🧠:Okay, let's try to figure out this problem. So we have two sets of points. The first set is all the points with x-coordinate -1 and y-coordinates from 1 to 2000. So they're like (-1, 1), (-1, 2), ..., (-1, 2000). The second set is similar but with x-coordinate 1: (1, 1), (1, 2), ..., (1, 2000). We need to connect each point in the first set to exactly one point in the second set with a line segment, and each point in the second set is connected to exactly one in the first set. So essentially, it's a perfect matching between the two sets. Then, we have to find the sum of the y-intercepts of all these 2000 line segments. Hmm, okay. Let's start by understanding what a y-intercept of a line segment is. The y-intercept is the point where the line crosses the y-axis, which is where x=0. For a line connecting two points, say (-1, a) from the first set and (1, b) from the second set, we can find the equation of the line and then find its y-intercept.So, let's recall the equation of a line given two points. If we have two points (x1, y1) and (x2, y2), the slope m is (y2 - y1)/(x2 - x1). Then, using point-slope form, the equation is y - y1 = m(x - x1). Then, to find the y-intercept, we set x=0 and solve for y.Let me do that for a general pair of points. Let's take a point (-1, a) from the first set and (1, b) from the second set. The slope m would be (b - a)/(1 - (-1)) = (b - a)/2. Then, using the point (-1, a), the equation of the line is:y - a = [(b - a)/2](x - (-1)) y - a = [(b - a)/2](x + 1)To find the y-intercept, set x=0:y - a = [(b - a)/2](0 + 1) y = a + (b - a)/2 y = (2a + b - a)/2 y = (a + b)/2So the y-intercept is (a + b)/2 for the line connecting (-1, a) and (1, b). Interesting. So regardless of the points chosen, the y-intercept is just the average of their y-coordinates. That simplifies things!Therefore, each line segment's y-intercept is the average of the y-coordinates of the two points it connects. Since we have to connect each point in the first set to exactly one in the second set, forming a perfect matching, the sum of all y-intercepts will be the sum over all pairs of (a + b)/2.Therefore, the total sum is (1/2) times the sum over all a from the first set plus the sum over all b from the second set. Wait, but hold on. Because each a is connected to exactly one b, and each b is connected to exactly one a. So the sum over all pairs is actually equal to the sum of all a's plus the sum of all b's. Because each a is included once and each b is included once. Then, multiplied by 1/2.Therefore, the total sum of y-intercepts is (1/2)(sum_{a} a + sum_{b} b). But the sum of all a's is the same as the sum of all b's because both sets have the same y-coordinates from 1 to 2000. So sum_{a} a = sum_{b} b = 1 + 2 + 3 + ... + 2000.Therefore, the total sum is (1/2)(2 * sum_{k=1}^{2000} k) = sum_{k=1}^{2000} k.Wait, so is it just the sum of 1 to 2000? That seems too straightforward. Let me verify.Let me take a small example. Suppose we have two sets with points (-1, 1), (-1, 2) and (1, 1), (1, 2). Then, there are two possible matchings: connecting (-1,1) to (1,1) and (-1,2) to (1,2), or connecting (-1,1) to (1,2) and (-1,2) to (1,1). Let's compute the sum of the y-intercepts in both cases.First matching:1. Line between (-1,1) and (1,1): y-intercept is (1+1)/2 = 1.2. Line between (-1,2) and (1,2): y-intercept is (2+2)/2 = 2.Sum: 1 + 2 = 3.Second matching:1. Line between (-1,1) and (1,2): y-intercept is (1+2)/2 = 1.5.2. Line between (-1,2) and (1,1): y-intercept is (2+1)/2 = 1.5.Sum: 1.5 + 1.5 = 3.In both cases, the sum is 3, which is equal to 1 + 2 = 3. So the sum of y-intercepts is the same regardless of the matching. So the total sum is indeed the sum of 1 to n, where n is the number of points in each set.Therefore, in the original problem, since each set has 2000 points with y-coordinates from 1 to 2000, the sum of the y-intercepts is the sum of 1 + 2 + ... + 2000.So now, we just need to compute the sum of the first 2000 positive integers. The formula for the sum of the first n integers is n(n + 1)/2. Therefore, substituting n = 2000:Sum = 2000 * 2001 / 2 = 1000 * 2001 = 2,001,000.Hence, the sum of the y-intercepts is 2,001,000.But wait, let me check the reasoning again. The key insight was that regardless of how we pair the points, the sum of the y-intercepts remains the same. Is this always true?In the small example, we saw that even when we swapped the pairings, the sum remained the same. That's because swapping two pairs (a, b) and (c, d) to (a, d) and (c, b) changes individual y-intercepts but keeps the total sum the same. Let me check that.Suppose we have two pairs: (a, b) and (c, d). Their y-intercepts are (a + b)/2 and (c + d)/2. If we swap to (a, d) and (c, b), the y-intercepts become (a + d)/2 and (c + b)/2. The total sum before and after:Before: (a + b + c + d)/2 After: (a + d + c + b)/2 = same as before.Therefore, swapping any two pairs doesn't change the total sum. Since any permutation can be broken down into a series of swaps, the total sum remains invariant under any permutation. Hence, no matter how you pair the points, the sum of the y-intercepts is always the sum of all (a_i + b_j)/2 where each a_i is paired with exactly one b_j.But since each a_i is paired with exactly one b_j, and vice versa, the total sum is (sum a_i + sum b_j)/2. But sum a_i and sum b_j are the same because both sets have y-coordinates from 1 to 2000. Therefore, sum a_i = sum b_j = sum_{k=1}^{2000} k. Therefore, total sum is (2 * sum_{k=1}^{2000} k)/2 = sum_{k=1}^{2000} k. Thus, the conclusion holds.Therefore, the answer is simply the sum from 1 to 2000, which is 2000*2001/2 = 2001000.I think that's solid. The key steps were realizing the y-intercept is the average of the paired y-coordinates, then recognizing that the sum of all averages is the average of the total sums, which for a perfect matching simplifies to the sum of all original y-coordinates. Since the pairing doesn't affect the total, it's just the sum from 1 to 2000.Final AnswerThe sum of the y-intercepts of the segments is boxed{2001000}.

✅:Given two sets of points in the coordinate plane: ({(-1,1), (-1,2), ldots, (-1,2000)}) and ({(1,1), (1,2), ldots, (1,2000)}), we need to find the sum of the (y)-intercepts of 2000 line segments connecting each point in the first set to exactly one point in the second set, and vice versa.1. Equation of the Line and (y)-intercept: - For a line segment connecting ((-1, a)) and ((1, b)), the slope (m) is (frac{b - a}{2}). - The equation of the line using point-slope form is (y - a = frac{b - a}{2}(x + 1)). - Setting (x = 0) to find the (y)-intercept, we get (y = frac{a + b}{2}).2. Sum of (y)-intercepts: - Each (y)-intercept is the average of the (y)-coordinates of the connected points, (frac{a + b}{2}). - The sum of all (y)-intercepts is (frac{1}{2}) times the sum of all (y)-coordinates from both sets. - Since both sets have the same (y)-coordinates (1 to 2000), the sum of all (y)-coordinates in each set is (sum_{k=1}^{2000} k).3. Sum of the First 2000 Positive Integers: - The sum (sum_{k=1}^{2000} k) is calculated using the formula (frac{n(n + 1)}{2}) with (n = 2000): [ sum_{k=1}^{2000} k = frac{2000 times 2001}{2} = 1000 times 2001 = 2001000 ]4. Conclusion: - The sum of the (y)-intercepts is (frac{1}{2} times 2 times 2001000 = 2001000).Thus, the sum of the (y)-intercepts of the segments is (boxed{2001000}).

🔑:1. Determine the general form of the y-intercept: Consider the line segment connecting point ((-1, a)) and ((1, b)). The general equation of a line through two points ((x_1, y_1)) and ((x_2, y_2)) is given by: [ y - y_1 = frac{y_2 - y_1}{x_2 - x_1}(x - x_1) ] Substituting (x_1 = -1, y_1 = a, x_2 = 1, y_2 = b), we get: [ y - a = frac{b - a}{1 - (-1)}(x + 1) = frac{b - a}{2}(x + 1) ] Solving for (y), we have: [ y = a + frac{b - a}{2}(x + 1) ]2. Calculate the (y)-intercept: The (y)-intercept is obtained by setting (x = 0): [ y = a + frac{b - a}{2}(0 + 1) = a + frac{b - a}{2} ] Simplifying, we get: [ text{Intercept} = frac{a + b}{2} ]3. Sum of (y)-intercepts: Consider the given points and their connections: - Points in set one: ((-1, 1), (-1, 2), ldots, (-1, 2000)) - Points in set two: ((1, 1), (1, 2), ldots, (1, 2000)) Suppose each point ((-1, k)) is connected to ((1, y_k)). Using our formula for the intercept: [ ytext{-intercept} = frac{k + y_k}{2} ] Summing over all points, we get: [ sum_{k=1}^{2000} frac{k + y_k}{2} ]4. Simplify the sum: Using the properties of summation, this can be split as: [ sum_{k=1}^{2000} frac{k + y_k}{2} = frac{1}{2} left( sum_{k=1}^{2000} k + sum_{k=1}^{2000} y_k right) ] Since (y_k) are simply a reordering of (1, 2, ldots, 2000), their sum is also the sum of an arithmetic series from 1 to 2000: [ sum_{k=1}^{2000} y_k = sum_{k=1}^{2000} k ] Therefore, [ sum_{k=1}^{2000} frac{k + y_k}{2} = frac{1}{2} left( sum_{k=1}^{2000} k + sum_{k=1}^{2000} k right) = frac{1}{2} left( 2 sum_{k=1}^{2000} k right) = sum_{k=1}^{2000} k ]5. Evaluate the sum of the arithmetic sequence: The formula for the sum of the first (n) natural numbers is: [ sum_{k=1}^{n} k = frac{n(n+1)}{2} ] Applying this to our sequence: [ sum_{k=1}^{2000} k = frac{2000 cdot 2001}{2} ]6. Final calculation: [ frac{2000 cdot 2001}{2} = 2001000 ]Conclusion:[boxed{2001000}]

🧠:Okay, so I need to find a new quadratic equation where one root is the reciprocal of the sum of the roots of the original equation 2x² -9x +8=0, and the other root is the square of the difference of the roots of the original equation. Hmm, let's break this down step by step.First, I remember that for a quadratic equation ax² + bx + c =0, the sum of the roots is -b/a and the product is c/a. Let me confirm that. So for the original equation, 2x² -9x +8=0, the coefficients are a=2, b=-9, c=8. Therefore, the sum of the roots, let's call them α and β, would be -b/a = 9/2. And the product αβ = c/a = 8/2 = 4.Wait, the first part of the problem says the reciprocal of the sum of the roots. The sum is 9/2, so its reciprocal is 2/9. That's straightforward. So one root of the new equation is 2/9.The other root is the square of the difference of the roots. The difference of the roots is α - β. But how do I find that? I know that (α - β)² = (α + β)² - 4αβ. Let me check. Expanding (α - β)² gives α² - 2αβ + β², and (α + β)² is α² + 2αβ + β². So subtracting 4αβ from (α + β)² would give α² + 2αβ + β² -4αβ = α² -2αβ + β², which is exactly (α - β)². So yes, (α - β)² = (α + β)² -4αβ.So substituting the values we have: (α - β)² = (9/2)² -4*(4). Let's compute that. (81/4) - 16. Wait, 16 is 64/4. So 81/4 -64/4 = 17/4. Therefore, the square of the difference of the roots is 17/4. So the other root of the new equation is 17/4.So now, the new quadratic equation has roots 2/9 and 17/4. To form the quadratic equation, we can use the sum and product of these roots. Let me recall that if the roots are r and s, the equation is x² - (r + s)x + rs =0. But wait, actually, the general form is ax² + bx + c =0, but the monic form (leading coefficient 1) would be x² - (sum)x + product. So if we use this, then the sum of the new roots is 2/9 +17/4. Let's calculate that. To add these fractions, find a common denominator. 9 and 4 have a common denominator of 36. So 2/9 is 8/36, and 17/4 is 153/36. Adding them gives 161/36. Then the product of the roots is (2/9)*(17/4) = (34)/36 = 17/18.Therefore, the quadratic equation would be x² - (161/36)x +17/18 =0. But usually, equations are written without fractions. So we can multiply through by 36 to eliminate denominators. Multiplying each term: 36x² -161x + 34 =0. Wait, 17/18 multiplied by 36 is 17*2=34. Correct. So the equation is 36x² -161x +34=0.Wait, let me double-check the calculations here. First, sum of roots 2/9 +17/4. Convert to decimal to check? 2/9 is approximately 0.222..., 17/4 is 4.25. Sum is about 4.472..., which is 161/36. Let me compute 161 divided by 36: 36*4=144, 161-144=17, so 4 and 17/36, which is approximately 4.472... Correct.Product is 2/9 *17/4 = (2*17)/(9*4)=34/36=17/18≈0.944..., which when multiplied by 36 gives 34. Correct.So the quadratic equation is 36x² -161x +34=0. But maybe they want the coefficients to be integers with no common factors. Let's check if 36, 161, 34 have a common divisor. 36 and 34 are both divisible by 2, but 161 is not (161 divided by 2 is 80.5). So no common divisor except 1. Therefore, the equation 36x² -161x +34=0 is the simplest form.Wait, but just to make sure, let me verify all steps again. Original equation: 2x² -9x +8=0. Sum of roots α + β = 9/2, product αβ =4. Then, reciprocal of sum is 2/9. Then difference squared: (α - β)² = (α + β)² -4αβ = (81/4) -16 = (81 -64)/4=17/4. So that's correct.Therefore, new roots: 2/9 and 17/4. Sum is 2/9 +17/4 = (8 +153)/36=161/36. Product is (2/9)(17/4)=34/36=17/18. So quadratic equation is x² - (161/36)x +17/18=0. Multiply by 36: 36x² -161x +34=0. Correct.I think that's it. Let me just check once more to make sure there's no arithmetic error. The difference squared: 9/2 squared is 81/4. Minus 4*4=16. 81/4 -16 =81/4 -64/4=17/4. Yes. Reciprocal of 9/2 is 2/9. Then sum and product of new roots: 2/9 +17/4. Correct. Product 2/9 *17/4=34/36=17/18. Yes. Multiply by 36: 36x² -161x +34. Correct. Coefficients 36, 161, 34 have no common factors. So yes, final answer is 36x² -161x +34=0.

✅:To solve the problem, we start by analyzing the original quadratic equation (2x^2 - 9x + 8 = 0). Using Vieta's formulas:- Sum of roots ((alpha + beta)): (frac{-b}{a} = frac{9}{2})- Product of roots ((alpha beta)): (frac{c}{a} = 4)First root of the new equation: The reciprocal of the sum of the roots:[frac{1}{alpha + beta} = frac{2}{9}]Second root of the new equation: The square of the difference of the roots. Using ((alpha - beta)^2 = (alpha + beta)^2 - 4alpha beta):[(alpha - beta)^2 = left(frac{9}{2}right)^2 - 4 cdot 4 = frac{81}{4} - 16 = frac{17}{4}]Forming the new quadratic equation with roots (frac{2}{9}) and (frac{17}{4}):- Sum of new roots: (frac{2}{9} + frac{17}{4} = frac{8 + 153}{36} = frac{161}{36})- Product of new roots: (frac{2}{9} cdot frac{17}{4} = frac{34}{36} = frac{17}{18})The quadratic equation is:[x^2 - left(frac{161}{36}right)x + frac{17}{18} = 0]Eliminating fractions by multiplying by 36:[36x^2 - 161x + 34 = 0]Final Answer:[boxed{36x^2 - 161x + 34 = 0}]

🔑:1. Consider the quadratic equation given in the problem: [ 2x^2 - 9x + 8 = 0 ] Let (x_1) and (x_2) be the roots of this equation.2. By Vieta's formulas, we know: [ x_1 + x_2 = frac{9}{2} quad text{and} quad x_1x_2 = 4 ]3. We need to construct a new quadratic equation with roots ( frac{1}{x_1 + x_2} ) and ( (x_1 - x_2)^2 ).4. First, calculate ( frac{1}{x_1 + x_2} ): [ x_1' = frac{1}{x_1 + x_2} = frac{1}{frac{9}{2}} = frac{2}{9} ]5. Next, calculate ( (x_1 - x_2)^2 ): [ (x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1x_2 ] Substituting the known values: [ (x_1 + x_2)^2 = left(frac{9}{2}right)^2 = frac{81}{4} ] and [ 4x_1x_2 = 4 times 4 = 16 ] Thus, [ x_2' = (x_1 - x_2)^2 = frac{81}{4} - 16 = frac{81}{4} - frac{64}{4} = frac{17}{4} ]6. We now have the roots (x_1' = frac{2}{9}) and (x_2' = frac{17}{4}).7. The quadratic equation with these roots can be written as: [ x^2 - (x_1' + x_2')x + x_1'x_2' = 0 ]8. Calculate the coefficients: [ p = -(x_1' + x_2')= -left(frac{2}{9} + frac{17}{4}right) ] Finding a common denominator: [ frac{2}{9} + frac{17}{4} = frac{8}{36} + frac{153}{36} = frac{161}{36} ] Hence, [ p = -frac{161}{36} ]9. Calculate (q = x_1'x_2'): [ q = frac{2}{9} times frac{17}{4} = frac{34}{36} = frac{17}{18} ]10. Therefore, the quadratic equation becomes: [ x^2 - left(frac{161}{36}right)x + frac{17}{18} = 0 ] To clear the fractions, multiply the entire equation by 36: [ 36x^2 - 161x + 34 = 0 ]Conclusion:[boxed{36x^2 - 161x + 34 = 0}]

🧠:Okay, so I need to figure out if the vectors a, b, and c are coplanar. Let me remember what coplanar vectors mean. If three vectors are coplanar, they all lie in the same plane, right? Which would mean that the volume of the parallelepiped formed by these three vectors is zero. The way to check that is by using the scalar triple product. If the scalar triple product of a, b, and c is zero, then they are coplanar. If not, then they aren't. So the scalar triple product is calculated as a · (b × c). Let me recall how to compute the cross product first, and then the dot product. First, let's find the cross product of vectors b and c. Vector b is {-2, -1, 0} and vector c is {5, 2, -1}. The cross product b × c can be found using the determinant of a matrix with the unit vectors i, j, k in the first row, the components of b in the second row, and the components of c in the third row. So writing that out:b × c = |i   j   k|        |-2 -1   0|        |5   2  -1|Calculating this determinant, the i component is (-1)(-1) - (0)(2) = 1 - 0 = 1. The j component is - [ (-2)(-1) - (0)(5) ] = - [2 - 0] = -2. The k component is (-2)(2) - (-1)(5) = -4 + 5 = 1. So the cross product b × c is {1, -2, 1}. Wait, let me verify that again. For the cross product, the formula for each component is:i component: (b2*c3 - b3*c2)j component: -(b1*c3 - b3*c1)k component: (b1*c2 - b2*c1)Wait, actually, the cross product of b × c is computed as:i*(b2*c3 - b3*c2) - j*(b1*c3 - b3*c1) + k*(b1*c2 - b2*c1)So plugging in the values:b2 is -1, c3 is -1; b3 is 0, c2 is 2. So i component: (-1)*(-1) - 0*2 = 1 - 0 = 1j component: - [ (-2)*(-1) - 0*5 ] = - [2 - 0] = -2k component: (-2)*2 - (-1)*5 = -4 +5 = 1Yes, so b × c = {1, -2, 1}. Okay, that checks out.Now, we need to take the dot product of vector a with this result. Vector a is {3, 1, -1}. So the dot product a · (b × c) is 3*1 + 1*(-2) + (-1)*1.Calculating each term:3*1 = 31*(-2) = -2-1*1 = -1Adding them up: 3 - 2 - 1 = 0. So the scalar triple product is zero. That means the volume of the parallelepiped formed by a, b, and c is zero, which implies that the vectors are coplanar. Wait a second, let me make sure I didn't make any calculation errors. Let me go through the cross product again step by step. Given b = {-2, -1, 0}, c = {5, 2, -1}Cross product formula:i*(b2*c3 - b3*c2) - j*(b1*c3 - b3*c1) + k*(b1*c2 - b2*c1)So:i*(-1*(-1) - 0*2) - j*(-2*(-1) - 0*5) + k*(-2*2 - (-1)*5)Calculates to:i*(1 - 0) - j*(2 - 0) + k*(-4 +5)Which is 1i -2j +1k. So {1, -2, 1}. Correct.Then a · (b × c) is 3*1 + 1*(-2) + (-1)*1 = 3 -2 -1 = 0. So yes, that's zero. Therefore, the scalar triple product is zero. Alternatively, another way to check if vectors are coplanar is to see if one of them can be expressed as a linear combination of the other two. Let's try that approach to confirm. Suppose we try to express vector c as a combination of a and b. So, does there exist scalars x and y such that c = x*a + y*b?Writing the equations component-wise:For the first component: 5 = 3x + (-2)ySecond component: 2 = 1x + (-1)yThird component: -1 = (-1)x + 0*ySo from the third component: -1 = -x + 0 => x = 1. Then plug x = 1 into the second component equation: 2 = 1*1 + (-1)y => 2 = 1 - y => -y = 1 => y = -1.Now check the first component with x =1 and y = -1: 3*1 + (-2)*(-1) = 3 + 2 = 5. Which matches the first component of c. So c = 1*a + (-1)*b. So c is indeed a linear combination of a and b. Therefore, the three vectors are linearly dependent, hence coplanar. Alternatively, if the determinant of the matrix formed by these vectors (as rows or columns) is zero, they are coplanar. Let me form the matrix with vectors a, b, c as rows and compute its determinant. Matrix M:3   1   -1-2 -1   05   2   -1Calculating determinant:3*[(-1)*(-1) - 0*2] -1*[(-2)*(-1) -0*5] + (-1)*[(-2)*2 - (-1)*5]Which is 3*(1 -0) -1*(2 -0) + (-1)*(-4 +5)= 3*1 -1*2 + (-1)*(1)= 3 -2 -1 = 0So determinant is zero, hence vectors are coplanar. Hmm, that's three different methods all leading to the same conclusion. The scalar triple product is zero, one vector is a linear combination of the other two, and the determinant of the matrix formed by them is zero. All confirm that the vectors are coplanar. I think that's thorough. Let me just check once more if I messed up any calculations, but all steps seem to check out. The cross product, the dot product, solving the linear system, and the determinant all give consistent results. So yes, the vectors a, b, and c are coplanar.

✅:To determine if vectors ( a = {3, 1, -1} ), ( b = {-2, -1, 0} ), and ( c = {5, 2, -1} ) are coplanar, we calculate the scalar triple product ( a cdot (b times c) ). If this product is zero, the vectors are coplanar.Step 1: Compute the cross product ( b times c ):[b times c = begin{vmatrix}mathbf{i} & mathbf{j} & mathbf{k} -2 & -1 & 0 5 & 2 & -1end{vmatrix}= mathbf{i}left((-1)(-1) - (0)(2)right) - mathbf{j}left((-2)(-1) - (0)(5)right) + mathbf{k}left((-2)(2) - (-1)(5)right)][= mathbf{i}(1) - mathbf{j}(2) + mathbf{k}(-4 + 5) = {1, -2, 1}]Step 2: Compute the dot product ( a cdot (b times c) ):[a cdot (b times c) = 3(1) + 1(-2) + (-1)(1) = 3 - 2 - 1 = 0]Since the scalar triple product equals zero, the vectors are coplanar. Additional verification includes expressing ( c ) as ( c = a - b ) and confirming the determinant of the matrix formed by ( a, b, c ) is zero.Final Answer: The vectors ( a, b, ) and ( c ) are coplanar. boxed{text{Yes}}

🔑:1. Problem Restatement: Given vectors (a, b,) and (c), [ a = {3, 1, -1}, quad b = {-2, -1, 0}, quad c = {5, 2, -1} ] we need to determine whether these vectors are coplanar.2. Coplanarity Criterion: Vectors (a, b,) and (c) are coplanar if and only if their scalar triple product is zero. The scalar triple product ((a, b, c)) is given by the determinant of the matrix formed by placing the vectors (a, b,) and (c) as rows or columns.3. Set up the Determinant: [ (a, b, c) = left| begin{array}{ccc} 3 & 1 & -1 -2 & -1 & 0 5 & 2 & -1 end{array} right| ]4. Expand the Determinant: To compute the determinant, we will expand along the first row: [ left| begin{array}{ccc} 3 & 1 & -1 -2 & -1 & 0 5 & 2 & -1 end{array} right| = 3 left| begin{array}{cc} -1 & 0 2 & -1 end{array} right| - 1 left| begin{array}{cc} -2 & 0 5 & -1 end{array} right| + (-1) left| begin{array}{cc} -2 & -1 5 & 2 end{array} right| ]5. Evaluate the 2x2 Determinants: [ 3 left( (-1)(-1) - (0)(2) right) = 3(1) = 3 ] [ -1 left( (-2)(-1) - (0)(5) right) = -1(2) = -2 ] [ (-1) left( (-2)(2) - (-1)(5) right) = (-1) left( -4 + 5 right) = (-1)(1) = -1 ]6. Sum the Results: [ left| begin{array}{ccc} 3 & 1 & -1 -2 & -1 & 0 5 & 2 & -1 end{array} right| = 3 - 2 - 1 = 0 ]7. Conclusion: Since the determinant is zero, the scalar triple product ((a, b, c) = 0), hence the vectors (a, b,) and (c) are coplanar.[boxed{{text{The vectors } a, b, text{ and } c text{ are coplanar.}}}]

🧠:Okay, so I need to show that the quadrilateral IJKL formed by these equilateral triangles on the sides of ABCD is a parallelogram. Hmm, let's start by recalling what defines a parallelogram. A quadrilateral is a parallelogram if both pairs of opposite sides are parallel and equal in length, or equivalently, if the midpoints of the diagonals coincide. Maybe using vectors or complex numbers could help here, since dealing with rotations and equilateral triangles might be more straightforward with those methods.First, let's try to visualize the problem. We have a quadrilateral ABCD. Then, on sides AB and CD, we construct outward equilateral triangles, resulting in points I and K. On sides BC and DA, we construct inward equilateral triangles, resulting in points J and L. The goal is to show that connecting these four points I, J, K, L forms a parallelogram.Since the problem involves constructing equilateral triangles on the sides of ABCD, which is a general quadrilateral, transformations like rotation might be key. Remember that constructing an equilateral triangle on a side involves rotating the side by 60 degrees. The direction of rotation (clockwise or counterclockwise) determines whether the triangle is outward or inward.Let me denote the rotation operator. Let’s define a rotation of a point around another point by 60 degrees. If we rotate point B around point A by 60 degrees, we get a new point. Similarly for other points. But in this case, the equilateral triangles are constructed on the sides AB, BC, CD, DA. So, for example, triangle ABJ is inward, so maybe the rotation is clockwise, whereas triangle ABJ is outward, so maybe counterclockwise? Wait, the problem says "outward" and "inward" with respect to the quadrilateral ABCD. Hmm, the direction of rotation might depend on the orientation of ABCD.Alternatively, perhaps using complex numbers to represent the points. Let me assign complex numbers to the points A, B, C, D. Let’s denote them as complex numbers a, b, c, d. Then, the points I, J, K, L can be expressed in terms of a, b, c, d with appropriate rotations.For an outward equilateral triangle on AB, the point I can be obtained by rotating vector AB by 60 degrees counterclockwise around point A or B? Wait, need to be precise. If the triangle is outward, the direction of rotation depends on the orientation. Let me think.Suppose we have segment AB. To construct an outward equilateral triangle, we need to rotate AB by 60 degrees in the plane such that the triangle does not intersect the quadrilateral. Similarly, for inward, it would be the opposite rotation.Assuming the quadrilateral is convex, but the problem states it's non-intersecting, which could be concave or convex. Hmm, maybe the direction of outward/inward is defined with respect to the orientation. Wait, perhaps "outward" means the normal direction away from the interior of the quadrilateral, but if the quadrilateral is not convex, the "interior" is ambiguous. Maybe the problem assumes a convex quadrilateral? Wait, no, it's a non-intersecting quadrilateral, so it's a simple quadrilateral, which can be convex or concave.Alternatively, maybe the outward and inward are defined with respect to the traversal direction. If the quadrilateral is traversed in order ABCD, then outward equilateral triangles on AB and CD would be on the same side relative to the traversal, while inward on BC and DA would be on the opposite side.This is getting a bit abstract. Let me try to formalize the construction.For the outward equilateral triangle on AB: Let's denote that as triangle AB I. To construct this, we can take vector AB and rotate it 60 degrees outward. Similarly, for CD K.For the inward equilateral triangles on BC and DA: Similarly, rotate the vector BC and DA 60 degrees inward.But how exactly to define outward and inward? Maybe outward is the rotation such that the triangle is on the "outside" of the quadrilateral when traversed in order, and inward is the opposite.Alternatively, perhaps using the right-hand rule. If we traverse the quadrilateral ABCD, the outward direction is to the left, and inward to the right? Not sure. Maybe another approach.Let me consider using complex numbers. Let's model the plane as the complex plane. Let’s assign complex numbers a, b, c, d to points A, B, C, D. Then, a rotation by 60 degrees can be represented by multiplying by e^{iπ/3} = cos(π/3) + i sin(π/3) = 0.5 + i (√3)/2. Similarly, rotation by -60 degrees is multiplying by e^{-iπ/3} = 0.5 - i (√3)/2.Now, to construct an equilateral triangle on AB outward. The standard way is to rotate vector AB by 60 degrees to get the third vertex. But depending on the direction, the rotation could be positive (counterclockwise) or negative (clockwise). However, "outward" and "inward" need to be consistent with the quadrilateral's orientation.Wait, maybe the key is that for outward equilateral triangles on AB and CD, the rotation is in the same direction, say counterclockwise, while for inward on BC and DA, the rotation is clockwise. But how to verify this?Alternatively, perhaps for each side, the direction of the equilateral triangle (outward or inward) is defined such that for AB and CD, the triangles are on the same side relative to the quadrilateral, and for BC and DA, they are on the opposite side.Alternatively, since the problem says "with respect to ABCD", perhaps the outward direction is determined by the orientation of ABCD. If ABCD is oriented counterclockwise, then outward equilateral triangles on AB and CD would be constructed by rotating the sides outward from the interior. Similarly, inward would be rotating into the interior.But since the quadrilateral is non-intersecting, it has a defined interior by the Jordan curve theorem. So perhaps "outward" equilateral triangles are those constructed outside the quadrilateral's interior, and "inward" are those constructed inside.If that's the case, then constructing an outward equilateral triangle on AB would involve rotating AB by 60 degrees away from the interior, and inward would be rotating towards the interior.But how do we model that? Let's try.Assume the quadrilateral is oriented counterclockwise, so the interior is on the left when traversing ABCD. Then, for the outward equilateral triangle on AB, moving from A to B, the outward direction would be to the left of the edge AB. Wait, but if the quadrilateral is counterclockwise, then the interior is to the left of each edge. So, outward would be to the right of each edge? Wait, maybe.Alternatively, if the quadrilateral is traversed counterclockwise, the outward normal direction for each edge is to the right. Hmm, no. In standard orientation, for a counterclockwise polygon, the interior is to the left of each edge. So, if you move from A to B, the interior is on the left, so outward would be on the right. Therefore, constructing an outward equilateral triangle on AB would be on the right side of AB, i.e., rotating AB by -60 degrees (clockwise) around point A or B.Wait, maybe I need to think in terms of rotating the edge to get the third vertex.Let me try to define point I such that triangle AB I is outward equilateral. To construct such a triangle, starting from AB, we can rotate vector AB by 60 degrees in the outward direction. If outward is to the right, then rotating AB by -60 degrees (clockwise) around point B would give point I. Alternatively, rotating AB by 60 degrees counterclockwise around point A. Wait, need to be precise.Suppose we want to construct an equilateral triangle on AB outward. There are two possible equilateral triangles on AB: one on one side and one on the other. The outward one is determined with respect to the quadrilateral. If the quadrilateral is oriented counterclockwise, then the outward direction for AB would be the side opposite to the interior. So, if the interior is on the left when going from A to B, then the outward direction is on the right. Therefore, to get point I, we can rotate vector AB by -60 degrees (clockwise) around point B, or rotate vector BA by 60 degrees (counterclockwise) around point A. Hmm, perhaps.Alternatively, let's use complex numbers. Let’s assume the quadrilateral is oriented counterclockwise. Let’s assign complex numbers a, b, c, d to points A, B, C, D. Then, to construct an outward equilateral triangle on AB, we can rotate vector AB by -60 degrees (clockwise) to get the third vertex. So, vector AB is b - a. Rotating this by -60 degrees gives (b - a) multiplied by e^{-iπ/3}. Then, adding this to point B would give the third vertex. Wait, let's check.Wait, to construct the equilateral triangle outward, starting from A to B, then the third vertex I can be obtained by rotating vector AB by -60 degrees around point B. So, the formula would be: I = B + (A - B) * e^{-iπ/3}. Wait, no. Wait, rotating vector AB around point B by -60 degrees. The vector from B to I is equal to the vector BA rotated by -60 degrees. Wait, perhaps better to use rotation matrices.Alternatively, the rotation of a point around another. To rotate point A around point B by -60 degrees, the formula is: I = B + (A - B) * e^{-iπ/3}. Similarly, rotating point B around point A by 60 degrees could also give another point. But the problem states that AB I is an outward equilateral triangle. Depending on the rotation direction, we might get different points.This is getting a bit confusing. Maybe we need a more systematic approach.Let me recall that constructing an equilateral triangle on a segment can be done by rotating one endpoint around the other by 60 degrees. So, for segment AB, rotating point B around point A by 60 degrees counterclockwise gives a point, say I1, and rotating point A around point B by 60 degrees clockwise gives another point, say I2. Depending on the desired direction (outward or inward), we choose between I1 and I2.Assuming the quadrilateral is oriented counterclockwise, the outward equilateral triangles on AB and CD would be constructed by rotating in the clockwise direction, i.e., using -60 degrees rotations, so that the triangles are on the exterior side of the quadrilateral. Similarly, inward equilateral triangles on BC and DA would be constructed by rotating in the counterclockwise direction, i.e., 60 degrees, but since they are inward, maybe it's the opposite.Wait, perhaps it's better to assign a rotation direction for each case. For outward triangles on AB and CD: Let's say we rotate the segment AB around A by 60 degrees clockwise to get point I. Similarly, rotate segment CD around D by 60 degrees clockwise to get point K. For inward triangles on BC and DA: rotate segment BC around B by 60 degrees counterclockwise to get point J, and rotate segment DA around D by 60 degrees counterclockwise to get point L. Wait, but this is arbitrary. The problem doesn't specify the exact rotation direction, just outward and inward with respect to the quadrilateral.Alternatively, perhaps all outward triangles are constructed by rotating the segments in one direction (say, clockwise) and inward in the other (counterclockwise), but relative to the quadrilateral's orientation.This is a bit of a stumbling block. Maybe instead of worrying about the exact direction, I can use vectors and express the points I, J, K, L in terms of A, B, C, D with rotation factors, then compute the vectors IJ and LK, and see if they are equal, and similarly for JK and IL.Let me try that.Let’s denote rotation by 60 degrees as multiplication by ω = e^{iπ/3} = 0.5 + i√3/2, and rotation by -60 degrees as ω̄ = e^{-iπ/3} = 0.5 - i√3/2.Assuming complex plane representation.For the outward equilateral triangle on AB: Let's suppose that point I is obtained by rotating vector AB by 60 degrees outward. If outward is defined as clockwise, then the rotation is by -60 degrees. So, vector from A to I: (B - A) rotated by -60 degrees. So, I = A + (B - A) * ω̄.Similarly, for the outward equilateral triangle on CD: point K = C + (D - C) * ω̄.For the inward equilateral triangles on BC and DA: Since they are inward, the rotation would be by 60 degrees (counterclockwise). So, point J = B + (C - B) * ω. Similarly, point L = D + (A - D) * ω.Wait, let me check. If we have segment BC, constructing an inward equilateral triangle. If the quadrilateral is oriented counterclockwise, then inward would be towards the interior, which would be to the left of the segment BC. Rotating vector BC by 60 degrees counterclockwise around point B would give the inward direction. So, starting at B, moving towards C, the inward direction is to the left. So, rotating BC by 60 degrees counterclockwise around B would give point J. So, J = B + (C - B) * ω.Similarly, for segment DA, inward equilateral triangle: rotating DA by 60 degrees counterclockwise around D. So, L = D + (A - D) * ω.Similarly, outward equilateral triangles on AB and CD: rotating AB by -60 degrees (clockwise) around A gives I = A + (B - A) * ω̄. Rotating CD by -60 degrees (clockwise) around C gives K = C + (D - C) * ω̄.Wait, but the problem statement says "outward equilateral triangles with respect to ABCD". If ABCD is oriented counterclockwise, then the outward direction for AB is to the right (clockwise rotation), and for CD, it's also to the right. Similarly, inward for BC and DA would be to the left (counterclockwise rotation). That seems consistent.So, with that, we can express the points I, J, K, L in complex numbers as:I = A + (B - A) * ω̄J = B + (C - B) * ωK = C + (D - C) * ω̄L = D + (A - D) * ωNow, to show that IJKL is a parallelogram, we need to show that vector IJ is equal to vector LK, and vector JK is equal to vector IL. Alternatively, that the midpoints of the diagonals IK and JL coincide, but vectors might be easier.First, compute vector IJ: J - ISimilarly, vector LK: K - LIf J - I = K - L, then sides IJ and LK are equal and parallel. Similarly, compute vector JK: K - J and vector IL: L - I. If K - J = L - I, then sides JK and IL are equal and parallel.Let’s compute these vectors step by step.First, express all points in complex numbers:I = A + (B - A) * ω̄ = A(1 - ω̄) + B ω̄Similarly,J = B + (C - B) * ω = B(1 - ω) + C ωK = C + (D - C) * ω̄ = C(1 - ω̄) + D ω̄L = D + (A - D) * ω = D(1 - ω) + A ωNow compute IJ = J - I:J - I = [B(1 - ω) + C ω] - [A(1 - ω̄) + B ω̄]= B(1 - ω - ω̄) + C ω - A(1 - ω̄)Similarly, LK = K - L:K - L = [C(1 - ω̄) + D ω̄] - [D(1 - ω) + A ω]= C(1 - ω̄) - D(1 - ω) + D ω̄ - A ω= C(1 - ω̄) + D(-1 + ω + ω̄) - A ωBut ω + ω̄ = e^{iπ/3} + e^{-iπ/3} = 2 cos(π/3) = 1. So, ω + ω̄ = 1.Therefore, 1 - ω̄ = ω, since 1 - ω̄ = 1 - (1 - ω) if ω + ω̄ = 1. Wait, hold on. Let's compute ω + ω̄:ω = e^{iπ/3} = cos(π/3) + i sin(π/3) = 0.5 + i (√3)/2ω̄ = e^{-iπ/3} = cos(π/3) - i sin(π/3) = 0.5 - i (√3)/2So ω + ω̄ = 0.5 + 0.5 = 1. Therefore, yes, ω + ω̄ = 1.Therefore, 1 - ω̄ = ω, because 1 - ω̄ = 1 - (0.5 - i (√3)/2) = 0.5 + i (√3)/2 = ωSimilarly, 1 - ω = ω̄.So, substituting back into IJ:J - I = B(1 - ω - ω̄) + C ω - A(1 - ω̄)But 1 - ω - ω̄ = 1 - (ω + ω̄) = 1 - 1 = 0. So the term with B disappears.Therefore, J - I = C ω - A(1 - ω̄) = C ω - A ω (since 1 - ω̄ = ω)Wait, wait. 1 - ω̄ = ω, because ω + ω̄ = 1. So 1 - ω̄ = ω.Therefore,J - I = C ω - A ω = ω (C - A)Similarly, compute LK = K - L:K - L = C(1 - ω̄) + D ω̄ - D(1 - ω) - A ω= C ω + D(ω̄ -1 + ω) - A ωBut ω̄ -1 + ω = (ω + ω̄) -1 = 1 -1 = 0Therefore, K - L = C ω - A ω = ω (C - A)So, IJ = J - I = ω (C - A) and LK = K - L = ω (C - A). Therefore, IJ = LK.Similarly, compute JK = K - J:K - J = [C(1 - ω̄) + D ω̄] - [B(1 - ω) + C ω]= C(1 - ω̄ - ω) + D ω̄ - B(1 - ω)Again, 1 - ω̄ - ω = 1 - (ω + ω̄) = 1 -1 = 0Thus, JK = D ω̄ - B(1 - ω) = D ω̄ - B ω̄ (since 1 - ω = ω̄)Therefore, JK = ω̄ (D - B)Now compute IL = L - I:L - I = [D(1 - ω) + A ω] - [A(1 - ω̄) + B ω̄]= D(1 - ω) + A ω - A(1 - ω̄) - B ω̄= D(1 - ω) + A(ω -1 + ω̄) - B ω̄Again, ω -1 + ω̄ = (ω + ω̄) -1 = 1 -1 = 0Therefore, IL = D(1 - ω) - B ω̄But 1 - ω = ω̄, so:IL = D ω̄ - B ω̄ = ω̄ (D - B)Therefore, IL = ω̄ (D - B) and JK = ω̄ (D - B), so IL = JK.Therefore, we have IJ = LK and JK = IL, meaning that opposite sides are equal and parallel. Therefore, quadrilateral IJKL is a parallelogram.Wait, this seems to check out. Let me recap to ensure I didn't make any miscalculations.Starting with the definitions of I, J, K, L using complex numbers with rotations by ω and ω̄. Then, computing the vectors IJ, LK, JK, IL. Through substitution and using the fact that ω + ω̄ = 1, the terms simplified, leading to IJ = LK = ω (C - A) and JK = IL = ω̄ (D - B). Therefore, the opposite sides are equal and parallel, hence IJKL is a parallelogram.I think this works. Let me consider a specific example to verify.Take a simple quadrilateral, say a square ABCD with A at (0,0), B at (1,0), C at (1,1), D at (0,1). Compute points I, J, K, L.First, construct outward equilateral triangles on AB and CD.For AB: from A(0,0) to B(1,0). Outward equilateral triangle would be below the square (assuming square is oriented counterclockwise). Rotating AB by -60 degrees (clockwise) around A.Vector AB is (1,0). Rotated by -60 degrees: the rotation matrix is [cos(-60) -sin(-60); sin(-60) cos(-60)] = [0.5, √3/2; -√3/2, 0.5]. Applied to vector (1,0):x' = 0.5*1 + √3/2*0 = 0.5y' = -√3/2*1 + 0.5*0 = -√3/2So point I is A + (0.5, -√3/2) = (0 + 0.5, 0 - √3/2) = (0.5, -√3/2)Similarly, outward equilateral triangle on CD. CD is from C(1,1) to D(0,1). Rotating CD by -60 degrees around C. Vector CD is (-1,0). Rotated by -60 degrees:x' = 0.5*(-1) + √3/2*0 = -0.5y' = -√3/2*(-1) + 0.5*0 = √3/2So point K is C + (-0.5, √3/2) = (1 - 0.5, 1 + √3/2) = (0.5, 1 + √3/2)Now, inward equilateral triangles on BC and DA.For BC: from B(1,0) to C(1,1). Inward equilateral triangle, which would be towards the interior of the square. Rotating BC by 60 degrees counterclockwise around B.Vector BC is (0,1). Rotated 60 degrees counterclockwise:x' = cos(60)*0 - sin(60)*1 = -√3/2y' = sin(60)*0 + cos(60)*1 = 0.5So point J is B + (-√3/2, 0.5) = (1 - √3/2, 0 + 0.5) = (1 - √3/2, 0.5)For DA: from D(0,1) to A(0,0). Inward equilateral triangle, towards the interior. Rotating DA by 60 degrees counterclockwise around D.Vector DA is (0,-1). Rotated 60 degrees counterclockwise:x' = cos(60)*0 - sin(60)*(-1) = √3/2y' = sin(60)*0 + cos(60)*(-1) = -0.5So point L is D + (√3/2, -0.5) = (0 + √3/2, 1 - 0.5) = (√3/2, 0.5)Now, we have points:I: (0.5, -√3/2)J: (1 - √3/2, 0.5)K: (0.5, 1 + √3/2)L: (√3/2, 0.5)Let's compute the coordinates of IJKL.First, compute vector IJ: J - IJ is (1 - √3/2, 0.5), I is (0.5, -√3/2)So, IJ = (1 - √3/2 - 0.5, 0.5 - (-√3/2)) = (0.5 - √3/2, 0.5 + √3/2)Similarly, vector LK: K - LK is (0.5, 1 + √3/2), L is (√3/2, 0.5)LK = (0.5 - √3/2, 1 + √3/2 - 0.5) = (0.5 - √3/2, 0.5 + √3/2)So IJ = LK. That's good.Now, compute vector JK: K - JK is (0.5, 1 + √3/2), J is (1 - √3/2, 0.5)JK = (0.5 - (1 - √3/2), 1 + √3/2 - 0.5) = (-0.5 + √3/2, 0.5 + √3/2)Vector IL: L - IL is (√3/2, 0.5), I is (0.5, -√3/2)IL = (√3/2 - 0.5, 0.5 - (-√3/2)) = (√3/2 - 0.5, 0.5 + √3/2)Comparing JK and IL:JK = (-0.5 + √3/2, 0.5 + √3/2)IL = (√3/2 - 0.5, 0.5 + √3/2)Indeed, -0.5 + √3/2 = √3/2 - 0.5, so JK = IL. Therefore, opposite sides are equal and parallel, so IJKL is a parallelogram.This example confirms the general proof. Therefore, the reasoning holds.Final Answerboxed{IJKL text{ is a parallelogram}}

✅:To show that (IJKL) is a parallelogram, we start by considering the construction of equilateral triangles on the sides of quadrilateral (ABCD). We use complex numbers to represent the points and rotations to construct the points (I), (J), (K), and (L).1. Defining Points with Rotations: - For the outward equilateral triangles on (AB) and (CD): - (I) is obtained by rotating (B) around (A) by (-60^circ) (clockwise): (I = A + (B - A) cdot overline{omega}) - (K) is obtained by rotating (D) around (C) by (-60^circ) (clockwise): (K = C + (D - C) cdot overline{omega}) - For the inward equilateral triangles on (BC) and (DA): - (J) is obtained by rotating (C) around (B) by (60^circ) (counterclockwise): (J = B + (C - B) cdot omega) - (L) is obtained by rotating (A) around (D) by (60^circ) (counterclockwise): (L = D + (A - D) cdot omega) Here, (omega = e^{ipi/3} = frac{1}{2} + ifrac{sqrt{3}}{2}) and (overline{omega} = e^{-ipi/3} = frac{1}{2} - ifrac{sqrt{3}}{2}).2. Calculating Vectors: - Compute vectors (IJ), (LK), (JK), and (IL) using complex numbers: - (IJ = J - I = omega (C - A)) - (LK = K - L = omega (C - A)) - (JK = K - J = overline{omega} (D - B)) - (IL = L - I = overline{omega} (D - B))3. Verifying Parallelogram Properties: - Show that opposite sides are equal and parallel: - (IJ = LK) since both are (omega (C - A)) - (JK = IL) since both are (overline{omega} (D - B))4. Example Verification: - Consider a square (ABCD) and compute the coordinates of (I), (J), (K), and (L). Verify that vectors (IJ) and (LK) are equal, and vectors (JK) and (IL) are equal, confirming the parallelogram structure.Thus, by constructing the points using rotations and verifying the properties of the vectors, we conclude that (IJKL) is a parallelogram.[boxed{IJKL text{ is a parallelogram}}]

🔑:1. Consider the Equilateral Triangles: - Let I be the point such that triangle ABI is an equilateral triangle. - Let K be the point such that triangle CDK is an equilateral triangle. - Let J be the point such that triangle BCJ is an equilateral triangle pointing inward. - Let L be the point such that triangle DAL is an equilateral triangle pointing inward.2. Utilize Rotations: - A rotation of 60^circ (or frac{pi}{3} radians) centered at B moves point (A) to (I) and point (C) to (J). Formally, if mathcal{R}_B^{pi/3} denotes the rotation by pi/3 radians about point B: [ mathcal{R}_B^{pi/3}(A) = I quad text{and} quad mathcal{R}_B^{pi/3}(C) = J ] - Similarly, a rotation of 60^circ centered at D moves point A to L and point C to K. Formally: [ mathcal{R}_D^{pi/3}(A) = L quad text{and} quad mathcal{R}_D^{pi/3}(C) = K ]3. Vectors and Rotational Symmetry: - The vector overrightarrow{IJ} represents the translation from I to J. This vector can be obtained by rotating the vector overrightarrow{AC} by pi/3 radians about B. - Similarly, the vector overrightarrow{LK} can be obtained by rotating overrightarrow{AC} by pi/3 radians about D: [ overrightarrow{IJ} = mathcal{R}_B^{pi/3}(overrightarrow{AC}) quad text{and} quad overrightarrow{LK} = mathcal{R}_D^{pi/3}(overrightarrow{AC}) ] - Since both overrightarrow{IJ} and overrightarrow{LK} are obtained by the same rotational transformation (though centered at different points), they are parallel and of equal magnitude: [ overrightarrow{IJ} = overrightarrow{LK} ]4. Conclusion: - Since overrightarrow{IJ} = overrightarrow{LK}, the quadrilateral IJKL must have opposite sides that are parallel and equal in length. - By definition, a quadrilateral with opposite sides that are both parallel and equal in length is a parallelogram.Therefore, IJKL is indeed a parallelogram.[boxed{IJKL text{ is a parallelogram}}]

🧠:Okay, let me try to work through this problem step by step. So, the problem is about a right triangle ABC, where AB and AC are the legs, and AC is longer than AB. Points E on AC and D on BC are chosen such that AB equals AE equals BD. We need to prove that triangle ADE is a right triangle if and only if the sides of triangle ABC are in the ratio 3:4:5. Hmm, right triangles with specific ratios... classic geometry problem. Let me start by visualizing the triangle.First, let me draw triangle ABC with right angle at A. So, AB and AC are the legs, and BC is the hypotenuse. Since AC > AB, that means AC is the longer leg. Let's denote AB as length 'a', AC as length 'b', so BC would be sqrt(a² + b²) by Pythagoras. Now, points E and D are chosen such that AE = AB = a, and BD = AB = a. So, AE is on AC, so starting from A, moving along AC a distance of a to point E. Similarly, BD is along BC, starting from B, moving along BC a distance of a to point D. Then we need to show that triangle ADE is right-angled if and only if the original triangle has sides in ratio 3:4:5.Alright, so first, let's set up coordinate axes to model this. Let me place point A at the origin (0,0). Then, since AB and AC are legs, let me have point B at (0, a) and point C at (b, 0). Then, BC would be the hypotenuse from (0,a) to (b,0). Now, point E is on AC such that AE = AB = a. Since AC has length b, and we're moving from A (0,0) to C (b,0), but wait, AC is along the x-axis from (0,0) to (b,0)? Wait, hold on. Wait, if A is at (0,0), and AB is a leg, which is vertical, then B should be at (0, a), and AC is the other leg, horizontal, so C is at (c, 0). Wait, but in the problem statement, AB and AC are the legs. So, AC is a leg, but since AC > AB, then AC is longer. So, let's define AB as length 'a', AC as length 'b', with b > a. Then BC is hypotenuse with length sqrt(a² + b²).So, coordinates: A(0,0), B(0,a), C(b,0). Then, point E is on AC such that AE = AB = a. Since AC is from (0,0) to (b,0), which is along the x-axis. So, moving a distance of a along AC from A would be point E at (a, 0), but wait, the length from A to E is a, but AC is of length b. So, if AC is along the x-axis from (0,0) to (b,0), then the distance from A to E is a, so E would be at (a, 0). But wait, but AC has length b, so unless a <= b, which it is since AC > AB. So E is at (a, 0).Then point D is on BC such that BD = a. So, BD is the distance from B to D along BC. Since BC is from (0,a) to (b,0), the length of BC is sqrt(b² + a²). So, we need to find point D on BC such that the distance from B to D is a. Let's parameterize BC. Let me parametrize BC as a line segment from B(0,a) to C(b,0). The parametric equations for BC would be x = bt, y = a(1 - t), where t ranges from 0 to 1. So, when t=0, we are at B(0,a), and t=1 at C(b,0). The length from B to any point D on BC is t * sqrt(b² + a²). Wait, but BD is supposed to be a. So, we can set the length BD = a = t * sqrt(b² + a²). Therefore, t = a / sqrt(b² + a²). Therefore, the coordinates of D would be (b * t, a(1 - t)) = ( (b * a)/sqrt(b² + a²), a(1 - a / sqrt(b² + a²)) ).Wait, but maybe there's another way to find coordinates of D. Alternatively, since BD = a, and BC has length sqrt(a² + b²), then DC = sqrt(a² + b²) - a. So, the ratio BD/BC is a / sqrt(a² + b²). Therefore, using similar triangles or parameterizing, D divides BC in the ratio BD:DC = a : (sqrt(a² + b²) - a). But maybe working with coordinates is better.Alternatively, perhaps using vectors. The vector from B to C is (b, -a). So, moving from B towards C by a distance a. The unit vector in the direction BC is (b, -a) / sqrt(b² + a²). Therefore, point D is B + a * unit vector. So, coordinates of D would be (0, a) + ( (b, -a) * a / sqrt(b² + a²) ) = ( (ab)/sqrt(b² + a²), a - (a²)/sqrt(b² + a²) )So, D has coordinates (ab / sqrt(b² + a²), a - a² / sqrt(b² + a²) )Similarly, point E is (a, 0). So now, we need to find the coordinates of points A(0,0), D, and E(a,0), and check when triangle ADE is a right triangle.Wait, triangle ADE. So, points A(0,0), D(ab / sqrt(b² + a²), a - a² / sqrt(b² + a²) ), and E(a, 0). To check if ADE is right-angled, we need to see if any of the angles at A, D, or E is 90 degrees.But since A is at (0,0), E is at (a,0), so AE is along the x-axis. Point D is somewhere in the plane. So, angles at A: between AD and AE. But AE is along x-axis, and AD is from A(0,0) to D(ab / sqrt(b² + a²), a - a² / sqrt(b² + a²) ). The angle at A would be between AD and AE. For it to be right-angled at A, the vectors AD and AE should be perpendicular, but AE is (a,0) and AD is (ab / sqrt(b² + a²), a - a² / sqrt(b² + a²) ). The dot product would be a * ab / sqrt(b² + a²) + 0 * (a - a² / sqrt(b² + a²)) = a² b / sqrt(b² + a²). Which is not zero unless a=0 or b=0, which they aren't. So angle at A isn't right.Angle at E: So, vectors ED and EA. Point E is (a,0). Vector ED is D - E = (ab / sqrt(b² + a²) - a, a - a² / sqrt(b² + a²) - 0 ) = ( a ( b / sqrt(b² + a²) - 1 ), a ( 1 - a / sqrt(b² + a²) ) )Vector EA is A - E = (-a, 0). The dot product of ED and EA should be zero if angle at E is right. Let's compute:ED ⋅ EA = [ a ( b / sqrt(b² + a²) - 1 ) ] * (-a) + [ a ( 1 - a / sqrt(b² + a²) ) ] * 0 = -a² ( b / sqrt(b² + a²) - 1 )Set this equal to zero:- a² ( b / sqrt(b² + a²) - 1 ) = 0Since a ≠ 0, then:b / sqrt(b² + a²) - 1 = 0 → b / sqrt(b² + a²) = 1 → sqrt(b² + a²) = b → sqrt(b² + a²) = b → b² + a² = b² → a² = 0 → a=0, which is impossible. So angle at E cannot be right.Angle at D: So, vectors DA and DE. DA is A - D = ( -ab / sqrt(b² + a²), -a + a² / sqrt(b² + a²) )DE is E - D = (a - ab / sqrt(b² + a²), -a + a² / sqrt(b² + a²) )Dot product DA ⋅ DE:[ -ab / sqrt(b² + a²) ] * [ a - ab / sqrt(b² + a²) ] + [ -a + a² / sqrt(b² + a²) ] * [ -a + a² / sqrt(b² + a²) ]Let me compute each term:First term:- ab / sqrt(b² + a²) * ( a - ab / sqrt(b² + a²) ) = -a² b / sqrt(b² + a²) + a² b² / (b² + a²)Second term:[ -a + a² / sqrt(b² + a²) ] * [ -a + a² / sqrt(b² + a²) ] = [ -a + a² / sqrt(b² + a²) ]²Let me expand that:= a² - 2a * (a² / sqrt(b² + a²)) ) + (a² / sqrt(b² + a²))²= a² - 2a³ / sqrt(b² + a²) + a⁴ / (b² + a²)So the total dot product is:(-a² b / sqrt(b² + a²) + a² b² / (b² + a²)) + (a² - 2a³ / sqrt(b² + a²) + a⁴ / (b² + a²))Let me combine terms:First, terms with 1/sqrt(b² + a²):- a² b / sqrt(b² + a²) - 2a³ / sqrt(b² + a²) = - a² (b + 2a) / sqrt(b² + a²)Terms with 1/(b² + a²):a² b² / (b² + a²) + a⁴ / (b² + a²) = a² (b² + a²) / (b² + a²) ) = a²Then the remaining term is + a²So total:- a² (b + 2a) / sqrt(b² + a²) + a² + a² = - a² (b + 2a) / sqrt(b² + a²) + 2a²So for angle at D to be right, this dot product must be zero:- a² (b + 2a) / sqrt(b² + a²) + 2a² = 0Divide both sides by a² (since a ≠ 0):- (b + 2a) / sqrt(b² + a²) + 2 = 0Move the term with sqrt to the other side:2 = (b + 2a)/sqrt(b² + a²)Multiply both sides by sqrt(b² + a²):2 sqrt(b² + a²) = b + 2aSquare both sides to eliminate sqrt:4 (b² + a²) = (b + 2a)²Expand the right side:4b² + 4a² = b² + 4a b + 4a²Subtract 4a² from both sides:4b² = b² + 4abSubtract b²:3b² = 4abDivide both sides by b (since b ≠ 0):3b = 4aTherefore, 3b = 4a → b = (4/3)aSo, the ratio of the legs is AC = b = (4/3)a, AB = a. Therefore, the sides are AB = a, AC = (4/3)a, BC = sqrt(a² + b²) = sqrt(a² + (16/9)a²) = sqrt(25/9 a²) = (5/3)a. Therefore, the ratios of AB:AC:BC are a : (4/3)a : (5/3)a = 3 : 4 : 5 when scaled by 3/a. So the sides are in the ratio 3:4:5. Therefore, triangle ADE is right-angled if and only if the original triangle ABC has sides in ratio 3:4:5. Hence proved.Wait, but the problem says "if and only if". So, we showed that if triangle ADE is right-angled, then the sides must be in ratio 3:4:5. But do we need to check the converse? That if the sides are in ratio 3:4:5, then triangle ADE is right-angled. Let's verify that.Suppose ABC has sides 3,4,5. So AB = 3, AC =4, BC=5. Then, AE = AB =3, so E is 3 units from A along AC. Since AC is 4, E is at (3,0) if we set coordinates as A(0,0), B(0,3), C(4,0). Then BD = AB =3. BD is along BC from B. The length BC is 5, so BD is 3, so DC = 5 -3 =2. So D divides BC in the ratio BD:DC =3:2. Coordinates of D can be found by section formula. From B(0,3) to C(4,0), coordinates of D are ( (3*4 + 2*0)/(3+2), (3*0 + 2*3)/(3+2) ) = (12/5, 6/5). So D is (12/5, 6/5). E is (3,0). Now, check if triangle ADE is right-angled.Points: A(0,0), D(12/5,6/5), E(3,0). Let's compute vectors:AD: (12/5,6/5)AE: (3,0)DE: (3 -12/5, 0 -6/5) = (3/5, -6/5)Compute the angles:First, check if any two vectors are perpendicular.AD ⋅ AE = (12/5)(3) + (6/5)(0) = 36/5 ≠0AD ⋅ DE = (12/5)(3/5) + (6/5)(-6/5) = (36/25) - (36/25) =0So the dot product of AD and DE is zero. So vectors AD and DE are perpendicular. Therefore, angle at D is right angle. Hence, triangle ADE is right-angled at D. Therefore, the converse holds. Hence, triangle ADE is right-angled if and only if ABC is a 3:4:5 triangle.Therefore, the proof is complete.Final Answerboxed{3:4:5}

✅:Given a right triangle (ABC) with (AB) and (AC) as the legs, where (AC > AB). Points (E) on (AC) and (D) on (BC) are chosen such that (AB = AE = BD). We need to prove that triangle (ADE) is a right triangle if and only if the sides of triangle (ABC) are in the ratio (3:4:5).1. Coordinate Setup: - Place (A) at the origin ((0,0)), (B) at ((0, a)), and (C) at ((b, 0)) where (b > a). - Point (E) is on (AC) such that (AE = AB = a), so (E) is at ((a, 0)). - Point (D) is on (BC) such that (BD = AB = a). Using parameterization, the coordinates of (D) are found to be (left(frac{ab}{sqrt{a^2 + b^2}}, a - frac{a^2}{sqrt{a^2 + b^2}}right)).2. Checking Right Angle in (ADE): - Coordinates of (A(0,0)), (Dleft(frac{ab}{sqrt{a^2 + b^2}}, a - frac{a^2}{sqrt{a^2 + b^2}}right)), and (E(a, 0)). - Vectors (DA) and (DE) are computed. The dot product of (DA) and (DE) is calculated to check if it is zero (indicating a right angle at (D)).3. Dot Product Calculation: - The dot product of vectors (DA) and (DE) is simplified to: [ - frac{a^2 (b + 2a)}{sqrt{a^2 + b^2}} + 2a^2 = 0 ] - Solving this equation leads to (3b = 4a), implying (b = frac{4}{3}a).4. Verification of Ratios: - Substituting (b = frac{4}{3}a) into the side lengths of (ABC) gives (AB = a), (AC = frac{4}{3}a), and (BC = frac{5}{3}a). These ratios simplify to (3:4:5).5. Converse Verification: - For a triangle (ABC) with sides (3:4:5), coordinates of (D) and (E) are calculated. Checking the dot product of vectors (AD) and (DE) confirms it is zero, verifying the right angle at (D).Thus, triangle (ADE) is a right triangle if and only if the sides of triangle (ABC) are in the ratio (3:4:5).[boxed{3:4:5}]

🔑:1. Analyzing the given triangle: Given a right triangle ( triangle ABC ) with legs ( AB ) and ( AC ), where ( AC > AB ). Point ( E ) is chosen on ( AC ) and point ( D ) is chosen on ( BC ) such that ( AB = AE = BD ).2. Considering a 3:4:5 triangle: Suppose ( triangle ABC ) is a right triangle where the sides are in the ratio ( 3:4:5 ). Let ( AB = 3 ), ( AC = 4 ), and the hypotenuse ( BC = 5 ).3. Lengths verification: We are given that ( AB = AE = BD ). Therefore, these lengths must be equal to 3, since ( AB = 3 ).4. Determining point ( E ): Since ( E ) is on ( AC ) and ( AE = AB = 3 ), and ( AC = 4 ): [ CE = AC - AE = 4 - 3 = 1 ]5. Determining point ( D ): Since ( BD = 3 ) and ( BC = 5 ): [ DC = BC - BD = 5 - 3 = 2 ]6. Coordinates Representation: Place ( A ) at the origin ((0, 0)), ( B ) at ((3, 0)), and ( C ) at ((0, 4)).7. Checking Triangle ( ADE ): - Find the coordinates of ( E ) on ( AC ): ( E ) is ((0, 3)) because it lies on the line ( AC ) and ( AE = 3 ). - Find the coordinates of ( D ) on ( BC ): The line ( BC ) is given by the equation ( y = -frac{4}{3}x + 4 ). Since ( B = D - 3 ): [ D text{ needs to satisfy } x^2 + y^2 = 9 text{ (as } BD = 3 text{)}. ]8. Proving right triangle ( ADE ): - Calculate the slopes of ( AD ) and ( DE ). - ( A ) is at ((0,0)), ( D ) is at ((1, 2)) and ( E ) is at ((0, 3)). [ text{slope of } AD = frac{2-0}{1-0} = 2 ] [ text{slope of } DE = frac{3-2}{0-1} = -1 ] - Check their product: [ 2 times (-1) = -2 neq -1 Rightarrow AD perp DE ] Thus, ( triangle ADE ) is indeed a right triangle.Conclusion:When the sides ( AB ), ( AC ), and ( BC ) of ( triangle ABC ) are in the ratio ( 3:4:5 ), we conclude that the triangle ( ADE ) is right-angled. [boxed{3:4:5}]

🧠:Okay, so I need to solve these two equations simultaneously. Let me write them down again to make sure I have them right:First equation: 3x² + xy - 2y² = -5Second equation: x² + 2xy + y² = 1Hmm, both equations are quadratic. That might make things a bit tricky, but let's see. Since they are both quadratic, maybe substitution is a way to go here. Or perhaps elimination? Let me think.Looking at the second equation, x² + 2xy + y² = 1. Wait a minute, that looks familiar. Isn't that a perfect square? Let me check: (x + y)² = x² + 2xy + y². Yes! So the second equation simplifies to (x + y)² = 1. That's helpful. So from the second equation, we can take the square root of both sides, which would give us x + y = ±1. That gives us two possibilities: x + y = 1 or x + y = -1. Okay, so maybe we can express x in terms of y (or vice versa) from these equations and substitute into the first equation. Let's try that. Let's take the two cases separately.Case 1: x + y = 1In this case, we can solve for x: x = 1 - y. Then substitute this expression for x into the first equation.So substituting x = 1 - y into 3x² + xy - 2y² = -5.First, compute x²: (1 - y)² = 1 - 2y + y²Then, compute xy: (1 - y)y = y - y²Now substitute these into the first equation:3(1 - 2y + y²) + (y - y²) - 2y² = -5Let me expand that step by step.First term: 3*(1 - 2y + y²) = 3*1 - 3*2y + 3*y² = 3 - 6y + 3y²Second term: (y - y²) remains as is.Third term: -2y²So combining all terms:3 - 6y + 3y² + y - y² - 2y² = -5Now combine like terms:The y² terms: 3y² - y² - 2y² = 0y²The y terms: -6y + y = -5yThe constant terms: 3So altogether: 0y² -5y +3 = -5Simplify:-5y + 3 = -5Subtract 3 from both sides:-5y = -8Divide by -5:y = (-8)/(-5) = 8/5 = 1.6So y = 8/5. Now, since x + y = 1, then x = 1 - y = 1 - 8/5 = (5/5 - 8/5) = -3/5 = -0.6So in Case 1, we have x = -3/5 and y = 8/5. Let me check if these values satisfy both equations.First, check the second equation (since we used it to get x + y =1):x + y = -3/5 + 8/5 = 5/5 =1. Good. Then, squaring that gives 1, which matches the second equation.Now check the first equation:3x² + xy -2y²Compute each term:x² = (-3/5)² = 9/25, so 3x² = 27/25xy = (-3/5)(8/5) = -24/25y² = (8/5)² = 64/25, so -2y² = -128/25Adding them all together: 27/25 -24/25 -128/25 = (27 -24 -128)/25 = (-125)/25 = -5. Which matches the first equation. Perfect, so Case 1 works. So (-3/5, 8/5) is a solution.Now onto Case 2: x + y = -1Similarly, solve for x: x = -1 - ySubstitute into the first equation: 3x² + xy -2y² = -5Compute x²: (-1 - y)² = (1 + 2y + y²)Compute xy: (-1 - y)y = -y - y²Substitute into the first equation:3*(1 + 2y + y²) + (-y - y²) -2y² = -5Expand:3*1 + 3*2y + 3*y² - y - y² -2y²Which is 3 + 6y + 3y² - y - y² -2y²Combine like terms:y² terms: 3y² - y² -2y² = 0y²y terms: 6y - y = 5yConstants: 3So altogether: 0y² +5y +3 = -5Simplify:5y + 3 = -5Subtract 3:5y = -8Divide by 5:y = -8/5 = -1.6Then x = -1 - y = -1 - (-8/5) = -1 +8/5 = (-5/5 +8/5)= 3/5 =0.6So x=3/5 and y= -8/5. Let me verify these in both equations.First, the second equation: x + y =3/5 + (-8/5)= -5/5 = -1. Squared is 1. So that checks out.Now the first equation:3x² +xy -2y²Compute each term:x² = (3/5)^2 =9/25, so 3x²=27/25xy = (3/5)(-8/5)= -24/25y²=(-8/5)^2=64/25, so -2y²= -128/25Adding them:27/25 -24/25 -128/25 = (27 -24 -128)/25= (-125)/25= -5. Perfect. So Case 2 also works. So (3/5, -8/5) is another solution.So far, we have two solutions from the cases where x + y =1 and x + y=-1. But since the original equations are quadratic, maybe there are more solutions? Wait, but the second equation is (x + y)^2 =1, which only gives two possibilities. So only two cases. So we might have two solutions. Let me check if there are any other solutions.But let me think again. The second equation is (x + y)^2=1, so x + y must be 1 or -1. So only two possibilities. Therefore, all solutions must come from these two cases. Therefore, the only solutions are (-3/5, 8/5) and (3/5, -8/5). But wait, let me check if these are the only solutions. Let me think if there's another way to approach this.Alternatively, maybe we can use substitution in another way, but since we already found two solutions that satisfy both equations, and since the system is quadratic, the maximum number of solutions is 4, but since we have only two solutions, perhaps that's all. Wait, but why? Let me think.Alternatively, let's consider solving the system by another method to confirm. Maybe using substitution or elimination.Let me try to use elimination. Let's denote the equations:Equation 1: 3x² + xy -2y² = -5Equation 2: x² +2xy + y² =1We can try to manipulate these equations to eliminate one variable. For instance, if we multiply Equation 2 by 3 to make the x² terms match:3*(Equation 2): 3x² +6xy +3y² =3Then subtract Equation 1 from this:(3x² +6xy +3y²) - (3x² +xy -2y²) =3 - (-5)Compute left side:3x² -3x² +6xy -xy +3y² - (-2y²) =0 +5xy +5y²Right side: 8Thus, 5xy +5y²=8Simplify by dividing both sides by 5:xy + y² =8/5Factor y:y(x + y)=8/5But from Equation 2, (x + y)^2=1, so x + y=±1. Therefore, in either case, x + y is either 1 or -1. Therefore, substituting into y(x + y)=8/5:Case1: x + y=1, then y*1=8/5 => y=8/5. Then x=1 -8/5=-3/5. Which matches our previous solution.Case2: x + y=-1, then y*(-1)=8/5 => y= -8/5. Then x= -1 -y= -1 -(-8/5)=3/5. Which also matches our previous solution. Therefore, there are only two solutions.Therefore, the solutions are (-3/5,8/5) and (3/5, -8/5). But wait, let me just check if there are any other possibilities. Suppose if x + y is not 1 or -1, but since Equation 2 is (x + y)^2=1, then x + y must be 1 or -1, so no other possibilities. Therefore, these are the only solutions.Alternatively, let me try another approach. Let's solve for one variable in terms of the other.From Equation 2: x² +2xy + y²=1. Let's write this as (x + y)^2=1, so x + y=±1 as before.Alternatively, if we didn't notice the perfect square, we could have solved for one variable. For example, solve Equation 2 for x in terms of y or vice versa. Let's try that.Equation 2: x² +2xy + y²=1. Let's treat this as a quadratic in x:x² +2xy + y² -1=0.This is quadratic in x: x² +2y x + (y² -1)=0We can solve for x using quadratic formula:x = [-2y ± sqrt{(2y)^2 -4*1*(y² -1)}]/2Compute discriminant:4y² -4*(y² -1)=4y² -4y² +4=4Therefore, sqrt{4}=2Thus, x = [-2y ±2]/2= [-2y +2]/2 or [-2y -2]/2= (-y +1) or (-y -1)Thus, x= -y +1 or x= -y -1Which is the same as x + y=1 or x + y=-1, which is what we had before. So that confirms that there are only two cases. Then substitution gives the same solutions.Therefore, the only solutions are x= -3/5, y=8/5 and x=3/5, y= -8/5.Wait, but just to make sure, let me check if there's any possibility of extraneous solutions or if we missed something. For example, when we squared or took square roots, but in this case, we didn't square both sides; instead, we took square roots, so the solutions are valid.Therefore, I think these are the only solutions. Let me just recap:1. From the second equation, (x + y)^2 =1, so x + y=1 or x + y=-1.2. For each case, solved for x in terms of y (or vice versa) and substituted into the first equation.3. Solved the resulting linear equation for y, found the corresponding x.4. Verified both solutions satisfy the original equations.Therefore, the solutions are (-3/5,8/5) and (3/5,-8/5). Let me write them as fractions to be precise:First solution: x = -3/5, y =8/5Second solution: x=3/5, y= -8/5Yes, that's correct. So these are the solutions.Final AnswerThe solutions to the simultaneous equations are boxed{left( -dfrac{3}{5}, dfrac{8}{5} right)} and boxed{left( dfrac{3}{5}, -dfrac{8}{5} right)}.

✅:To solve the simultaneous equations:[begin{aligned}3 x^{2}+x y-2 y^{2} & =-5 x^{2}+2 x y+y^{2} & =1end{aligned}]First, recognize that the second equation is a perfect square:[(x + y)^2 = 1]This gives two cases:1. ( x + y = 1 )2. ( x + y = -1 )Case 1: ( x + y = 1 )Express ( x ) in terms of ( y ):[x = 1 - y]Substitute ( x = 1 - y ) into the first equation:[3(1 - y)^2 + (1 - y)y - 2y^2 = -5]Expand and simplify:[3(1 - 2y + y^2) + y - y^2 - 2y^2 = -5][3 - 6y + 3y^2 + y - y^2 - 2y^2 = -5][-5y + 3 = -5][-5y = -8][y = frac{8}{5}]Then,[x = 1 - frac{8}{5} = -frac{3}{5}]Case 2: ( x + y = -1 )Express ( x ) in terms of ( y ):[x = -1 - y]Substitute ( x = -1 - y ) into the first equation:[3(-1 - y)^2 + (-1 - y)y - 2y^2 = -5]Expand and simplify:[3(1 + 2y + y^2) - y - y^2 - 2y^2 = -5][3 + 6y + 3y^2 - y - y^2 - 2y^2 = -5][5y + 3 = -5][5y = -8][y = -frac{8}{5}]Then,[x = -1 - left(-frac{8}{5}right) = frac{3}{5}]Verify both solutions satisfy the original equations. The solutions are:[left( -frac{3}{5}, frac{8}{5} right) quad text{and} quad left( frac{3}{5}, -frac{8}{5} right)] Final AnswerThe solutions to the simultaneous equations are boxed{left( -dfrac{3}{5}, dfrac{8}{5} right)} and boxed{left( dfrac{3}{5}, -dfrac{8}{5} right)}.

🔑:Given the simultaneous equations:[begin{aligned}3 x^2 + x y - 2 y^2 & = -5 quad quad (1) x^2 + 2 x y + y^2 & = 1 quad quad (2)end{aligned}]1. Multiply equation ((2)) by 5: [ 5(x^2 + 2 x y + y^2) = 5 ] This simplifies to: [ 5x^2 + 10 xy + 5y^2 = 5 quad quad (3) ]2. Add equation ((1)) to equation ((3)): [ (3x^2 + xy - 2y^2) + (5x^2 + 10xy + 5y^2) = -5 + 5 ] Simplifying the left-hand side: [ 8x^2 + 11xy + 3y^2 = 0 ]3. Factorize the quadratic expression: [ 8x^2 + 11xy + 3y^2 = 0 ] We can rewrite it as: [ (8x + 3y)(x + y) = 0 ] Thus, either: [ 8x + 3y = 0 quad quad(4) quad text{or} quad x + y = 0 quadquad(5) ] for (8x + 3y = 0):4. Express ( y ) in terms of ( x ) using equation ((4)): [ 8x + 3y = 0 implies y = -frac{8x}{3} ]5. Substitute ( y ) in equation ((2)): [ x^2 + 2xleft(-frac{8x}{3}right) + left(-frac{8x}{3}right)^2 = 1 ] Simplify the expression: [ x^2 - frac{16x^2}{3} + frac{64x^2}{9} = 1 ] Combine like terms: [ x^2left(1 - frac{16}{3} + frac{64}{9}right) = 1 ] Find a common denominator: [ frac{9 - 48 + 64}{9}x^2 = 1 ] Simplify the fraction: [ frac{25x^2}{9} = 1 ] Solve for ( x ): [ 25x^2 = 9 implies x^2 = frac{9}{25} implies x = pmfrac{3}{5} ] If ( x = frac{3}{5} ), then: [ y = -frac{8}{5} ] If ( x = -frac{3}{5} ), then: [ y = frac{8}{5} ] for ( x + y = 0 ):6. Express ( y ) in terms of ( x ) using equation ( (5) ): [ x + y = 0 implies y = -x ]7. Substitute ( y ) in equation ((2)): [ x^2 + 2x(-x) + (-x)^2 = 1 ] Simplify the expression: [ x^2 - 2x^2 + x^2 = 1 ] Combine like terms: [ 0 = 1 ] This gives a contradiction, so there are no solutions for ( x + y = 0 ). Conclusion:The solutions to the given simultaneous equations are:[x = frac{3}{5}, y = -frac{8}{5}; quad x = -frac{3}{5}, y = frac{8}{5}](boxed{x = frac{3}{5}, y = -frac{8}{5}; quad x = -frac{3}{5}, y = frac{8}{5}})