🧠:[<think>Alright, let me try to work through this problem step by step. It's about a parabola with some given conditions, and then a geometry problem involving points on the parabola. Let's take it part by part.Part (1): Find the equation of the parabola C.First, the problem states that the parabola has its vertex at the origin, and the focus is on the y-axis. Since the vertex is at (0,0) and the focus is on the y-axis, this parabola must open either upward or downward. The standard form of such a parabola is ( x^2 = 4py ), where p is the distance from the vertex to the focus. If p is positive, it opens upward; if negative, downward.We are given a point Q(a, 2) on the parabola C, and the distance from Q to the focus is 3. Let's use this information to find p.First, since Q(a, 2) is on the parabola, it must satisfy the equation ( a^2 = 4p(2) ), so ( a^2 = 8p ).Next, the focus of the parabola is at (0, p). The distance from Q(a, 2) to the focus (0, p) is given as 3. The distance formula between (a, 2) and (0, p) is:( sqrt{(a - 0)^2 + (2 - p)^2} = 3 )Squaring both sides:( a^2 + (2 - p)^2 = 9 )But from earlier, we know ( a^2 = 8p ). Substitute that into the equation:( 8p + (2 - p)^2 = 9 )Let's expand (2 - p)^2:( 8p + 4 - 4p + p^2 = 9 )Combine like terms:( p^2 + 4p + 4 = 9 )Wait, hold on. Let me check the expansion again. (2 - p)^2 is 4 - 4p + p^2. Then adding 8p:8p + 4 - 4p + p^2 = p^2 + 4p + 4. That's correct.So:( p^2 + 4p + 4 = 9 )Subtract 9:( p^2 + 4p - 5 = 0 )Factor the quadratic equation:Looking for two numbers that multiply to -5 and add to +4. Those numbers are +5 and -1.( (p + 5)(p - 1) = 0 )Therefore, p = -5 or p = 1.But wait, let's recall that the standard parabola equation ( x^2 = 4py ). If p is positive, the parabola opens upward; if negative, downward. However, since the point Q(a, 2) is on the parabola, the y-coordinate is 2. Let's check both possibilities.Case 1: p = 1. Then the equation is ( x^2 = 4(1)y ), so ( x^2 = 4y ). Then for point Q(a, 2), substituting y=2, we have ( a^2 = 8 ), so a is ±2√2. Then the distance from Q to focus (0,1) is √[(2√2)^2 + (2 - 1)^2] = √[8 + 1] = √9 = 3, which matches.Case 2: p = -5. Then the equation is ( x^2 = 4(-5)y ), so ( x^2 = -20y ). For point Q(a, 2), substituting y=2, we get ( a^2 = -40 ). But a^2 can't be negative. So this is impossible. Therefore, p = -5 is invalid.Therefore, the correct value is p = 1, so the equation of the parabola is ( x^2 = 4y ).So part (1) answer is ( x^2 = 4y ).Part (2): Find the minimum area of triangle ABD under given conditions.Alright, this part is more complex. Let me parse the problem again.We have points A(x1, y1) and B(x2, y2) on the parabola C (which is x² = 4y). So, these points satisfy x1² = 4y1 and x2² = 4y2.There exists a point D(x3, y3) on the parabola C, with x3 < x1 < x2. So, the x-coordinate of D is less than that of A, which is less than that of B.Triangle ABD is an isosceles right triangle with angle A being the right angle. We need to find the minimum value of the area of triangle ABD.First, let's recall that in an isosceles right triangle with right angle at A, the legs (from A to B and from A to D) are equal in length, and the hypotenuse is BD, with length √2 times the legs.Moreover, the area of such a triangle is (1/2)*(AB)^2, since each leg is AB, so (1/2)*leg*leg.But let's formalize this.Since angle A is the right angle, vectors AB and AD must be perpendicular and of equal length.Let's express coordinates:Point A: (x1, y1)Point B: (x2, y2)Point D: (x3, y3)All on parabola C: x² = 4y, so y1 = x1²/4, y2 = x2²/4, y3 = x3²/4.Given x3 < x1 < x2.First, vectors:Vector AB = (x2 - x1, y2 - y1) = (x2 - x1, (x2² - x1²)/4)Vector AD = (x3 - x1, y3 - y1) = (x3 - x1, (x3² - x1²)/4)Since angle at A is right angle, AB ⋅ AD = 0.Also, since it's isosceles, |AB| = |AD|.Therefore, two conditions:1. Dot product AB ⋅ AD = 0.2. |AB| = |AD|.Let me write these out.First, the dot product:(x2 - x1)(x3 - x1) + [(x2² - x1²)/4][(x3² - x1²)/4] = 0Second, the magnitudes:√[(x2 - x1)^2 + ((x2² - x1²)/4)^2] = √[(x3 - x1)^2 + ((x3² - x1²)/4)^2]Squaring both sides to remove the square roots:(x2 - x1)^2 + [(x2² - x1²)/4]^2 = (x3 - x1)^2 + [(x3² - x1²)/4]^2This seems complicated. Maybe we can simplify.Alternatively, since y = x²/4, perhaps parameterize the points.Let me parameterize points A, B, D as follows:Let’s let A be (2a, a²), since x² = 4y ⇒ y = x²/4. Let’s set x = 2a, so y = (2a)^2 / 4 = a². Similarly, point B can be (2b, b²), and point D can be (2c, c²). This substitution might simplify the algebra.Given that x3 < x1 < x2, so 2c < 2a < 2b ⇒ c < a < b.So, points:A: (2a, a²)B: (2b, b²)D: (2c, c²)With c < a < b.Now, the vectors AB and AD are:AB = (2b - 2a, b² - a²)AD = (2c - 2a, c² - a²)First condition: AB ⋅ AD = 0.Compute the dot product:(2b - 2a)(2c - 2a) + (b² - a²)(c² - a²) = 0Factor 2 out of the x-components:[2(b - a)][2(c - a)] + (b² - a²)(c² - a²) = 0So 4(b - a)(c - a) + (b² - a²)(c² - a²) = 0Second condition: |AB| = |AD|Compute |AB|² and |AD|²:|AB|² = (2b - 2a)^2 + (b² - a²)^2 = 4(b - a)^2 + (b² - a²)^2|AD|² = (2c - 2a)^2 + (c² - a²)^2 = 4(c - a)^2 + (c² - a²)^2Set them equal:4(b - a)^2 + (b² - a²)^2 = 4(c - a)^2 + (c² - a²)^2Hmm, these equations look quite involved. Let's see if we can factor some terms.First, note that (b² - a²) = (b - a)(b + a), and similarly (c² - a²) = (c - a)(c + a).So let's rewrite the dot product equation:4(b - a)(c - a) + [(b - a)(b + a)][(c - a)(c + a)] = 0Factor out (b - a)(c - a):(b - a)(c - a)[4 + (b + a)(c + a)] = 0Since c < a < b, then (b - a) > 0, (c - a) < 0. So their product is negative. Therefore, the term in brackets must be zero to satisfy the equation:4 + (b + a)(c + a) = 0So:(b + a)(c + a) = -4Similarly, for the second condition:4(b - a)^2 + (b - a)^2(b + a)^2 = 4(c - a)^2 + (c - a)^2(c + a)^2Let’s factor out (b - a)^2 and (c - a)^2:(b - a)^2[4 + (b + a)^2] = (c - a)^2[4 + (c + a)^2]From the first equation, we have (b + a)(c + a) = -4. Let’s denote (b + a) = m, so (c + a) = -4/m. Then:From (b + a) = m and (c + a) = -4/m.But let's also express b and c in terms of m and a.From (b + a) = m ⇒ b = m - aFrom (c + a) = -4/m ⇒ c = -4/m - aTherefore, b and c can be expressed in terms of m and a.Let’s substitute these into the second condition.First, compute (b - a) = (m - a - a) = m - 2aWait, no:Wait, b = m - a, so (b - a) = m - a - a = m - 2aSimilarly, c = -4/m - a, so (c - a) = (-4/m - a) - a = -4/m - 2aThen, let's substitute into the second equation:(b - a)^2[4 + (b + a)^2] = (c - a)^2[4 + (c + a)^2]Substitute (b - a) = m - 2a, (b + a) = mSimilarly, (c - a) = -4/m - 2a, (c + a) = -4/mSo:(m - 2a)^2[4 + m^2] = (-4/m - 2a)^2[4 + (-4/m)^2]Simplify both sides.Left side:(m - 2a)^2 (4 + m²)Right side:[(-4/m - 2a)^2] [4 + 16/m²]First, compute (-4/m - 2a)^2. Let's factor out the negative sign:(-4/m - 2a)^2 = (4/m + 2a)^2 = [2a + 4/m]^2 = 4a² + 16a/m + 16/m²Similarly, 4 + 16/m² = 4 + (16)/m²Therefore, right side becomes:[4a² + 16a/m + 16/m²] [4 + 16/m²]This is getting quite complicated. Maybe we need a smarter approach.Alternatively, let's recall that we have from the first equation:(b + a)(c + a) = -4But also, perhaps we can find a relation between b and c.Alternatively, let's try to express everything in terms of m, as we started.Let’s set m = b + a. Then c + a = -4/m.So c = -4/m - a.Then, we can express (c - a) = -4/m - 2a.So, we can write the second equation as:(m - 2a)^2 [4 + m^2] = (-4/m - 2a)^2 [4 + ( -4/m)^2]But (-4/m)^2 = 16/m², so 4 + 16/m².Let’s compute both sides.Left side:(m - 2a)^2 (4 + m^2) = [m² -4a m +4a²][4 + m²]Right side:[(-4/m - 2a)^2][4 + 16/m²] = [16/m² + 16a/m +4a²][4 + 16/m²]Hmm. This seems messy. Maybe instead of trying to work with variables m and a, we can look for a substitution or consider a as a parameter.Alternatively, perhaps there's a symmetry or substitution that can simplify this.Wait, maybe instead of parameterizing the points as (2a, a²), which introduces the 2, we can use a different parameter. Let me check:Original parabola is x² = 4y. So, if we parameterize using t where x = 2t, then y = (2t)^2 /4 = t². So points can be represented as (2t, t²). So, that's the same as before, with t = a, b, c. So maybe that's the standard way.Alternatively, let's use parameter t where x = t, then y = t²/4. Then points are (t, t²/4). Maybe that's simpler? Let's try.Let’s define points:A: (t1, t1²/4)B: (t2, t2²/4)D: (t3, t3²/4)With t3 < t1 < t2.Then vectors AB and AD:AB = (t2 - t1, (t2² - t1²)/4)AD = (t3 - t1, (t3² - t1²)/4)First condition: AB ⋅ AD = 0.So:(t2 - t1)(t3 - t1) + [(t2² - t1²)/4][(t3² - t1²)/4] = 0Second condition: |AB| = |AD|So:√[(t2 - t1)^2 + ((t2² - t1²)/4)^2] = √[(t3 - t1)^2 + ((t3² - t1²)/4)^2]Squaring both sides:(t2 - t1)^2 + [(t2² - t1²)/4]^2 = (t3 - t1)^2 + [(t3² - t1²)/4]^2This seems similar to before. Let's factor.Note that t2² - t1² = (t2 - t1)(t2 + t1)Similarly, t3² - t1² = (t3 - t1)(t3 + t1)So let's substitute into the first equation:(t2 - t1)(t3 - t1) + [(t2 - t1)(t2 + t1)/4][(t3 - t1)(t3 + t1)/4] = 0Factor out (t2 - t1)(t3 - t1):(t2 - t1)(t3 - t1)[1 + (t2 + t1)(t3 + t1)/16] = 0Since t2 > t1 and t3 < t1, (t2 - t1) > 0 and (t3 - t1) < 0, so their product is negative. Therefore, the term in brackets must be zero:1 + (t2 + t1)(t3 + t1)/16 = 0So:(t2 + t1)(t3 + t1) = -16Similarly, in the second condition:(t2 - t1)^2 + [(t2 - t1)(t2 + t1)/4]^2 = (t3 - t1)^2 + [(t3 - t1)(t3 + t1)/4]^2Factor out (t2 - t1)^2 and (t3 - t1)^2:(t2 - t1)^2[1 + (t2 + t1)^2/16] = (t3 - t1)^2[1 + (t3 + t1)^2/16]Let me denote (t2 - t1) = p, (t3 - t1) = q. Then since t2 > t1 and t3 < t1, p > 0 and q < 0.From the first equation:(t2 + t1)(t3 + t1) = -16But t2 = t1 + p, t3 = t1 + q.So substituting:(t1 + p + t1)(t1 + q + t1) = -16(2t1 + p)(2t1 + q) = -16Similarly, the second equation:p²[1 + ( (t1 + p) + t1 )² /16 ] = q²[1 + ( (t1 + q) + t1 )² /16 ]Simplify inside the brackets:(2t1 + p)² /16 and (2t1 + q)^2 /16.So:p²[1 + (2t1 + p)^2 /16] = q²[1 + (2t1 + q)^2 /16 ]But from the first equation, (2t1 + p)(2t1 + q) = -16. Let's denote S = 2t1 + p and T = 2t1 + q, so ST = -16.Then the equation becomes:p²[1 + S² /16] = q²[1 + T² /16 ]But since S * T = -16, perhaps we can express T = -16/S.So substituting T = -16/S:p²[1 + S² /16] = q²[1 + (256 / S²)/16 ]Simplify:p²[1 + S² /16] = q²[1 + 16 / S² ]But we need to relate p, q, and S.From S = 2t1 + p and T = 2t1 + q, and T = -16/S.So:2t1 + q = -16/SBut S = 2t1 + p ⇒ 2t1 = S - pSubstitute into above:(S - p) + q = -16/STherefore:q = -16/S - (S - p) = -16/S - S + pBut also, p = t2 - t1, q = t3 - t1.This is getting too abstract. Maybe another approach.Alternatively, let's use the first equation:(2t1 + p)(2t1 + q) = -16Let’s denote u = 2t1. Then:(u + p)(u + q) = -16Expand:u² + u(p + q) + pq = -16But perhaps this is not helpful.Alternatively, let's consider that in the first equation, (t2 + t1)(t3 + t1) = -16. Let's denote t1 as fixed, and express t2 and t3 in terms of t1.But maybe we need to find a relation between t2 and t3. Let's suppose that given t1, we can express t2 and t3 accordingly.Alternatively, let's think of the problem as follows: for a given point A(t1), we can find points B(t2) and D(t3) such that triangle ABD is isosceles right-angled at A. We need to find the minimum area over all possible such triangles.The area of triangle ABD is (1/2)*|AB|^2, since it's an isosceles right-angled triangle with legs AB and AD.Wait, since |AB| = |AD|, and the area is (1/2)*|AB|*|AD| = (1/2)*|AB|².Therefore, minimizing the area is equivalent to minimizing |AB|².Therefore, we need to find the minimum value of |AB|², given the constraints.Thus, if we can express |AB|² in terms of variables and then minimize it, that would give the minimal area.So, let's try to find |AB|².From earlier, |AB|² = (t2 - t1)^2 + [(t2² - t1²)/4]^2Factor (t2 - t1)^2:= (t2 - t1)^2 [1 + (t2 + t1)^2 /16 ]Similarly, from the first equation, (t2 + t1)(t3 + t1) = -16.So, (t3 + t1) = -16 / (t2 + t1)Therefore, t3 = -16 / (t2 + t1) - t1Let’s substitute t3 into the expression for |AD|², which should equal |AB|².But since |AB| = |AD|, and we are to minimize |AB|², perhaps we can express |AB|² in terms of t2 and t1, then find its minimum.Alternatively, let's express everything in terms of t2 and t1.Let me set s = t2 + t1. Then, from the first equation, t3 + t1 = -16 / s, so t3 = -16/s - t1.Then, t2 = s - t1.Therefore, t2 - t1 = s - 2t1.Similarly, (t2 + t1) = s.Thus, |AB|² = (s - 2t1)^2 [1 + s² /16 ]But we need to express this in terms of s and t1.But we also have t3 = -16/s - t1.We need to ensure that t3 < t1 < t2.Given that t3 < t1 < t2, and t2 = s - t1.So, t3 = -16/s - t1 < t1 ⇒ -16/s - t1 < t1 ⇒ -16/s < 2t1 ⇒ -8/s < t1.Also, t1 < t2 = s - t1 ⇒ t1 < s - t1 ⇒ 2t1 < s ⇒ t1 < s/2.Therefore, combining the inequalities:-8/s < t1 < s/2.Moreover, since t3 must be a real number, s ≠ 0.But we need to see the possible values of s.Alternatively, maybe we can set t1 as a variable and s as another variable, but this might not simplify things.Alternatively, perhaps express t1 in terms of s.Alternatively, let's consider choosing s as a parameter and express t1 in terms of s such that the inequalities hold.But this seems complicated. Maybe another approach.Let’s try to use calculus to minimize |AB|².Express |AB|² as a function of s and t1.But with the constraints.Alternatively, perhaps fix t1 and express |AB|² in terms of s.Wait, s = t2 + t1, and t3 = -16/s - t1.But t2 = s - t1.Therefore, in terms of s and t1:|AB|² = (s - 2t1)^2 [1 + s² /16 ]But we need to ensure that t3 < t1 < t2.Given t3 = -16/s - t1 < t1.So:-16/s - t1 < t1 ⇒ -16/s < 2t1 ⇒ t1 > -8/s.And t1 < t2 = s - t1 ⇒ t1 < s/2.Therefore, the domain of t1 is (-8/s, s/2). Since s is t2 + t1, and t2 > t1, s = t2 + t1 > t1 + t1 = 2t1. Therefore, s > 2t1 ⇒ t1 < s/2, which is already considered.But s is related to t1 and t2. Perhaps it's better to consider s as a variable independent of t1, but that complicates things.Alternatively, suppose we set u = t1. Then, we have:s = t2 + ut3 = -16/s - uBut t2 = s - u.So, |AB|² = (s - 2u)^2 [1 + s² /16 ]We need to express this in terms of s and u, with the constraints:-8/s < u < s/2But how do we relate u and s? If s is given, then u is constrained between -8/s and s/2. But s can be any real number except zero. However, s = t2 + u, and since t2 > u, s = t2 + u > u + u = 2u. But since u < s/2 ⇒ 2u < s. Therefore, consistent.Alternatively, perhaps treat u as a function of s, or vice versa.Alternatively, let's make a substitution to reduce variables. Let’s set v = u/s. Then u = v s, where v is a scalar.Given that u < s/2 ⇒ v s < s/2 ⇒ v < 1/2 (if s > 0). If s < 0, the inequality would reverse, but let's consider s > 0 first.Similarly, from -8/s < u ⇒ -8/s < v s ⇒ -8 < v s². Since s > 0, this becomes -8/s² < v. But if s > 0 and v < 1/2, but also -8/s² < v. So v is in (-8/s², 1/2).But this substitution might not help much.Alternatively, consider that s and u are variables. Perhaps take partial derivatives with respect to u and s to find minima. But this might be complicated.Alternatively, set s as a parameter and find the minimal |AB|² over u for each s, then minimize over s.But this is getting too involved. Maybe try specific values.Alternatively, consider that the expression for |AB|² is (s - 2u)^2 (1 + s² /16 ). To minimize this, we can see for a given s, the minimal value occurs when (s - 2u)^2 is minimized. Since (s - 2u) can vary depending on u. The term (s - 2u) is linear in u. The minimal value of (s - 2u)^2 occurs when derivative with respect to u is zero. But within the domain of u.But u is in (-8/s, s/2). Let's consider (s - 2u)^2 as a function of u. Expanding, it's quadratic in u with minimum at u = s/2. But u < s/2, so the closer u is to s/2, the smaller (s - 2u)^2 becomes. However, u must be less than s/2, so the minimal value is approached as u approaches s/2 from below. But then (s - 2u) approaches zero. Therefore, the minimal |AB|² for a given s is approaching zero. But this can't be, because when u approaches s/2, t1 approaches s/2, and t2 = s - u approaches s/2. So points A and B would coincide, which isn't allowed.Wait, but the problem states that AB is a segment with endpoints A and B on the parabola. So, A and B must be distinct points, hence t2 > t1. If u approaches s/2, then t2 = s - u approaches s - s/2 = s/2, so t1 approaches s/2 from below. Thus, t2 approaches s/2 from above. So A and B approach each other.But in that case, the area approaches zero. But the problem states "there exists a point D", so perhaps the minimum area is zero? But the problem says "find the minimum value of the area", which might not be zero because maybe such a triangle can't have zero area. Or perhaps there's a lower bound.But this approach might not be correct. Alternatively, maybe there's a specific configuration where the area is minimized.Alternatively, think geometrically. The parabola is x² = 4y. Points A, B, D are on the parabola. Triangle ABD is a right isosceles triangle with right angle at A. So, AB and AD are equal in length and perpendicular.Given that all points are on the parabola, maybe we can use reflection properties or geometric properties of parabolas.Alternatively, use coordinates. Suppose point A is (2a, a²). Then, vectors AB and AD must satisfy the perpendicular and equal length conditions.Let’s attempt to write these conditions.Let’s let A = (2a, a²). Then, points B and D are also on the parabola: B = (2b, b²), D = (2c, c²), with c < a < b.Vector AB = (2(b - a), b² - a²)Vector AD = (2(c - a), c² - a²)Condition 1: AB ⋅ AD = 0So:2(b - a)*2(c - a) + (b² - a²)(c² - a²) = 0Which simplifies to:4(b - a)(c - a) + (b² - a²)(c² - a²) = 0Factor:(b - a)(c - a)[4 + (b + a)(c + a)] = 0As before, leading to (b + a)(c + a) = -4.Condition 2: |AB| = |AD|Which gives:√[4(b - a)^2 + (b² - a²)^2] = √[4(c - a)^2 + (c² - a²)^2]Square both sides:4(b - a)^2 + (b² - a²)^2 = 4(c - a)^2 + (c² - a²)^2Let’s divide both sides by 4:(b - a)^2 + [(b² - a²)^2]/4 = (c - a)^2 + [(c² - a²)^2]/4But since (b² - a²) = (b - a)(b + a) and similarly for c:= (b - a)^2 + [(b - a)^2(b + a)^2]/4= (b - a)^2 [1 + (b + a)^2 /4]Similarly for c.Therefore, we have:(b - a)^2 [1 + (b + a)^2 /4 ] = (c - a)^2 [1 + (c + a)^2 /4 ]From condition 1, (b + a)(c + a) = -4. Let’s denote (b + a) = m ⇒ (c + a) = -4/m.Then, substitute into the above equation:(b - a)^2 [1 + m² /4 ] = (c - a)^2 [1 + (16/m²)/4 ]Simplify:(b - a)^2 [1 + m² /4 ] = (c - a)^2 [1 + 4/m² ]But we need to express (b - a) and (c - a) in terms of m.Since b + a = m ⇒ b = m - aSo, (b - a) = m - 2aSimilarly, c + a = -4/m ⇒ c = -4/m - aTherefore, (c - a) = -4/m - 2aSubstitute into the equation:(m - 2a)^2 [1 + m²/4 ] = (-4/m - 2a)^2 [1 + 4/m² ]Let’s compute each side.Left side:(m - 2a)^2 (1 + m²/4 )Right side:(-4/m - 2a)^2 (1 + 4/m² )First, compute (-4/m - 2a)^2:= (4/m + 2a)^2 (since squared, negative doesn't matter)= (2a + 4/m)^2= 4a² + 16a/m + 16/m²Similarly, 1 + 4/m² = (m² + 4)/m²Therefore, right side becomes:(4a² + 16a/m + 16/m²) * (m² + 4)/m²= [4a² + 16a/m + 16/m²] * (m² + 4)/m²Similarly, left side:(m - 2a)^2 (1 + m²/4 )Compute (m - 2a)^2 = m² - 4a m + 4a²Then multiplied by (1 + m²/4 ):= (m² - 4a m + 4a²)(1 + m²/4 )Expand this:= m²(1) + m²(m²/4) - 4a m(1) -4a m(m²/4) + 4a²(1) + 4a²(m²/4 )= m² + m^4/4 -4a m - a m^3 + 4a² + a² m²Now, the right side:[4a² + 16a/m + 16/m²] * (m² + 4)/m²First, expand the numerator:(4a²)(m² + 4) + (16a/m)(m² + 4) + (16/m²)(m² + 4)= 4a² m² + 16a² + 16a m + 64a/m + 16 + 64/m²Then divide by m²:= [4a² m² + 16a² + 16a m + 64a/m + 16 + 64/m²] / m²= 4a² + 16a²/m² + 16a/m + 64a/m^3 + 16/m² + 64/m^4So, equating left side and right side:Left side:m² + m^4/4 -4a m - a m^3 + 4a² + a² m²Right side:4a² + 16a²/m² + 16a/m + 64a/m^3 + 16/m² + 64/m^4This seems extremely complex. Perhaps there is a symmetry or substitution that can simplify this.Alternatively, let's assume that a = 0. Let's see if that's possible.If a = 0, then point A is (0, 0), the vertex. Then:From condition (b + 0)(c + 0) = -4 ⇒ b * c = -4From condition 2:(m - 0)^2 [1 + m²/4 ] = (c - 0)^2 [1 + c²/4 ]But m = b + 0 = b. So:b² (1 + b²/4 ) = c² (1 + c²/4 )But b * c = -4 ⇒ c = -4/bSubstitute c = -4/b into the equation:b² (1 + b²/4 ) = (16/b²) (1 + (16/b²)/4 )Simplify:b² (1 + b²/4 ) = (16/b²)(1 + 4/b² )Multiply both sides by b²:b^4 (1 + b²/4 ) = 16 (1 + 4/b² )Expand left side:b^4 + b^6/4Right side:16 + 64/b²Multiply both sides by 4 to eliminate denominators:4b^6 + b^8 = 64 + 256/b²Multiply both sides by b²:4b^8 + b^{10} = 64b² + 256Rearranged:b^{10} +4b^8 -64b² -256 =0This is a tenth-degree equation, which is very difficult to solve. So, a =0 might not be a good assumption.Alternatively, maybe there's a case where a is not zero but the equation simplifies.Alternatively, maybe take specific values. Suppose m = 2. Then (b + a) = 2, and (c + a) = -4/2 = -2. So c = -2 -a.Then, b = 2 - a.Compute (b - a) = 2 - 2a.Compute (c - a) = -2 - 2a.Left side:(2 - 2a)^2 [1 + (2)^2 /4 ] = (4 -8a +4a²)(1 + 1) = (4 -8a +4a²)(2) = 8 -16a +8a²Right side:(-2 - 2a)^2 [1 + (-2)^2 /4 ] = (4 +8a +4a²)(1 + 1) = (4 +8a +4a²)(2) =8 +16a +8a²Set equal:8 -16a +8a² =8 +16a +8a²Simplify:-16a =16a ⇒ -32a =0 ⇒ a=0.So, when m=2, a=0. Then b=2-0=2, c=-2-0=-2.Points:A: (0,0)B: (4, 4) (since x=2b=4, y=b²=4)D: (-4, 4) (x=2c=-4, y=c²=4)Then vectors AB=(4,4), AD=(-4,4)Check if AB ⋅ AD = 4*(-4) +4*4= -16 +16=0 ✔️Check |AB|=√(16+16)=√32, |AD|=√(16+16)=√32 ✔️Area= (1/2)*(√32)^2=(1/2)*32=16.So, area 16.Is this the minimal area? Let's check another m.Suppose m=4. Then c + a = -1.Let’s go through similar steps.But this might take forever. Alternatively, maybe the minimal area is 16, achieved at a=0, b=2, c=-2.But let's check another case. Let's pick m= sqrt( something ). Wait, when a=0, the area is 16. Let's see if there's a case where a≠0 with smaller area.Suppose a=1. Then, (b +1)(c +1) = -4.Assume b +1 = 2 ⇒ b=1. Then c +1 = -2 ⇒ c=-3.Check if this works.So, b=1, c=-3.Points:A: (2*1,1²)=(2,1)B: (2*1,1²)=(2,1) Wait, no. Wait, if a=1, then A is (2a, a²)=(2,1). Then b=1, so point B is (2*1,1²)=(2,1). Same as A. Not allowed.Oops, mistake. If a=1, then (b +1)(c +1) = -4. Let's pick b +1=4, then c +1= -1 ⇒ b=3, c=-2.Then:A: (2,1)B: (6,9)D: (-4,4)Check vectors AB=(6-2,9-1)=(4,8)AD=(-4-2,4-1)=(-6,3)Check AB ⋅ AD=4*(-6) +8*3= -24 +24=0 ✔️Check |AB|=√(16+64)=√80=4√5|AD|=√(36 +9)=√45=3√5But 4√5 ≠3√5. So, this does not satisfy |AB|=|AD|. So invalid.Alternatively, maybe other values.Alternatively, let's find m and a such that the equations are satisfied.But this seems too time-consuming. Perhaps 16 is indeed the minimal area.Alternatively, suppose a≠0 but find a case where the area is smaller.Let’s try with m= something else.Suppose m= sqrt(8). Then (b +a)=sqrt(8), (c +a)= -4/sqrt(8)= -sqrt(2)/2.But this might not help.Alternatively, let's use calculus.Let’s denote that from the first condition, we have (b + a)(c + a) = -4, so c = -4/(b + a) -a.Then, we can write the area in terms of a and b.The area is (1/2)*|AB|².Compute |AB|² = (2(b -a))² + (b² -a²)^2 = 4(b -a)^2 + (b² -a²)^2.Express this in terms of b and a.But since c is dependent on a and b via c = -4/(b + a) -a, we need to ensure that t3 = c < a < b.But we can consider the area as a function of variables a and b, with the constraint (b +a)(c +a) = -4.This is quite complex, but perhaps by substituting c, we can express the area solely in terms of a and b.But even then, taking derivatives would be complicated.Alternatively, notice that when a=0, we get area 16, and when a increases, the area might increase. Let me test with a=1.Wait, when a=1, from earlier, with b +1=2, c=-2 -1=-3. Then points A=(2,1), B=(2b, b²), c=-3. But we need to solve for b.From (b +1)(c +1)= -4, if a=1, then c = -4/(b +1) -1.So, pick a=1, then:c = -4/(b +1) -1And c < a=1 < b.So, c <1 <b.But c= -4/(b +1) -1. Let's solve for b.Let’s suppose b=3. Then c= -4/4 -1= -1 -1= -2. So c=-2 <1 <3.Then compute |AB|².Point A=(2,1), B=(6,9), D=(-4,4).AB vector=(4,8), |AB|²=16+64=80.Area= (1/2)*80=40, which is larger than 16.If we take a=0.5, then:From (b +0.5)(c +0.5)= -4.Let’s assume b +0.5=8 ⇒ b=7.5. Then c +0.5= -4/8= -0.5 ⇒ c= -1.Points A=(1,0.25), B=(15,56.25), D=(-2,1).Compute AB vector=(14,56.0), |AB|²=14² +56²=196 +3136=3332. Area=3332/2=1666, which is way bigger.Alternatively, take b +a= m, c +a= -4/m.Assume m is small. Let’s take m=1. Then c +a= -4/1= -4.Let’s set a=0. Then b=1, c=-4 -0= -4.Points:A=(0,0), B=(2,1), D=(-8,16)AB vector=(2,1), |AB|²=4+1=5. Area=5/2=2.5. Wait, but is this valid?Wait, check if angle at A is right angle.AB=(2,1), AD=(-8,16)Dot product: 2*(-8) +1*16= -16 +16=0. ✔️Check |AB| and |AD|.|AB|=√(4 +1)=√5.|AD|=√(64 +256)=√320=8√5.Not equal. So, not an isosceles triangle. Therefore, invalid.Wait, so this case doesn't satisfy the second condition.Thus, my previous assumption was wrong. When a=0, m=1, we do not have |AB|=|AD|.Earlier, when m=2, a=0, we had |AB|=|AD|=√32, which worked. So that's a valid case. So maybe that's the minimal case.But when m=2, a=0, the area is 16. If there's a case where a≠0 with a smaller area, then 16 isn't the minimum. But so far, I can't find one.Alternatively, perhaps 16 is the minimal area.Alternatively, let's consider using Lagrange multipliers to minimize |AB|² with the given constraints.But this is getting too advanced for an initial problem-solving approach. Given that when a=0, we can achieve area 16, and other configurations give larger areas, perhaps 16 is indeed the minimal area.But let's confirm.Suppose we have points A(0,0), B(4,4), D(-4,4). The triangle ABD.Coordinates:A(0,0), B(4,4), D(-4,4)AB vector: (4,4), AD vector: (-4,4)They are perpendicular, and equal in length. The legs are each √(32), so area is 1/2*(√32)^2=1/2*32=16.This seems to satisfy all conditions. Also, D(-4,4) is on the parabola x²=4y, since (-4)^2=16=4*4=16.Yes, valid.Is there a triangle with smaller area? Let's think.Suppose we take points closer to the vertex. For example, A at (2,1), B and D somewhere else.But earlier attempts showed that areas get larger. For example, A(2,1), B( something ), D( something ). The area was 40 or larger.Alternatively, take A closer to the vertex but not at the vertex.Let’s try A(1,0.25). Then, suppose we have vectors AB and AD perpendicular and equal.Let’s suppose AB = (h, k), AD = (-k, h) to be perpendicular and same length.But since points B and D are on the parabola.So, point B = A + (h, k) = (1 + h, 0.25 + k)This must satisfy (1 + h)^2 =4*(0.25 + k)Similarly, point D = A + (-k, h) = (1 - k, 0.25 + h)This must satisfy (1 - k)^2 =4*(0.25 + h)Also, |AB|² = h² +k² = |AD|² =k² +h², which is equal.So, we have two equations:(1 + h)^2 =4*(0.25 + k) ⇒1 +2h +h²=1 +4k ⇒2h +h²=4kAnd(1 -k)^2 =4*(0.25 +h) ⇒1 -2k +k²=1 +4h ⇒-2k +k²=4hNow, from first equation: 4k=2h +h² ⇒k=(2h +h²)/4Substitute into second equation:-2*( (2h +h²)/4 ) + ( (2h +h²)/4 )² =4hSimplify:- ( (4h +2h²)/4 ) + ( (2h +h²)^2 )/16 =4h= - (h +0.5h²) + (4h² +4h^3 +h^4)/16 =4hMultiply everything by 16 to eliminate denominators:-16h -8h² +4h² +4h^3 +h^4 =64hSimplify:h^4 +4h^3 -4h² -16h -64h=0h^4 +4h^3 -4h² -80h=0Factor:h(h^3 +4h² -4h -80)=0So, h=0 is a solution, but h=0 would mean AB=(0,k), which would imply point B=(1,0.25+k), but then since B is on parabola, (1)^2=4*(0.25 +k) ⇒1=1 +4k ⇒k=0. So points A and B coincide. Not allowed.Other solutions come from h^3 +4h² -4h -80=0.Try h=4: 64 +64 -16 -80= 12≠0h=3:27 +36 -12 -80= -29≠0h=5:125 +100 -20 -80=125≠0h= -4: -64 +64 +16 -80= -64≠0h=2:8 +16 -8 -80= -64≠0h=1:1 +4 -4 -80= -79≠0Not factorable easily. So, no real solutions here? Maybe this cubic has one real solution.Using rational root theorem, possible roots are ±1,±2,±4,±5,±8,±10,±16,±20,±40,±80. None seem to work.Therefore, no real solutions except h=0. Therefore, no such triangle exists with A at (1,0.25). Therefore, the minimal area might indeed be 16.Therefore, after all these trials and considerations, the minimal area is 16.Final Answer(1) The equation of the parabola is boxed{x^2 = 4y}.(2) The minimum value of the area of triangle ABD is boxed{16}.</think>Given a parabola ( C ) with its vertex at the origin and its focus on the ( y )-axis, we need to find the equation of the parabola and the minimum value of the area of triangle ( triangle ABD ) under certain conditions.(1) Find the equation of the parabola ( C ):1. The parabola has its vertex at the origin and focus on the ( y )-axis, so its equation is ( x^2 = 4py ).2. A point ( Q(a, 2) ) on the parabola satisfies ( a^2 = 8p ).3. The distance from ( Q(a, 2) ) to the focus ( (0, p) ) is 3. Using the distance formula: [ sqrt{a^2 + (2 - p)^2} = 3 ]4. Substituting ( a^2 = 8p ) into the distance equation: [ 8p + (2 - p)^2 = 9 ]5. Solving the quadratic equation: [ p^2 + 4p - 5 = 0 implies (p + 5)(p - 1) = 0 ]6. Since ( p = -5 ) leads to a contradiction, we use ( p = 1 ). Thus, the equation is ( x^2 = 4y ).The equation of the parabola ( C ) is (boxed{x^2 = 4y}).(2) Find the minimum value of the area of ( triangle ABD ):1. Points ( A ), ( B ), and ( D ) are on the parabola ( x^2 = 4y ).2. The triangle ( triangle ABD ) is an isosceles right triangle with the right angle at ( A ).3. Using parameterization, let ( A = (2a, a^2) ), ( B = (2b, b^2) ), and ( D = (2c, c^2) ).4. Vectors ( overrightarrow{AB} ) and ( overrightarrow{AD} ) must be perpendicular and equal in length.5. Solving the conditions for perpendicularity and equal length, we find that the minimum area occurs when ( A ) is at the vertex ((0, 0)), ( B = (4, 4) ), and ( D = (-4, 4) ).6. The area of the triangle is calculated as: [ text{Area} = frac{1}{2} times |AB|^2 = frac{1}{2} times 32 = 16 ]The minimum value of the area of ( triangle ABD ) is (boxed{16}).,<think>Okay, let's tackle this problem step by step. First, part (1) asks for the equation of the parabola C. The vertex is at the origin, and the focus is on the y-axis. So, since the vertex is at (0,0) and the focus is on the y-axis, the parabola must open either upwards or downwards. The standard form for such a parabola is ( x^2 = 4py ), where p is the distance from the vertex to the focus. If p is positive, it opens upwards; if negative, downwards.We are given a point Q(a, 2) on the parabola C. Since Q is on the parabola, its coordinates must satisfy the equation ( a^2 = 4p(2) ), so ( a^2 = 8p ).Additionally, the distance from Q to the focus is 3. The focus is at (0, p) because for the parabola ( x^2 = 4py ), the focus is at (0, p). So, the distance between Q(a, 2) and the focus (0, p) is given by the distance formula:( sqrt{(a - 0)^2 + (2 - p)^2} = 3 )Squaring both sides:( a^2 + (2 - p)^2 = 9 )But we already know that ( a^2 = 8p ), so substitute that into the equation:( 8p + (2 - p)^2 = 9 )Let's expand (2 - p)^2:( 8p + 4 - 4p + p^2 = 9 )Combine like terms:( p^2 + 4p + 4 = 9 )Wait, that seems off. Let me check:Wait, 8p + (4 - 4p + p^2) = 9So 8p + 4 -4p + p^2 = 9Combine 8p -4p = 4p, so:p^2 + 4p + 4 = 9Yes, that's correct.Then, subtract 9:p^2 + 4p + 4 -9 = 0 → p^2 + 4p -5 = 0Factor this quadratic equation:Looking for two numbers that multiply to -5 and add to 4. That would be 5 and -1.So, (p + 5)(p -1) = 0Thus, p = -5 or p = 1.But since the parabola has vertex at the origin and the focus is on the y-axis, if p is positive, the parabola opens upwards, if negative, downwards. Now, let's check if both solutions are valid.First, if p = 1:Then, the equation of the parabola is ( x^2 = 4(1)y = 4y ). Then, point Q(a,2) is on it, so ( a^2 = 8(1) = 8 ). Then, the distance from Q(±√8, 2) to focus (0,1) is √[(√8)^2 + (2 -1)^2] = √(8 +1) = √9 = 3, which matches the given condition. So p=1 is valid.If p = -5:The equation of the parabola is ( x^2 = 4*(-5)y = -20y ). Then, point Q(a, 2) is on the parabola, so ( a^2 = -20*2 = -40 ). But ( a^2 ) can't be negative. So this is invalid. Therefore, p=1 is the only valid solution.Therefore, the equation of the parabola is ( x^2 = 4y ).So part (1) answer is ( x^2 = 4y ).Moving to part (2). The problem states that there's a point D(x3, y3) on the parabola C, with x3 < x1 < x2. So points A(x1,y1) and B(x2,y2) are on the parabola, and D is another point on the parabola with x-coordinate less than x1. The triangle ABD is an isosceles right triangle with angle A being the right angle. We need to find the minimum value of the area of triangle ABD.First, since all points are on the parabola C: ( x^2 = 4y ). So for any point (x, y) on C, y = x²/4.So points A, B, D can be parameterized as (x1, x1²/4), (x2, x2²/4), (x3, x3²/4) respectively.Given that triangle ABD is an isosceles right triangle with right angle at A. So angle at A is 90 degrees, and two legs from A (i.e., AB and AD) should be equal in length if it's isosceles. Wait, but in a right-angled triangle, the two legs are the sides forming the right angle. So since angle A is the right angle, the legs are AB and AD, and the hypotenuse is BD. If it's isosceles, then legs AB and AD must be equal. Therefore, AB = AD.So the triangle is right-angled and isosceles at A, so AB = AD, and angle at A is 90 degrees.Therefore, we need two conditions:1. Vectors AB and AD are perpendicular.2. The lengths of AB and AD are equal.But perhaps it's easier to use coordinate geometry. Let's denote points:A: (x1, x1²/4)B: (x2, x2²/4)D: (x3, x3²/4)Given that x3 < x1 < x2.First, the coordinates:Vector AB is (x2 - x1, x2²/4 - x1²/4) = (x2 - x1, (x2² - x1²)/4)Vector AD is (x3 - x1, x3²/4 - x1²/4) = (x3 - x1, (x3² - x1²)/4)Since angle at A is right angle, vectors AB and AD must be perpendicular. Therefore, their dot product is zero:AB ⋅ AD = 0So,(x2 - x1)(x3 - x1) + [(x2² - x1²)/4][(x3² - x1²)/4] = 0Also, since the triangle is isosceles with legs AB and AD equal in length:|AB| = |AD|So,√[(x2 - x1)^2 + ((x2² - x1²)/4)^2] = √[(x3 - x1)^2 + ((x3² - x1²)/4)^2]Squaring both sides:(x2 - x1)^2 + [(x2² - x1²)/4]^2 = (x3 - x1)^2 + [(x3² - x1²)/4]^2So these are the two equations we need to satisfy.This looks quite complicated. Maybe we can simplify by parametrizing the points.Alternatively, since all points are on the parabola, perhaps we can use parameters t1, t2, t3 such that:A: (2t1, t1²)B: (2t2, t2²)D: (2t3, t3²)Wait, since the parabola is ( x^2 = 4y ), so for a general parameterization, we can let x = 2t, then y = (2t)^2 /4 = t². So parametric equations: (2t, t²). So maybe using parameters t1, t2, t3 for points A, B, D respectively. Then x1 = 2t1, y1 = t1²; x2 = 2t2, y2 = t2²; x3 = 2t3, y3 = t3². Given that x3 < x1 < x2, so 2t3 < 2t1 < 2t2 → t3 < t1 < t2.Then, vectors AB and AD:AB = (2t2 - 2t1, t2² - t1²) = (2(t2 - t1), (t2 - t1)(t2 + t1))AD = (2t3 - 2t1, t3² - t1²) = (2(t3 - t1), (t3 - t1)(t3 + t1))Dot product of AB and AD must be zero:[2(t2 - t1)][2(t3 - t1)] + [(t2 - t1)(t2 + t1)][(t3 - t1)(t3 + t1)] = 0Factor out (t2 - t1)(t3 - t1):(t2 - t1)(t3 - t1)[4 + (t2 + t1)(t3 + t1)] = 0Since t2 > t1 and t3 < t1, (t2 - t1) >0, (t3 - t1) <0, so the product (t2 - t1)(t3 - t1) ≠ 0. Therefore, the term in brackets must be zero:4 + (t2 + t1)(t3 + t1) = 0So,(t2 + t1)(t3 + t1) = -4Also, the lengths of AB and AD must be equal:|AB| = |AD|Compute |AB|:√[ (2(t2 - t1))^2 + ( (t2² - t1²) )^2 ]= √[4(t2 - t1)^2 + (t2 - t1)^2(t2 + t1)^2 ]= |t2 - t1| √[4 + (t2 + t1)^2 ]Similarly, |AD|:√[ (2(t3 - t1))^2 + ( (t3² - t1²) )^2 ]= √[4(t3 - t1)^2 + (t3 - t1)^2(t3 + t1)^2 ]= |t3 - t1| √[4 + (t3 + t1)^2 ]Since |AB| = |AD|:|t2 - t1| √[4 + (t2 + t1)^2 ] = |t3 - t1| √[4 + (t3 + t1)^2 ]But t2 - t1 >0, t3 - t1 <0, so |t2 - t1| = t2 - t1 and |t3 - t1| = t1 - t3Therefore:(t2 - t1) √[4 + (t2 + t1)^2 ] = (t1 - t3) √[4 + (t3 + t1)^2 ]Let’s denote s = t1 + t2 and q = t1 + t3. But maybe not. Alternatively, let's let’s denote variables.Let’s denote u = t2 + t1 and v = t3 + t1.From the previous equation: (t2 + t1)(t3 + t1) = -4 → u*v = -4Also, note that t2 = u - t1, and t3 = v - t1.But maybe substituting variables is messy. Let's see.Alternatively, from the first equation:u*v = -4.From the length equation:(t2 - t1) √(4 + u²) = (t1 - t3) √(4 + v²)But t2 - t1 = (u - t1) - t1 = u - 2t1 (Wait, no: t2 = u - t1 (since u = t1 + t2), so t2 = u - t1, so t2 - t1 = u - 2t1Similarly, t3 = v - t1, so t1 - t3 = t1 - (v - t1) = 2t1 - vTherefore, substituting into the length equation:(u - 2t1) √(4 + u²) = (2t1 - v) √(4 + v²)But since u*v = -4, so v = -4/u.Substituting v = -4/u into the equation:(u - 2t1) √(4 + u²) = (2t1 - (-4/u)) √(4 + (-4/u)^2 )Simplify RHS:(2t1 + 4/u) √(4 + 16/u² )= (2t1 + 4/u) √( (4u² +16)/u² )= (2t1 + 4/u) * (√(4u² +16)/|u| )Since u is a variable here, but we need to see if u is positive or negative. Let's think about the coordinates. Since x3 < x1 < x2, and x = 2t, so t3 < t1 < t2. Then, t3 < t1 < t2. Therefore, t2 is greater than t1, so u = t1 + t2. Since t2 > t1, u could be positive or negative depending on the values of t1 and t2. Similarly, v = t3 + t1. Since t3 < t1, v could be positive or negative as well.But given that u*v = -4, the product is negative. So if u is positive, v is negative, and vice versa.Let's suppose u is positive. Then v = -4/u is negative. Then, in the RHS expression:√(4u² +16)/|u| = √(4(u² +4))/|u| = 2√(u² +4)/|u| = 2√(u² +4)/u (since u is positive)Therefore, RHS becomes:(2t1 + 4/u) * (2√(u² +4)/u ) = (2t1 + 4/u)*(2√(u² +4)/u )Similarly, LHS is:(u - 2t1)√(4 + u² )So equating LHS and RHS:(u - 2t1)√(u² +4 ) = (2t1 + 4/u)*(2√(u² +4)/u )Simplify RHS:(2t1 + 4/u)*(2√(u² +4)/u ) = [ (2t1 u + 4)/u ] * [ 2√(u² +4)/u ] = (2t1 u +4) * 2√(u² +4)/u²Therefore, equation:(u - 2t1)√(u² +4 ) = (2t1 u +4) * 2√(u² +4)/u²Divide both sides by √(u² +4 ) (assuming √(u² +4 ) ≠0, which it isn't):u - 2t1 = (2t1 u +4)*2 / u²Multiply both sides by u²:u³ - 2t1 u² = 4t1 u + 8Bring all terms to left-hand side:u³ -2t1 u² -4t1 u -8 =0This equation relates u and t1. Hmm. This seems complicated, but maybe we can find a relation between u and t1 here.Alternatively, perhaps we can express t1 in terms of u. Let's try:Let’s rearrange the equation:u³ -2t1 u² -4t1 u -8 =0Group terms with t1:- t1 (2u² +4u) + u³ -8 =0Therefore,t1 = (u³ -8)/(2u² +4u )Simplify denominator:2u² +4u = 2u(u +2 )So,t1 = (u³ -8)/(2u(u +2)) )Notice that u³ -8 = (u -2)(u² +2u +4 )Therefore,t1 = (u -2)(u² +2u +4 )/(2u(u +2)) )Simplify numerator and denominator:But this might not simplify easily.Alternatively, perhaps we can substitute specific values to test, but since we need the minimal area, maybe we need a general expression.Wait, but perhaps we can express t1 in terms of u, and then express other variables in terms of u, and then express the area in terms of u, and then find its minimum.So let's proceed.We have t1 = (u³ -8)/(2u(u +2))Now, recall that u = t1 + t2, and t2 = u - t1.But t2 = u - t1 = u - (u³ -8)/(2u(u +2)) )This seems messy. Let's see if we can find expressions for t3 and other variables.From u*v = -4, and v = t1 + t3, so t3 = v - t1 = (-4/u) - t1So t3 = (-4/u) - t1 = (-4/u) - (u³ -8)/(2u(u +2)) )Combine the terms:= [ -8 - (u³ -8) ] / [2u(u +2) ]Wait, let me compute:First, express (-4/u) as (-8)/(2u)So,t3 = (-8)/(2u) - (u³ -8)/(2u(u +2)) )= [ -8(u +2) - (u³ -8) ] / [2u(u +2) ]Expand numerator:-8(u +2) -u³ +8 = -8u -16 -u³ +8 = -u³ -8u -8So numerator is - (u³ +8u +8 )Therefore,t3 = - (u³ +8u +8 ) / [2u(u +2) ]Hmm. Let's check if this can be factored:u³ +8u +8 = u³ +8u +8. Let's try rational roots. Possible roots are ±1, ±2, ±4, ±8. Testing u=-1: (-1)^3 +8*(-1) +8 = -1 -8 +8 = -1 ≠0. u=-2: -8 + (-16) +8 = -16 ≠0. u=1: 1 +8 +8=17≠0. Not factorable easily. So probably can't factor.So, t3 is expressed as above.Now, to compute the area of triangle ABD.The area of a triangle with coordinates can be calculated via determinant formula:Area = (1/2)| (x2(y3 - y1) + x3(y1 - y2) + x1(y2 - y3)) | Alternatively, since it's a right-angled triangle at A, the legs are AB and AD, so area is (1/2)*|AB|*|AD|But since it's isosceles, |AB|=|AD|, so area is (1/2)*|AB|²But wait, since AB and AD are perpendicular and equal in length, the area would indeed be (1/2)*|AB|².But perhaps using coordinates is safer.Alternatively, since angle at A is right angle, vectors AB and AD are perpendicular, the area is ½ |AB × AD|, which is ½ |AB_x * AD_y - AB_y * AD_x|But since AB and AD are perpendicular, the area is ½ |AB| |AD|, and since |AB|=|AD|, it's ½ |AB|².But since they are equal, it's (1/2)*|AB|².Therefore, Area = (1/2)*( |AB| )²But |AB| can be expressed as:From earlier, |AB| = |t2 - t1| √[4 + (t2 + t1)^2 ]But since t2 = u - t1, and u = t1 + t2, so t2 = u - t1.Wait, actually, we have parametrized the points in terms of u and t1, so maybe we need to express |AB| in terms of u.Wait, but we have expressions for t1 and t3 in terms of u.Alternatively, maybe express the area in terms of u.Given that area = (1/2)*|AB|²Compute |AB|²:From earlier:|AB|² = (t2 - t1)^2 [4 + (t2 + t1)^2 ]But t2 = u - t1, so t2 - t1 = u - 2t1And t2 + t1 = uTherefore, |AB|² = (u - 2t1)^2 [4 + u² ]Therefore, area = (1/2)*(u - 2t1)^2 [4 + u² ]But t1 is expressed in terms of u: t1 = (u³ -8)/(2u(u +2))Therefore, compute (u - 2t1):u - 2t1 = u - 2*(u³ -8)/(2u(u +2)) ) = u - (u³ -8)/(u(u +2)) )= [ u²(u +2) - (u³ -8) ] / [u(u +2) ]Compute numerator:u²(u +2) - u³ +8 = u³ +2u² -u³ +8 = 2u² +8Therefore,u -2t1 = (2u² +8)/(u(u +2)) ) = 2(u² +4)/(u(u +2)) )Thus,|AB|² = [2(u² +4)/(u(u +2)) ]² * (4 + u² )= 4(u² +4)^2 / [u²(u +2)^2 ] * (u² +4 )= 4(u² +4)^3 / [u²(u +2)^2 ]Therefore, area = (1/2)*[4(u² +4)^3 / (u²(u +2)^2 ) ] = 2(u² +4)^3 / [u²(u +2)^2 ]So area is 2(u² +4)^3 / [u²(u +2)^2 ]Our goal is to minimize this expression with respect to u, considering the constraints from the problem.But what is the domain of u? Remember that u = t1 + t2, and t3 < t1 < t2.From x3 < x1 < x2, which translates to t3 < t1 < t2. Also, t3 is related to u via v = t3 + t1 = -4/u.But since u is in the denominator, u cannot be zero. Also, since t3 < t1, and v = t3 + t1, then v = t3 + t1 < t1 + t1 = 2t1. But v = -4/u, so -4/u < 2t1.But t1 = (u³ -8)/(2u(u +2))So substituting:-4/u < 2*(u³ -8)/(2u(u +2)) → -4/u < (u³ -8)/(u(u +2)) )Multiply both sides by u(u +2). But need to consider the sign of u(u +2). Let's see.From the earlier assumption that u >0 (since if u is positive, v = -4/u is negative, which might align with t3 < t1. Alternatively, if u is negative, v = -4/u is positive. But with t3 < t1, we need to check.Wait, since x3 < x1, and x = 2t, so t3 < t1. Also, u = t1 + t2, and t2 > t1. If u is positive, then t1 + t2 >0. If u is negative, then t1 + t2 <0.But t2 > t1. If u is positive, then even if t1 is negative, t2 is greater than t1. If u is negative, then t1 + t2 is negative, but since t2 > t1, t1 must be negative enough such that their sum is negative.But perhaps we can consider both cases u >0 and u <0. Let's first consider u >0.Case 1: u >0Then v = -4/u <0. Since v = t3 + t1 <0, and t1 is expressed as (u³ -8)/(2u(u +2)).But with u >0, let's check if t1 is positive or negative.t1 = (u³ -8)/(2u(u +2))For u >0, denominator is positive. Numerator: u³ -8.So when u³ -8 >0 → u >2, then t1 is positive.When u <2, u³ -8 <0, so t1 is negative.But we have t3 < t1 < t2.If u >2, then t1 is positive, t2 = u - t1. If u >2 and t1 positive, then t2 = u - t1. Since u = t1 + t2, but u is a separate variable here. Wait, actually, we need to ensure that t2 > t1. Given that u = t1 + t2, and t2 > t1, so u = t1 + t2 > t1 + t1 = 2t1. If t1 is positive, then u > 2t1. If t1 is negative, u could be greater or less than 2t1.This is getting complicated. Maybe instead of splitting into cases, we can look for possible u values.But perhaps it's better to proceed with the expression for the area and find its minimum.So we need to minimize 2(u² +4)^3 / [u²(u +2)^2 ] over real numbers u ≠0, u ≠-2 (since denominator cannot be zero), and considering the constraints from the problem (e.g., t3 < t1 < t2). Let's consider u >0 first.Let’s set f(u) = 2(u² +4)^3 / [u²(u +2)^2 ]We need to find the minimum of f(u) for u >0.To minimize f(u), take derivative f’(u) and set to zero.Compute f’(u):Let’s write f(u) = 2 * (u² +4)^3 * [u²(u +2)^2 ]^{-1}Take natural logarithm to differentiate more easily:ln f(u) = ln2 + 3 ln(u² +4) - 2 ln u - 2 ln(u +2)Differentiate both sides:f’(u)/f(u) = 3*(2u)/(u² +4) - 2*(1/u) - 2*(1/(u +2))Therefore,f’(u) = f(u) * [ 6u/(u² +4) - 2/u - 2/(u +2) ]Set f’(u) =0:6u/(u² +4) - 2/u - 2/(u +2) =0Multiply through by u(u +2)(u² +4) to eliminate denominators:6u * u(u +2) - 2(u +2)(u² +4) - 2u(u² +4) =0Expand each term:First term: 6u * u(u +2) =6u²(u +2) =6u³ +12u²Second term: -2(u +2)(u² +4) = -2[u(u² +4) +2(u² +4)] = -2[u³ +4u +2u² +8] = -2u³ -8u -4u² -16Third term: -2u(u² +4) = -2u³ -8uCombine all terms:6u³ +12u² -2u³ -8u -4u² -16 -2u³ -8u =0Combine like terms:6u³ -2u³ -2u³ =2u³12u² -4u² =8u²-8u -8u =-16u-16So equation becomes:2u³ +8u² -16u -16=0Divide both sides by 2:u³ +4u² -8u -8=0Now, solve u³ +4u² -8u -8=0Try rational roots: possible roots are ±1, ±2, ±4, ±8.Test u=2: 8 +16 -16 -8=0. Yes! u=2 is a root.Factor out (u -2):Using polynomial division or synthetic division.Divide u³ +4u² -8u -8 by (u -2):Wait, but wait, u=2 is a root, so factor is (u -2). Wait, but when u=2, the value is 8 +16 -16 -8=0. Correct.So, using synthetic division:Coefficients: 1 | 4 | -8 | -8Divide by (u -2):Bring down 1.Multiply by 2: 1*2=2. Add to next coefficient: 4 +2=6.Multiply by 2:6*2=12. Add to next coefficient: -8 +12=4.Multiply by 2:4*2=8. Add to last coefficient: -8 +8=0.So quotient polynomial is u² +6u +4.Thus,u³ +4u² -8u -8=(u -2)(u² +6u +4)=0Set to zero:(u -2)(u² +6u +4)=0Solutions: u=2, or roots of u² +6u +4=0 → u = [-6 ±√(36 -16)]/2 = [-6 ±√20]/2 = [-6 ±2√5]/2 = -3 ±√5So roots are u=2, u= -3 +√5 ≈ -3 +2.236≈-0.764, u= -3 -√5≈-5.236.But we are considering u >0, so only u=2 is valid here.Therefore, critical point at u=2.Check if this is a minimum.Second derivative test or check intervals.For u>0, let's check values around u=2.Take u approaching 0+:f(u) =2(u² +4)^3 / [u²(u +2)^2 ] ~ 2*(4)^3 / [u²*(2)^2 ] = 2*64 / (4u²) =32/u² → ∞ as u→0+At u=2, compute f(2):f(2)=2*(4 +4)^3 / [4*(4)^2 ]=2*(8)^3 / (4*16)=2*512 /64= 1024/64=16At u approaching ∞:f(u) ~2*(u²)^3 / [u²*u² ]=2u^6 /u^4=2u² → ∞ as u→inftyTherefore, the function has a critical point at u=2, and since it goes to infinity as u approaches 0+ and infty, it's likely that u=2 is a minimum.Therefore, the minimum area is 16.But we need to check if u=2 satisfies the original constraints.At u=2:t1=(8 -8)/(2*2*(2 +2))=0/(16)=0Wait, t1=(u³ -8)/(2u(u +2))=(8 -8)/(2*2*(4))=0. So t1=0.Then, t2 = u -t1=2 -0=2.t3= (-4/u) -t1=(-4/2)-0=-2.So points:A: (2*0, 0²)=(0,0)B: (2*2, 2²)=(4,4)D: (2*(-2), (-2)^2)=(-4,4)Wait, but check if x3 <x1 <x2: x3=-4, x1=0, x2=4: yes, -4 <0 <4. Good.Check triangle ABD with points A(0,0), B(4,4), D(-4,4)Compute vectors AB=(4,4), AD=(-4,4)Dot product AB ⋅ AD=4*(-4) +4*4= -16 +16=0. Correct, perpendicular.Lengths |AB|=√(16 +16)=√32=4√2, |AD|=√(16 +16)=√32=4√2. Equal lengths. So it is an isosceles right triangle at A. Area is (1/2)*(4√2)^2= (1/2)*32=16. Which matches our calculation.So this seems valid.But wait, the problem states "find the minimum value of the area of triangle ABD". So according to this, the minimum area is 16. But let's check if there are other critical points for u <0.Case 2: u <0If u is negative, then v = -4/u is positive. Then, t3 + t1 = v = -4/u. Since u <0, v = positive.But t3 < t1, so t3 = v - t1 = (-4/u) - t1. Given that t3 < t1, so (-4/u) - t1 < t1 → -4/u <2t1. But u is negative, let's denote u = -k where k>0. Then, v= -4/u=4/k.Then, t1 = (u³ -8)/(2u(u +2)) = ( (-k)^3 -8 )/(2*(-k)(-k +2)) = ( -k³ -8 )/( -2k(2 -k) ) = ( - (k³ +8 ) )/( -2k(2 -k ) ) = (k³ +8 )/(2k(2 -k )).So t1 = (k³ +8)/(2k(2 -k )) with k>0 and k≠2.Also, t3 = (-4/u) - t1 = (4/k) - t1.Given that t3 < t1, so (4/k) - t1 < t1 →4/k <2t1 →2/k <t1.But t1 = (k³ +8)/(2k(2 -k )). So:2/k < (k³ +8)/(2k(2 -k ))Multiply both sides by 2k(2 -k ) (but need to consider the sign of 2 -k )If 2 -k >0 →k <2, then inequality remains:2*2(2 -k ) <k³ +8Wait, no:Wait, starting again:2/k < (k³ +8)/(2k(2 -k ))Multiply both sides by 2k(2 -k ). Since u <0, so k>0, and 2 -k could be positive or negative.But if k <2, then 2 -k >0, so inequality direction remains:2*2(2 -k ) <k³ +8Wait, no:Wait, left side: (2/k)*2k(2 -k ) =4(2 -k )Right side: (k³ +8)/(2k(2 -k )) *2k(2 -k ) =k³ +8Thus, inequality becomes 4(2 -k ) <k³ +8Simplify:8 -4k <k³ +8 → -4k <k³ → since k>0, divide both sides by k: -4 <k² → which is always true as k² >0.So for k <2, the inequality holds.If k >2, then 2 -k <0, so multiplying both sides by 2k(2 -k ) reverses inequality:4(2 -k ) >k³ +8 →8 -4k >k³ +8 → -4k >k³ → which is impossible since k>0, left side negative, right side positive. So no solution for k>2.Therefore, for u <0 (k>0), possible only when k <2.So in case u <0, k∈(0,2). Now, compute the area expression in terms of k.Given that u=-k, k>0.Area=2(u² +4)^3 / [u²(u +2)^2 ]=2(k² +4)^3 / [k²(-k +2)^2 ]So Area=2(k² +4)^3 / [k²(2 -k )² ]To find critical points, similar to previous case, set derivative to zero.But this might be symmetric, but let's check.Compute derivative of Area w.r. to k:Let’s denote g(k) =2(k² +4)^3 / [k²(2 -k )² ]Take ln g(k)= ln2 +3 ln(k² +4) -2 lnk -2 ln(2 -k )Differentiate:g’(k)/g(k)= 3*(2k)/(k² +4) -2*(1/k) -2*( -1/(2 -k ) )Thus,g’(k)=g(k)*[6k/(k² +4) -2/k +2/(2 -k ) ]Set g’(k)=0:6k/(k² +4) -2/k +2/(2 -k )=0Multiply through by k(2 -k )(k² +4):6k *k(2 -k ) -2(2 -k )(k² +4) +2k(k² +4)=0Expand each term:First term:6k^2(2 -k )=12k² -6k³Second term:-2(2 -k )(k² +4)= -2[2(k² +4) -k(k² +4)]= -2[2k² +8 -k³ -4k ]= -4k² -16 +2k³ +8kThird term:2k(k² +4)=2k³ +8kCombine all terms:12k² -6k³ -4k² -16 +2k³ +8k +2k³ +8k=0Combine like terms:(-6k³ +2k³ +2k³)=(-2k³)12k² -4k²=8k²8k +8k=16k-16Thus:-2k³ +8k² +16k -16=0Multiply by -1:2k³ -8k² -16k +16=0Divide by 2:k³ -4k² -8k +8=0Try rational roots: ±1, ±2, ±4, ±8.Test k=2:8 -16 -16 +8= -16≠0k=1:1 -4 -8 +8= -3≠0k=4:64 -64 -32 +8= -24≠0k= -2: Not applicable since k>0.Maybe factor by grouping:k³ -4k² -8k +8= k²(k -4) -8(k -1). Doesn't help.Alternatively, use rational root theorem. Since no rational roots, maybe this cubic has real roots.Use derivative test. Let h(k)=k³ -4k² -8k +8h’(k)=3k² -8k -8Set to zero:3k² -8k -8=0 →k=(8 ±√(64 +96))/6=(8 ±√160)/6=(8 ±4√10)/6=(4 ±2√10)/3≈Positive root: (4 +2√10)/3≈(4 +6.324)/3≈10.324/3≈3.44Negative root ignored.Thus, h(k) has critical points at k≈3.44. Since we are looking for k∈(0,2), and h(2)=8 -16 -16 +8= -16, h(0)=0 -0 -0 +8=8>0. So between k=0 and k=2, h(k) decreases from 8 to -16, so crosses zero once. Therefore, there's a real root in (0,2). Let's approximate.At k=1: h(1)=1 -4 -8 +8= -3At k=1.5: h(1.5)=3.375 -9 -12 +8= -9.625Wait, wait:Wait, k=1.5:k³=3.375-4k²= -4*(2.25)= -9-8k= -12+8=8Total:3.375 -9 -12 +8=3.375 -13= -9.625k=1: h= -3; k=1.2:k=1.2:k³=1.728-4k²= -4*(1.44)= -5.76-8k= -9.6+8=8Total:1.728 -5.76 -9.6 +8= (1.728 +8) - (5.76 +9.6)=9.728 -15.36= -5.632k=0.5:k³=0.125-4k²= -1-8k= -4+8=8Total:0.125 -1 -4 +8=3.125So between k=0.5 and k=1, h(k) changes from 3.125 to -3, crossing zero. Let's try k=0.8:k=0.8:k³=0.512-4k²= -4*0.64= -2.56-8k= -6.4+8=8Total:0.512 -2.56 -6.4 +8=0.512 -8.96 +8= -0.448k=0.7:k³=0.343-4k²= -4*0.49= -1.96-8k= -5.6+8=8Total:0.343 -1.96 -5.6 +8=0.343 -7.56 +8≈0.783So between k=0.7 and k=0.8, h(k) crosses zero. Use linear approximation:Between k=0.7 (0.783) and k=0.8 (-0.448). The root is around k=0.7 + (0 -0.783)*(-0.448 -0.783)^(-1)*(0.8 -0.7)But this is tedious. Alternatively, we can use Newton-Raphson method.Let’s take k0=0.75h(0.75)=0.421875 -4*(0.5625) -8*(0.75) +8=0.421875 -2.25 -6 +8=0.421875 -8.25 +8≈0.421875 -0.25≈0.171875h’(0.75)=3*(0.75)^2 -8*(0.75) -8=3*0.5625 -6 -8=1.6875 -14= -12.3125Next iteration: k1=0.75 - h(k)/h’(k)=0.75 -0.171875/(-12.3125)=0.75 +0.01395≈0.76395Compute h(0.76395):k³≈0.76395^3≈0.445-4k²≈-4*(0.76395²)= -4*0.583≈-2.332-8k≈-6.1116+8≈8Total≈0.445 -2.332 -6.1116 +8≈0.445 -8.4436 +8≈0.445 -0.4436≈0.0014Very close to zero. So root≈0.764Thus, there's a critical point at k≈0.764, which is in (0,2). Compute the area at this point.But this is complicated, and given that in the positive case we already found a valid minimum area of 16, which is achievable, we need to check if this critical point gives a lower area.But given the complexity, perhaps the minimal area is 16, as found in the positive u case.Additionally, in the negative u case, the minimal area might be larger, but to confirm, let's compute the area at k≈0.764:g(k)=2(k² +4)^3 / [k²(2 -k )² ]Compute k≈0.764:k²≈0.5832 -k≈1.236So numerator:2*(0.583 +4)^3≈2*(4.583)^3≈2*(96.2)≈192.4Denominator:0.583*(1.236)^2≈0.583*1.527≈0.890Thus, area≈192.4 /0.890≈216.2, which is much larger than 16.Therefore, the minimal area is indeed 16, achieved at u=2.Hence, the minimum value of the area of triangle ABD is 16.Final Answer(1) The equation of the parabola is boxed{x^2 = 4y}.(2) The minimum value of the area of (triangle ABD) is boxed{16}.</think>Given a parabola ( C ) with its vertex at the origin and its focus on the ( y )-axis, we need to find the equation of the parabola and the minimum value of the area of an isosceles right triangle ( triangle ABD ) with (angle A) being the right angle. Part (1): Finding the Equation of the Parabola1. The standard form of a parabola with vertex at the origin and focus on the ( y )-axis is ( x^2 = 4py ), where ( p ) is the distance from the vertex to the focus.2. Given point ( Q(a, 2) ) on the parabola, we have ( a^2 = 8p ).3. The distance from ( Q(a, 2) ) to the focus ( (0, p) ) is 3. Using the distance formula: [ sqrt{a^2 + (2 - p)^2} = 3 ]4. Substituting ( a^2 = 8p ) into the distance equation: [ 8p + (2 - p)^2 = 9 ]5. Solving the quadratic equation ( p^2 + 4p - 5 = 0 ), we find ( p = 1 ) (since ( p = -5 ) leads to a negative ( a^2 )).6. The equation of the parabola is ( x^2 = 4y ). Part (2): Finding the Minimum Area of (triangle ABD)1. Points ( A ), ( B ), and ( D ) on the parabola can be parameterized as ( (2t_1, t_1^2) ), ( (2t_2, t_2^2) ), and ( (2t_3, t_3^2) ) respectively.2. For (triangle ABD) to be an isosceles right triangle with (angle A) as the right angle, vectors ( overrightarrow{AB} ) and ( overrightarrow{AD} ) must be perpendicular and equal in length.3. Using parametric coordinates and solving the equations for perpendicularity and equal lengths, we find that the critical point occurs at ( u = 2 ) (where ( u = t_1 + t_2 )).4. Substituting ( u = 2 ) into the coordinates of ( A ), ( B ), and ( D ), we find the points ( A(0, 0) ), ( B(4, 4) ), and ( D(-4, 4) ).5. Calculating the area of (triangle ABD) using the determinant formula, we find the area to be 16. Final Answers(1) The equation of the parabola is (boxed{x^2 = 4y}).(2) The minimum value of the area of (triangle ABD) is (boxed{16}).]

🔑:Solution:(1) Let the equation of the parabola C be x^2=2py(p > 0). Then, the focus F is at (0, frac{p}{2}), and the equation of the directrix is y=-frac{p}{2}.Drawing a perpendicular line from point Q to the directrix l, and let the foot of the perpendicular be Q_1.By the definition of a parabola, we have |QF|=|QQ_1|,therefore 2-(-frac{p}{2})=3, p=2,therefore The equation of the parabola is x^2=4y;(2) Let the slope of line AB be k, k > 0, and its equation be y-y_1=k(x-x_1), i.e., y=kx+y_1-kx_1,therefore begin{cases}y=kx+y_1-kx_1 x^2=4yend{cases}, simplifying gives: x^2-4kx+4kx_1-4y_1=0,therefore x_1+x_2=4k, x_2=-x_1+4k,|AB|^2=(1+k^2)[(x_1+x_2)^2-4x_1x_2]=(1+k^2)[(x_1+x_2)^2-4x_1x_2],=4(1+k^2)(x_1^2-4kx_1+4k^2),Similarly, |AD|^2=4[1+(-frac{1}{k})^2][x_1^2-4(-frac{1}{k})x_1+4(-frac{1}{k})^2],=4(1+frac{1}{k^2})(x_1^2+frac{4}{k}x_1+frac{4}{k^2}),Since |AB|=|AD|, then |AB|^2=|AD|^2, 4(1+k^2)(x_1^2-4kx_1+4k^2)=4(1+frac{1}{k^2})(x_1^2+frac{4}{k}x_1+frac{4}{k^2}),Simplifying gives: x_1=frac{k^2-1}{k}=k-frac{1}{k},Then |AB|^2=4(1+k^2)[(k-frac{1}{k})^2-4k(k-frac{1}{k})+4k^2]=4(1+k^2)(k+frac{1}{k})^2, |AB|=2sqrt{1+k^2}(k+frac{1}{k}),|AD|^2=4(1+frac{1}{k^2})[(k-frac{1}{k})^2+frac{4}{k}(k-frac{1}{k})+frac{4}{k^2}]=4(1+frac{1}{k^2})(k+frac{1}{k})^2, |AD|=2sqrt{1+frac{1}{k^2}}(k+frac{1}{k}),therefore The area of triangle ABD, S=frac{1}{2}times|AB|times|AD|=frac{1}{2}times2sqrt{1+k^2}(k+frac{1}{k})times2sqrt{1+frac{1}{k^2}}(k+frac{1}{k}),=frac{2(1+k^2)(k+frac{1}{k})^2}{k}=2(k+frac{1}{k})^3geqslant 2(2sqrt{ktimesfrac{1}{k}})^3=16,Equality holds if and only if k=frac{1}{k}, i.e., k^2=1, k=1,therefore The minimum value of the area of triangle ABD is boxed{16}.

🧠:[<think>Okay, let's tackle these problems one by one. Hmm, starting with the first one: Calculate (4 - (-28) + (-2)). Alright, subtracting a negative is like adding a positive, right? So (4 - (-28)) becomes (4 + 28). That should be 32. Then adding (-2) would be subtracting 2. So 32 - 2 is 30. Wait, let me check that again. Yeah, subtracting a negative turns into addition. So yes, 4 plus 28 is 32, then minus 2 gives 30. That seems right.Moving on to the second problem: ((-3) times [(- frac{2}{5}) div (- frac{1}{4})]). Okay, let's work inside the brackets first. Dividing two negatives. Remember that dividing negatives gives a positive. So (frac{-2/5}{-1/4}). Dividing fractions is the same as multiplying by the reciprocal, so it becomes (-2/5 times -4/1). Multiply the numerators: (-2)(-4) is 8, and denominators 5*1=5. So that's 8/5. Then multiply by (-3). So (-3) times 8/5. Hmm, that's -24/5. Which is -4.8 or as a fraction, -24/5. Let me confirm. Negative times positive is negative, so yes, -24/5. Alternatively, if we need to write it as a mixed number, that's -4 and 4/5, but I think the question just wants it as an improper fraction. So -24/5.Third problem: ((-42) div (-7) - (-6) times 4). Let's break it down. First, divide -42 by -7. Dividing two negatives gives a positive. 42 divided by 7 is 6. Then subtract (-6) times 4. Wait, no, the operation is minus (-6) times 4. So that part is - (-6)*4. Let me parse the order. According to PEMDAS, multiplication comes before subtraction. So first compute (-6) * 4, which is -24. Then the expression becomes 6 - (-24). Subtracting a negative is adding a positive. So 6 + 24 = 30. Let me double-check. Original problem: (-42)/(-7) is 6. Then minus (-6)*4. So (-6)*4 is -24, so minus (-24) is +24. 6 +24=30. Yep, that's right.Fourth problem: (-3^{2} div (-3)^{2} + 3 times (-2) + |-4|). This one looks tricky. Let's handle each term step by step. First, (-3^{2}). Wait, exponentiation takes precedence over the negative sign unless there's a parenthesis. So, is it (-(3^2)) or (-3)^2? The problem is written as -3^2, which typically is interpreted as -(3^2) = -9. Whereas (-3)^2 is 9. Then divided by (-3)^2. So that's (-3)^2 which is 9. Then the next terms: 3 * (-2) is -6. And absolute value of -4 is 4. So putting it all together: -9 ÷ 9 + (-6) + 4. Let's compute each part: -9 divided by 9 is -1. Then -1 plus (-6) is -7. Then -7 plus 4 is -3. Wait, let me go through again. First term: -3^2 is -9. Second term: (-3)^2 is 9. So -9 divided by 9 is -1. Then 3*(-2) is -6. Then |-4| is 4. So total: -1 + (-6) + 4. That's (-1 -6) +4 = -7 +4 = -3. That seems right.Fifth problem: ((-24) times (frac{3}{4} - frac{5}{6} + frac{7}{12})). Okay, need to compute the expression inside the parentheses first. Let's find a common denominator. The denominators are 4, 6, and 12. The least common denominator is 12. So convert each fraction:3/4 = 9/12,5/6 = 10/12,7/12 stays the same.So the expression becomes: 9/12 - 10/12 + 7/12. Combine them: (9 -10 +7)/12 = (6)/12 = 1/2. Then multiply by -24. So (-24)*(1/2) = -12. Let me verify. 3/4 is 0.75, 5/6 is approximately 0.8333, 7/12 is ~0.5833. 0.75 -0.8333 +0.5833 ≈ 0.75 -0.8333 is -0.0833 +0.5833=0.5. Which is 1/2. So -24*(1/2)= -12. Correct.Last problem: (-1^{4} - (1 - 0.5) div frac{5}{2} times frac{1}{5}). Let's handle each part. First term: -1^4. Similar to the fourth problem, exponentiation before the negative sign. So -(1^4) = -1. Then the second part: (1 - 0.5) is 0.5. Then divide by 5/2 and multiply by 1/5. Let's parse the order of operations. Division and multiplication from left to right. So 0.5 ÷ (5/2) × (1/5). Dividing by 5/2 is multiplying by 2/5. So 0.5 * 2/5 = (0.5 * 2)/5 = 1/5. Then multiply by 1/5: (1/5) * (1/5) = 1/25. So the entire expression is -1 - 1/25. Which is -1.04 or as a fraction, -26/25. Let me check again. 1 -0.5=0.5. 0.5 ÷ 5/2 = 0.5 * 2/5 = 0.2. Then 0.2 *1/5 =0.04. So total is -1 -0.04 = -1.04. Converting 0.04 to fraction is 1/25. So total is -1 -1/25 = -26/25. Yes, that's correct.So summarizing the answers:1. 302. -24/53. 304. -35. -126. -26/25Final Answer(1) boxed{30}(2) boxed{-dfrac{24}{5}}(3) boxed{30}(4) boxed{-3}(5) boxed{-12}(6) boxed{-dfrac{26}{25}}</think>(1) Calculate (4 - (-28) + (-2)):[4 - (-28) + (-2) = 4 + 28 - 2 = 32 - 2 = 30]Result: (boxed{30})(2) Calculate ((-3) times left[ left( -frac{2}{5} right) div left( -frac{1}{4} right) right]):[left( -frac{2}{5} right) div left( -frac{1}{4} right) = left( -frac{2}{5} right) times left( -4 right) = frac{8}{5}][(-3) times frac{8}{5} = -frac{24}{5}]Result: (boxed{-dfrac{24}{5}})(3) Calculate ((-42) div (-7) - (-6) times 4):[(-42) div (-7) = 6][(-6) times 4 = -24][6 - (-24) = 6 + 24 = 30]Result: (boxed{30})(4) Calculate (-3^{2} div (-3)^{2} + 3 times (-2) + |-4|):[-3^{2} = -9][(-3)^{2} = 9][-9 div 9 = -1][3 times (-2) = -6][|-4| = 4][-1 + (-6) + 4 = -3]Result: (boxed{-3})(5) Calculate ((-24) times left( frac{3}{4} - frac{5}{6} + frac{7}{12} right)):[frac{3}{4} - frac{5}{6} + frac{7}{12} = frac{9}{12} - frac{10}{12} + frac{7}{12} = frac{6}{12} = frac{1}{2}][(-24) times frac{1}{2} = -12]Result: (boxed{-12})(6) Calculate (-1^{4} - (1 - 0.5) div frac{5}{2} times frac{1}{5}):[-1^{4} = -1][1 - 0.5 = 0.5][0.5 div frac{5}{2} = 0.5 times frac{2}{5} = 0.2][0.2 times frac{1}{5} = 0.04][-1 - 0.04 = -1.04 = -frac{26}{25}]Result: (boxed{-dfrac{26}{25}}),<think>Okay, let's see. I have these six math problems to solve. Let me take them one by one. I need to be careful with the signs and the order of operations. Let's start with problem (1).Problem (1): Calculate (4 - (-28) + (-2)).Hmm, subtraction of a negative is like adding a positive. So, (4 - (-28)) becomes (4 + 28). Then adding (-2) would be subtracting 2. Let me compute that step by step.First, (4 - (-28)) is (4 + 28 = 32). Then adding (-2) is (32 - 2 = 30). So, the answer for (1) should be 30. Let me check again. Subtracting a negative adds, so yes, that part is right. Then 32 minus 2 is 30. Yeah, that seems correct.Problem (2): Calculate ((-3) times [(- frac{2}{5}) div (- frac{1}{4})]).Alright, this involves division of fractions and multiplication. Let me handle the division inside the brackets first. Dividing by a fraction is multiplying by its reciprocal. So, ((- frac{2}{5}) div (- frac{1}{4}) = (- frac{2}{5}) times (-4)). Multiplying two negatives gives a positive. So, (frac{2}{5} times 4 = frac{8}{5}). Then multiply by (-3). Wait, the original problem is (-3) times that result. So, (-3 times frac{8}{5} = -frac{24}{5}). Let me confirm.First, division part: (-frac{2}{5} div -frac{1}{4}). Flip the second fraction: -2/5 * -4/1 = 8/5. Then multiply by -3: -3 * 8/5 = -24/5. So, that's -24/5. Which can be written as (-4 frac{4}{5}) but probably should leave it as an improper fraction. So, answer is -24/5.Problem (3): Calculate ((-42) div (-7) - (-6) times 4).Okay, division and multiplication first. Let's compute each part step by step.First, (-42) divided by (-7) is positive 6 because negative divided by negative is positive. Then, (-6) multiplied by 4 is -24. Now the expression becomes 6 - (-24). Subtracting a negative is adding a positive, so 6 + 24 = 30. Let me verify: -42 ÷ -7 is 6. Then -6 ×4 is -24. Then 6 - (-24) is 6 +24 =30. Correct.Problem (4): Calculate (-3^{2} div (-3)^{2} + 3 times (-2) + |-4|).This one looks tricky with exponents and absolute value. Let me recall order of operations: exponents, then multiplication/division left to right, then addition/subtraction. Also, note that (-3^2) is different from ((-3)^2). The first is - (3^2) = -9, the second is 9. So:First term: (-3^2 = -9). Second term: ((-3)^2 = 9). So, division is -9 ÷ 9 = -1.Third term: 3 × (-2) = -6.Fourth term: |-4| = 4.Now add them all together: -1 + (-6) + 4. That is (-1 -6) +4 = -7 +4 = -3.Wait, let me check again. First term: (-3^2) is indeed -9. Then divided by (-3)^2 which is 9. So -9 ÷9 = -1. Then 3×(-2)= -6. Then absolute value of -4 is 4. Then combining: -1 + (-6) +4. So, -1 -6 is -7, then +4 is -3. Correct. So answer is -3.Problem (5): Calculate ((-24) times (frac{3}{4} - frac{5}{6} + frac{7}{12})).This involves distributing -24 over the parentheses. Maybe first compute the terms inside the parentheses. Let me find a common denominator. Denominators are 4,6,12. The least common denominator is 12.Convert each fraction:3/4 = 9/12,-5/6 = -10/12,7/12 stays as is.So, adding them together: 9/12 -10/12 +7/12 = (9 -10 +7)/12 = (6)/12 = 1/2.So inside the parentheses simplifies to 1/2. Then multiply by -24: -24 × 1/2 = -12. Let me check:First, 3/4 -5/6 +7/12. Converting to twelfths: 9 -10 +7 = 6, which is 6/12=1/2. Then -24*(1/2) = -12. Correct.Problem (6): Calculate (-1^{4} - (1 - 0.5) div frac{5}{2} times frac{1}{5}).Hmm, order of operations again. Let me parse this carefully.First, compute exponents. (-1^4) is -(1^4) = -1.Then, compute the terms in the parenthesis: 1 -0.5 = 0.5.Then, division and multiplication from left to right. So, 0.5 ÷ (5/2) × (1/5).First, 0.5 ÷ (5/2) is the same as 0.5 × (2/5) = (0.5 ×2)/5 = 1/5.Then multiply by 1/5: 1/5 × 1/5 = 1/25.So the entire expression is -1 - 1/25. Which is -1.04 or as a fraction, -26/25.Wait, let me verify step by step.Original problem: (-1^4 - (1 -0.5) ÷ 5/2 × 1/5).First term: (-1^4). Since exponentiation precedes negation, this is -(1^4) = -1.Second part: (1 -0.5) = 0.5.Then, 0.5 ÷ (5/2) × (1/5). Let's convert 0.5 to 1/2 for easier fractions.So, 1/2 ÷5/2 = (1/2) × (2/5) = (1×2)/(2×5) = 2/10 = 1/5.Then, 1/5 ×1/5 =1/25.So, the entire expression becomes -1 -1/25. To combine, -1 is -25/25, so total is -26/25. Which is -1.04. But as a fraction, -26/25 is the answer. Let me check again:Yes, 1 -0.5 is 0.5. Then 0.5 divided by 5/2 is 0.5 *2/5=0.2. Then 0.2 times 1/5 is 0.04. So, 0.04. Then the expression is -1 -0.04 = -1.04. Which is -26/25. So, correct.So, the six answers:1) 302) -24/53) 304) -35) -126) -26/25Wait, let me check problem 2 again to ensure I didn't make a mistake with signs. The division was between two negative fractions: (-2/5) divided by (-1/4). So that's positive 8/5. Then multiplied by -3: -3 *8/5 = -24/5. Yes, correct. So problem 2 is -24/5.Problem 6: The calculation of 0.5 divided by (5/2). 0.5 is 1/2, so 1/2 ÷5/2 = (1/2)*(2/5) =1/5. Then 1/5 *1/5=1/25. So total expression: -1 -1/25= -26/25. Correct.Alright, I think all answers are correct. Let me just recap each problem briefly:1. Subtracting negatives becomes addition, straightforward.2. Division of two negatives gives positive, then multiplied by a negative.3. Division and multiplication first, then subtraction of a negative.4. Careful with exponents: -3² vs (-3)². Then combining terms.5. Common denominator inside the parentheses simplifies to 1/2, multiplied by -24.6. Order of operations with exponents, parentheses, division, and multiplication. Converting decimals to fractions helps.Yes, all steps check out.Final Answer(1) boxed{30}(2) boxed{-dfrac{24}{5}}(3) boxed{30}(4) boxed{-3}(5) boxed{-12}(6) boxed{-dfrac{26}{25}}</think>(1) Calculate (4 - (-28) + (-2)):[4 - (-28) + (-2) = 4 + 28 - 2 = 32 - 2 = 30](boxed{30})(2) Calculate ((-3) times [(- frac{2}{5}) div (- frac{1}{4})]):[(-3) times left[ left( -frac{2}{5} right) div left( -frac{1}{4} right) right] = (-3) times left[ left( -frac{2}{5} right) times left( -4 right) right] = (-3) times left( frac{8}{5} right) = -frac{24}{5}](boxed{-dfrac{24}{5}})(3) Calculate ((-42) div (-7) - (-6) times 4):[(-42) div (-7) - (-6) times 4 = 6 - (-24) = 6 + 24 = 30](boxed{30})(4) Calculate (-3^{2} div (-3)^{2} + 3 times (-2) + |-4|):[-3^{2} div (-3)^{2} + 3 times (-2) + |-4| = -9 div 9 + (-6) + 4 = -1 - 6 + 4 = -3](boxed{-3})(5) Calculate ((-24) times (frac{3}{4} - frac{5}{6} + frac{7}{12})):[(-24) times left( frac{9}{12} - frac{10}{12} + frac{7}{12} right) = (-24) times left( frac{6}{12} right) = (-24) times frac{1}{2} = -12](boxed{-12})(6) Calculate (-1^{4} - (1 - 0.5) div frac{5}{2} times frac{1}{5}):[-1^{4} - (0.5) div frac{5}{2} times frac{1}{5} = -1 - left( frac{1}{2} div frac{5}{2} times frac{1}{5} right) = -1 - left( frac{1}{5} times frac{1}{5} right) = -1 - frac{1}{25} = -frac{26}{25}](boxed{-dfrac{26}{25}})]

🔑:(1) First, we simplify the expression by changing subtraction to addition: 4 + 28 + (-2). Next, we add the numbers: 4 + 28 = 32. Finally, we add the result to the last number: 32 + (-2) = boxed{30}.(2) We start by simplifying the expression inside the brackets: (- frac{2}{5}) div (- frac{1}{4}) = frac{2}{5} times 4 = frac{8}{5}. Now, we multiply this result by (-3): (-3) times frac{8}{5} = - frac{24}{5} = boxed{-frac{24}{5}}.(3) We start by dividing -42 by -7: (-42) div (-7) = 6. Next, we multiply -6 by 4: -6 times 4 = -24. Then, subtract -24 from 6: 6 - (-24) = 6 + 24 = boxed{30}.(4) First, we square -3 and -3: -3^{2} = -9 and (-3)^{2} = 9. Then, we divide -9 by 9: -9 div 9 = -1. Next, we multiply 3 by -2: 3 times (-2) = -6. Then, add the absolute value of -4 to the result: |-4| = 4, so -1 + (-6) + 4 = -3 = boxed{-3}.(5) First, we find the common denominator of the fractions inside the parentheses: frac{3}{4} - frac{5}{6} + frac{7}{12} = frac{9 - 10 + 7}{12} = frac{6}{12} = frac{1}{2}. Now, we multiply this result by (-24): (-24) times frac{1}{2} = -12 = boxed{-12}.(6) First, we calculate 1 - 0.5 = 0.5. Then, we divide 0.5 by frac{5}{2}: 0.5 div frac{5}{2} = frac{1}{2}. Now, we multiply this result by frac{1}{5}: frac{1}{2} times frac{1}{5} = frac{1}{10}. Finally, subtract this result from -1: -1 - frac{1}{10} = -frac{11}{10} = boxed{-frac{26}{25}}.

🧠:[<think>Okay, so I have this geometry problem here: In a right triangle ABC, the bisector CL is drawn. It's given that point L is equidistant from the vertex of the right angle B and the midpoint of the hypotenuse AC. I need to find the angle BAC. Hmm, let me try to visualize this first.First, let me draw a right triangle ABC, with the right angle at B. So, angle B is 90 degrees. Then, CL is the angle bisector. Since it's called CL, I think that must mean the bisector of angle C. So point C is one of the other vertices, either A or the other one. Wait, in a triangle, the vertices are labeled A, B, C. Since B is the right angle, then the other two angles are at A and C, each of which is acute. So, angle BAC is one of the acute angles, which we need to find.CL is the bisector of angle C, so angle ACL is equal to angle LCB. Then, point L is on side AB because the bisector from angle C would go to side AB. Wait, but in a triangle, an angle bisector goes from a vertex to the opposite side. So, angle C is being bisected by CL, which would meet AB at point L. So, point L is on AB.Now, the problem states that point L is equidistant from vertex B and the midpoint of hypotenuse AC. The hypotenuse is AC because in a right triangle, the hypotenuse is opposite the right angle. Since the right angle is at B, the hypotenuse is AC, and the midpoint of AC would be, let's call it M. So, midpoint M of AC is equidistant from A and C. The problem says that L is equidistant from B and M. So, distance from L to B equals distance from L to M.Our goal is to find angle BAC. Let's denote angle BAC as θ. Then, angle ABC is 90 degrees, and angle BCA would be 90 - θ degrees. Since CL is the angle bisector of angle C (which is 90 - θ), then angle ACL and angle LCB are each (90 - θ)/2.I need to relate this information to the fact that L is equidistant from B and M. Let me think about coordinate geometry. Maybe setting up coordinates would help. Let me assign coordinates to the triangle.Let's place point B at the origin (0, 0). Since it's a right triangle at B, let's let point A be on the x-axis and point C on the y-axis. Wait, but then hypotenuse AC would be from (x, 0) to (0, y). Alternatively, maybe point A is at (a, 0), point B at (0, 0), and point C at (0, b). Then hypotenuse AC would be from (a, 0) to (0, b), midpoint M at (a/2, b/2).Then, angle BAC is θ, which is the angle at point A between BA and AC. Wait, no. Wait, angle BAC is the angle at point A. Wait, no. Wait, angle BAC is the angle at point A between BA and CA? Wait, no. In triangle ABC, angle BAC is the angle at point A between BA and CA. Wait, but if we have coordinates with B at (0,0), A at (a,0), and C at (0,b), then angle BAC is the angle at A between points B, A, and C. So, BA is from A to B, which is along the x-axis from (a,0) to (0,0), and AC is from A to C, which is from (a,0) to (0,b). So angle BAC is the angle at A between BA (which is along the negative x-axis) and AC (which goes up to (0,b)). So, that angle can be calculated using coordinates.Alternatively, maybe using trigonometry. In triangle ABC, with legs AB and BC, but wait, in this coordinate system, AB is along the x-axis from (0,0) to (a,0), BC is along the y-axis from (0,0) to (0,b), and AC is the hypotenuse from (a,0) to (0,b).Wait, in this case, angle at B is 90 degrees. Then, angle at A is angle BAC, which is θ, and angle at C is angle BCA, which is 90 - θ. So, tan(theta) would be opposite over adjacent. In triangle ABC, angle at A: the side opposite to angle A is BC, which has length b, and the adjacent side is AB, which has length a. So, tan(theta) = opposite/adjacent = BC/AB = b/a. So, theta = arctan(b/a). So, if we can find the ratio of b/a, we can find theta.Alternatively, maybe using coordinate geometry, since we can assign coordinates and then find the position of point L on AB such that CL bisects angle C, and L is equidistant from B and M.First, let's set up coordinates. Let me proceed step by step.Let me place point B at (0, 0), point A at (a, 0), and point C at (0, b). Then, hypotenuse AC is between (a, 0) and (0, b), and its midpoint M is at (a/2, b/2).Now, angle bisector CL. The angle bisector of angle C (which is at (0, b)) will meet AB at point L. So, we need to find the coordinates of point L on AB such that CL bisects angle C.In coordinate terms, AB is the line from (a, 0) to (0, 0). So, AB is along the x-axis from (0,0) to (a,0). Wait, no: point A is at (a,0), point B is at (0,0), so AB is from (0,0) to (a,0). Wait, no: if B is at (0,0), and ABC is a right triangle with right angle at B, then AB and BC are the legs. So, point A is at (a,0), point C is at (0,c), for some a and c. Then, hypotenuse AC is from (a,0) to (0,c), midpoint M is at (a/2, c/2).Now, CL is the angle bisector from point C to AB. So, point L is on AB. Let's find coordinates for L.The angle bisector theorem states that the angle bisector divides the opposite side into segments proportional to the adjacent sides. So, in triangle ABC, angle bisector CL divides AB into segments AL and LB such that AL / LB = AC / BC.Wait, angle bisector theorem: in triangle ABC, if CL is the bisector of angle C, then AL / LB = AC / BC.AC is the length from A to C, which is sqrt(a² + c²). BC is the length from B to C, which is c.So, AL / LB = sqrt(a² + c²) / c.But AB is the length from A to B, which is a. So, AB = AL + LB = a.Therefore, AL = [sqrt(a² + c²) / (sqrt(a² + c²) + c)] * aSimilarly, LB = [c / (sqrt(a² + c²) + c)] * aBut maybe working with coordinates is better.Since point L is on AB, which is the x-axis from (0,0) to (a,0). Let's denote point L as (l, 0), where 0 < l < a.We need to find l such that CL bisects angle C.Alternatively, using the angle bisector formula. The coordinates of L can be determined using the angle bisector theorem ratio.From angle bisector theorem:AL / LB = AC / BCAC = sqrt(a² + c²), BC = cTherefore, AL / LB = sqrt(a² + c²) / cBut AL = a - l, since AB is from (0,0) to (a,0), so AL is the distance from A (a,0) to L (l,0), which is a - l.Wait, no. Wait, AB is from A (a,0) to B (0,0). So, AL is the length from A to L, which is a - l, and LB is the length from L to B, which is l.Wait, but angle bisector theorem states that AL / LB = AC / BC. Wait, but AL is the segment adjacent to AC, and LB is the segment adjacent to BC. Wait, let me confirm.In triangle ABC, angle at C is being bisected by CL, meeting AB at L. Then, according to the angle bisector theorem, AL / LB = AC / BC. So, AL is adjacent to AC, and LB is adjacent to BC. So yes, AL / LB = AC / BC.So, AC = sqrt(a² + c²), BC = c. So, AL / LB = sqrt(a² + c²)/c.But AL is the distance from A to L, which is a - l, and LB is the distance from L to B, which is l. Therefore, (a - l)/l = sqrt(a² + c²)/cSolving for l:(a - l)/l = sqrt(a² + c²)/cCross-multiplying:c(a - l) = l sqrt(a² + c²)Expanding:ca - cl = l sqrt(a² + c²)Bring all terms to one side:ca = l sqrt(a² + c²) + clFactor out l on the right:ca = l (sqrt(a² + c²) + c)Therefore:l = ca / (sqrt(a² + c²) + c)Okay, so coordinates of L are (l, 0) = ( ca / (sqrt(a² + c²) + c ), 0 )Now, the problem states that L is equidistant from B and M.Point B is at (0,0), midpoint M is at (a/2, c/2). So, distance from L to B is LB = l, since L is on the x-axis at (l,0), so distance to B (0,0) is l.Distance from L to M: coordinates of L are (l,0), coordinates of M are (a/2, c/2). So, distance LM is sqrt( (a/2 - l)^2 + (c/2 - 0)^2 )Set LB = LM:l = sqrt( (a/2 - l)^2 + (c/2)^2 )Square both sides:l² = (a/2 - l)^2 + (c/2)^2Expand the right-hand side:(a/2 - l)^2 + (c/2)^2 = (a²/4 - a l + l²) + c²/4Therefore, l² = a²/4 - a l + l² + c²/4Subtract l² from both sides:0 = a²/4 - a l + c²/4Rearrange:a l = a²/4 + c²/4Multiply both sides by 4:4 a l = a² + c²So, 4 a l = a² + c²But from earlier, we had l = ca / (sqrt(a² + c²) + c )So, substitute l into 4 a l = a² + c²:4 a * [ ca / (sqrt(a² + c²) + c ) ] = a² + c²Simplify left-hand side:4 a * [ ca / (sqrt(a² + c²) + c ) ] = 4 a² c / (sqrt(a² + c²) + c )Set equal to a² + c²:4 a² c / (sqrt(a² + c²) + c ) = a² + c²Multiply both sides by (sqrt(a² + c²) + c ):4 a² c = (a² + c²)(sqrt(a² + c²) + c )Let me denote sqrt(a² + c²) as d. Then, we have:4 a² c = (d²)(d + c )But d² = a² + c², so:4 a² c = (a² + c²)(d + c )But let's keep it in terms of a and c. Let me expand the right-hand side:(a² + c²)(sqrt(a² + c²) + c ) = (a² + c²)sqrt(a² + c²) + c(a² + c²)So,4 a² c = (a² + c²)^(3/2) + c(a² + c²)Let me factor out (a² + c²):4 a² c = (a² + c²)[ sqrt(a² + c²) + c ]Let me divide both sides by (a² + c²):4 a² c / (a² + c²) = sqrt(a² + c²) + cLet me denote k = a / c. Then, a = k c. Let me substitute this into the equation.So, 4 (k c)² c / ( (k c)^2 + c² ) = sqrt( (k c)^2 + c² ) + cSimplify numerator and denominator:Numerator: 4 k² c² * c = 4 k² c³Denominator: k² c² + c² = c²(k² + 1)So, left-hand side becomes 4 k² c³ / [ c²(k² + 1) ] = 4 k² c / (k² + 1)Right-hand side: sqrt( k² c² + c² ) + c = sqrt( c²(k² + 1) ) + c = c sqrt(k² + 1) + c = c( sqrt(k² + 1 ) + 1 )So, equation becomes:4 k² c / (k² + 1 ) = c( sqrt(k² + 1 ) + 1 )Divide both sides by c:4 k² / (k² + 1 ) = sqrt(k² + 1 ) + 1Let me let t = sqrt(k² + 1). Then, t² = k² + 1, so k² = t² - 1.Substitute into equation:4 (t² - 1 ) / ( t² ) = t + 1Simplify left-hand side:4(t² - 1)/t² = 4 - 4/t²So, equation becomes:4 - 4/t² = t + 1Bring all terms to one side:4 - 4/t² - t - 1 = 0Simplify:3 - t - 4/t² = 0Multiply both sides by t² to eliminate denominators:3 t² - t³ - 4 = 0Rearrange:-t³ + 3 t² - 4 = 0Multiply by -1:t³ - 3 t² + 4 = 0So, we have the cubic equation:t³ - 3 t² + 4 = 0Let me try to factor this. Let's look for rational roots using Rational Root Theorem. Possible roots are ±1, ±2, ±4.Testing t=1: 1 - 3 + 4 = 2 ≠0t= -1: -1 - 3 +4=0. Wait, t= -1: (-1)^3 -3*(-1)^2 +4 = -1 -3 +4=0. So t=-1 is a root.Therefore, we can factor (t + 1)(quadratic) = t³ -3t² +4.Let me perform polynomial division or use synthetic division.Divide t³ -3t² +4 by (t +1).Using synthetic division:-1 | 1 -3 0 4Multiply -1 by 1, add to -3: -1*1= -1; -3 + (-1)= -4Multiply -1 by -4, add to 0: (-1)*(-4)=4; 0 +4=4Multiply -1 by 4, add to 4: (-1)*4= -4; 4 + (-4)=0.So, the cubic factors as (t +1)(t² -4t +4). Then, t² -4t +4 factors as (t -2)^2.Therefore, t³ -3t² +4= (t +1)(t -2)^2=0Solutions are t=-1, t=2 (double root). But t = sqrt(k² +1) ≥1, since k is a real number. So t=-1 is discarded, t=2 is the solution. Therefore, t=2.Recall that t= sqrt(k² +1)=2, so:sqrt(k² +1)=2Square both sides:k² +1=4Thus, k²=3, so k= sqrt(3) or k=-sqrt(3). Since k= a/c is a ratio of lengths, it must be positive. Therefore, k= sqrt(3). So a/c= sqrt(3), so a= sqrt(3) c.Therefore, in triangle ABC, the legs are AB = a = sqrt(3)c and BC = c. Then, hypotenuse AC= sqrt(a² + c²)= sqrt(3c² +c²)=sqrt(4c²)=2c.Therefore, triangle ABC is a 30-60-90 triangle. Because in a right triangle where the legs are in ratio sqrt(3) :1, the angles are 30°, 60°, and 90°. Specifically, angle opposite the shorter leg (BC =c) is 30°, and angle opposite the longer leg (AB= sqrt(3)c) is 60°.But wait, angle BAC is at point A. Let's confirm. In triangle ABC with right angle at B, AB is the leg adjacent to angle BAC. Wait, let's see:Point A is at (a,0)= (sqrt(3)c, 0), point B at (0,0), point C at (0,c). Then, angle BAC is the angle at point A between BA and AC.But BA is along the x-axis from A to B, and AC is the hypotenuse from A to C. So, in triangle ABC, angle at A (BAC) can be found using trigonometry.In triangle ABC, tan(angle BAC) = opposite/adjacent. The side opposite angle BAC is BC, which is length c, and the adjacent side is AB, which is length sqrt(3)c. Therefore, tan(theta)= c / (sqrt(3)c )=1/sqrt(3), so theta=30°.Alternatively, since the hypotenuse AC is 2c, and in a 30-60-90 triangle, the sides are 1, sqrt(3), 2. So, angle opposite the side of length 1 is 30°, opposite sqrt(3) is 60°, so angle at A is 30°, angle at C is 60°.Therefore, angle BAC is 30 degrees, which is π/6 radians.But let me verify this conclusion. Let's check if with a/c= sqrt(3), the conditions of the problem are satisfied.So, let's take c=1, so a=sqrt(3). Then, coordinates:A: (sqrt(3),0)B: (0,0)C: (0,1)Midpoint M of AC: (sqrt(3)/2, 0.5)Point L is on AB. Let's compute l using l=ca / (sqrt(a² + c²) + c )With c=1, a=sqrt(3):l= (1 * sqrt(3)) / (sqrt( (sqrt(3))² +1² ) +1 )= sqrt(3) / (sqrt(3 +1 ) +1 )= sqrt(3) / (2 +1 )= sqrt(3)/3 ≈0.577So coordinates of L are (sqrt(3)/3,0)Now, distance from L to B is sqrt( (sqrt(3)/3 -0)^2 + (0 -0)^2 )=sqrt(3)/3 ≈0.577Distance from L to M:Coordinates of M: (sqrt(3)/2, 0.5)Coordinates of L: (sqrt(3)/3, 0)Distance LM= sqrt( (sqrt(3)/2 - sqrt(3)/3)^2 + (0.5 -0)^2 )Compute differences:x-coordinate: sqrt(3)/2 - sqrt(3)/3 = sqrt(3)(3/6 -2/6)=sqrt(3)/6y-coordinate: 0.5 -0=0.5So, distance LM= sqrt( (sqrt(3)/6)^2 + (0.5)^2 )= sqrt( (3/36) + 0.25 )= sqrt( 1/12 + 1/4 )= sqrt( (1 +3)/12 )= sqrt(4/12 )= sqrt(1/3 )= 1/sqrt(3 ) ≈0.577Which is equal to sqrt(3)/3, which is the same as distance LB. So yes, in this case, LB=LM. Therefore, the conditions hold. Therefore, angle BAC is 30 degrees.Hence, the answer should be 30 degrees, which is π/6 radians, so in boxed form, boxed{30^circ}.Final Answerboxed{30^circ}</think>In a right triangle (ABC) with the right angle at (B), the angle bisector (CL) is drawn. It is given that point (L) is equidistant from vertex (B) and the midpoint (M) of the hypotenuse (AC). We need to find the angle (BAC).1. Assign Coordinates: - Place (B) at ((0, 0)), (A) at ((a, 0)), and (C) at ((0, b)). - Midpoint (M) of (AC) is (left(frac{a}{2}, frac{b}{2}right)).2. Angle Bisector Theorem: - Using the angle bisector theorem, the coordinates of (L) on (AB) are found. The ratio ( frac{AL}{LB} = frac{AC}{BC} ). - Solving for (L), we get ( l = frac{ab}{sqrt{a^2 + b^2} + b} ).3. Equidistant Condition: - Distance from (L) to (B) is ( l ). - Distance from (L) to (M) is ( sqrt{left(frac{a}{2} - lright)^2 + left(frac{b}{2}right)^2} ). - Setting these distances equal and solving, we find the relationship ( 4al = a^2 + b^2 ).4. Substitute and Solve: - Substitute ( l = frac{ab}{sqrt{a^2 + b^2} + b} ) into the equation ( 4al = a^2 + b^2 ). - Let ( k = frac{a}{b} ), leading to the cubic equation ( t^3 - 3t^2 + 4 = 0 ) where ( t = sqrt{k^2 + 1} ). - Solving, we find ( t = 2 ), leading to ( k = sqrt{3} ).5. Conclusion: - The ratio ( frac{a}{b} = sqrt{3} ), indicating a 30-60-90 triangle. - Angle ( BAC ) is ( 30^circ ).Thus, the angle ( BAC ) is (boxed{30^circ}).,<think>Okay, so I have this geometry problem here. Let me try to parse it step by step. It says in a right triangle ABC, the bisector CL is drawn. Point L is equidistant from vertex B (the right angle) and the midpoint of hypotenuse AC. We need to find the angle BAC. Hmm. Let me visualize this first.First, let me sketch the triangle mentally. Since it's a right triangle ABC with the right angle at B. So, ABC is right-angled at B. Let me note that: triangle ABC, right-angled at B. Then, CL is the bisector. Wait, angle bisector? Which angle? It says the bisector CL. Since CL is the bisector, point L must be on side AB or BC? Wait, in a triangle, angle bisectors go from a vertex to the opposite side. So, CL is an angle bisector. So, point C is one of the vertices. So, angle at C is being bisected by CL, and L is the point where the bisector meets the opposite side. But since ABC is right-angled at B, the sides are AB, BC, and hypotenuse AC.Wait, let me make sure. If ABC is right-angled at B, then AB and BC are the legs, and AC is the hypotenuse. So, angle at C is angle BCA, which is one of the acute angles. So, CL is the bisector of angle C, so it starts at C and splits angle BCA into two equal angles. Then, L is the point where this bisector meets the opposite side, which would be AB? Wait, no. Wait, in a triangle, the angle bisector from vertex C would go to the opposite side, which is AB. So, CL is the angle bisector from C to AB, meeting AB at L. So, point L is on AB.But the problem says that point L is equidistant from vertex B and the midpoint of hypotenuse AC. So, point L is equidistant to B and to the midpoint of AC. Let me denote the midpoint of AC as M. So, M is the midpoint of AC. Then, the distance from L to B is equal to the distance from L to M. So, LB = LM. We need to find angle BAC, which is the angle at vertex A. Let's denote angle BAC as α. Since ABC is a right-angled triangle at B, angle BAC and angle BCA are complementary, so angle BCA is 90° - α.Given that CL is the angle bisector of angle BCA, so angle LCB = angle LCA = (90° - α)/2. Hmm. Maybe we can use some properties of angle bisectors here, or coordinate geometry. Let me think about possible approaches.One approach is coordinate geometry. Let's place the triangle in coordinate system. Let me assign coordinates to the points. Let me put point B at the origin (0, 0). Then, since ABC is right-angled at B, let me put point A on the x-axis and point C on the y-axis. So, coordinates: B (0, 0), A (a, 0), C (0, c), where a and c are positive real numbers. Then, hypotenuse AC connects (a, 0) to (0, c). The midpoint M of AC would be at ((a/2), (c/2)).Now, angle bisector CL from point C (0, c) to side AB. Wait, side AB is from (a, 0) to (0, 0). So, AB is along the x-axis from (0,0) to (a,0). So, point L is somewhere on AB. Let me find coordinates of point L.By the angle bisector theorem, the angle bisector from C to AB divides AB into segments proportional to the adjacent sides. So, in triangle ABC, angle bisector CL divides AB into AL and LB such that AL / LB = AC / BC. Let me compute that.First, AC is the hypotenuse, so AC = sqrt(a² + c²). BC is the leg from B to C, which is length c. So, AL / LB = sqrt(a² + c²) / c. Therefore, AL = (sqrt(a² + c²)/c) * LB. But AL + LB = AB = a. So, substituting:Let me denote LB = x. Then AL = (sqrt(a² + c²)/c) * x. So, AL + LB = (sqrt(a² + c²)/c + 1) * x = a. Solving for x:x = a / (1 + sqrt(a² + c²)/c) = (a c) / (c + sqrt(a² + c²))Thus, AL = (sqrt(a² + c²)/c) * x = (sqrt(a² + c²)/c) * (a c)/(c + sqrt(a² + c²)) ) = (a sqrt(a² + c²)) / (c + sqrt(a² + c²))Therefore, the coordinates of point L can be determined as follows. Since AB is from (0,0) to (a, 0), the coordinates of L will be (LB, 0) since starting from B at (0,0), moving LB along AB which is the x-axis. Wait, actually, if LB is the length from L to B, then since AB is along the x-axis from (0,0) to (a,0), point L is at (a - AL, 0)? Wait, no. Wait, if AL is the length from A to L, then since AB is length a, AL + LB = a. So, if AL is as computed above, then the coordinate of L would be (AL, 0) measured from A? Wait, no. Let me clarify.Point A is at (a, 0), point B is at (0,0). So, AB is from (a,0) to (0,0). So, the distance from A to L is AL, and from L to B is LB. Therefore, the coordinate of L is (LB, 0) because starting from B at (0,0), moving LB along AB (which is the x-axis) towards A. Wait, no. If we move from B towards A, the x-coordinate increases from 0 to a. So, point L is at (LB, 0), but LB is the length from L to B. Wait, no. Wait, LB is the length from L to B, so if AB has length a, then LB is the distance from L to B, which would mean that the coordinate of L is (a - AL, 0). Wait, this is confusing. Let me better parameterize AB.Since AB is from (a,0) to (0,0), the coordinates of any point on AB can be written as (a - t, 0) where t is the distance from A. Alternatively, since AB is along the x-axis, moving from B (0,0) to A (a,0), the coordinates can be parameterized as (x, 0) where x ranges from 0 to a. So, point L is somewhere on AB, so its coordinates are (x, 0) for some x between 0 and a. Then, AL is the distance from A (a,0) to L (x,0), which is |a - x|. Similarly, LB is the distance from L (x,0) to B (0,0), which is |x - 0| = x. So, since AL / LB = AC / BC, as per angle bisector theorem, we have (a - x)/x = AC / BC.Compute AC: length from A (a,0) to C (0,c): sqrt(a² + c²). BC is length from B (0,0) to C (0,c): c. So, (a - x)/x = sqrt(a² + c²)/c. Solving for x:(a - x)/x = sqrt(a² + c²)/cMultiply both sides by x:a - x = (sqrt(a² + c²)/c) xBring all terms with x to one side:a = x (sqrt(a² + c²)/c + 1)Therefore:x = a / (1 + sqrt(a² + c²)/c) = (a c) / (c + sqrt(a² + c²))So, x = (a c) / (c + sqrt(a² + c²))Therefore, coordinates of L are (x, 0) = ( (a c)/(c + sqrt(a² + c²)), 0 )Now, midpoint M of AC is at ((a/2), (c/2)).We are told that L is equidistant from B and M. So, distance from L to B equals distance from L to M.Compute LB: already defined as x, which is (a c)/(c + sqrt(a² + c²)).Distance LM: coordinates of L are (x, 0), coordinates of M are (a/2, c/2). So, LM is sqrt( (a/2 - x)^2 + (c/2 - 0)^2 )Set LB = LM:x = sqrt( (a/2 - x)^2 + (c/2)^2 )Square both sides:x² = (a/2 - x)^2 + (c/2)^2Expand right-hand side:= (a²/4 - a x + x²) + c²/4Therefore:x² = a²/4 - a x + x² + c²/4Subtract x² from both sides:0 = a²/4 - a x + c²/4Rearranged:a x = (a² + c²)/4Therefore:x = (a² + c²)/(4 a)But earlier, we have x = (a c)/(c + sqrt(a² + c²))Therefore, setting them equal:(a c)/(c + sqrt(a² + c²)) = (a² + c²)/(4 a)Multiply both sides by (c + sqrt(a² + c²)) and 4a:4 a * a c = (a² + c²)(c + sqrt(a² + c²))Simplify left side: 4 a² cRight side: (a² + c²)c + (a² + c²)sqrt(a² + c²) = c(a² + c²) + (a² + c²)^{3/2}So, equation becomes:4 a² c = c(a² + c²) + (a² + c²)^{3/2}Let me factor out (a² + c²) on the right side:= (a² + c²)[c + sqrt(a² + c²)]Therefore:4 a² c = (a² + c²)[c + sqrt(a² + c²)]Divide both sides by [c + sqrt(a² + c²)]:4 a² c / [c + sqrt(a² + c²)] = a² + c²But note that from earlier, x = (a c)/(c + sqrt(a² + c²)) = (a² + c²)/(4 a)Wait, perhaps substitute t = sqrt(a² + c²). Let me denote t = sqrt(a² + c²). Then, t² = a² + c².Then, the equation becomes:4 a² c = (t²)(c + t)But t² = a² + c². So:4 a² c = (a² + c²)(c + t)Hmm. Let's write this as:4 a² c = (a² + c²)c + (a² + c²) tBut (a² + c²) t = t³, since t = sqrt(a² + c²), so t³ = (a² + c²)^{3/2}Wait, but maybe we can express a² in terms of t. Let me see.Alternatively, let's divide both sides by c:4 a² = (a² + c²) [1 + t/c]But t = sqrt(a² + c²). Let me write t/c = sqrt( (a² + c²)/c² ) = sqrt( (a/c)^2 + 1 )Let me set k = a/c, so that a = k c. Then, t = sqrt( (k² c² + c² ) ) = c sqrt(k² + 1 )So, substitute into the equation:4 a² = (a² + c²) [1 + t/c ]Expressed in terms of k:4 (k c)^2 = ( (k c)^2 + c^2 ) [1 + (c sqrt(k² + 1 )) / c ]Simplify:4 k² c² = (k² c² + c²)(1 + sqrt(k² + 1 ))Factor c² on both sides:4 k² c² = c² (k² + 1)(1 + sqrt(k² + 1 ))Divide both sides by c²:4 k² = (k² + 1)(1 + sqrt(k² + 1 ))Let me denote s = sqrt(k² + 1 ). Then, s² = k² + 1, so k² = s² - 1. Substitute into equation:4 (s² - 1) = (s²)(1 + s )Expand left side:4 s² - 4 = s² + s³Bring all terms to left side:4 s² - 4 - s² - s³ = 0Simplify:- s³ + 3 s² - 4 = 0Multiply both sides by -1:s³ - 3 s² + 4 = 0So, we have the cubic equation s³ - 3 s² + 4 = 0.Let me try to factor this cubic. Let's check possible rational roots using Rational Root Theorem. Possible roots are factors of 4 over factors of 1: ±1, ±2, ±4.Test s = 1: 1 - 3 + 4 = 2 ≠ 0.s = -1: -1 - 3 + 4 = 0. Wait, s = -1: (-1)^3 - 3*(-1)^2 +4 = -1 -3 +4=0. So, s = -1 is a root. But since s = sqrt(k² + 1), which is positive, s cannot be negative. So, s = -1 is a root but not acceptable here. Let's factor out (s + 1):Using polynomial division or synthetic division:Divide s³ - 3 s² + 4 by (s + 1). Coefficients: 1 | -3 | 0 | 4 (Wait, original polynomial is s³ -3 s² +0 s +4)Using synthetic division for root s = -1:-1 | 1 -3 0 4 -1 4 -4 1 -4 4 0So, the cubic factors as (s + 1)(s² - 4 s + 4) = 0.Then, s² - 4 s + 4 factors as (s - 2)^2. So, the cubic is (s + 1)(s - 2)^2 = 0. So, roots are s = -1 and s = 2 (double root). Since s must be positive, s = 2 is the only valid root. Therefore, s = 2.Recall that s = sqrt(k² + 1) = 2, so sqrt(k² + 1) = 2. Then, k² + 1 = 4 => k² = 3 => k = sqrt(3). Since k = a/c, so a/c = sqrt(3) => a = c sqrt(3).Therefore, the ratio of the legs is a : c = sqrt(3) : 1. So, in triangle ABC, right-angled at B, the legs AB and BC are in the ratio sqrt(3) : 1.Therefore, angle BAC can be found by tan(α) = opposite/adjacent = BC / AB. Wait, in triangle ABC, angle at A is α = angle BAC. The sides opposite to α is BC, and adjacent is AB. So, tan(α) = BC / AB.Wait, BC is the length from B to C, which is c, and AB is the length from A to B, which is a = c sqrt(3). So, tan(α) = c / (c sqrt(3)) = 1/sqrt(3). Therefore, α = arctan(1/sqrt(3)) = 30°. So, angle BAC is 30 degrees.Wait, let me check that again. If angle BAC is α, then tan(α) = opposite / adjacent = BC / AB. But BC is the side opposite angle A, right? Wait, in triangle ABC, angle at A is α. The side opposite angle A is BC, which has length c. The side adjacent to angle A is AB, which has length a. So, tan(α) = BC / AB = c / a. But since a = c sqrt(3), then tan(α) = 1 / sqrt(3), so α = 30°, yes.Alternatively, since in a right-angled triangle, if the legs are in ratio 1 : sqrt(3), then the angles are 30°, 60°, 90°. So, yes, angle opposite the shorter leg (which is BC = c, if a = c sqrt(3)) is 30°, which is angle BAC.Therefore, the angle BAC is 30 degrees. So, the answer should be 30°, which in boxed form is boxed{30^circ}.But wait, let me confirm all steps once more to be sure.1. Placed triangle in coordinate system with B at (0,0), A at (a,0), C at (0,c).2. Midpoint M of AC is (a/2, c/2).3. Angle bisector CL intersects AB at L. Found coordinates of L using angle bisector theorem.4. Expressed the condition that LB = LM.5. Derived equation relating a and c, eventually leading to the ratio a/c = sqrt(3).6. Concluded that angle BAC is 30°, since tan(α) = 1/sqrt(3).All steps seem consistent. Also, checked the cubic equation solution, got s = 2, leading to a/c = sqrt(3). So, the angle is indeed 30 degrees. So, I think that's correct.Final AnswerThe angle ( BAC ) is boxed{30^circ}.</think>In a right triangle (ABC) with the right angle at (B), the angle bisector (CL) is drawn. Point (L) is equidistant from vertex (B) and the midpoint (M) of the hypotenuse (AC). We need to find the angle (BAC).1. Coordinate System Setup: - Place (B) at the origin ((0, 0)), (A) at ((a, 0)), and (C) at ((0, c)). - The midpoint (M) of (AC) is (left(frac{a}{2}, frac{c}{2}right)).2. Angle Bisector Theorem: - Using the angle bisector theorem, the coordinates of (L) on (AB) are determined. The ratio ( frac{AL}{LB} = frac{AC}{BC} ). - Calculations show the coordinates of (L) are (left(frac{ac}{c + sqrt{a^2 + c^2}}, 0right)).3. Equidistant Condition: - The distance (LB = frac{ac}{c + sqrt{a^2 + c^2}}). - The distance (LM) is calculated using the coordinates of (L) and (M), leading to the equation: [ frac{ac}{c + sqrt{a^2 + c^2}} = sqrt{left(frac{a}{2} - frac{ac}{c + sqrt{a^2 + c^2}}right)^2 + left(frac{c}{2}right)^2} ]4. Solving the Equation: - Simplifying the equation leads to solving a cubic equation in terms of (s = sqrt{k^2 + 1}) where (k = frac{a}{c}). - The valid solution (s = 2) gives (k = sqrt{3}), meaning (a = csqrt{3}).5. Determining the Angle: - The ratio ( frac{BC}{AB} = frac{c}{a} = frac{1}{sqrt{3}} ). - Therefore, (tan(angle BAC) = frac{1}{sqrt{3}}), leading to (angle BAC = 30^circ).The angle (BAC) is (boxed{30^circ}).]

🔑:square Answer: 30^{circ}.Let M be the midpoint of AC. Drop a perpendicular LN from point L to the hypotenuse AC. Since L lies on the bisector of angle BCA, then LB = LN. By the condition, LB = LM. Therefore, points M and N coincide. Fig. 8. Then LM is both a median and an altitude in triangle ALC, i.e., this triangle is isosceles. Therefore, angle LAC = angle ACL = angle BCL. At the same time, angle LAC + angle ACL + angle BCL = 90^{circ}. Thus, angle BAC = angle LAC = 30^{circ}.(F. Nilov)

🧠:[<think>Okay, let's tackle this geometry problem step by step. First, let me read it again to make sure I understand all the given conditions.We have triangle ABC with two equations:1. (2sin^{2}frac{A}{2} = sqrt{3}sin A)2. (sin(B - C) = 2cos B sin C)And we need to find the value of (frac{AC}{AB}).Hmm, so we need to find the ratio of sides AC to AB. In triangle terms, sides opposite angles are proportional to the sines of their opposite angles. So, by the Law of Sines, (frac{AC}{AB} = frac{sin B}{sin C}). Therefore, if I can find the ratio of (sin B) to (sin C), that would give me the answer. So maybe the key is to find angles B and C first?Let's start with the first equation: (2sin^{2}frac{A}{2} = sqrt{3}sin A). Maybe we can solve for angle A first. Let me recall some trigonometric identities that might help here.The double-angle identity for sine: (sin A = 2sinfrac{A}{2}cosfrac{A}{2}). Also, the identity for (sin^{2}frac{A}{2}) is related to the cosine of A. Specifically, (sin^{2}frac{A}{2} = frac{1 - cos A}{2}). Let me substitute that into the equation.So substituting into the left side:(2 times frac{1 - cos A}{2} = sqrt{3}sin A)Simplify:(1 - cos A = sqrt{3}sin A)Hmm, that's a simpler equation. Let me write that as:(1 = sqrt{3}sin A + cos A)This looks like a linear combination of sine and cosine. Maybe I can express this as a single sine or cosine function. The standard approach is to write it as (Rsin(A + phi)) or (Rcos(A - theta)), where R is the amplitude and (phi) or (theta) is the phase shift.Let's see. Let me recall that (asin x + bcos x = Rsin(x + phi)), where (R = sqrt{a^2 + b^2}) and (tanphi = frac{b}{a}). Alternatively, it can also be written as (Rcos(x - theta)), where (R = sqrt{a^2 + b^2}) and (tantheta = frac{a}{b}).In this case, we have (sqrt{3}sin A + cos A = 1). Let's compute R:(R = sqrt{(sqrt{3})^2 + 1^2} = sqrt{3 + 1} = 2).So, we can write this as:(2sin(A + phi) = 1), where (tanphi = frac{1}{sqrt{3}}). Since (tanphi = frac{1}{sqrt{3}}), that means (phi = 30^circ) or (frac{pi}{6}) radians.Alternatively, since the coefficients are (sqrt{3}) (for sin) and 1 (for cos), perhaps it's better to write it as:(sqrt{3}sin A + cos A = 2 times frac{sqrt{3}}{2}sin A + 2 times frac{1}{2}cos A = 2sinleft(A + 30^circright)). Wait, is that correct?Wait, let me check:Using the identity (asin x + bcos x = Rsin(x + phi)), here (a = sqrt{3}), (b = 1), so R is 2 as calculated. Then, (sinphi = frac{b}{R} = frac{1}{2}), so (phi = 30^circ). Therefore, (sqrt{3}sin A + cos A = 2sin(A + 30^circ)). Therefore, the equation becomes:(2sin(A + 30^circ) = 1)Hence,(sin(A + 30^circ) = frac{1}{2})The solutions to this equation are:(A + 30^circ = 30^circ + 360^circ k) or (A + 30^circ = 150^circ + 360^circ k), where k is any integer.Therefore,Case 1: (A + 30^circ = 30^circ + 360^circ k) ⇒ (A = 0^circ + 360^circ k). But angle A in a triangle can't be 0°, so discard.Case 2: (A + 30^circ = 150^circ + 360^circ k) ⇒ (A = 120^circ + 360^circ k). Since angles in a triangle are less than 180°, the only valid solution is A = 120°.So angle A is 120°. That's one piece of information.Now, moving on to the second equation: (sin(B - C) = 2cos B sin C). Let's try to simplify this equation.First, let's recall that (sin(B - C) = sin B cos C - cos B sin C). So substituting into the left side:(sin B cos C - cos B sin C = 2cos B sin C)Let's bring all terms to one side:(sin B cos C - cos B sin C - 2cos B sin C = 0)Combine like terms:(sin B cos C - 3cos B sin C = 0)Hmm, so:(sin B cos C = 3cos B sin C)Divide both sides by (cos B cos C) (assuming neither cos B nor cos C is zero, which they can't be in a triangle with angles B and C that are less than 90°, maybe? Wait, since A is 120°, then B and C must add up to 60°, so each is less than 60°, so cos B and cos C are positive and not zero. So division is okay.So:(frac{sin B}{cos B} = 3frac{sin C}{cos C})Which simplifies to:(tan B = 3 tan C)So we have that (tan B = 3 tan C). That's a useful relation.Since in triangle ABC, angles A + B + C = 180°, and we already know that A = 120°, so B + C = 60°. Let me denote angle B as x and angle C as y. So x + y = 60°, and tan x = 3 tan y.We need to solve for x and y. Let's write y = 60° - x. Then:tan x = 3 tan(60° - x)Let me compute tan(60° - x) using the tangent subtraction formula:tan(60° - x) = (frac{tan 60° - tan x}{1 + tan 60° tan x}) = (frac{sqrt{3} - tan x}{1 + sqrt{3} tan x})Therefore, substituting into tan x = 3 tan(60° - x):tan x = 3 * (frac{sqrt{3} - tan x}{1 + sqrt{3} tan x})Let me set t = tan x for simplicity. Then the equation becomes:t = 3 * (frac{sqrt{3} - t}{1 + sqrt{3} t})Multiply both sides by (1 + √3 t):t(1 + √3 t) = 3(√3 - t)Expand left side:t + √3 t² = 3√3 - 3tBring all terms to left side:t + √3 t² + 3t - 3√3 = 0Combine like terms:4t + √3 t² - 3√3 = 0Let's write this as:√3 t² + 4t - 3√3 = 0Multiply both sides by √3 to rationalize:3 t² + 4√3 t - 9 = 0Now, this is a quadratic equation in t. Let's use the quadratic formula:t = [-4√3 ± √( (4√3)^2 - 4*3*(-9) ) ] / (2*3)Calculate discriminant:(4√3)^2 - 4*3*(-9) = 16*3 + 108 = 48 + 108 = 156So sqrt(156) = sqrt(4*39) = 2*sqrt(39)Therefore,t = [ -4√3 ± 2√39 ] / 6 = [ -2√3 ± √39 ] / 3Now, since x is an angle between 0° and 60°, tan x must be positive. So we discard the negative root.Compute both roots:First root: [ -2√3 + √39 ] / 3Second root: [ -2√3 - √39 ] / 3 (negative, discard)So t = [ -2√3 + √39 ] / 3Let me approximate the value to check:√3 ≈ 1.732, √39 ≈ 6.245So t ≈ [ -2*1.732 + 6.245 ] / 3 ≈ [ -3.464 + 6.245 ] /3 ≈ 2.781 /3 ≈ 0.927So tan x ≈ 0.927 ⇒ x ≈ arctan(0.927) ≈ 43° (since tan 40° ≈ 0.839, tan 45°=1, so approx 43°)But since x + y = 60°, y ≈ 17°, let's check tan y:tan y = tan(60° - x) ≈ tan(17°) ≈ 0.305, and 3 tan y ≈ 0.916, which is close to tan x ≈0.927. Close enough considering approximation.But we need the exact value. Let's see if we can find an exact angle.Alternatively, maybe the exact value relates to some special angles? But given the quadratic solution, it's probably not a standard angle. Therefore, perhaps we can find sin B and sin C using the relation tan B = 3 tan C, and B + C = 60°, then compute the ratio sin B / sin C.Alternatively, since we need to find AC / AB = sin B / sin C, as per the Law of Sines.Let me denote this ratio as k = sin B / sin C. Our goal is to find k.Given that tan B = 3 tan C, and B + C = 60°, let's express tan B and tan C in terms of k.But first, let's express sin B and sin C in terms of angles. Let me consider B and C such that B + C = 60°, tan B = 3 tan C.Let me let C = θ, so B = 60° - θ.Then tan(60° - θ) = 3 tan θAgain, same equation as before. Let's use this substitution.tan(60° - θ) = [tan 60° - tan θ] / [1 + tan 60° tan θ] = [√3 - tan θ] / [1 + √3 tan θ] = 3 tan θSo:[√3 - tan θ] / [1 + √3 tan θ] = 3 tan θMultiply both sides by denominator:√3 - tan θ = 3 tan θ (1 + √3 tan θ )Expand RHS:3 tan θ + 3√3 tan² θBring all terms to left:√3 - tan θ - 3 tan θ - 3√3 tan² θ = 0Simplify:√3 - 4 tan θ - 3√3 tan² θ = 0Multiply both sides by -1:-√3 + 4 tan θ + 3√3 tan² θ = 0Rearranged:3√3 tan² θ + 4 tan θ - √3 = 0This is similar to the quadratic we had before. Let me let u = tan θ.Then:3√3 u² + 4 u - √3 = 0Quadratic in u:a = 3√3, b = 4, c = -√3Discriminant D = b² - 4ac = 16 - 4*(3√3)*(-√3) = 16 - 4*(-3*3) = 16 + 36 = 52Solutions:u = [ -4 ± √52 ] / (2*3√3 ) = [ -4 ± 2√13 ] / (6√3 ) = [ -2 ± √13 ] / (3√3 )Again, since θ is between 0° and 60°, tan θ is positive. So take the positive root:u = [ -2 + √13 ] / (3√3 )Rationalize the denominator:Multiply numerator and denominator by √3:u = [ (-2 + √13 )√3 ] / (3*3 ) = [ (-2√3 + √39 ) ] / 9So tan θ = [ √39 - 2√3 ] / 9Hmm, that's messy. Maybe not the most useful.Alternatively, since we need sin B / sin C = k.Given that B = 60° - C, so k = sin(60° - C)/sin CUsing sine subtraction formula:sin(60° - C) = sin 60° cos C - cos 60° sin C = (√3/2)cos C - (1/2)sin CTherefore,k = [ (√3/2)cos C - (1/2)sin C ] / sin C = (√3/2) cot C - 1/2But from tan B = 3 tan C, and B = 60° - C, so tan(60° - C) = 3 tan CLet me denote t = tan C. Then tan(60° - C) = [√3 - t]/[1 + √3 t] = 3 tSo:[√3 - t] = 3 t (1 + √3 t )As before, leading to the quadratic equation. However, perhaps instead of solving for the angle, we can express cot C in terms of t.If t = tan C, then cot C = 1/t. So, let's see:From the previous equation, we have tan(60° - C) = 3 tan C, which gives:[√3 - t]/[1 + √3 t] = 3 tCross-multiplying:√3 - t = 3 t (1 + √3 t )Which gives:√3 - t = 3 t + 3√3 t²Bring all terms to left:√3 - t - 3 t - 3√3 t² = 0 ⇒ √3 -4 t - 3√3 t² = 0Multiply by -1:-√3 +4 t +3√3 t² = 0 ⇒ 3√3 t² +4 t -√3 =0This is the same quadratic as before. Solving for t:t = [ -4 ± √(16 + 4*3√3*√3 ) ] / (2*3√3 )Wait, discriminant D = 16 + 4*3√3*√3 = 16 + 4*3*3 = 16 + 36 = 52. So same as before.Therefore, t = [ -4 ± √52 ] / (6√3 ) = [ -4 ± 2√13 ] / (6√3 ) = [ -2 ± √13 ] / (3√3 )Again, taking the positive root:t = [ -2 + √13 ] / (3√3 )So tan C = [ -2 + √13 ] / (3√3 )Therefore, cot C = 1/t = 3√3 / (-2 + √13 )Multiply numerator and denominator by (√13 + 2) to rationalize:cot C = 3√3 (√13 + 2 ) / [ (-2 + √13)(√13 + 2 ) ] = 3√3 (√13 + 2 ) / [ (√13)^2 - (2)^2 ] = 3√3 (√13 + 2 ) / (13 - 4 ) = 3√3 (√13 + 2 ) / 9 = √3 (√13 + 2 ) / 3Therefore, cot C = [√3 (√13 + 2 )]/3Plugging back into k:k = (√3/2) cot C - 1/2 = (√3/2)( [√3 (√13 + 2 )]/3 ) - 1/2Compute first term:(√3/2)( [√3 (√13 + 2 )]/3 ) = (3/2)(√13 + 2 ) /3 ) = ( (√13 + 2 ) / 2 )Then subtract 1/2:k = (√13 + 2 ) /2 - 1/2 = (√13 + 2 -1)/2 = (√13 +1)/2Therefore, the ratio AC / AB = sin B / sin C = k = (√13 +1)/2But let me check this because the steps were a bit involved.Wait, starting from:k = (√3/2)cot C - 1/2We found cot C = √3 (√13 + 2 ) /3So:k = (√3/2)( √3 (√13 + 2 ) /3 ) - 1/2Simplify first term:(√3 * √3 ) / (2*3 ) * (√13 + 2 ) = (3 /6 ) (√13 +2 ) = (1/2)(√13 + 2 )Then subtract 1/2:(1/2)(√13 +2 ) - 1/2 = (1/2)(√13 +2 -1 ) = (1/2)(√13 +1 )Therefore, k = (√13 +1)/2 ≈ (3.605 +1)/2 ≈ 4.605/2 ≈2.3025But earlier, when we approximated angles, we had angle B ≈43°, angle C≈17°, so sin B / sin C ≈ sin43 / sin17 ≈0.682 /0.292≈2.336, which is close to 2.3025. The discrepancy is due to approximation errors. Therefore, the exact value is (√13 +1)/2. So that's the ratio AC/AB.Alternatively, maybe there's a better way. Let me try another approach.Given that in triangle ABC, angle A =120°, and tan B =3 tan C, with B + C =60°, and want to find AC/AB = sin B / sin C.Let me use the sine rule. Let me denote sides as follows: AB = c, BC = a, AC = b. Then by Law of Sines:a / sin A = b / sin B = c / sin C.We need to find b/c = sin B / sin C.Given that angle A =120°, so angles B + C =60°, and tan B =3 tan C.Let me set angle C = θ, so angle B=60° - θ.Then, tan(60° - θ) =3 tan θ.As before. Let's express sin B / sin C = sin(60° - θ)/sin θ.Expressed as:sin(60° - θ)/sin θ = [ sin60 cosθ - cos60 sinθ ] / sinθ = sin60 cotθ - cos60Which is (√3/2)cotθ -1/2If we can find cotθ, then we can compute this.From tan B =3 tan C, which is tan(60° - θ)=3 tanθ.Let me set t=tanθ, then tan(60° - θ)= [√3 - t]/[1 +√3 t] =3 tAs before. Which leads to:√3 - t =3 t (1 +√3 t )Which gives √3 - t =3 t +3√3 t²Bring all terms to left:3√3 t² +4t -√3=0Which is the same quadratic as before. So same solution for t=tanθ= [ -2 +√13 ] / (3√3 )Then, cotθ=1/t= 3√3 / ( -2 +√13 )Multiply numerator and denominator by (√13 +2 ):cotθ= 3√3 (√13 +2 ) / [ (-2 +√13 )(√13 +2 ) ] = 3√3 (√13 +2 ) / (13 -4 )= 3√3 (√13 +2 ) /9= √3 (√13 +2 ) /3Therefore, cotθ=√3 (√13 +2 ) /3Plug back into expression for sin B / sin C:(√3/2)(√3 (√13 +2 ) /3 ) -1/2= (3/2)( (√13 +2 ) /3 ) -1/2= ( (√13 +2 ) /2 ) -1/2= (√13 +2 -1)/2= (√13 +1)/2So yes, same result. Therefore, AC/AB=(√13 +1)/2Therefore, the answer is boxed{dfrac{1 + sqrt{13}}{2}}.Wait, but usually, the order is to write the number in the numerator in ascending order, but since 1 + √13 is already simplified, it's fine. Alternatively, it can be written as (sqrt{13} +1)/2, which is the same thing.Therefore, the final answer is boxed{dfrac{1 + sqrt{13}}{2}}.Final Answerboxed{dfrac{1 + sqrt{13}}{2}}</think>Given the problem in (triangle ABC) with the conditions (2sin^{2}frac{A}{2} = sqrt{3}sin A) and (sin(B - C) = 2cos B sin C), we need to find the value of (frac{AC}{AB}).First, we solve for angle (A) using the given equation (2sin^{2}frac{A}{2} = sqrt{3}sin A). Using the identity (sin^{2}frac{A}{2} = frac{1 - cos A}{2}), we substitute and simplify:[2 times frac{1 - cos A}{2} = sqrt{3}sin A implies 1 - cos A = sqrt{3}sin A]Rewriting this as a linear combination of sine and cosine:[1 = sqrt{3}sin A + cos A]We recognize this as (2sin(A + 30^circ) = 1), leading to:[sin(A + 30^circ) = frac{1}{2} implies A + 30^circ = 150^circ implies A = 120^circ]Next, we use the second condition (sin(B - C) = 2cos B sin C). Expanding and simplifying using trigonometric identities:[sin B cos C - cos B sin C = 2cos B sin C implies sin B cos C = 3cos B sin C implies tan B = 3 tan C]Given (B + C = 60^circ), we let (B = 60^circ - C). Substituting into the tangent condition:[tan(60^circ - C) = 3 tan C]Using the tangent subtraction formula and solving the resulting quadratic equation, we find:[tan C = frac{-2 + sqrt{13}}{3sqrt{3}}]Using the Law of Sines, the ratio (frac{AC}{AB} = frac{sin B}{sin C}). Expressing (sin B) and (sin C) in terms of angles and solving, we find:[frac{sin B}{sin C} = frac{sqrt{13} + 1}{2}]Thus, the value of (frac{AC}{AB}) is:[boxed{dfrac{1 + sqrt{13}}{2}}],<think>Okay, so I have this geometry problem here about triangle ABC. It says two things: first, that 2 times the square of sine of half angle A equals the square root of 3 times sine of angle A. Second, that the sine of (B minus C) equals 2 times cosine B times sine C. Then I need to find the value of AC divided by AB. Alright, let's break this down step by step.Starting with the first equation: 2 sin²(A/2) = √3 sin A. Hmm. Maybe I can use some trigonometric identities to simplify this. I remember that sin²(theta) can be written using the double-angle identity. Specifically, sin²(theta) = (1 - cos 2theta)/2. So if I apply that here, replacing theta with A/2, then sin²(A/2) would be (1 - cos A)/2. Let me try that.So substituting into the left-hand side: 2 * [(1 - cos A)/2] = (1 - cos A). So the left-hand side simplifies to 1 - cos A. The right-hand side is √3 sin A. So now the equation becomes:1 - cos A = √3 sin A.Hmm, okay. Now, maybe I can use another identity here. I recall that expressions of the form a cos A + b sin A can be written as R sin(A + phi) or R cos(A - phi) or something like that. Maybe this equation can be rearranged into such a form. Let's see. Let's move all terms to one side:1 = √3 sin A + cos A.Yes, so that's 1 = √3 sin A + cos A. Now, if I can express the right-hand side as a single sine or cosine function, maybe I can solve for A.The general formula is that a sin x + b cos x = R sin(x + phi), where R = sqrt(a² + b²) and phi is some phase shift. Let's compute R here. Here, a is √3 and b is 1. So R would be sqrt( (√3)² + 1² ) = sqrt(3 + 1) = sqrt(4) = 2. So R is 2. Then, phi is arctan(b/a) or something? Wait, actually, the formula is a sin x + b cos x = R sin(x + phi), where phi = arctan(b/a) if I remember correctly. Wait, let me check:Wait, actually, the formula can be written as R sin(x + phi) = R sin x cos phi + R cos x sin phi. Comparing this to a sin x + b cos x, we have:a = R cos phib = R sin phiSo, tan phi = b/a. Therefore, phi = arctan(b/a). Here, a is the coefficient of sin x, which is √3, and b is the coefficient of cos x, which is 1. So tan phi = 1/√3, which means phi is 30 degrees, or pi/6 radians.Therefore, √3 sin A + cos A = 2 sin(A + 30°). Therefore, the equation becomes:2 sin(A + 30°) = 1So sin(A + 30°) = 1/2The solutions to sin(theta) = 1/2 are theta = 30° + 360°n or theta = 150° + 360°n, where n is an integer. Since A is an angle in a triangle, it's between 0° and 180°, so theta = A + 30° must be between 30° and 210°. Therefore, possible solutions are theta = 30° or 150°, but 30° would lead to A = 0°, which is impossible. So theta must be 150°, so:A + 30° = 150° => A = 120°.So angle A is 120 degrees. That's good progress.Now, moving on to the second equation: sin(B - C) = 2 cos B sin C.Hmm, let's see. Maybe I can expand the left-hand side using the sine subtraction formula: sin(B - C) = sin B cos C - cos B sin C. So substituting into the equation:sin B cos C - cos B sin C = 2 cos B sin CLet's bring all terms to one side:sin B cos C - cos B sin C - 2 cos B sin C = 0Combine like terms:sin B cos C - 3 cos B sin C = 0Hmm, so sin B cos C = 3 cos B sin C.Let me divide both sides by cos B cos C, assuming that cos B and cos C are not zero. Since B and C are angles in a triangle, and angle A is 120°, then angles B and C must sum to 60°, so each of them is less than 60°, so their cosines should be positive, right? Because angles less than 90° have positive cosine. So dividing by cos B cos C is safe.So:(sin B / cos B) = 3 (sin C / cos C)Which simplifies to:tan B = 3 tan CSo tan B = 3 tan C. Alright, so that relates angles B and C. Since angles in a triangle sum to 180°, and angle A is 120°, we have B + C = 60°. Let me note that down:B + C = 60°tan B = 3 tan CSo perhaps we can set up variables here. Let me let angle C = x, then angle B = 60° - x.Therefore, tan(60° - x) = 3 tan x.Now, let's expand tan(60° - x). Using the tangent subtraction formula:tan(60° - x) = (tan 60° - tan x) / (1 + tan 60° tan x)We know that tan 60° is √3, so:tan(60° - x) = (√3 - tan x) / (1 + √3 tan x)Set this equal to 3 tan x:(√3 - tan x) / (1 + √3 tan x) = 3 tan xMultiply both sides by (1 + √3 tan x):√3 - tan x = 3 tan x (1 + √3 tan x)Expand the right-hand side:√3 - tan x = 3 tan x + 3√3 tan² xBring all terms to one side:√3 - tan x - 3 tan x - 3√3 tan² x = 0Combine like terms:√3 - 4 tan x - 3√3 tan² x = 0Let me rearrange this:-3√3 tan² x -4 tan x + √3 = 0Multiply both sides by -1 to make the leading coefficient positive:3√3 tan² x +4 tan x - √3 = 0Now, let's let t = tan x. Then the equation becomes:3√3 t² + 4 t - √3 = 0This is a quadratic equation in t. Let's solve for t using the quadratic formula.The quadratic formula says t = [-b ± sqrt(b² - 4ac)] / (2a)Here, a = 3√3, b = 4, c = -√3Compute discriminant D:D = b² - 4ac = 16 - 4*(3√3)*(-√3) = 16 - 4*(-3√3*√3) = 16 - 4*(-3*3) = 16 - 4*(-9) = 16 +36=52So D=52. Then,t = [-4 ± sqrt(52)] / (2*3√3) = [-4 ± 2*sqrt(13)] / (6√3) = [ -4 ± 2√13 ] / (6√3 )Simplify numerator and denominator:Factor numerator: 2[ -2 ± √13 ] / (6√3 ) = [ -2 ± √13 ] / (3√3 )So tan x = [ -2 ± √13 ] / (3√3 )Now, since x is angle C, which is between 0° and 60°, so tan x should be positive. Let's check the two solutions:First solution: [ -2 + √13 ] / (3√3 )Compute -2 + √13: Since √13 ≈ 3.6055, so -2 + 3.6055 ≈ 1.6055, which is positive.Second solution: [ -2 - √13 ] / (3√3 ), which is negative, so we can discard that.So tan x = [ -2 + √13 ] / (3√3 )Let me rationalize the denominator:Multiply numerator and denominator by √3:tan x = [ (-2 + √13 ) * √3 ] / (3√3 * √3 ) = [ (-2√3 + √39 ) ] / (9 )So tan x = (√39 - 2√3)/9But maybe I don't need to compute the exact value. Alternatively, since tan x = [ -2 + √13 ] / (3√3 ), maybe there's a way to find x or to relate sides?Alternatively, perhaps we can use the Law of Sines to find the ratio AC/AB.In triangle ABC, AC is the side opposite angle B, and AB is the side opposite angle C. So by the Law of Sines:AC / AB = sin B / sin CSo if we can find sin B / sin C, that would give the ratio. Alternatively, since we have tan B = 3 tan C, and B + C = 60°, maybe we can find sin B / sin C.Let me think. Let's denote angle C = x, so angle B = 60° - x. Then, tan(60° - x) = 3 tan x. From earlier steps, we found tan x = [ -2 + √13 ] / (3√3 )But maybe instead of calculating tan x, we can use the relation tan B = 3 tan C, which is equivalent to (sin B / cos B) = 3 (sin C / cos C )So, sin B / sin C = 3 cos B / cos CTherefore, sin B / sin C = 3 cos B / cos CBut from the Law of Sines, sin B / sin C = AC / AB, which is the ratio we need. So:AC / AB = 3 cos B / cos CSo if we can find cos B / cos C, then multiply by 3 to get the ratio.Alternatively, maybe we can find cos B and cos C in terms of angle x, since angle C = x and angle B = 60° - x.Wait, so cos B = cos(60° - x) and cos C = cos x.So cos B / cos C = cos(60° - x)/cos xUsing the cosine subtraction formula:cos(60° - x) = cos 60° cos x + sin 60° sin x = (0.5 cos x) + ( (√3)/2 sin x )So cos(60° - x)/cos x = 0.5 + ( (√3)/2 ) (sin x / cos x ) = 0.5 + (√3 / 2 ) tan xBut we know that tan x = [ -2 + √13 ] / (3√3 )Therefore, let's compute that term:(√3 / 2 ) tan x = (√3 / 2 ) * [ (-2 + √13 ) / (3√3 ) ] = [ (-2 + √13 ) / (6 ) ]So, cos B / cos C = 0.5 + [ (-2 + √13 ) / 6 ]Compute 0.5 = 3/6, so:3/6 + (-2 + √13)/6 = (3 -2 + √13 ) /6 = (1 + √13 ) /6Therefore, cos B / cos C = (1 + √13 ) /6Thus, AC / AB = 3 * (1 + √13 ) /6 = (1 + √13 ) /2Hence, the value of AC/AB is (1 + √13)/2.Wait, but let me verify that again. Because I had AC / AB = 3 cos B / cos C, and cos B / cos C = (1 + √13)/6, so 3 times that is (1 + √13)/2. That seems correct.But let me check if there's a possible miscalculation. Let's retrace:From AC / AB = sin B / sin C = 3 cos B / cos C.Then, cos B / cos C was calculated as (1 + √13)/6, so 3*(1 + √13)/6 is (1 + √13)/2. Yes, that's right.Alternatively, maybe there's another approach. Let me check with Law of Sines.Given that angle A is 120°, and angles B and C sum to 60°, with tan B = 3 tan C.Alternatively, using the Law of Sines:AC / AB = sin B / sin C.But since angle B = 60° - C, sin B = sin(60° - C).So AC / AB = sin(60° - C) / sin CExpand sin(60° - C) as sin60 cos C - cos60 sin C:= [ (√3/2 cos C ) - (1/2 sin C ) ] / sin C= (√3/2 cos C ) / sin C - (1/2 sin C ) / sin C= (√3/2 ) cot C - 1/2So, AC / AB = (√3 / 2 ) cot C - 1/2But we know that tan B = 3 tan C. Since B = 60° - C, tan(60° - C) = 3 tan CLet me write that as:tan(60° - C) = 3 tan CAgain, using tan(60° - C) = (tan60 - tan C)/(1 + tan60 tan C) = (√3 - tan C)/(1 + √3 tan C) = 3 tan CWhich is the same equation as before. So solving that gives tan C = [ -2 + √13 ] / (3√3 )But maybe instead of going through that, perhaps express cot C as 1/tan C.From earlier, we have tan C = [ -2 + √13 ] / (3√3 )Therefore, cot C = 1 / tan C = 3√3 / ( -2 + √13 )Multiply numerator and denominator by ( -2 - √13 ) to rationalize the denominator:cot C = 3√3 ( -2 - √13 ) / [ ( -2 + √13 )( -2 - √13 ) ] = 3√3 ( -2 - √13 ) / (4 - 13 ) = 3√3 ( -2 - √13 ) / (-9 ) = 3√3 (2 + √13 ) /9 = ( √3 (2 + √13 ) ) /3Therefore, cot C = ( √3 (2 + √13 ) ) /3Therefore, AC / AB = (√3 / 2 ) * ( √3 (2 + √13 ) /3 ) - 1/2Simplify:First term: (√3 / 2 ) * ( √3 (2 + √13 ) /3 ) = (3 (2 + √13 ) ) / (6 ) ) = (2 + √13 ) / 2Second term: -1/2Therefore, AC / AB = (2 + √13 ) /2 - 1/2 = (2 + √13 -1 ) /2 = (1 + √13 ) /2So that matches the earlier result. Therefore, the ratio is (1 + √13)/2.So, after working through two different approaches, both lead to the same conclusion. Therefore, the value of AC divided by AB is (1 + √13)/2.Final AnswerThe value of frac{AC}{AB} is boxed{dfrac{1 + sqrt{13}}{2}}.</think>Given the problem in triangle ABC with the conditions 2sin^2 frac{A}{2} = sqrt{3}sin A and sin(B - C) = 2cos B sin C, we need to find the value of frac{AC}{AB}.First, we solve the equation 2sin^2 frac{A}{2} = sqrt{3}sin A. Using the identity sin^2 frac{A}{2} = frac{1 - cos A}{2}, we rewrite the equation as:[2 left(frac{1 - cos A}{2}right) = sqrt{3}sin A implies 1 - cos A = sqrt{3}sin A]Rearranging and using the identity for linear combinations of sine and cosine, we get:[1 = sqrt{3}sin A + cos A implies 1 = 2sinleft(A + 30^circright)]Solving for A, we find:[sinleft(A + 30^circright) = frac{1}{2} implies A + 30^circ = 150^circ implies A = 120^circ]Next, we use the given condition sin(B - C) = 2cos B sin C. Expanding using the sine subtraction formula:[sin B cos C - cos B sin C = 2cos B sin C implies sin B cos C = 3cos B sin C]Dividing both sides by cos B cos C gives:[tan B = 3 tan C]Since B + C = 60^circ, let C = x and B = 60^circ - x. Then:[tan(60^circ - x) = 3 tan x]Using the tangent subtraction formula and solving the resulting quadratic equation, we find tan x = frac{-2 + sqrt{13}}{3sqrt{3}}.Using the Law of Sines, frac{AC}{AB} = frac{sin B}{sin C}. Since B = 60^circ - x and C = x, we have:[sin B = sin(60^circ - x) text{ and } sin C = sin x]Using the sine subtraction formula and solving, we find:[frac{sin(60^circ - x)}{sin x} = frac{sqrt{3}}{2} cot x - frac{1}{2}]Rationalizing and simplifying, we find cot x = frac{sqrt{3}(2 + sqrt{13})}{3}, leading to:[frac{AC}{AB} = frac{1 + sqrt{13}}{2}]Thus, the value of frac{AC}{AB} is boxed{dfrac{1 + sqrt{13}}{2}}.]

🔑:Given 2sin ^{2} frac{A}{2}= sqrt{3}sin A,We can rewrite this as 1-cos A= sqrt{3}sin A,Which implies sin (A+ frac{pi}{6})= frac{1}{2},Given 0 < A < pi, we have A= frac{2pi}{3}.By the Cosine Rule, we have a^{2}=b^{2}+c^{2}+bc (1),Expanding sin (B-C)=2cos Bsin C, we get sin Bcos C=3cos Bsin C,Angle-ifying the sides, we get b cdot frac{a^{2}+b^{2}-c^{2}}{2ab}=3 cdot frac{a^{2}+c^{2}-b^{2}}{2ac} cdot c, which simplifies to 2b^{2}-2c^{2}=a^{2} (2),Substituting (1) into (2), we get b^{2}-3c^{2}-bc=0, dividing both sides by c^{2}, we get (frac{b}{c})^{2}-frac{b}{c}-3=0 (3),Solving (3), we get frac{b}{c}= frac{1+sqrt{13}}{2},Hence, frac{AC}{AB}= frac{1+sqrt{13}}{2}.Thus, the answer is boxed{frac{1+sqrt{13}}{2}}.This problem tests the application of the Cosine Rule and Sine Rule, and requires some computational skills. It is a moderate difficulty problem.

🧠:[<think>Okay, so I need to figure out which of the given sets of vectors are not coplanar. The problem states that {a, b, c} form a basis for the space, which means they are linearly independent and span the entire space, so they are not coplanar themselves. The question is asking which of the options A, B, C, D are sets of vectors that are not coplanar, which I think means they are linearly independent, forming a basis as well. Wait, but coplanar vectors lie in the same plane, so if three vectors are coplanar, they are linearly dependent. Therefore, the question is asking which set of three vectors are not coplanar, i.e., linearly independent. So, the answer would be the set that is linearly independent.So, I need to check each option to see if the three vectors in each are linearly independent. If they are, then they are not coplanar; otherwise, they are coplanar. Let's go through each option one by one.Starting with Option A: vectors are b + c, b, b - c. Let me write them as:v1 = b + cv2 = bv3 = b - cTo check if they are linearly independent, we can set up a linear combination αv1 + βv2 + γv3 = 0 and see if the only solution is α=β=γ=0.So:α(b + c) + βb + γ(b - c) = 0Let's distribute the coefficients:αb + αc + βb + γb - γc = 0Combine like terms:(α + β + γ)b + (α - γ)c = 0Since {a, b, c} are a basis, which implies that b and c are linearly independent (as part of the basis). Therefore, the coefficients of b and c must each be zero. Wait, but the original basis has three vectors, a, b, c. However, in this equation, there's no component of a. So, the equation is in the subspace spanned by b and c. But the vectors v1, v2, v3 are all in the subspace spanned by b and c. Therefore, since they are three vectors in a two-dimensional subspace (since b and c are linearly independent), they must be linearly dependent. Therefore, Option A is coplanar. So A is not the answer.Wait, but actually, the problem says "form a basis for a space," which is probably three-dimensional since there are three vectors. So {a, b, c} is a basis for a 3-dimensional space. Therefore, any three vectors that are coplanar (i.e., lie in the same plane) would be linearly dependent. So if the three vectors can be expressed with only two basis vectors, then they lie in a plane and are coplanar.But in Option A, all vectors are combinations of b and c, so they lie in the b-c plane, hence coplanar. Therefore, Option A is coplanar. So A is out.Moving to Option B: vectors are a, a + b, a - b.So v1 = av2 = a + bv3 = a - bCheck linear independence:αa + β(a + b) + γ(a - b) = 0Expanding:αa + βa + βb + γa - γb = 0Combine like terms:(α + β + γ)a + (β - γ)b = 0Again, since {a, b, c} are a basis, coefficients must be zero:For a: α + β + γ = 0For b: β - γ = 0For c: 0 = 0 (since there are no c components in the vectors)So, equations:1. α + β + γ = 02. β - γ = 0From equation 2: β = γSubstitute into equation 1: α + β + β = α + 2β = 0 → α = -2βSo, we can choose β = 1, then γ = 1, α = -2. Therefore, non-trivial solution exists. For example, -2a + 1*(a + b) + 1*(a - b) = -2a + a + b + a - b = 0. So, yes, linear combination equals zero with non-zero coefficients. Therefore, Option B is linearly dependent, hence coplanar. So B is out.Option C: vectors are a + b, a - b, c.So v1 = a + bv2 = a - bv3 = cCheck linear independence:α(a + b) + β(a - b) + γc = 0Expanding:αa + αb + βa - βb + γc = 0Combine like terms:(α + β)a + (α - β)b + γc = 0Since {a, b, c} are linearly independent, all coefficients must be zero:1. α + β = 02. α - β = 03. γ = 0From equation 1 and 2:Adding equations 1 and 2: 2α = 0 → α = 0Then from equation 1: 0 + β = 0 → β = 0And equation 3: γ = 0So the only solution is α = β = γ = 0. Therefore, vectors in Option C are linearly independent. Hence, they are not coplanar. So C is a candidate for the correct answer.Check Option D: vectors are a + b, a + b + c, c.So v1 = a + bv2 = a + b + cv3 = cCheck linear independence:α(a + b) + β(a + b + c) + γc = 0Expanding:αa + αb + βa + βb + βc + γc = 0Combine like terms:(α + β)a + (α + β)b + (β + γ)c = 0Again, coefficients must be zero:1. α + β = 02. α + β = 03. β + γ = 0So equations 1 and 2 are the same. From equation 1: α = -βFrom equation 3: γ = -βSo, set β = 1, then α = -1, γ = -1Check the combination:-1*(a + b) + 1*(a + b + c) -1*c = (-a - b) + (a + b + c) - c = 0Yes, non-trivial solution. Therefore, Option D is linearly dependent, hence coplanar. So D is out.Therefore, the only set that is not coplanar is Option C. So the answer is C.Final Answerboxed{C}</think>Given that ({overrightarrow{a}, overrightarrow{b}, overrightarrow{c}}) form a basis for a space, we need to determine which set of vectors is not coplanar (i.e., linearly independent).Option A: (overrightarrow{b} + overrightarrow{c}), (overrightarrow{b}), (overrightarrow{b} - overrightarrow{c})These vectors can be written as:[begin{aligned}overrightarrow{v_1} &= overrightarrow{b} + overrightarrow{c}, overrightarrow{v_2} &= overrightarrow{b}, overrightarrow{v_3} &= overrightarrow{b} - overrightarrow{c}.end{aligned}]Setting up the linear combination (alpha overrightarrow{v_1} + beta overrightarrow{v_2} + gamma overrightarrow{v_3} = 0):[alpha (overrightarrow{b} + overrightarrow{c}) + beta overrightarrow{b} + gamma (overrightarrow{b} - overrightarrow{c}) = 0.]Combining terms:[(alpha + beta + gamma) overrightarrow{b} + (alpha - gamma) overrightarrow{c} = 0.]Since (overrightarrow{b}) and (overrightarrow{c}) are linearly independent, the coefficients must be zero. However, solving the equations shows non-trivial solutions, indicating linear dependence. Hence, coplanar.Option B: (overrightarrow{a}), (overrightarrow{a} + overrightarrow{b}), (overrightarrow{a} - overrightarrow{b})These vectors can be written as:[begin{aligned}overrightarrow{v_1} &= overrightarrow{a}, overrightarrow{v_2} &= overrightarrow{a} + overrightarrow{b}, overrightarrow{v_3} &= overrightarrow{a} - overrightarrow{b}.end{aligned}]Setting up the linear combination (alpha overrightarrow{v_1} + beta overrightarrow{v_2} + gamma overrightarrow{v_3} = 0):[alpha overrightarrow{a} + beta (overrightarrow{a} + overrightarrow{b}) + gamma (overrightarrow{a} - overrightarrow{b}) = 0.]Combining terms:[(alpha + beta + gamma) overrightarrow{a} + (beta - gamma) overrightarrow{b} = 0.]Solving the equations shows non-trivial solutions, indicating linear dependence. Hence, coplanar.Option C: (overrightarrow{a} + overrightarrow{b}), (overrightarrow{a} - overrightarrow{b}), (overrightarrow{c})These vectors can be written as:[begin{aligned}overrightarrow{v_1} &= overrightarrow{a} + overrightarrow{b}, overrightarrow{v_2} &= overrightarrow{a} - overrightarrow{b}, overrightarrow{v_3} &= overrightarrow{c}.end{aligned}]Setting up the linear combination (alpha overrightarrow{v_1} + beta overrightarrow{v_2} + gamma overrightarrow{v_3} = 0):[alpha (overrightarrow{a} + overrightarrow{b}) + beta (overrightarrow{a} - overrightarrow{b}) + gamma overrightarrow{c} = 0.]Combining terms:[(alpha + beta) overrightarrow{a} + (alpha - beta) overrightarrow{b} + gamma overrightarrow{c} = 0.]Solving the equations shows only trivial solutions, indicating linear independence. Hence, not coplanar.Option D: (overrightarrow{a} + overrightarrow{b}), (overrightarrow{a} + overrightarrow{b} + overrightarrow{c}), (overrightarrow{c})These vectors can be written as:[begin{aligned}overrightarrow{v_1} &= overrightarrow{a} + overrightarrow{b}, overrightarrow{v_2} &= overrightarrow{a} + overrightarrow{b} + overrightarrow{c}, overrightarrow{v_3} &= overrightarrow{c}.end{aligned}]Setting up the linear combination (alpha overrightarrow{v_1} + beta overrightarrow{v_2} + gamma overrightarrow{v_3} = 0):[alpha (overrightarrow{a} + overrightarrow{b}) + beta (overrightarrow{a} + overrightarrow{b} + overrightarrow{c}) + gamma overrightarrow{c} = 0.]Combining terms:[(alpha + beta) overrightarrow{a} + (alpha + beta) overrightarrow{b} + (beta + gamma) overrightarrow{c} = 0.]Solving the equations shows non-trivial solutions, indicating linear dependence. Hence, coplanar.Thus, the set of vectors that are not coplanar is (boxed{C}).,<think>Okay, let's see. The problem is about determining which set of vectors is not coplanar, given that {a, b, c} form a basis for the space. Since they form a basis, they are linearly independent and span the entire space, which I think is three-dimensional because there are three vectors. The question is asking which of the options A, B, C, D consists of vectors that are not coplanar. So, if vectors are coplanar, they lie in the same plane, which would mean they are linearly dependent in a three-dimensional space. Therefore, if a set of three vectors is not coplanar, they must be linearly independent, forming a basis themselves. Wait, but actually, in three-dimensional space, three vectors are coplanar if they lie in the same plane, which would mean their span is two-dimensional. So, if they are not coplanar, their span is three-dimensional, hence they are linearly independent. Therefore, the question is equivalent to asking which set of vectors is linearly independent. So, we need to check for each option whether the three vectors are linearly independent.Alright, let's go through each option one by one.Option A: b + c, b, b - c.Let me check if these vectors are linearly independent. Let's denote the vectors as v1 = b + c, v2 = b, v3 = b - c.To check linear independence, we can form a matrix with these vectors as columns (or rows) and compute the determinant. If the determinant is non-zero, they are linearly independent. Since the original vectors a, b, c are a basis, we can express everything in terms of coordinates relative to this basis. Let's see.Expressing each vector in terms of the basis {a, b, c}:v1 = 0a + 1b + 1cv2 = 0a + 1b + 0cv3 = 0a + 1b - 1cSo, writing them as column vectors in the basis {a, b, c}, the matrix would be:[0, 0, 0;1, 1, 1;1, 0, -1]Wait, actually, each vector is a column in the matrix. Wait, v1 is (0,1,1), v2 is (0,1,0), v3 is (0,1,-1). So the matrix is:First column: [0; 1; 1]Second column: [0; 1; 0]Third column: [0; 1; -1]So, arranging them:0 0 01 1 11 0 -1Wait, no. Wait, each vector is a column vector. So the coordinates are in rows? Wait, maybe I need to be careful here.In standard linear algebra, when you write vectors as columns, each vector is a column. So if the basis is {a, b, c}, then the coordinates of v1 = 0a + 1b + 1c would be [0, 1, 1]^T, right? So the matrix would have these as columns. So columns are:First column: [0, 1, 1]^TSecond column: [0, 1, 0]^TThird column: [0, 1, -1]^TSo the matrix is:0 0 01 1 11 0 -1Wait, no. Wait, if each vector is a column, then the first column is [0, 1, 1], the second is [0, 1, 0], the third is [0, 1, -1]. So stacking them as columns:Row 1: 0, 0, 0Row 2: 1, 1, 1Row 3: 1, 0, -1Wait, that can't be right because row 1 is all zeros. So the determinant of this matrix would be zero, since the first row is all zeros. Therefore, the determinant is zero, which means the vectors are linearly dependent. Therefore, the vectors in option A are coplanar. So A is not the answer.Moving on to option B: a, a + b, a - bVectors are v1 = a, v2 = a + b, v3 = a - bExpressed in the basis:v1 = 1a + 0b + 0cv2 = 1a + 1b + 0cv3 = 1a - 1b + 0cSo the matrix formed by these columns would be:1 1 10 1 -10 0 0Wait, similar to before. Let me check again. Each vector is a column:v1: [1, 0, 0]^Tv2: [1, 1, 0]^Tv3: [1, -1, 0]^TTherefore, the matrix is:1 1 10 1 -10 0 0Again, the third row is all zeros, so determinant is zero. Hence, these vectors are linearly dependent as well, meaning they are coplanar. So option B is coplanar. Not the answer.Option C: a + b, a - b, cVectors are v1 = a + b, v2 = a - b, v3 = cExpressed in the basis:v1 = 1a + 1b + 0cv2 = 1a - 1b + 0cv3 = 0a + 0b + 1cSo forming the matrix with these columns:First column: [1, 1, 0]^TSecond column: [1, -1, 0]^TThird column: [0, 0, 1]^TTherefore, the matrix is:1 1 01 -1 00 0 1Now, let's compute the determinant. For a 3x3 matrix:det = 1*(-1*1 - 0*0) - 1*(1*1 - 0*0) + 0*(1*0 - (-1)*0)= 1*(-1) - 1*(1) + 0*(0)= -1 -1 + 0 = -2Since determinant is -2, which is non-zero. Therefore, these vectors are linearly independent. Hence, they are not coplanar. So option C is a candidate.Option D: a + b, a + b + c, cVectors are v1 = a + b, v2 = a + b + c, v3 = cExpressed in the basis:v1 = 1a + 1b + 0cv2 = 1a + 1b + 1cv3 = 0a + 0b + 1cMatrix columns:First column: [1, 1, 0]^TSecond column: [1, 1, 1]^TThird column: [0, 0, 1]^TSo the matrix is:1 1 01 1 00 1 1Wait, wait, no. Let me write that again:v1: [1,1,0]v2: [1,1,1]v3: [0,0,1]Therefore, the matrix is:1 1 01 1 00 1 1Wait, hold on. No. Wait, the components are (a, b, c). So each vector is expressed as (coefficient of a, coefficient of b, coefficient of c). Therefore:v1: (1, 1, 0)v2: (1, 1, 1)v3: (0, 0, 1)Therefore, the matrix is:Row 1 (coefficients of a): 1, 1, 0Row 2 (coefficients of b): 1, 1, 0Row 3 (coefficients of c): 0, 1, 1Wait, that's not correct. Wait, each vector is a column. So columns are:v1: [1; 1; 0]v2: [1; 1; 1]v3: [0; 0; 1]Therefore, the matrix is:1 1 01 1 00 1 1Wait, no, that's incorrect. Wait, if v1 is (1,1,0), that's 1 in a, 1 in b, 0 in c. So as a column vector, it's [1; 1; 0]. Similarly, v2 is [1; 1; 1], and v3 is [0; 0; 1].Therefore, the matrix is:First column: [1,1,0]Second column: [1,1,1]Third column: [0,0,1]Therefore, the matrix is:1 1 01 1 00 1 1Wait, no. Wait, rows are a, b, c components. So:Row 1: a components: 1 (from v1), 1 (from v2), 0 (from v3)Row 2: b components: 1,1,0Row 3: c components: 0,1,1Therefore, the matrix is:1 1 01 1 00 1 1Now, compute the determinant. Let's calculate it step by step.The determinant of a 3x3 matrix:|1 1 0||1 1 0||0 1 1|Calculating this using expansion by minors.First, expand along the first row:1 * det( [1 0; 1 1] ) - 1 * det( [1 0; 0 1] ) + 0 * det( [1 1; 0 1] )Compute each minor:First minor: determinant of [[1,0],[1,1]] = (1)(1) - (0)(1) = 1Second minor: determinant of [[1,0],[0,1]] = (1)(1) - (0)(0) = 1Third minor: doesn't matter because multiplied by 0.So determinant = 1*1 - 1*1 + 0 = 1 - 1 = 0Therefore, determinant is zero. Hence, vectors are linearly dependent, so they are coplanar. Thus, option D is coplanar.Therefore, the only option that is not coplanar is option C.Wait, but just to confirm, since the answer is C, let me just double-check.For option C, vectors are a + b, a - b, and c. So in matrix form, as columns:[1,1,0; 1,-1,0; 0,0,1]. The determinant was calculated as -2, which is non-zero. So yes, determinant is non-zero, hence linearly independent, hence not coplanar. So option C is correct.Therefore, the answer is C.Final Answerboxed{C}</think>Given that ({overrightarrow{a}, overrightarrow{b}, overrightarrow{c}}) form a basis for a three-dimensional space, we need to determine which set of vectors is not coplanar. Non-coplanar vectors are linearly independent.Option A: (overrightarrow{b} + overrightarrow{c}), (overrightarrow{b}), (overrightarrow{b} - overrightarrow{c})- Vectors in terms of the basis: (0overrightarrow{a} + 1overrightarrow{b} + 1overrightarrow{c}), (0overrightarrow{a} + 1overrightarrow{b} + 0overrightarrow{c}), (0overrightarrow{a} + 1overrightarrow{b} - 1overrightarrow{c})- Matrix form: [ begin{pmatrix} 0 & 0 & 0 1 & 1 & 1 1 & 0 & -1 end{pmatrix} ]- The first row is all zeros, determinant is zero, so vectors are linearly dependent (coplanar).Option B: (overrightarrow{a}), (overrightarrow{a} + overrightarrow{b}), (overrightarrow{a} - overrightarrow{b})- Vectors in terms of the basis: (1overrightarrow{a} + 0overrightarrow{b} + 0overrightarrow{c}), (1overrightarrow{a} + 1overrightarrow{b} + 0overrightarrow{c}), (1overrightarrow{a} - 1overrightarrow{b} + 0overrightarrow{c})- Matrix form: [ begin{pmatrix} 1 & 1 & 1 0 & 1 & -1 0 & 0 & 0 end{pmatrix} ]- The third row is all zeros, determinant is zero, so vectors are linearly dependent (coplanar).Option C: (overrightarrow{a} + overrightarrow{b}), (overrightarrow{a} - overrightarrow{b}), (overrightarrow{c})- Vectors in terms of the basis: (1overrightarrow{a} + 1overrightarrow{b} + 0overrightarrow{c}), (1overrightarrow{a} - 1overrightarrow{b} + 0overrightarrow{c}), (0overrightarrow{a} + 0overrightarrow{b} + 1overrightarrow{c})- Matrix form: [ begin{pmatrix} 1 & 1 & 0 1 & -1 & 0 0 & 0 & 1 end{pmatrix} ]- Determinant calculation: (1(-1 cdot 1 - 0 cdot 0) - 1(1 cdot 1 - 0 cdot 0) + 0 = -1 - 1 = -2), which is non-zero. Vectors are linearly independent (not coplanar).Option D: (overrightarrow{a} + overrightarrow{b}), (overrightarrow{a} + overrightarrow{b} + overrightarrow{c}), (overrightarrow{c})- Vectors in terms of the basis: (1overrightarrow{a} + 1overrightarrow{b} + 0overrightarrow{c}), (1overrightarrow{a} + 1overrightarrow{b} + 1overrightarrow{c}), (0overrightarrow{a} + 0overrightarrow{b} + 1overrightarrow{c})- Matrix form: [ begin{pmatrix} 1 & 1 & 0 1 & 1 & 0 0 & 1 & 1 end{pmatrix} ]- Determinant calculation: (1(1 cdot 1 - 0 cdot 1) - 1(1 cdot 1 - 0 cdot 0) + 0 = 1 - 1 = 0), determinant is zero, so vectors are linearly dependent (coplanar).The correct answer is (boxed{C}).]

🔑:To determine which set of vectors is not coplanar, we analyze each option based on the necessary and sufficient conditions for vectors to be coplanar.Option A: overrightarrow{b}+overrightarrow{c}, overrightarrow{b}, overrightarrow{b}-overrightarrow{c}We can express overrightarrow{b} as a linear combination of overrightarrow{b}+overrightarrow{c} and overrightarrow{b}-overrightarrow{c}:[overrightarrow{b} = frac{1}{2}(overrightarrow{b}+overrightarrow{c}) + frac{1}{2}(overrightarrow{b}-overrightarrow{c})]This shows that the vectors in option A are coplanar since one of them can be represented as a linear combination of the other two.Option B: overrightarrow{a}, overrightarrow{a}+overrightarrow{b}, overrightarrow{a}-overrightarrow{b}Similarly, we can see that these vectors are also coplanar because they can be related through linear combinations.Option D: overrightarrow{a}+overrightarrow{b}, overrightarrow{a}+overrightarrow{b}+overrightarrow{c}, overrightarrow{c}Here, overrightarrow{a}+overrightarrow{b}+overrightarrow{c} can be expressed as the sum of overrightarrow{a}+overrightarrow{b} and overrightarrow{c}:[overrightarrow{a}+overrightarrow{b}+overrightarrow{c} = (overrightarrow{a}+overrightarrow{b}) + overrightarrow{c}]This indicates that the vectors in option D are coplanar because one vector is the sum of the other two.Option C: overrightarrow{a}+overrightarrow{b}, overrightarrow{a}-overrightarrow{b}, overrightarrow{c}For option C, there is no immediate linear combination that shows one vector can be represented as a combination of the other two without assuming additional relationships beyond those given. This suggests that these vectors are not necessarily coplanar, making option C the correct choice.Therefore, the set of vectors that are not coplanar is given by option C: overrightarrow{a}+overrightarrow{b}, overrightarrow{a}-overrightarrow{b}, overrightarrow{c}.So, the correct choice is boxed{text{C}}.

🧠:[<think>Okay, so I need to find the sum of all integer values of x where the function y doesn't exceed 8. The function given is:y = x² + x(log₂18 - log₃12) - log₃16 - 4 log₂3First, I should simplify the logarithmic terms to make the equation easier to handle. Let me start by breaking down each logarithm.Starting with log₂18. Since 18 is 2 * 3², I can write log₂(2*3²) = log₂2 + log₂3² = 1 + 2 log₂3. Similarly, log₃12. 12 is 3 * 4, which is 3 * 2². So log₃(3*2²) = log₃3 + log₃2² = 1 + 2 log₃2. Then, log₃16. 16 is 2⁴, so log₃2⁴ = 4 log₃2. Lastly, 4 log₂3 is already simplified.So substituting these back into the original equation:y = x² + x[ (1 + 2 log₂3) - (1 + 2 log₃2) ] - (4 log₃2) - 4 log₂3Simplify the terms inside the brackets:(1 + 2 log₂3) - (1 + 2 log₃2) = 1 - 1 + 2 log₂3 - 2 log₃2 = 2(log₂3 - log₃2)So the equation becomes:y = x² + 2x(log₂3 - log₃2) - 4 log₃2 - 4 log₂3Hmm. Now, log₂3 and log₃2 are reciprocals of each other. Remember that log_b a = 1 / log_a b. So log₂3 = 1 / log₃2. Let me denote log₃2 as t. Then log₂3 = 1/t. Let's substitute that.So:y = x² + 2x(1/t - t) - 4t - 4/tThat's a bit more manageable, perhaps. Let's see if we can find the numerical value of t. Since t = log₃2, we can calculate its approximate value. Let's recall that log₃2 is the exponent to which 3 must be raised to get 2. So 3^t = 2. Taking natural logarithm, t ln3 = ln2, so t = ln2 / ln3 ≈ 0.6309. Similarly, 1/t ≈ 1.58496.But maybe instead of approximating, we can work with exact expressions. Let me see. Let's note that 1/t - t = (1 - t²)/t. So:2x(1/t - t) = 2x(1 - t²)/tSimilarly, the constants:-4t - 4/t = -4(t + 1/t)So now:y = x² + 2x(1 - t²)/t - 4(t + 1/t)But I'm not sure if this helps. Alternatively, perhaps express everything in terms of log₂3, since we can compute log₂3 numerically. Let me compute log₂3. log₂3 is approximately 1.58496, as above. Similarly, log₃2 is approximately 0.6309. Let me note these down:log₂3 ≈ 1.58496log₃2 ≈ 0.6309So log₂3 - log₃2 ≈ 1.58496 - 0.6309 ≈ 0.95406Then 2*(log₂3 - log₃2) ≈ 1.90812Similarly, the constant terms:- log₃16 -4 log₂3log₃16 = log₃2⁴ = 4 log₃2 ≈ 4*0.6309 ≈ 2.52364 log₂3 ≈ 4*1.58496 ≈ 6.33984Therefore, the constant term is -2.5236 -6.33984 ≈ -8.86344So substituting these approximated values into the equation, we have:y ≈ x² + 1.90812x -8.86344So the inequality y ≤ 8 becomes:x² + 1.90812x -8.86344 ≤ 8Bring 8 to the left:x² + 1.90812x -16.86344 ≤ 0So now we need to solve the quadratic inequality x² + 1.90812x -16.86344 ≤ 0To find the values of x where this holds, we first find the roots of the equation x² + 1.90812x -16.86344 = 0Using the quadratic formula:x = [-b ± √(b² - 4ac)]/(2a)Here, a = 1, b = 1.90812, c = -16.86344Discriminant D = b² - 4ac = (1.90812)^2 - 4*1*(-16.86344)Calculate (1.90812)^2 ≈ 3.6409Then, 4*1*16.86344 ≈ 67.45376So D ≈ 3.6409 + 67.45376 ≈ 71.09466Square root of D ≈ √71.09466 ≈ 8.43So the roots are:x = [-1.90812 ± 8.43]/2First root: (-1.90812 +8.43)/2 ≈ (6.52188)/2 ≈ 3.26094Second root: (-1.90812 -8.43)/2 ≈ (-10.33812)/2 ≈ -5.16906So the quadratic is less than or equal to zero between x ≈ -5.16906 and x ≈ 3.26094Since x must be an integer, the integer values of x are from -5 up to 3, inclusive.Wait, check the endpoints. -5.16906 is between -6 and -5, so the smallest integer x is -5. The upper root is 3.26, so the largest integer x is 3.Therefore, integer x values are: -5, -4, -3, -2, -1, 0, 1, 2, 3Now, sum these integers:Sum = (-5) + (-4) + (-3) + (-2) + (-1) + 0 + 1 + 2 + 3Let's compute:Add the negative numbers: -5 -4 -3 -2 -1 = -15Add the non-negative numbers: 0 +1 +2 +3 = 6Total sum: -15 + 6 = -9Wait, but I need to confirm if all these integers actually satisfy the original inequality. Because we approximated the coefficients, there might be a case where the actual roots are slightly different, which might affect whether the endpoints (like x=3 or x=-5) actually satisfy the original inequality.So perhaps we need to compute the exact values instead of approximating. Let me try to do that.First, let's go back to the original equation:y = x² + x(log₂18 - log₃12) - log₃16 -4 log₂3Let me compute each logarithmic term exactly.log₂18: 18 = 2*3², so log₂18 = log₂2 + log₂3² = 1 + 2 log₂3log₃12: 12 = 3*2², so log₃12 = log₃3 + log₃2² = 1 + 2 log₃2log₃16: 16 = 2⁴, so log₃16 = log₃2⁴ = 4 log₃24 log₂3 is as is.So substituting back:y = x² + x[(1 + 2 log₂3) - (1 + 2 log₃2)] - (4 log₃2 +4 log₂3)Simplify:The x terms: (1 -1) + 2 log₂3 - 2 log₃2 = 2(log₂3 - log₃2)The constants: -4(log₃2 + log₂3)So y = x² + 2x(log₂3 - log₃2) -4(log₃2 + log₂3)Now, note that log₂3 = 1 / log₃2 (change of base formula). Let's denote a = log₃2. Then log₂3 = 1/a. Therefore, the equation becomes:y = x² + 2x(1/a - a) -4(a + 1/a)So y = x² + 2x( (1 - a²)/a ) -4(a + 1/a)Alternatively, we can write this as:y = x² + 2x*(1 - a²)/a -4(a + 1/a)But since a = log₃2, which is approximately 0.6309 as before, but we need to work with exact expressions.Alternatively, perhaps we can express all terms in terms of log₂3. Let me denote b = log₂3, so that log₃2 = 1/b. Then:y = x² + 2x(b - 1/b) -4(1/b + b)So y = x² + 2x(b - 1/b) -4(b + 1/b)This might be helpful. Let's compute this expression.Let me compute the coefficient of x: 2(b - 1/b). Let me compute b - 1/b:If b = log₂3, then b ≈ 1.58496, so 1/b ≈ 0.6309, so b - 1/b ≈ 0.95406, which matches the earlier approximation.Similarly, the constant term: -4(b + 1/b). b +1/b = log₂3 + log₃2. Again, since b = log₂3, this is b + 1/b. Let's compute:If b ≈ 1.58496, then 1/b ≈ 0.6309, so b + 1/b ≈ 2.21586. Multiply by -4: ≈ -8.86344, which matches the earlier constant term.So, the quadratic equation is y = x² + 2*(b -1/b)x -4*(b +1/b) ≤8.So, moving 8 to the left:x² + 2*(b -1/b)x -4*(b +1/b) -8 ≤0Let me denote the quadratic as:x² + [2(b -1/b)]x + [ -4(b +1/b) -8 ] ≤0So, to solve the inequality, we need to find the roots of the quadratic equation:x² + [2(b -1/b)]x + [ -4(b +1/b) -8 ] =0But solving this exactly is complicated because of the log terms. However, maybe there is a clever substitution or relation between these logarithmic terms that can help simplify the equation.Alternatively, perhaps notice that 4(b +1/b) +8 can be written as 4(b +1/b + 2), but not sure.Alternatively, perhaps there is a way to write the quadratic in terms of (x + something)^2.Alternatively, let's compute the discriminant of this quadratic equation to find exact roots.Let me denote:A = 1B = 2(b -1/b)C = -4(b +1/b) -8Then discriminant D = B² -4ACCompute D:[2(b -1/b)]² -4*1*(-4(b +1/b) -8)First term: 4(b -1/b)^2Second term: -4*1*(-4(b +1/b) -8) = 16(b +1/b) +32So D = 4(b -1/b)^2 +16(b +1/b) +32Let me expand (b -1/b)^2:= b² - 2*(b)*(1/b) + (1/b)^2 = b² - 2 + 1/b²Thus, 4(b -1/b)^2 = 4b² -8 + 4/b²So D = 4b² -8 +4/b² +16(b +1/b) +32Combine like terms:4b² +4/b² +16b +16/b + (-8 +32)=4b² +4/b² +16b +16/b +24This seems complicated, but maybe factor terms:Factor 4 from the first two terms: 4(b² +1/b²) +16(b +1/b) +24Note that (b +1/b)^2 = b² + 2 + 1/b², so b² +1/b² = (b +1/b)^2 -2Therefore,4[(b +1/b)^2 -2] +16(b +1/b) +24=4(b +1/b)^2 -8 +16(b +1/b) +24=4(b +1/b)^2 +16(b +1/b) +16Let me factor this expression:Let’s set z = b +1/b. Then the expression becomes:4z² +16z +16 = 4(z² +4z +4) =4(z +2)^2Thus, D =4(z +2)^2, where z = b +1/bTherefore, D is a perfect square, which is nice.Thus, sqrt(D) = 2(z +2) =2(z +2) =2(b +1/b +2)Therefore, the roots are:x = [-B ± sqrt(D)]/(2A) = [ -2(b -1/b) ± 2(b +1/b +2) ] / 2Simplify numerator:First root: -2(b -1/b) +2(b +1/b +2) = -2b +2/b +2b +2/b +4 = ( -2b +2b ) + (2/b +2/b ) +4 = 4/b +4Second root: -2(b -1/b) -2(b +1/b +2) = -2b +2/b -2b -2/b -4 = (-4b) -4Therefore, roots:First root: (4/b +4)/2 = 2/b +2Second root: (-4b -4)/2 = -2b -2Thus, the roots are x = -2b -2 and x = 2/b +2Recall that b = log₂3 ≈1.58496, and 1/b ≈0.6309Therefore, the exact roots are:x = -2 log₂3 -2 and x = 2 / log₂3 +2But 2 / log₂3 = 2 log₃2, since 1 / log₂3 = log₃2Therefore, x = 2 log₃2 + 2So the roots are x = -2 log₂3 -2 and x = 2 log₃2 +2So the quadratic is less than or equal to zero between these two roots.Therefore, the solution set is x ∈ [ -2 log₂3 -2 , 2 log₃2 +2 ]But we need integer x within this interval.Therefore, we need to find integer x such that:-2 log₂3 -2 ≤x ≤ 2 log₃2 +2Compute the numerical values.First, compute -2 log₂3 -2:log₂3 ≈1.58496Multiply by -2: ≈-3.16992Subtract 2: ≈-5.16992Similarly, 2 log₃2 +2:log₃2 ≈0.6309Multiply by 2: ≈1.2618Add 2:≈3.2618So, the interval is approximately [-5.16992, 3.2618]Therefore, integer x values from -5 to 3 inclusive.Thus, the integers are: -5, -4, -3, -2, -1, 0, 1, 2, 3Now, sum these integers:Sum = (-5) + (-4) + (-3) + (-2) + (-1) + 0 +1 +2 +3Combine terms:Negative numbers: -5 -4 -3 -2 -1 = -15Non-negative numbers:0 +1 +2 +3 =6Total sum: -15 +6= -9However, before finalizing, we need to check whether x=-5 and x=3 actually satisfy the original inequality y ≤8, because sometimes endpoints might be just over due to approximation.So let's compute y when x=-5 and x=3 to verify.First, compute y at x=-5:Original function:y = (-5)^2 + (-5)(log₂18 - log₃12) - log₃16 -4 log₂3Compute each term:(-5)^2 =25Compute log₂18 - log₃12: previously simplified to 2(log₂3 - log₃2) ≈2*(1.58496 -0.6309)=2*(0.95406)=1.90812Thus, (-5)*(1.90812)= -9.5406log₃16=4 log₃2≈4*0.6309≈2.52364 log₂3≈4*1.58496≈6.33984Thus, constants: -2.5236 -6.33984≈-8.86344So total y≈25 -9.5406 -8.86344≈25 -18.404≈6.596Which is less than 8. So x=-5 is okay.Now, check x=3:y=3² +3*(log₂18 - log₃12) -log₃16 -4 log₂3Compute:3²=93*(1.90812)=5.72436log₃16≈2.5236, 4 log₂3≈6.33984, so constants≈-8.86344Thus, total y≈9 +5.72436 -8.86344≈9 +5.72436=14.72436 -8.86344≈5.86092, which is also less than 8.Wait, that's strange. Wait, according to my previous calculation, when x=3, y≈5.86, which is less than 8. But according to the quadratic inequality, the upper root was approx 3.26, so x=3 is within the interval, so it should satisfy the inequality.But wait, in the exact roots, x=2 log₃2 +2. Let's compute exactly:log₃2 = ln2/ln3≈0.6309, so 2 log₃2≈1.2618, plus 2 is≈3.2618, so x=3.2618 is the upper root, so x=3 is less than that, so indeed x=3 is within the interval. Therefore, y at x=3 is below 8.Therefore, all integers from -5 to 3 inclusive are valid.Therefore, their sum is -9.But let me check another value. For example, x=4. Since the upper root is≈3.26, x=4 is outside, so y(4) should be greater than 8. Let's check:y=4² +4*(log₂18 - log₃12) -log₃16 -4 log₂3≈16 +4*1.90812 -8.86344≈16 +7.63248 -8.86344≈23.63248 -8.86344≈14.769, which is greater than 8. So x=4 is invalid.Similarly, x=-6:y=(-6)^2 + (-6)*(log₂18 - log₃12) -log₃16 -4 log₂3≈36 -6*1.90812 -8.86344≈36 -11.44872 -8.86344≈36 -20.31216≈15.68784, which is greater than 8. So x=-6 is invalid.Therefore, indeed the integer solutions are x=-5 to x=3 inclusive. Sum is -9.Thus, the final answer is boxed{-9}Final Answerboxed{-9}</think>To find the sum of all integer values of ( x ) for which the corresponding values of the function[y = x^2 + xleft(log_2 18 - log_3 12right) - log_3 16 - 4 log_2 3]do not exceed 8, we start by simplifying the logarithmic terms.First, we express the logarithms in terms of (log_2 3) and (log_3 2):- (log_2 18 = 1 + 2 log_2 3)- (log_3 12 = 1 + 2 log_3 2)- (log_3 16 = 4 log_3 2)Substituting these into the function, we get:[y = x^2 + xleft((1 + 2 log_2 3) - (1 + 2 log_3 2)right) - 4 log_3 2 - 4 log_2 3]Simplifying further:[y = x^2 + 2x(log_2 3 - log_3 2) - 4(log_3 2 + log_2 3)]Recognizing that (log_2 3 = frac{1}{log_3 2}), let ( a = log_3 2 ). Then (log_2 3 = frac{1}{a}). Substituting these, we get:[y = x^2 + 2xleft(frac{1}{a} - aright) - 4left(a + frac{1}{a}right)]To solve ( y leq 8 ), we convert this into a quadratic inequality:[x^2 + 2xleft(frac{1}{a} - aright) - 4left(a + frac{1}{a}right) leq 8][x^2 + 2xleft(frac{1 - a^2}{a}right) - 4left(a + frac{1}{a}right) - 8 leq 0]Using the quadratic formula, we find the roots of the quadratic equation:[x = -2 log_2 3 - 2 quad text{and} quad x = 2 log_3 2 + 2]Approximating the roots:- (-2 log_2 3 - 2 approx -5.16992)- (2 log_3 2 + 2 approx 3.2618)The integer values of ( x ) within the interval ([-5.16992, 3.2618]) are (-5, -4, -3, -2, -1, 0, 1, 2, 3). Summing these integers:[(-5) + (-4) + (-3) + (-2) + (-1) + 0 + 1 + 2 + 3 = -15 + 6 = -9]Thus, the sum of all integer values of ( x ) for which ( y leq 8 ) is (boxed{-9}).,<think>Okay, so I need to find the sum of all integer values of x where the function y doesn't exceed 8. The function is given by:y = x² + x*(log₂18 - log₃12) - log₃16 - 4 log₂3First, let me make sure I understand the problem. I need to find all integer x such that y ≤ 8, then sum those x's. Alright, so the steps are probably: simplify the logarithmic terms, rewrite the inequality y ≤ 8, solve for x, find the integer solutions, and then sum them up.Let me start by simplifying the coefficients involving logarithms. The function has two logarithmic expressions: log₂18 - log₃12 and -log₃16 - 4 log₂3.Starting with log₂18 - log₃12. Let me compute each term separately.log₂18: 18 is 2*9, so log₂(2*9) = log₂2 + log₂9 = 1 + log₂9. Since 9 is 3², log₂9 = 2 log₂3. So log₂18 = 1 + 2 log₂3.Similarly, log₃12: 12 is 3*4, so log₃(3*4) = log₃3 + log₃4 = 1 + log₃4. And 4 is 2², so log₃4 = 2 log₃2. So log₃12 = 1 + 2 log₃2.Therefore, log₂18 - log₃12 = [1 + 2 log₂3] - [1 + 2 log₃2] = 2 log₂3 - 2 log₃2.Hmm, so that simplifies to 2(log₂3 - log₃2). Let me note that log₃2 is 1 / log₂3, right? Because log_b a = 1 / log_a b. So if log₃2 = 1 / log₂3.Therefore, log₂3 - log₃2 = log₂3 - 1 / log₂3. Let me denote log₂3 as t for a moment. Then this expression becomes t - 1/t. So the coefficient of x is 2(t - 1/t). Let me compute t numerically? Wait, maybe I can express this in terms of known logarithms or find a relationship. Alternatively, maybe we can compute the numerical value.But before that, let's check the other logarithmic terms: -log₃16 -4 log₂3.log₃16: 16 is 2^4, so log₃2^4 = 4 log₃2. Therefore, -log₃16 = -4 log₃2.So the entire constant term is -4 log₃2 -4 log₂3. Which can be written as -4(log₃2 + log₂3). Again, log₃2 is 1 / log₂3, so this is -4(1 / log₂3 + log₂3). Let me denote log₂3 as t again. So this becomes -4(t + 1/t).Therefore, the original function y can be rewritten in terms of t:y = x² + 2(t - 1/t)x -4(t + 1/t)But maybe instead of substituting t, I can compute these logarithmic terms numerically. Since log₂3 is approximately 1.58496 (because 2^1.58496 ≈ 3). Let me check:2^1 = 2, 2^1.58496 ≈ 3. So log₂3 ≈ 1.58496.Therefore, log₃2 = 1 / 1.58496 ≈ 0.63093.So let's compute the coefficient for x: 2(log₂3 - log₃2) ≈ 2(1.58496 - 0.63093) ≈ 2(0.95403) ≈ 1.90806.So approximately 1.90806. Let me note that.The constant term: -4(log₃2 + log₂3) ≈ -4(0.63093 + 1.58496) ≈ -4(2.21589) ≈ -8.86356.Therefore, the function y can be approximated as:y ≈ x² + 1.90806x -8.86356But we need to solve y ≤ 8. So:x² + 1.90806x -8.86356 ≤ 8Which becomes:x² + 1.90806x -16.86356 ≤ 0But solving this quadratic inequality with approximate coefficients might not be precise. Maybe there's an exact way.Alternatively, let's try to express all logs in terms of log₂3 and log₃2, and see if they can be simplified.Let me denote a = log₂3. Then log₃2 = 1/a.So the coefficient of x is 2(a - 1/a). The constant term is -4(a + 1/a).Therefore, the function becomes:y = x² + 2(a - 1/a)x -4(a + 1/a)Then we need to solve:x² + 2(a - 1/a)x -4(a + 1/a) ≤ 8Bring 8 to the left:x² + 2(a - 1/a)x -4(a + 1/a) -8 ≤ 0So, x² + 2(a - 1/a)x - [4(a + 1/a) +8] ≤ 0Let me compute 4(a + 1/a) +8. That's 4a + 4/a +8.Alternatively, perhaps factor this expression? Hmm.Alternatively, let's note that a = log₂3. Maybe express everything in terms of a.But I might be overcomplicating. Alternatively, let's use change of base formula to express all logs in terms of natural logs or common logs, to see if something cancels.Alternatively, maybe there's a relationship between log₂3 and log₃2. Since log₃2 = 1 / log₂3, as mentioned before.Let me see. So:The coefficient of x is 2(a - 1/a), where a = log₂3.The constant term is -4(a + 1/a) -8.But perhaps we can write the quadratic in terms of a.Alternatively, maybe note that a = log₂3, so 2^a = 3.But I don't know if that helps.Alternatively, maybe we can compute the exact value of the coefficients. Let's try.Let me see:Let me compute 2(a - 1/a):= 2a - 2/aSimilarly, the constant term is -4(a + 1/a) -8= -4a -4/a -8So the quadratic inequality is:x² + (2a - 2/a)x -4a -4/a -8 ≤ 0But without knowing a's exact value, this seems difficult. Maybe there's a way to write this quadratic in terms of a common variable.Alternatively, perhaps multiply through by a to eliminate denominators. Let's try that.Multiply both sides by a (assuming a ≠ 0, which it is not since a = log₂3 ≈1.585):a*x² + (2a² - 2)x -4a² -4 -8a ≤ 0So, a*x² + (2a² - 2)x - (4a² +4 +8a) ≤0But a is log₂3, which is irrational, so this might not help. Maybe not the best approach.Alternatively, perhaps note that a is log₂3, and 2a = log₂3^2 = log₂9, but not sure if that's helpful.Alternatively, let's check if there's a relationship between the coefficients.Wait, perhaps think of the quadratic in x as x² + bx + c ≤0, where b = 2(a - 1/a) and c = -4(a +1/a) -8.Alternatively, maybe factor the quadratic? Let me check.But the quadratic is x² + (2a - 2/a)x -4(a +1/a) -8.Alternatively, maybe we can complete the square? Let's see.Completing the square for x² + bx + c:x² + bx + c = (x + b/2)^2 - (b²)/4 + cSo let's compute that:Here, b = 2(a - 1/a), so b/2 = a -1/aTherefore,(x + a -1/a)^2 - ( (2(a -1/a))² ) /4 + cCompute the second term:(4(a -1/a)^2)/4 = (a -1/a)^2Therefore, the expression becomes:(x + a -1/a)^2 - (a -1/a)^2 + cThen c is -4(a +1/a) -8Therefore, total expression:(x + a -1/a)^2 - (a -1/a)^2 -4(a +1/a) -8 ≤ 0Let me compute the constants:First, let me compute (a -1/a)^2:= a² - 2*(a)*(1/a) + (1/a)^2 = a² -2 + 1/a²Therefore, the expression becomes:(x + a -1/a)^2 - (a² -2 + 1/a²) -4(a +1/a) -8= (x + a -1/a)^2 -a² +2 -1/a² -4a -4/a -8Combine constants:= (x + a -1/a)^2 -a² -1/a² -4a -4/a +2 -8= (x + a -1/a)^2 -a² -1/a² -4a -4/a -6Hmm, not sure if this helps. Maybe not. Perhaps another approach.Alternatively, maybe notice that the quadratic equation in x can be written as:x² + (log₂18 - log₃12)x - log₃16 -4 log₂3 ≤8But perhaps convert all logarithms to the same base. Let me try converting all logs to base 2.First, log₂18 is already base 2. log₃12 can be written as log₂12 / log₂3. Similarly, log₃16 is log₂16 / log₂3, and log₃2 is 1 / log₂3.Let me denote log₂3 as t again.So:log₃12 = log₂12 / tlog₃16 = log₂16 / tSo rewrite the original function y:y = x² + x*(log₂18 - log₃12) - log₃16 -4 log₂3= x² + x*(log₂18 - (log₂12)/t) - (log₂16)/t -4tNow, compute each log term:log₂18: log₂(2*3²) = 1 + 2 log₂3 = 1 + 2tlog₂12: log₂(4*3) = log₂4 + log₂3 = 2 + tlog₂16: log₂16 = 4Therefore, substituting back:y = x² + x*( (1 + 2t) - (2 + t)/t ) - 4/t -4tSimplify each term:First, the coefficient of x:(1 + 2t) - (2 + t)/t = 1 + 2t - (2/t + 1)= 1 + 2t -1 -2/t= 2t -2/tWhich matches the previous result. So that's 2(t - 1/t)Similarly, the constant term:-4/t -4tSo the function is:y = x² + 2(t -1/t)x -4(t +1/t)Again, same as before. So even after converting all logs to base 2, we still end up with the same expression. So perhaps this is as simplified as it can get.So, to solve y ≤8, we have:x² + 2(t -1/t)x -4(t +1/t) ≤8Bring 8 to the left:x² + 2(t -1/t)x -4(t +1/t) -8 ≤0Let me denote the quadratic as:x² + Bx + C ≤0, where B = 2(t -1/t) and C = -4(t +1/t) -8.To find the solutions, we can compute the roots of the quadratic equation x² + Bx + C =0, then the inequality holds between the roots if the quadratic opens upwards (which it does, since the coefficient of x² is 1 >0). Therefore, the solution set is the interval between the two roots, inclusive.Therefore, we need to compute the roots:x = [-B ± sqrt(B² -4*1*C)] /2Compute B and C:B = 2(t -1/t) = 2t - 2/tC = -4(t +1/t) -8 = -4t -4/t -8Therefore, discriminant D = B² -4*1*CLet me compute D:D = (2t -2/t)^2 -4*(-4t -4/t -8)First, expand (2t -2/t)^2:= 4t² - 2*2t*(2/t) + 4/t²Wait, wait, (a - b)^2 = a² -2ab +b². So here a=2t, b=2/t.Therefore:(2t -2/t)^2 = (2t)^2 -2*(2t)*(2/t) + (2/t)^2= 4t² -8 + 4/t²So D = 4t² -8 +4/t² -4*(-4t -4/t -8)Compute the second part:-4*(-4t -4/t -8) = 16t +16/t +32Therefore, D = 4t² -8 +4/t² +16t +16/t +32Combine like terms:4t² +4/t² +16t +16/t + (-8 +32)=4t² +4/t² +16t +16/t +24Hmm, this seems complicated, but maybe factor out 4:=4(t² +1/t² +4t +4/t +6)Wait, t² +1/t² can be written as (t +1/t)^2 -2. Similarly, 4t +4/t =4(t +1/t). Let's check:Let me write t² +1/t² +4t +4/t +6= (t² +1/t²) +4(t +1/t) +6= [(t +1/t)^2 -2] +4(t +1/t) +6= (t +1/t)^2 +4(t +1/t) +4So that becomes:(t +1/t)^2 +4(t +1/t) +4Which is equal to:[(t +1/t) +2]^2Let me check:Let me expand [(t +1/t) +2]^2 = (t +1/t)^2 +4(t +1/t) +4. Yes, exactly. So:D =4*[(t +1/t) +2]^2Therefore, the discriminant D is a perfect square. Therefore, sqrt(D) =2*[(t +1/t) +2]Therefore, the roots are:x = [-B ± sqrt(D)] /2We have B=2t -2/tSo:x = [ -2t +2/t ± 2(t +1/t +2) ] /2Simplify numerator:First, take the '+' case:-2t +2/t +2(t +1/t +2) = -2t +2/t +2t +2/t +4 = ( -2t +2t ) + (2/t +2/t ) +4 = 4/t +4Divide by 2:(4/t +4)/2 =2/t +2Similarly, the '-' case:-2t +2/t -2(t +1/t +2) = -2t +2/t -2t -2/t -4 = (-2t -2t) + (2/t -2/t) -4 = -4t -4Divide by 2:(-4t -4)/2 = -2t -2Therefore, the roots are x=2/t +2 and x=-2t -2Therefore, the quadratic inequality x² + Bx + C ≤0 is satisfied for x between -2t -2 and 2/t +2, inclusive.Therefore, the integer solutions x must satisfy:-2t -2 ≤x ≤2/t +2Since t = log₂3 ≈1.58496, let's compute the numerical values of the bounds.Compute left bound: -2t -2 ≈ -2*1.58496 -2 ≈ -3.16992 -2 ≈ -5.16992Compute right bound: 2/t +2 ≈2 /1.58496 +2 ≈1.26186 +2≈3.26186Therefore, x must be between approximately -5.16992 and 3.26186. Since x must be integers, the integer values of x are: -5, -4, -3, -2, -1, 0, 1, 2, 3Now, check if x=3 is included. The upper bound is approximately3.26186, so 3 is less than that. x=4 would be 4 which is higher than 3.26186, so 4 is excluded. Similarly, lower bound is approx -5.16992, so integers from -5 upwards. x=-6 would be less than -5.16992, so -6 is excluded.Therefore, integer solutions: x=-5,-4,-3,-2,-1,0,1,2,3Now, sum these integers.Sum from -5 to3 inclusive.Sum = (-5) + (-4) + (-3) + (-2) + (-1) +0 +1 +2 +3Let me compute step by step:Adding from left to right:Start with -5.-5 + (-4) = -9-9 + (-3) = -12-12 + (-2) = -14-14 + (-1) = -15-15 +0 = -15-15 +1 = -14-14 +2 = -12-12 +3 = -9So total sum is -9.Wait, but let me check again:List of numbers: -5, -4, -3, -2, -1, 0,1,2,3Sum them:(-5 +3) = -2(-4 +2)= -2(-3 +1)= -2(-2 +0)= -2-1 remains.Total sum: (-2)*4 + (-1) = -8 -1 = -9. Yes, same result. So the sum is -9.But before finalizing, we need to ensure that all these integers indeed satisfy the original inequality y ≤8. Because sometimes when we use approximate values for the roots, the endpoints might be close to an integer, and due to approximation errors, we might include an integer that shouldn't be included or exclude one that should be included.Therefore, it's essential to verify each integer x from -5 to3 by plugging them into the original function y and checking if y ≤8.Let me compute y for each x in {-5, -4, -3, -2, -1,0,1,2,3}First, compute the exact value of the coefficients. Let me note that:The coefficient for x is log₂18 - log₃12Earlier, we found that:log₂18 =1 +2 log₂3log₃12=1 +2 log₃2=1 +2*(1 / log₂3)Let me denote log₂3 as t≈1.58496Thus, coefficient of x is:(1 +2t) - [1 + 2*(1/t)] =2t -2*(1/t)=2(t -1/t)= approximately2*(1.58496 -0.63093)=2*0.95403≈1.90806, as before.Constant term: -log₃16 -4 log₂3log₃16= log₃(2^4)=4 log₃2=4*(1/t)Thus, constant term: -4*(1/t) -4t≈-4/1.58496 -4*1.58496≈-2.523 -6.339≈-8.862So y ≈x² +1.90806x -8.862Therefore, approximate y values:For x=-5:y≈(-5)^2 +1.90806*(-5) -8.862≈25 -9.5403 -8.862≈25 -18.4023≈6.5977 ≤8: yesx=-4:y≈16 +1.90806*(-4) -8.862≈16 -7.63224 -8.862≈16 -16.49424≈-0.49424 ≤8: yesx=-3:y≈9 +1.90806*(-3) -8.862≈9 -5.72418 -8.862≈9 -14.58618≈-5.58618 ≤8: yesx=-2:y≈4 +1.90806*(-2) -8.862≈4 -3.81612 -8.862≈4 -12.67812≈-8.67812 ≤8: yesx=-1:y≈1 +1.90806*(-1) -8.862≈1 -1.90806 -8.862≈1 -10.77006≈-9.77006 ≤8: yesx=0:y≈0 +0 -8.862≈-8.862 ≤8: yesx=1:y≈1 +1.90806*1 -8.862≈1 +1.90806 -8.862≈2.90806 -8.862≈-5.95394 ≤8: yesx=2:y≈4 +1.90806*2 -8.862≈4 +3.81612 -8.862≈7.81612 -8.862≈-1.04588 ≤8: yesx=3:y≈9 +1.90806*3 -8.862≈9 +5.72418 -8.862≈14.72418 -8.862≈5.86218 ≤8: yesSo all these x values indeed satisfy y ≤8. Therefore, the integer solutions are indeed from -5 to3 inclusive, and their sum is -9.Therefore, the answer is boxed{-9}Final Answerboxed{-9}</think>To find the sum of all integer values of ( x ) for which the corresponding values of the function[y = x^2 + xleft(log_2 18 - log_3 12right) - log_3 16 - 4 log_2 3]do not exceed 8, we start by simplifying the logarithmic terms.First, we convert the logarithms to a common base and simplify:- (log_2 18 = 1 + 2 log_2 3)- (log_3 12 = 1 + 2 log_3 2)- (log_3 16 = 4 log_3 2)- (log_3 2 = frac{1}{log_2 3})Let ( t = log_2 3 ). Then:- (log_3 2 = frac{1}{t})- The coefficient of ( x ) becomes ( 2(t - frac{1}{t}) )- The constant term becomes ( -4(t + frac{1}{t}) - 8 )The function ( y ) can be rewritten as:[y = x^2 + 2(t - frac{1}{t})x - 4(t + frac{1}{t}) - 8]We need to solve the inequality ( y leq 8 ). Substituting the simplified terms, we get:[x^2 + 2(t - frac{1}{t})x - 4(t + frac{1}{t}) - 8 leq 0]This is a quadratic inequality in ( x ). Solving the quadratic equation ( x^2 + 2(t - frac{1}{t})x - 4(t + frac{1}{t}) - 8 = 0 ), we find the roots using the quadratic formula. The roots are approximately ( -5.16992 ) and ( 3.26186 ).The integer values of ( x ) that satisfy the inequality are ( -5, -4, -3, -2, -1, 0, 1, 2, 3 ). Summing these integers:[-5 + (-4) + (-3) + (-2) + (-1) + 0 + 1 + 2 + 3 = -9]Thus, the sum of all integer values of ( x ) for which ( y leq 8 ) is (boxed{-9}).]

🔑:Solution. Let a=log _{2} 3. Then the condition of the problem will turn into the inequalityx^{2}+2left(a-frac{1}{a}right) x-4left(a+frac{1}{a}+2right) leqslant 0Considering that a inleft(frac{3}{2}, 2right), we get x inleft[-2 a-2, frac{2}{a}+2right]. Since -6<-2 a-2<-5.3<frac{2}{a}+2<4, the integer solutions will be the numbers -5,-4,-3,-2,-1,0,1,2,3.Answer: -9 . (A)