So, factors would be ( (x - 1/2)(x^2 + 0x + 1/2) ), which is ( (x - 1/2)(x^2 + 1/2) ). Multiply out:( x^3 + (1/2)x - (1/2)x^2 - (1/2)*(1/2) = x^3 - (1/2)x^2 + (1/2)x - 1/4 ), which matches. So, roots are ( x = 1/2 ) and ( x = pm sqrt{-1/2} ), which are complex. So, only real root is ( x = 1/2 ). Therefore, when ( a = 1/2 ), the equation has exactly one real root (and two complex roots). So, that seems like a valid case.However, when ( a = 1 ), the equation had two real roots, one of which was a double root. Therefore, ( a = 1 ) is not a valid solution for the problem. So, perhaps when ( a = 1/2 ), we have exactly one real root, and in other cases, maybe two or three. Therefore, maybe the answer is ( a = 1/2 ). But the problem asks for a range of values for ( a ), so perhaps there's a range where the discriminant is negative, leading to one real root. But when we tried factoring, we saw that for ( a = 1/2 ), the cubic factors into a linear term and an irreducible quadratic, which gives one real root. So, perhaps there's a range around ( a = 1/2 ) where the discriminant is negative, leading to one real root. Alternatively, maybe only ( a = 1/2 ).Wait, but we need to check if there are other values of ( a ) where the cubic has exactly one real root. To do that, we need to analyze the discriminant of the cubic.The discriminant ( D ) of a general cubic equation ( ax^3 + bx^2 + cx + d = 0 ) is given by:( D = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2 )In our case, the cubic is:( x^3 + (1 - 3a)x^2 + (2a^2 - 2a + 1)x + (a^2 - a) = 0 )So, coefficients:- ( a_3 = 1 )- ( a_2 = 1 - 3a )- ( a_1 = 2a^2 - 2a + 1 )- ( a_0 = a^2 - a )Plugging into the discriminant formula:( D = 18(1)(1 - 3a)(2a^2 - 2a + 1)(a^2 - a) - 4(1 - 3a)^3(a^2 - a) + (1 - 3a)^2(2a^2 - 2a + 1)^2 - 4(1)(2a^2 - 2a + 1)^3 - 27(1)^2(a^2 - a)^2 )This looks extremely complicated. Maybe expanding this is not feasible by hand. Perhaps there's a better way.Alternatively, since we have the cubic equation and we want to know when it has exactly one real root, which is when the discriminant is negative. So, we can set ( D < 0 ).Alternatively, perhaps we can compute the derivative of the cubic and analyze its critical points to determine when the cubic is monotonic (which would imply only one real root) or when it has local maxima and minima but they don't cross the x-axis (so the cubic still has only one real root). That might be a better approach.Let me recall that if a cubic function is strictly increasing or decreasing (i.e., its derivative has no real roots), then it will have exactly one real root. Alternatively, if the derivative has real roots (i.e., local maxima and minima), then the cubic will have three real roots if the local maximum is above the x-axis and the local minimum is below the x-axis, or one real root if both extrema are on the same side of the x-axis.So, first, let's compute the derivative of the cubic:( f(x) = x^3 + (1 - 3a)x^2 + (2a^2 - 2a + 1)x + (a^2 - a) )The derivative is:( f'(x) = 3x^2 + 2(1 - 3a)x + (2a^2 - 2a + 1) )To find critical points, set ( f'(x) = 0 ):( 3x^2 + 2(1 - 3a)x + (2a^2 - 2a + 1) = 0 )This is a quadratic equation in ( x ). The discriminant of this quadratic is:( Delta = [2(1 - 3a)]^2 - 4*3*(2a^2 - 2a + 1) )Compute:( Delta = 4(1 - 6a + 9a^2) - 12(2a^2 - 2a + 1) )Expand:( 4 - 24a + 36a^2 - 24a^2 + 24a - 12 )Combine like terms:- ( 36a^2 - 24a^2 = 12a^2 )- ( -24a + 24a = 0 )- ( 4 - 12 = -8 )So, ( Delta = 12a^2 - 8 )Therefore, the discriminant of the derivative is ( 12a^2 - 8 ). So, when is this discriminant positive, zero, or negative?- If ( 12a^2 - 8 > 0 ), then ( a^2 > 8/12 = 2/3 ), so ( |a| > sqrt{2/3} approx 0.8165 ). In this case, the derivative has two real roots, so the cubic has a local maximum and minimum.- If ( 12a^2 - 8 = 0 ), then ( a = pm sqrt{2/3} ). The derivative has a repeated real root, so the cubic has a point of inflection.- If ( 12a^2 - 8 < 0 ), then ( |a| < sqrt{2/3} ). The derivative has no real roots, so the cubic is strictly increasing (since the leading coefficient is positive). Therefore, if the cubic is strictly increasing, it will cross the x-axis exactly once. Hence, in this case, the equation has exactly one real root.Therefore, when ( |a| < sqrt{2/3} ), the cubic is strictly increasing and has exactly one real root. However, we need to check if for ( |a| geq sqrt{2/3} ), the cubic might still have only one real root if the local maximum and minimum are both above or both below the x-axis.So, for ( |a| geq sqrt{2/3} ), the cubic has a local maximum and minimum. To check whether the cubic has only one real root in this case, we need to ensure that the local maximum and the local minimum are both above the x-axis or both below the x-axis. This can be checked by evaluating the cubic at the critical points (i.e., substituting the x-values of the local maxima and minima into the original cubic) and seeing if both are positive or both are negative.Therefore, let's denote the critical points as ( x_1 ) and ( x_2 ), which are the roots of the derivative. Then, compute ( f(x_1) ) and ( f(x_2) ). If both are positive or both are negative, then the cubic has only one real root. Otherwise, it has three.But since computing ( x_1 ) and ( x_2 ) in terms of ( a ) is complicated, perhaps there's a better way. Alternatively, we can use the concept of discriminant: if the discriminant of the cubic is negative, then there is one real root and two complex conjugate roots. If the discriminant is positive, three real roots. If discriminant is zero, multiple roots.But since calculating the discriminant is complicated, maybe we can use the following approach: for the cubic to have only one real root when it has two critical points (i.e., when ( |a| geq sqrt{2/3} )), then both critical points must lie on the same side of the x-axis. So, the cubic evaluated at both critical points must have the same sign.Therefore, to find when ( f(x_1) ) and ( f(x_2) ) have the same sign. To compute ( f(x_1) ) and ( f(x_2) ), we can use the fact that if ( x_1 ) and ( x_2 ) are critical points, then ( f(x_i) ) can be expressed in terms of the coefficients of the cubic and the derivative. However, this might still be complicated.Alternatively, perhaps we can use the resultant or other methods. But maybe it's easier to use the discriminant of the cubic.Alternatively, let me recall that the discriminant of the cubic is related to the resultant of the cubic and its derivative. The discriminant is proportional to the resultant, which can be computed using the Sylvester matrix.But this might be time-consuming. Alternatively, maybe there's a shortcut. Wait, in our earlier attempt to factor the cubic, we found that when ( a = 1/2 ), the cubic factors into a linear term and an irreducible quadratic, leading to exactly one real root. Similarly, maybe when ( a = 1 ), the cubic factors into a linear term and a quadratic with a real root, leading to two real roots. So, perhaps for certain values of ( a ), the cubic has a multiple root.Wait, if the cubic has a multiple root, then the discriminant is zero, which is the boundary between three real roots and one real root. So, when the discriminant is zero, the cubic has a multiple root. Therefore, the values of ( a ) that make the discriminant zero are the bifurcation points between one and three real roots.Therefore, perhaps solving for when the discriminant is zero will give critical values of ( a ), and we can test intervals between these critical values to see when the discriminant is negative (one real root) or positive (three real roots).But since computing the discriminant is complicated, let's try to proceed step by step.First, let me try to compute the discriminant of the cubic:Given the cubic ( x^3 + bx^2 + cx + d ), discriminant ( D ) is:( D = 18bcd - 4b^3d + b^2c^2 - 4c^3 - 27d^2 )Wait, actually, in some sources, the discriminant is given by:( D = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2 )But in our case, ( a = 1 ), so:( D = 18(1)bcd - 4b^3d + b^2c^2 - 4(1)c^3 - 27(1)^2d^2 )Where in the cubic ( x^3 + bx^2 + cx + d ), so in our case:( b = 1 - 3a )( c = 2a^2 - 2a + 1 )( d = a^2 - a )So, substituting:( D = 18(1)(1 - 3a)(2a^2 - 2a + 1)(a^2 - a) - 4(1 - 3a)^3(a^2 - a) + (1 - 3a)^2(2a^2 - 2a + 1)^2 - 4(2a^2 - 2a + 1)^3 - 27(a^2 - a)^2 )This is going to be a very long computation. Let's tackle each term one by one.First term: ( 18(1)(1 - 3a)(2a^2 - 2a + 1)(a^2 - a) )Let me compute this term step by step. Let's factor out common terms:First, compute ( (1 - 3a)(a^2 - a) ):( (1 - 3a)(a(a - 1)) = a(a - 1)(1 - 3a) )Then, multiply by ( 2a^2 - 2a + 1 ):( a(a - 1)(1 - 3a)(2a^2 - 2a + 1) )Then multiply by 18:( 18a(a - 1)(1 - 3a)(2a^2 - 2a + 1) )This term is quite complex. Let's hold onto this for a moment.Second term: ( -4(1 - 3a)^3(a^2 - a) )Factor out ( (1 - 3a)^3(a^2 - a) ):= ( -4(1 - 3a)^3a(a - 1) )Third term: ( (1 - 3a)^2(2a^2 - 2a + 1)^2 )Fourth term: ( -4(2a^2 - 2a + 1)^3 )Fifth term: ( -27(a^2 - a)^2 )= ( -27a^2(a - 1)^2 )This seems really tedious. Maybe there's a smarter way. Alternatively, maybe we can use the fact that when ( a = 1/2 ), the discriminant is zero (as we saw that the cubic factors into a linear term and an irreducible quadratic). Similarly, we can check when the cubic has a multiple root by checking if there exists some ( x ) such that ( f(x) = 0 ) and ( f'(x) = 0 ).So, suppose there exists an ( x ) such that:1. ( x^3 + (1 - 3a)x^2 + (2a^2 - 2a + 1)x + (a^2 - a) = 0 )2. ( 3x^2 + 2(1 - 3a)x + (2a^2 - 2a + 1) = 0 )We can solve this system for ( x ) and ( a ).From equation 2, we can express ( x ) in terms of ( a ), then substitute into equation 1.Let me write equation 2:( 3x^2 + 2(1 - 3a)x + (2a^2 - 2a + 1) = 0 )Let me solve for ( x ):Quadratic in ( x ):( 3x^2 + 2(1 - 3a)x + (2a^2 - 2a + 1) = 0 )Solutions:( x = frac{ -2(1 - 3a) pm sqrt{4(1 - 3a)^2 - 12(2a^2 - 2a + 1)} }{ 6 } )Simplify discriminant inside sqrt:( 4(1 - 6a + 9a^2) - 12(2a^2 - 2a + 1) )= ( 4 - 24a + 36a^2 - 24a^2 + 24a - 12 )= ( (36a^2 - 24a^2) + (-24a + 24a) + (4 - 12) )= ( 12a^2 - 8 )Which is the same discriminant we found earlier. So, the solutions are:( x = frac{ -2(1 - 3a) pm sqrt{12a^2 - 8} }{ 6 } )= ( frac{ -2 + 6a pm sqrt{12a^2 - 8} }{ 6 } )= ( frac{ -1 + 3a pm sqrt{3a^2 - 2} }{ 3 } )Therefore, the critical points are:( x = frac{ -1 + 3a + sqrt{3a^2 - 2} }{ 3 } ) and ( x = frac{ -1 + 3a - sqrt{3a^2 - 2} }{ 3 } )Note that ( sqrt{12a^2 - 8} = 2sqrt{3a^2 - 2} ), which is why we can write the sqrt term as ( sqrt{3a^2 - 2} ).Now, these critical points exist only when ( 3a^2 - 2 geq 0 ), i.e., when ( |a| geq sqrt{2/3} ), as before.So, for ( |a| geq sqrt{2/3} ), there are two critical points. Let's denote them as ( x_1 ) and ( x_2 ).Now, to check if the cubic has a multiple root, we need to find if there exists an ( a ) such that ( f(x_i) = 0 ) for either ( x_1 ) or ( x_2 ).Therefore, substitute ( x = x_i ) into the cubic equation and solve for ( a ).This is going to be complicated, but let's try.Take ( x = x_1 = frac{ -1 + 3a + sqrt{3a^2 - 2} }{ 3 } )Plugging into the cubic equation ( f(x) = 0 ):This seems very tedious. Maybe there's a smarter way. Alternatively, since we already found that when ( a = 1/2 ), the cubic factors into a linear term and an irreducible quadratic, which implies that at ( a = 1/2 ), the cubic has a multiple root? Wait, no. If it factors into a linear term and an irreducible quadratic, then there's one real root and two complex roots, so it's not a multiple root. But earlier, when we factored for ( a = 1 ), we had a double root, which is a multiple root. So, maybe ( a = 1 ) is a case of a multiple root.Wait, when ( a = 1 ), the cubic becomes ( x^3 - 2x^2 + x ), which factors into ( x(x - 1)^2 ). So, there's a double root at ( x = 1 ), so yes, multiple root.Therefore, ( a = 1 ) is a value where the cubic has a multiple root. Similarly, maybe ( a = 1/2 ) is not a multiple root case but an irreducible quadratic factor.Therefore, to find the values of ( a ) where the cubic has multiple roots, we need to solve the system ( f(x) = 0 ) and ( f'(x) = 0 ).So, let's set up the equations:From ( f'(x) = 0 ):( 3x^2 + 2(1 - 3a)x + (2a^2 - 2a + 1) = 0 )From ( f(x) = 0 ):( x^3 + (1 - 3a)x^2 + (2a^2 - 2a + 1)x + (a^2 - a) = 0 )Let me denote equation 1 as ( f'(x) = 0 ), equation 2 as ( f(x) = 0 ).We can solve this system by expressing ( a ) in terms of ( x ) from equation 1 and substituting into equation 2.From equation 1:( 3x^2 + 2(1 - 3a)x + (2a^2 - 2a + 1) = 0 )Let me rearrange this equation:( 2a^2 - 2a + 1 + 2(1 - 3a)x + 3x^2 = 0 )Let's collect terms by ( a ):Quadratic in ( a ):( 2a^2 + (-2 - 6x)a + (1 + 2x + 3x^2) = 0 )So, quadratic equation:( 2a^2 - (2 + 6x)a + (1 + 2x + 3x^2) = 0 )Let me solve for ( a ):( a = frac{ (2 + 6x) pm sqrt{(2 + 6x)^2 - 8(1 + 2x + 3x^2)} }{4} )Compute discriminant:( (2 + 6x)^2 - 8(1 + 2x + 3x^2) )= ( 4 + 24x + 36x^2 - 8 - 16x - 24x^2 )= ( (36x^2 - 24x^2) + (24x - 16x) + (4 - 8) )= ( 12x^2 + 8x - 4 )Factor out 4:= ( 4(3x^2 + 2x - 1) )Therefore,( a = frac{2 + 6x pm 2sqrt{3x^2 + 2x - 1}}{4} )Simplify:( a = frac{1 + 3x pm sqrt{3x^2 + 2x - 1}}{2} )So, these are the possible values of ( a ) in terms of ( x ). Now, substitute this into equation 2, which is ( f(x) = 0 ):( x^3 + (1 - 3a)x^2 + (2a^2 - 2a + 1)x + (a^2 - a) = 0 )But since ( a ) is expressed in terms of ( x ), substituting the expression for ( a ) into this equation would allow us to solve for ( x ), but this seems very complicated.Alternatively, maybe we can find a relationship between ( a ) and ( x ) using the above expression for ( a ), and then substitute back into equation 2. Let's see.Alternatively, perhaps we can use polynomial division. Since ( f(x) ) and ( f'(x) ) have a common root when there's a multiple root, the greatest common divisor (GCD) of ( f(x) ) and ( f'(x) ) would be a polynomial of degree at least 1. To find the GCD, we can perform the Euclidean algorithm on ( f(x) ) and ( f'(x) ).Let me attempt polynomial division of ( f(x) ) by ( f'(x) ).Divide ( f(x) ) by ( f'(x) ):Dividend: ( x^3 + (1 - 3a)x^2 + (2a^2 - 2a + 1)x + (a^2 - a) )Divisor: ( 3x^2 + 2(1 - 3a)x + (2a^2 - 2a + 1) )First term: ( x^3 / 3x^2 = (1/3)x )Multiply divisor by (1/3)x:( (1/3)x*(3x^2 + 2(1 - 3a)x + (2a^2 - 2a + 1)) = x^3 + (2/3)(1 - 3a)x^2 + ( (2a^2 - 2a + 1)/3 )x )Subtract this from dividend:( [x^3 + (1 - 3a)x^2 + (2a^2 - 2a + 1)x + (a^2 - a)] - [x^3 + (2/3)(1 - 3a)x^2 + ( (2a^2 - 2a + 1)/3 )x ] )= ( [0x^3] + [ (1 - 3a - 2/3(1 - 3a) )x^2 ] + [ (2a^2 - 2a + 1 - (2a^2 - 2a + 1)/3 )x ] + [a^2 - a] )Simplify each coefficient:Quadratic term:( 1 - 3a - (2/3)(1 - 3a) = (1 - 3a)(1 - 2/3) = (1 - 3a)(1/3) = (1/3 - a) )Linear term:( (2a^2 - 2a + 1) - (2a^2 - 2a + 1)/3 = (2a^2 - 2a + 1)(1 - 1/3) = (2a^2 - 2a + 1)(2/3) = (4a^2 - 4a + 2)/3 )Constant term remains ( a^2 - a )Therefore, the remainder after division is:( (1/3 - a)x^2 + (4a^2 - 4a + 2)/3 x + (a^2 - a) )Now, continue the Euclidean algorithm. Next step: divide the divisor ( f'(x) ) by the remainder.But this is getting too involved. Maybe another approach is better. Let me recall that when a polynomial ( f(x) ) has a multiple root ( r ), then ( r ) is also a root of ( f'(x) ). Therefore, the multiple root ( r ) must satisfy both equations:1. ( f(r) = 0 )2. ( f'(r) = 0 )Therefore, we can set up the system:From ( f'(r) = 0 ):( 3r^2 + 2(1 - 3a)r + (2a^2 - 2a + 1) = 0 )From ( f(r) = 0 ):( r^3 + (1 - 3a)r^2 + (2a^2 - 2a + 1)r + (a^2 - a) = 0 )Let me subtract ( r ) times ( f'(r) ) from ( f(r) ). This is a technique used to eliminate higher-degree terms.Compute ( f(r) - r f'(r) ):( [r^3 + (1 - 3a)r^2 + (2a^2 - 2a + 1)r + (a^2 - a)] - r[3r^2 + 2(1 - 3a)r + (2a^2 - 2a + 1)] )Compute each term:- ( r^3 - 3r^3 = -2r^3 )- ( (1 - 3a)r^2 - 2(1 - 3a)r^2 = (1 - 3a - 2 + 6a)r^2 = (-1 + 3a)r^2 )- ( (2a^2 - 2a + 1)r - (2a^2 - 2a + 1)r = 0 )- ( a^2 - a )Therefore, ( f(r) - r f'(r) = -2r^3 + (-1 + 3a)r^2 + (a^2 - a) )But since ( f(r) = 0 ) and ( f'(r) = 0 ), this expression equals zero:( -2r^3 + (-1 + 3a)r^2 + (a^2 - a) = 0 )So now we have two equations:1. ( 3r^2 + 2(1 - 3a)r + (2a^2 - 2a + 1) = 0 ) (from ( f'(r) = 0 ))2. ( -2r^3 + (-1 + 3a)r^2 + (a^2 - a) = 0 ) (from ( f(r) - r f'(r) = 0 ))Let me denote equation 1 as Eq1 and equation 2 as Eq2.From Eq1, we can express ( 3r^2 = -2(1 - 3a)r - (2a^2 - 2a + 1) )Then, substitute ( 3r^2 ) into Eq2. Let me see:Eq2: ( -2r^3 + (-1 + 3a)r^2 + (a^2 - a) = 0 )Express ( r^3 ) in terms of ( r^2 ) using Eq1. From Eq1:( 3r^2 = -2(1 - 3a)r - (2a^2 - 2a + 1) )Multiply both sides by r:( 3r^3 = -2(1 - 3a)r^2 - (2a^2 - 2a + 1)r )Therefore,( r^3 = frac{ -2(1 - 3a)r^2 - (2a^2 - 2a + 1)r }{ 3 } )Substitute this into Eq2:( -2*[ (-2(1 - 3a)r^2 - (2a^2 - 2a + 1)r ) / 3 ] + (-1 + 3a)r^2 + (a^2 - a) = 0 )Simplify term by term:First term:( -2*(-2(1 - 3a)r^2 - (2a^2 - 2a + 1)r ) / 3 = [4(1 - 3a)r^2 + 2(2a^2 - 2a + 1)r ] / 3 )So, equation becomes:( [4(1 - 3a)r^2 + 2(2a^2 - 2a + 1)r ] / 3 + (-1 + 3a)r^2 + (a^2 - a) = 0 )Multiply all terms by 3 to eliminate denominator:( 4(1 - 3a)r^2 + 2(2a^2 - 2a + 1)r + 3(-1 + 3a)r^2 + 3(a^2 - a) = 0 )Combine like terms:Quadratic terms:( 4(1 - 3a)r^2 + 3(-1 + 3a)r^2 = [4 - 12a - 3 + 9a]r^2 = (1 - 3a)r^2 )Linear terms:( 2(2a^2 - 2a + 1)r = (4a^2 - 4a + 2)r )Constant terms:( 3(a^2 - a) )Thus, the equation becomes:( (1 - 3a)r^2 + (4a^2 - 4a + 2)r + 3a^2 - 3a = 0 )Now, we have two equations involving ( r ) and ( a ). Let me recall Eq1:( 3r^2 + 2(1 - 3a)r + (2a^2 - 2a + 1) = 0 )Let me denote this as Eq1: ( 3r^2 + 2(1 - 3a)r + (2a^2 - 2a + 1) = 0 )And the modified Eq2: ( (1 - 3a)r^2 + (4a^2 - 4a + 2)r + 3a^2 - 3a = 0 )Let me call the modified Eq2 as Eq3.Now, we can try to eliminate ( r^2 ) from Eq1 and Eq3.From Eq1:Solve for ( r^2 ):( 3r^2 = -2(1 - 3a)r - (2a^2 - 2a + 1) )=> ( r^2 = [ -2(1 - 3a)r - (2a^2 - 2a + 1) ] / 3 )Substitute this into Eq3:( (1 - 3a)*[ (-2(1 - 3a)r - (2a^2 - 2a + 1)) / 3 ] + (4a^2 - 4a + 2)r + 3a^2 - 3a = 0 )Multiply out:First term:( (1 - 3a)*(-2(1 - 3a)r - (2a^2 - 2a + 1)) / 3 )= ( [ -2(1 - 3a)^2 r - (1 - 3a)(2a^2 - 2a + 1) ] / 3 )Second term: ( (4a^2 - 4a + 2)r )Third term: ( 3a^2 - 3a )Combine all terms:[ -2(1 - 3a)^2 r - (1 - 3a)(2a^2 - 2a + 1) ] / 3 + (4a^2 - 4a + 2)r + 3a^2 - 3a = 0Multiply all terms by 3 to eliminate denominator:-2(1 - 3a)^2 r - (1 - 3a)(2a^2 - 2a + 1) + 3(4a^2 - 4a + 2)r + 9a^2 - 9a = 0Now, expand each term:First term: -2(1 - 6a + 9a^2)r = -2r + 12a r - 18a^2 rSecond term: -(1 - 3a)(2a^2 - 2a + 1) = -(2a^2 - 2a + 1 - 6a^3 + 6a^2 - 3a) = -2a^2 + 2a -1 + 6a^3 -6a^2 +3a = 6a^3 -8a^2 +5a -1Third term: 3*(4a^2 -4a +2)r = 12a^2 r -12a r +6rFourth term: 9a^2 -9aNow, combine all terms:Terms with r:(-2r + 12a r -18a^2 r) + (12a^2 r -12a r +6r) = [(-2r +6r) + (12a r -12a r) + (-18a^2 r +12a^2 r)] = 4r -6a^2 rTerms without r:6a^3 -8a^2 +5a -1 +9a^2 -9a = 6a^3 + ( -8a^2 +9a^2 ) + (5a -9a ) + (-1) = 6a^3 +a^2 -4a -1Therefore, overall equation:(4 -6a^2)r +6a^3 +a^2 -4a -1 =0Now, from this equation, solve for r:(4 -6a^2)r = -6a^3 -a^2 +4a +1=> r = [ -6a^3 -a^2 +4a +1 ] / (4 -6a^2 )Simplify numerator and denominator:Factor numerator:Let me see if -6a^3 -a^2 +4a +1 can be factored.Let me factor by grouping:-6a^3 -a^2 +4a +1 = (-6a^3 -a^2) + (4a +1) = -a^2(6a +1) +1*(4a +1). Not obvious.Alternatively, use rational root theorem. Possible roots are ( pm1, pm1/2, pm1/3, pm1/6 ).Test a = 1:-6(1)^3 -1^2 +4(1) +1 = -6 -1 +4 +1 = -2 ≠ 0a = -1:-6(-1)^3 -(-1)^2 +4*(-1) +1 = 6 -1 -4 +1 = 2 ≠ 0a = 1/2:-6*(1/8) -1/4 +4*(1/2) +1 = -3/4 -1/4 +2 +1 = (-1) +3 = 2 ≠ 0a = -1/2:-6*(-1/2)^3 -(-1/2)^2 +4*(-1/2) +1 = -6*(-1/8) -1/4 -2 +1 = 6/8 -1/4 -1 = 3/4 -1/4 -1 = -1/2 ≠0a=1/3:-6*(1/27) -1/9 +4*(1/3) +1 = -2/9 -1/9 +4/3 +1 = (-3/9) + (12/9 +9/9) = (-1/3) +21/9 = (-1/3 +7/3) = 6/3=2≠0a= -1/3:-6*(-1/3)^3 -(-1/3)^2 +4*(-1/3) +1 = -6*(-1/27) -1/9 -4/3 +1 = 2/9 -1/9 -4/3 +1 = (1/9) -4/3 +1 = (1 -12 +9)/9 = (-2)/9 ≠0a=1/6:-6*(1/6)^3 - (1/6)^2 +4*(1/6) +1 = -6*(1/216) -1/36 +2/3 +1 = -1/36 -1/36 +2/3 +1 = (-2/36) + (24/36 +36/36)= (-1/18) +60/36= (-1/18)+5/3≈ (-0.055)+1.666≈1.611≠0So, no rational roots. Therefore, the numerator doesn't factor nicely. Therefore, perhaps we need to express r in terms of a as above and substitute back into Eq1.But this seems very complex. Let's recall that we also have from Eq1:3r^2 + 2(1 - 3a)r + (2a^2 - 2a +1)=0Let me denote this as:3r^2 = -2(1 -3a)r - (2a^2 -2a +1)So, substitute r from the previous expression into this.But this seems too involved. Maybe another approach is better.Alternatively, notice that when a=1, the cubic has a double root. When a=1/2, it has one real root and two complex roots. Therefore, perhaps the discriminant D is zero at a=1 and a=1/2, and negative between them or outside. Wait, but we need to verify.Wait, when a=1, we have a double root, so D=0. When a=1/2, we have one real root and two complex roots, so D<0. But earlier factorization showed that at a=1/2, the cubic factors into (x - 1/2)(x^2 + 1/2). The discriminant of the quadratic x^2 + 1/2 is negative, so the cubic discriminant would be related to that.But perhaps the discriminant D is zero when a=1 and a=1/2. Wait, but when a=1/2, the cubic has one real root and two complex roots, which would imply D<0. When a=1, it's a double root and a simple root, which would imply D=0. So, discriminant D=0 at a=1 and another value? Or maybe there's another value.Alternatively, perhaps the discriminant is zero at a=1 and a=sqrt(2/3), but this is just guessing.Alternatively, let's compute the discriminant for specific values of a:Case 1: a=0The original equation becomes:x^3 + x^2 + x +0 = x^3 + x^2 + x = x(x^2 +x +1)=0Roots: x=0, and roots of x^2 +x +1=0, which are complex. So, one real root. Therefore, for a=0, the equation has exactly one real root. Therefore, the discriminant must be negative at a=0.Case 2: a=1/2As before, factors into (x -1/2)(x^2 +1/2)=0, which has one real root. So discriminant D<0.Case 3: a=1Double root at x=1 and simple root at x=0. Therefore, discriminant D=0.Case 4: a=2Original equation:x^3 + (1 -6)x^2 + (8 -4 +1)x + (4 -2) =x^3 -5x^2 +5x +2=0Not sure how many real roots this has. Let me check by evaluating the cubic at several points:f(0)=2>0f(1)=1 -5 +5 +2=3>0f(2)=8 -20 +10 +2=0. So, x=2 is a root. Then factor (x-2):Using polynomial division or synthetic division:Divide x^3 -5x^2 +5x +2 by (x-2):Coefficients: 1 | -5 | 5 | 2Bring down 1Multiply by 2: 2Add to next coefficient: -5 +2=-3Multiply by2: -6Add to next coefficient:5 + (-6)= -1Multiply by2: -2Add to last coefficient:2 + (-2)=0. So, quotient is x^2 -3x -1.So, factors to (x -2)(x^2 -3x -1). The quadratic has discriminant 9 +4=13>0, so two real roots. So, total three real roots. Therefore, when a=2, equation has three real roots. So, discriminant D>0.Therefore, when a=0 and a=1/2, discriminant is negative (one real root). When a=1, discriminant is zero (double root). When a=2, discriminant is positive (three real roots). Therefore, this suggests that the discriminant changes sign at a=1 and another point between a=1/2 and a=1? Wait, but when a=1/2, discriminant is negative, at a=1 it's zero, and for a>1, at least at a=2, it's positive. Similarly, for a<1/2, like a=0, it's negative. So, maybe the discriminant is negative when a <1 and positive when a>1. But at a=1/2, which is less than 1, it's negative, and at a=1, zero, and a=2, positive. Wait, but that conflicts with the previous case of a=1. Wait, maybe the discriminant is negative for all a <1 and positive for a>1, but at a=1, it's zero. But when a=1/2, it's negative, which is less than 1. At a=0, negative, at a=1, zero, and at a=2, positive. So, possibly the discriminant transitions from negative to positive at a=1. So, the equation has exactly one real root when a <1, and three real roots when a >1. But this contradicts the case when a=1/2, which is less than 1, but when a=1, it has a double root.Wait, but earlier when we considered the derivative's discriminant, we found that for |a| < sqrt(2/3) ≈0.816, the cubic is strictly increasing and thus has exactly one real root. For |a| > sqrt(2/3), the cubic has local maxima and minima. However, the existence of critical points does not automatically mean three real roots. It depends on the behavior at those critical points.But from the specific examples:- When a=0 (which is |a| < sqrt(2/3)), the cubic is strictly increasing (as the derivative discriminant is negative), so one real root.- When a=1/2 (which is |a| < sqrt(2/3) since sqrt(2/3)≈0.816), but in this case, the cubic is not strictly increasing because when a=1/2, the derivative discriminant is 12*(1/2)^2 -8= 12*(1/4) -8=3 -8=-5<0, which contradicts our earlier conclusion. Wait, no: wait, when a=1/2, the derivative discriminant is 12a^2 -8=12*(1/4)-8=3-8=-5<0. Therefore, the derivative has no real roots, which means the cubic is strictly increasing. But earlier, we saw that when a=1/2, the cubic factors into (x -1/2)(x^2 +1/2). So, the cubic is not strictly increasing, since it has a local maximum and minimum?Wait, there's a contradiction here. Wait, if the derivative discriminant is negative, then the cubic is strictly increasing. But in the case of a=1/2, we have a cubic that factors into (x -1/2)(x^2 +1/2). The quadratic term x^2 +1/2 has no real roots, but the cubic itself has a real root at x=1/2 and two complex roots. The graph of the cubic should cross the x-axis only once, and if the derivative has no real roots, then it's strictly increasing. But wait, how can a cubic with a strictly increasing derivative have only one real root? Because if it's strictly increasing, it will cross the x-axis exactly once. So, that must be the case. However, when we factored the cubic for a=1/2 into (x -1/2)(x^2 +1/2), the cubic should indeed be strictly increasing. Let me check by taking the derivative:For a=1/2, the cubic is x^3 - (1/2)x^2 + (1/2)x -1/4. Its derivative is 3x^2 - x +1/2. The discriminant of the derivative is (-1)^2 -4*3*(1/2)=1 -6= -5 <0. So, derivative has no real roots, meaning the cubic is strictly increasing. Therefore, even though it factors into a linear term and a quadratic, the fact that the quadratic has no real roots is consistent with the cubic being strictly increasing and crossing the x-axis once. So, all is good.But then, when a=1/2, even though the cubic is strictly increasing, it factors into a linear and quadratic term. That's okay, because the quadratic term doesn't contribute any real roots. So, the cubic crosses the x-axis once, even though it's factorable.Therefore, the key point is that for |a| < sqrt(2/3), the cubic is strictly increasing and thus has exactly one real root. For |a| >= sqrt(2/3), the cubic has local maxima and minima, but may still have only one real root if both extrema are on the same side of the x-axis.So, we need to check for |a| >= sqrt(2/3), whether the cubic has one or three real roots. To do that, we need to evaluate the cubic at its local maximum and minimum and check their signs.But since computing the critical points is complicated, we can use the following method:The cubic will have three real roots if and only if there is a sign change between the local maximum and minimum. So, if the local maximum is positive and the local minimum is negative (or vice versa), then the cubic crosses the x-axis three times. If both extrema are positive or both are negative, then the cubic has only one real root.Therefore, to have exactly one real root for |a| >= sqrt(2/3), the cubic's local maximum and minimum must both be positive or both be negative.To compute this, we can use the fact that the cubic at the critical points can be expressed in terms of the coefficients. However, this is still complicated.Alternatively, recall that if a cubic has local maximum and local minimum, and both are on the same side of the x-axis, then the cubic will have only one real root. Otherwise, three real roots.To determine when both extrema are positive or both are negative, we can use the following approach:Let f(x) be the cubic. Let x1 and x2 be the critical points (x1 < x2). Then:- If f(x1) > 0 and f(x2) > 0, then the cubic has one real root.- If f(x1) < 0 and f(x2) < 0, then the cubic has one real root.- If f(x1) and f(x2) have opposite signs, then three real roots.So, we need to find when f(x1) and f(x2) have the same sign.But calculating f(x1) and f(x2) in terms of a is complicated. Alternatively, perhaps we can use the concept of "discriminant sequences" or other methods, but this might be beyond the scope.Alternatively, let's consider that for large |a|, the cubic will behave like x^3, so as x approaches infinity, it goes to infinity, and as x approaches negative infinity, it goes to negative infinity. But if the leading coefficient is positive, the cubic tends to +infty as x approaches +infty and -infty as x approaches -infty. Therefore, if there's a local maximum and minimum both above the x-axis, the cubic will start from -infty, rise to the local maximum (above the x-axis), then decrease to the local minimum (still above the x-axis), and then increase to +infty. Therefore, crossing the x-axis only once. Similarly, if both extrema are below the x-axis, the cubic will cross the x-axis once.Therefore, to find when both extrema are above or both below the x-axis, we need to ensure that the minimum value of the cubic is positive or the maximum value is negative.But calculating these extrema is still complicated.Alternatively, let's use the fact that for the cubic to have exactly one real root when it has two critical points, the cubic must satisfy f(x1) * f(x2) > 0, where x1 and x2 are the critical points.Therefore, the product of the cubic evaluated at the critical points must be positive. If this product is positive, the cubic has one real root; if negative, three real roots.Therefore, let's compute f(x1) * f(x2).This can be done using the resultant or other methods, but it's complicated. However, there's a formula for the product of the cubic evaluated at the critical points in terms of the coefficients.Let me recall that for a cubic polynomial ( f(x) = x^3 + bx^2 + cx + d ), with critical points x1 and x2, the product f(x1)f(x2) can be computed as follows:First, note that x1 and x2 are roots of f’(x) = 3x^2 + 2bx + c = 0.Therefore, x1 + x2 = -2b/3, x1x2 = c/3.Then, f(x1) = x1^3 + bx1^2 + cx1 + dBut since x1 is a critical point, f’(x1) = 0 => 3x1^2 + 2bx1 + c =0 => x1^2 = (-2bx1 -c)/3Therefore, x1^3 = x1 * x1^2 = x1*(-2bx1 -c)/3 = (-2b x1^2 -c x1)/3But x1^2 = (-2bx1 -c)/3, so substitute:x1^3 = (-2b*(-2bx1 -c)/3 -c x1)/3 = (4b^2 x1 + 2bc - 3c x1)/9= [ (4b^2 -3c) x1 + 2bc ] /9Therefore, f(x1) = x1^3 + b x1^2 +c x1 +d= [ (4b^2 -3c)x1 + 2bc ] /9 + b x1^2 +c x1 +dBut x1^2 = (-2b x1 -c)/3, so:= [ (4b^2 -3c)x1 + 2bc ] /9 + b*(-2b x1 -c)/3 +c x1 +d= [ (4b^2 -3c)x1 + 2bc ] /9 + (-2b^2 x1 - bc)/3 +c x1 +dConvert all terms to ninths:= [ (4b^2 -3c)x1 + 2bc ] /9 + [ -6b^2 x1 -3bc ] /9 + [9c x1]/9 + [9d]/9Combine terms:Numerator:(4b^2 -3c)x1 + 2bc -6b^2 x1 -3bc +9c x1 +9dGroup like terms:x1 terms:(4b^2 -3c -6b^2 +9c)x1 = (-2b^2 +6c)x1Constants:2bc -3bc +9d = (-bc +9d)Therefore, f(x1) = [ (-2b^2 +6c)x1 - bc +9d ] /9Similarly, f(x2) = [ (-2b^2 +6c)x2 - bc +9d ] /9Therefore, the product f(x1)f(x2) is:[ (-2b^2 +6c)x1 - bc +9d ] [ (-2b^2 +6c)x2 - bc +9d ] /81Let me denote K = (-2b^2 +6c)Then, f(x1)f(x2) = [ K x1 - bc +9d ] [ K x2 - bc +9d ] /81Expand the numerator:K^2 x1x2 + K(- bc +9d)(x1 +x2) + (- bc +9d)^2From earlier, x1 +x2 = -2b/3, x1x2 = c/3Substitute:= K^2 (c/3) + K(- bc +9d)(-2b/3) + (- bc +9d)^2Therefore,= (K^2 c)/3 + (2bK(bc -9d))/3 + (bc -9d)^2This is very complicated, but let's substitute K = -2b^2 +6c:= [ (-2b^2 +6c)^2 c ] /3 + [2b(-2b^2 +6c)(bc -9d)] /3 + (bc -9d)^2This expression is quite involved, but perhaps we can substitute the coefficients for our specific cubic.In our case:b =1 -3ac =2a^2 -2a +1d =a^2 -aPlugging these into the expression:First, compute K = -2b^2 +6c = -2(1 -3a)^2 +6(2a^2 -2a +1)Expand:= -2(1 -6a +9a^2) +12a^2 -12a +6= -2 +12a -18a^2 +12a^2 -12a +6= (-18a^2 +12a^2) + (12a -12a) + (-2 +6)= (-6a^2) +0 +4= -6a^2 +4Next, compute bc -9d:= b*c -9d = (1 -3a)(2a^2 -2a +1) -9(a^2 -a)Expand (1 -3a)(2a^2 -2a +1):= 2a^2 -2a +1 -6a^3 +6a^2 -3a= -6a^3 +8a^2 -5a +1Subtract 9a^2 -9a:= -6a^3 +8a^2 -5a +1 -9a^2 +9a= -6a^3 -a^2 +4a +1So, bc -9d = -6a^3 -a^2 +4a +1Now, compute each term:First term:[ (-2b^2 +6c)^2 c ] /3 = K^2 c /3 = (-6a^2 +4)^2 (2a^2 -2a +1) /3Second term:[2bK(bc -9d)] /3 = 2*(1 -3a)*(-6a^2 +4)*(-6a^3 -a^2 +4a +1) /3Third term:(bc -9d)^2 = (-6a^3 -a^2 +4a +1)^2This is extremely complex, but perhaps there is cancellation or factorization.Given the time constraints, maybe it's better to abandon this approach and consider the specific cases and examples we computed earlier.We know that:- For a=0: D<0 (one real root)- For a=1/2: D<0 (one real root)- For a=1: D=0 (double root)- For a=2: D>0 (three real roots)Therefore, it appears that the discriminant is negative for a <1, zero at a=1, and positive for a>1. However, when a=1/2, which is less than 1, the discriminant is also negative. This suggests that the discriminant is negative for all a <1 and positive for a>1. But when a=1, it's zero.But wait, when a=1/2, which is less than 1, the discriminant is negative. When a=0, it's negative. When a=1, zero. When a=2, positive. Therefore, the discriminant transitions from negative to zero at,<think>Okay, so I need to find the range of values for 'a' such that the given cubic equation has exactly one root. The equation is:x³ + (1 - 3a)x² + 2a²x - 2ax + x + a² - a = 0First, I should probably simplify the equation by combining like terms. Let me look at each term:Starting with the x³ term: that's just x³, nothing to combine.Next, the x² term: (1 - 3a)x². That seems straightforward.Now, the x terms: 2a²x - 2ax + x. Let me combine those. Let's factor out x:x(2a² - 2a + 1). So combining those coefficients:2a² - 2a + 1. Hmm, that might be important later.Then the constant term: a² - a.So simplifying the entire equation:x³ + (1 - 3a)x² + (2a² - 2a + 1)x + (a² - a) = 0So, we have a cubic equation in standard form. Now, the problem states that this equation has exactly one root. But wait, a cubic equation can have one or three real roots (counting multiplicities). So, for it to have exactly one real root, it must mean that the other two roots are complex conjugates. Therefore, the equation has one real root and two non-real complex roots. So, the condition here is that the cubic equation has only one real root, which would require its discriminant to be negative. But maybe there's another way to approach this.Alternatively, maybe the equation has a triple root? But a triple root would technically be one root with multiplicity three. But in that case, the discriminant would be zero. But the problem says "exactly one root," which might mean exactly one distinct root. So, in that case, it would need to have a triple root. But I need to check what the question is exactly asking. It says "exactly one root," which usually in math problems means exactly one real root (with the other two being complex), unless specified otherwise. So perhaps I need to check when the cubic has one real root and two complex roots.But let me confirm. If a cubic equation has three real roots, but two of them are equal (i.e., a double root and a simple root), then technically, it has two distinct roots. If all three roots are equal, it's a triple root, so one distinct root. But the problem says "exactly one root," so maybe it's considering multiplicity. Hmm, but the wording is ambiguous. Wait, in the problem statement, when it says "has exactly one root," does that mean exactly one real root (regardless of multiplicity) or exactly one distinct real root (counting multiplicity)? This is crucial.But in standard terminology, when they say a polynomial equation has "exactly one root," without specifying, it usually means exactly one real root, with the others being non-real. So, the equation crosses the x-axis once, and the other two roots are complex. Therefore, we need the cubic to have one real root and two complex roots. For that to happen, the discriminant of the cubic must be negative.Alternatively, maybe the cubic can be factored, and then we can find conditions on 'a' so that two factors have no real roots. Let me check if the cubic can be factored.Looking at the simplified equation:x³ + (1 - 3a)x² + (2a² - 2a + 1)x + (a² - a) = 0Let me try to factor this. Maybe factor by grouping. Let me see:Group terms as [x³ + (1 - 3a)x²] + [(2a² - 2a + 1)x + (a² - a)]Factor x² from the first group:x²(x + 1 - 3a) + [(2a² - 2a + 1)x + (a² - a)]Hmm, not sure if that helps. Alternatively, perhaps trying to find a rational root. By Rational Root Theorem, possible roots are factors of the constant term (a² - a) divided by factors of the leading coefficient, which is 1. So possible rational roots are ±(a² - a). Let me test x = 1. Let me plug x = 1 into the equation:1 + (1 - 3a) + (2a² - 2a + 1) + (a² - a) = 0Calculate:1 + 1 - 3a + 2a² - 2a + 1 + a² - a = 0Combine like terms:1 + 1 + 1 = 3-3a -2a -a = -6a2a² + a² = 3a²So total equation: 3a² -6a + 3 = 0Divide by 3: a² - 2a + 1 = 0 → (a - 1)² = 0 → a = 1So when a = 1, x = 1 is a root. Let me check if x =1 is a root when a =1:Original equation:x³ + (1 - 3*1)x² + 2*1²x - 2*1x + x + 1² - 1 = x³ -2x² + 2x -2x + x +0 = x³ -2x² + x = x(x² -2x +1) = x(x-1)^2. So, when a=1, the equation factors as x(x-1)^2=0, which has roots x=0 and x=1 (double root). So two distinct real roots. But the problem says "exactly one root," so a=1 would not satisfy the condition. So maybe that approach isn't helpful.Alternatively, maybe x = a is a root? Let me try plugging x = a into the equation:a³ + (1 - 3a)a² + 2a²*a - 2a*a + a + a² - a = ?Calculate each term:a³ + (a² - 3a³) + 2a³ - 2a² + a + a² - aCombine like terms:a³ -3a³ + 2a³ = 0a² -2a² + a² = 0a -a =0So total is 0. Therefore, x = a is always a root of the equation, regardless of the value of a. Wait, that's a key insight!So, the cubic can be factored as (x - a)(quadratic) = 0. Let me perform polynomial division or factorization.Given that x = a is a root, we can factor (x - a) out of the cubic polynomial.Let me write the cubic polynomial as P(x) = x³ + (1 - 3a)x² + (2a² - 2a + 1)x + (a² - a)Divide P(x) by (x - a). Let's use polynomial long division or synthetic division.Alternatively, write P(x) = (x - a)(x² + bx + c). Let me expand this:(x - a)(x² + bx + c) = x³ + bx² + cx - a x² - abx - ac = x³ + (b - a)x² + (c - ab)x - acSet this equal to the original polynomial:x³ + (1 - 3a)x² + (2a² - 2a + 1)x + (a² - a)So equate coefficients:1. Coefficient of x³: 1 = 1 (okay)2. Coefficient of x²: b - a = 1 - 3a ⇒ b = 1 - 3a + a = 1 - 2a3. Coefficient of x: c - ab = 2a² - 2a + 14. Constant term: -ac = a² - a ⇒ -a c = a² - a ⇒ c = (-a² + a)/a = -a + 1 (assuming a ≠ 0)So let's check:From the constant term, c = -a + 1.From the coefficient of x², b = 1 - 2a.Now, check the coefficient of x:c - ab = ( -a +1 ) - a*(1 - 2a ) = (-a +1) - a + 2a² = -2a +1 + 2a²But according to the original polynomial, the coefficient of x is 2a² -2a +1. So that's exactly equal. So yes, the factorization works.Therefore, the cubic equation factors as:(x - a)(x² + (1 - 2a)x + (-a +1)) = 0Simplify the quadratic term:x² + (1 - 2a)x + (-a +1) = x² + (1 - 2a)x + (1 - a)So, the equation is (x - a)(x² + (1 - 2a)x + (1 - a)) = 0Therefore, the roots are x = a and the roots of the quadratic x² + (1 - 2a)x + (1 - a) = 0For the original equation to have exactly one real root, the quadratic must have no real roots. Because if the quadratic has real roots, then the cubic would have three real roots (counting x = a), but if the quadratic has no real roots, then the only real root is x = a. So, the problem reduces to finding the values of 'a' such that the quadratic equation x² + (1 - 2a)x + (1 - a) = 0 has no real roots.A quadratic equation ax² + bx + c = 0 has no real roots when its discriminant is negative. The discriminant D is D = b² - 4ac.So, compute the discriminant of the quadratic:D = (1 - 2a)² - 4 * 1 * (1 - a)Expand and simplify:First, expand (1 - 2a)²: 1 - 4a + 4a²Then subtract 4*(1 - a): 1 - 4a + 4a² - 4 + 4aSimplify term by term:1 - 4a + 4a² -4 +4a = (1 -4) + (-4a +4a) +4a² = -3 + 0 +4a² = 4a² -3So discriminant D = 4a² -3Wait, so D = 4a² -3. Wait, that seems too simple. Let me check again.Original quadratic discriminant:D = (1 - 2a)^2 - 4*(1)*(1 - a)Compute (1 - 2a)^2 = 1 - 4a + 4a²4*1*(1 - a) = 4 -4aTherefore, D = (1 -4a +4a²) - (4 -4a) = 1 -4a +4a² -4 +4a = (1 -4) + (-4a +4a) +4a² = -3 +0 +4a² = 4a² -3. Yes, correct.Therefore, the discriminant of the quadratic is D = 4a² -3.For the quadratic to have no real roots, D < 0:4a² -3 < 0 ⇒ 4a² <3 ⇒ a² < 3/4 ⇒ |a| < sqrt(3)/2 ≈0.866Therefore, the quadratic has no real roots when -sqrt(3)/2 < a < sqrt(3)/2.Therefore, the original cubic equation will have exactly one real root (x = a) when the quadratic factor has no real roots, which occurs when a is in (-sqrt(3)/2, sqrt(3)/2).But wait, hold on! Wait, we need to check if there are any exceptions. For example, when a is such that x = a is also a root of the quadratic. If the quadratic has x = a as a root, then the cubic would have a double root at x = a, which would mean that the cubic has two real roots (x = a and the other root from the quadratic). Wait, but in our previous factorization, we factored out (x - a), so the quadratic cannot have x = a as a root unless (x - a) is a factor of the quadratic, which would mean the quadratic is (x - a)(x - b). But in our case, the quadratic is x² + (1 - 2a)x + (1 - a). Let's check if x = a is a root of the quadratic.Plug x = a into the quadratic:a² + (1 - 2a)a + (1 - a) = a² + a - 2a² +1 -a = (a² -2a²) + (a -a) +1 = -a² +0 +1 = 1 -a²So, 1 -a² =0 ⇒ a² =1 ⇒ a=1 or a=-1. Therefore, when a=1 or a=-1, x=a is a root of the quadratic. Which would make the cubic have a double root at x=a and another root from the quadratic. But wait, if a=1, then as we saw earlier, the equation becomes x(x -1)^2=0, which has roots x=0 and x=1 (double root). So two distinct real roots, which contradicts the "exactly one root" condition. Similarly, if a=-1, let's check:Original cubic: factors as (x - (-1))(quadratic). So (x +1)(quadratic). Let's compute the quadratic when a=-1:Quadratic becomes x² + (1 - 2*(-1))x + (1 - (-1)) = x² + (1 +2)x + (1 +1) = x² +3x +2. Which factors as (x+1)(x+2). So the cubic becomes (x +1)(x +1)(x +2) = (x+1)^2(x+2). So roots at x=-1 (double) and x=-2. So two distinct real roots. Therefore, when a=-1 or a=1, the cubic equation has two distinct real roots. But in our discriminant condition above, when a is in (-sqrt(3)/2, sqrt(3)/2), the quadratic has no real roots. However, when a=1 or a=-1, those a's are outside the interval (-sqrt(3)/2 ≈-0.866, 0.866). Since sqrt(3)/2 ≈0.866, and 1 and -1 are outside that interval, so the discriminant is positive there, so quadratic has two real roots. But even so, in those cases, even though the quadratic has real roots, one of them coincides with x=a, leading to a multiple root.But in our earlier analysis, the discriminant D =4a² -3. When a=1, D=4*1 -3=1>0, so quadratic has two real roots, but one of them is x=a=1, leading to a double root. Similarly for a=-1.But the problem states that the equation has "exactly one root." If the equation has a double root and another distinct root, then it's two distinct roots. Hence, even though the discriminant is positive, if one of the roots coincides with x=a, then the cubic would have two real roots (one single, one double). Therefore, in order for the cubic to have exactly one real root, the quadratic must have no real roots, which is when D<0 ⇒ |a| < sqrt(3)/2. However, we must also check that when D=0 (i.e., discriminant zero), which is when |a|=sqrt(3)/2. In that case, the quadratic would have a double root. So if a=sqrt(3)/2, then the quadratic would have a double root. But we need to check if that double root is equal to x=a or not.Wait, if a=sqrt(3)/2, then let's check if the quadratic's double root is x=a.Quadratic equation: x² + (1 - 2a)x + (1 -a) =0If discriminant D=0, the quadratic has a double root at x = [-(1 - 2a)]/(2*1) = (2a -1)/2So the double root is x=(2a -1)/2. Let's check if this equals a:(2a -1)/2 =a ⇒ 2a -1 =2a ⇒ -1=0, which is impossible. Therefore, when D=0, the double root of the quadratic is different from x=a. Therefore, when a= sqrt(3)/2, the cubic equation would have two distinct real roots: x=a and the double root from the quadratic. Similarly, when a=-sqrt(3)/2, the quadratic would have a double root, which is different from x=a. Therefore, in those cases, the cubic would have two distinct real roots. Therefore, even when a= sqrt(3)/2 or a=-sqrt(3)/2, the equation has two real roots, so those endpoints are excluded.Therefore, the range of values for 'a' such that the equation has exactly one real root is when the quadratic has no real roots, i.e., when |a| < sqrt(3)/2.But wait, let me confirm with a test value. Let me pick a=0, which is inside the interval. Then the equation factors as (x -0)(x² + (1 -0)x + (1 -0))=x(x² +x +1)=0. The quadratic x² +x +1 has discriminant 1 -4= -3 <0, so indeed only real root is x=0. So that works.Another test: a=1/2. Then sqrt(3)/2≈0.866, so 1/2=0.5 is inside. Quadratic becomes x² + (1 -2*(1/2))x + (1 -1/2)=x² +0x +1/2= x² +1/2=0. Which has roots x=±sqrt(-1/2), imaginary. So the cubic equation has only x=1/2 as a real root. Correct.Another test: a=sqrt(3)/2. Then the quadratic discriminant D=4*(3/4) -3=3 -3=0. Quadratic equation has a double root at x=(2a -1)/2. Let's compute a=sqrt(3)/2:x=(2*(sqrt(3)/2) -1)/2=(sqrt(3)-1)/2≈(1.732 -1)/2≈0.366. So double root at x≈0.366, which is different from a≈0.866. Therefore, the cubic has two real roots: x≈0.866 and x≈0.366 (double). So two distinct real roots. Therefore, a=sqrt(3)/2 is excluded.Similarly, for a=-sqrt(3)/2, quadratic discriminant D=4*(3/4)-3=0. Quadratic equation: x² + (1 -2*(-sqrt(3)/2))x + (1 -(-sqrt(3)/2))=x² + (1 +sqrt(3))x +1 +sqrt(3)/2. The double root is at x=(2a -1)/2= (2*(-sqrt(3)/2)-1)/2= (-sqrt(3)-1)/2≈(-1.732 -1)/2≈-1.366, which is different from a≈-0.866. Therefore, cubic has two distinct real roots. Hence, these endpoints are excluded.Therefore, the range is all real numbers a such that -sqrt(3)/2 < a < sqrt(3)/2.But let me check another edge case. Suppose a=0. The equation is x(x² +x +1)=0, which has one real root x=0. Correct.Another case: a=1/2, which we checked. So yes, seems consistent.Therefore, the answer should be a is in (-sqrt(3)/2, sqrt(3)/2). So written as:The range of values for a is boxed{left( -dfrac{sqrt{3}}{2}, dfrac{sqrt{3}}{2} right)}.Final AnswerThe range of values for ( a ) is boxed{left( -dfrac{sqrt{3}}{2}, dfrac{sqrt{3}}{2} right)}.</think>To determine the range of values for ( a ) such that the given cubic equation has exactly one real root, we start by simplifying the equation:[x^3 + (1 - 3a)x^2 + (2a^2 - 2a + 1)x + (a^2 - a) = 0]We factor out ( x - a ) from the cubic polynomial, knowing that ( x = a ) is a root. This gives us:[(x - a)(x^2 + (1 - 2a)x + (1 - a)) = 0]For the cubic equation to have exactly one real root, the quadratic factor ( x^2 + (1 - 2a)x + (1 - a) ) must have no real roots. This occurs when the discriminant of the quadratic is negative. The discriminant ( D ) of the quadratic ( x^2 + (1 - 2a)x + (1 - a) ) is:[D = (1 - 2a)^2 - 4 cdot 1 cdot (1 - a)]Simplifying the discriminant:[D = 1 - 4a + 4a^2 - 4 + 4a = 4a^2 - 3]For the quadratic to have no real roots, we require ( D < 0 ):[4a^2 - 3 < 0 implies 4a^2 < 3 implies a^2 < frac{3}{4} implies |a| < frac{sqrt{3}}{2}]Thus, the range of values for ( a ) is:[boxed{left( -dfrac{sqrt{3}}{2}, dfrac{sqrt{3}}{2} right)}]]

🔑:Since x^3 + (1-3a)x^2 + 2ax - 2ax + x + a^2 - a = 0, we have x^3 - 3ax^2 + 2a^2x + x^2 - (2a-1)x + a^2 - a = 0, thus x(x-a)(x-2a) + (x-a)[x-(a-1)] = 0, which simplifies to (x-a)[x^2 - (2a-1)x - (a-1)] = 0, Given that the equation x^3 + (1-3a)x^2 + 2a^2x - 2ax + x + a^2 - a = 0 has exactly one root, it implies that x^2 - (2a-1)x - (a-1) = 0 has no solution, Therefore, the discriminant Delta = (2a-1)^2 + 4(a-1) < 0, which leads to 4a^2 - 3 < 0, Thus, boxed{frac{-sqrt{3}}{2} < a < frac{sqrt{3}}{2}}.