🧠:[<think>Okay, so I have this problem where Ilya takes a triplet of numbers {70, 61, 20} and transforms it by replacing each number with the sum of the other two. I need to find the difference between the largest and smallest numbers after 1989 applications of this rule. Hmm, let's see how to approach this.First, let me understand the transformation step. If the triplet is (a, b, c), then after one step, it becomes (b + c, a + c, a + b). Right? Each number is replaced by the sum of the other two. Let me test this with the initial numbers to see what happens.Starting with {70, 61, 20}:First step:- New a = 61 + 20 = 81- New b = 70 + 20 = 90- New c = 70 + 61 = 131So the new triplet is {81, 90, 131}Difference between largest and smallest: 131 - 81 = 50.Second step:- New a = 90 + 131 = 221- New b = 81 + 131 = 212- New c = 81 + 90 = 171Triplet: {221, 212, 171}Difference: 221 - 171 = 50.Third step:- New a = 212 + 171 = 383- New b = 221 + 171 = 392- New c = 221 + 212 = 433Triplet: {383, 392, 433}Difference: 433 - 383 = 50.Wait a minute, the difference after each step is 50? That's interesting. Let me check again.Wait, first step: difference was 131 - 81 = 50.Second step: 221 - 171 = 50.Third step: 433 - 383 = 50.Huh, so the difference between the largest and smallest numbers is staying at 50 each time? Or maybe not. Let me do one more step to confirm.Fourth step:New a = 392 + 433 = 825New b = 383 + 433 = 816New c = 383 + 392 = 775Triplet: {825, 816, 775}Difference: 825 - 775 = 50. Yeah, same difference. So, is the difference always 50? Then regardless of how many steps, even 1989 steps, the difference remains 50? That seems to be the pattern here.But wait, let me check with another example to see if this is a general rule. Suppose I have a different starting triplet. Let's say {1, 2, 3}.First step: {5, 4, 3}. Difference is 5 - 3 = 2.Second step: {7, 8, 9}. Difference is 9 - 7 = 2.Third step: {17, 16, 15}. Difference is 17 - 15 = 2.Fourth step: {31, 32, 33}. Difference is 33 - 31 = 2.Wait, so even here, the difference remains the same? Starting with {1, 2, 3}, difference is 2, and after each step, the difference remains 2. So maybe this transformation preserves the difference between the largest and smallest numbers? But wait, in the first example, the difference was 50, which was maintained. In the second example, difference was 2, which was also maintained. So maybe the difference is invariant under this transformation? If that's the case, then regardless of the number of applications, the difference remains the same as the original.But wait, let's check another example where the initial difference isn't between the first and last numbers. Let's take {1, 3, 2}. The initial difference is 3 - 1 = 2.First step: 3 + 2 = 5, 1 + 2 = 3, 1 + 3 = 4. So {5, 3, 4}. The difference here is 5 - 3 = 2.Second step: 3 + 4 = 7, 5 + 4 = 9, 5 + 3 = 8. So {7, 9, 8}. Difference is 9 - 7 = 2.Third step: 9 + 8 = 17, 7 + 8 = 15, 7 + 9 = 16. {17, 15, 16}. Difference is 17 - 15 = 2.Yep, so even if the initial triplet is not ordered, after the first step, the difference seems to stabilize as the difference between the largest and smallest in the first step. Wait, in this case, the initial difference was 2, but the first step also had a difference of 2. Wait, but if the initial triplet had a different difference, would that difference stay?Wait, let me test with another triplet. Let's take {1, 1, 3}. Initial difference is 3 - 1 = 2.First step: 1 + 3 = 4, 1 + 3 = 4, 1 + 1 = 2. Triplet {4, 4, 2}. Difference is 4 - 2 = 2.Second step: 4 + 2 = 6, 4 + 2 = 6, 4 + 4 = 8. Triplet {6, 6, 8}. Difference is 8 - 6 = 2.Third step: 6 + 8 = 14, 6 + 8 = 14, 6 + 6 = 12. Triplet {14, 14, 12}. Difference 14 - 12 = 2.So again, difference is preserved. Wait, so maybe the difference between the maximum and minimum is an invariant here? That is, it remains constant no matter how many times we apply the transformation. If that's the case, then in the original problem, since after the first step, the difference is 50, and even in the original triplet, {70, 61, 20}, the difference is 70 - 20 = 50. Wait, original difference was already 50. So maybe that's why in all subsequent steps, the difference is 50. So perhaps the initial difference is preserved.Wait, let me check the original triplet. Original triplet: {70, 61, 20}. The maximum is 70, the minimum is 20. So difference is 50. After first transformation, triplet is {81, 90, 131}. Max 131, min 81, difference 50. Then next step, 221, 212, 171. Max 221, min 171, difference 50. So even though the numbers themselves are changing, the difference between the largest and smallest is staying the same.Therefore, if this is an invariant, then regardless of how many transformations we apply, the difference will remain 50. So even after 1989 steps, the difference is still 50. Therefore, the answer is 50. So I should check if this is the case.But wait, let me think again. Maybe this is a cycle? Wait, the problem says "difference between the largest and smallest". So perhaps the difference is preserved each time. Let me think of the transformation algebraically.Suppose we have triplet (a, b, c). Let's assume a ≥ b ≥ c. Then the difference is a - c.After one transformation, the triplet becomes (b + c, a + c, a + b). Let's order them again. Let's see:If a ≥ b ≥ c, then:- a + b is the largest, since a is the largest, b is next, so a + b is bigger than a + c (since b ≥ c) and bigger than b + c.Then, a + c is next, since a + c ≥ b + c (because a ≥ b).So ordered triplet would be (a + b, a + c, b + c). Wait, but original ordering was a ≥ b ≥ c, so the new numbers would be:First number: b + cSecond number: a + cThird number: a + bBut when we order them, the largest is a + b, then a + c, then b + c.Therefore, the new maximum is a + b, and the new minimum is b + c.Difference after transformation: (a + b) - (b + c) = a - c. Which is the same as the original difference. So the difference is preserved.Therefore, regardless of the number of transformations, the difference between the maximum and minimum remains equal to the original difference, which in this case is 70 - 20 = 50.Therefore, even after 1989 steps, the difference is still 50. So the answer is 50.But wait, let me check in the first step of the initial problem. Original triplet {70, 61, 20}. The maximum is 70, minimum 20, difference 50. After first transformation, {81, 90, 131}, ordered as {81, 90, 131}, difference 50. Then next step, {171, 212, 221}, difference 50. So each time, the difference remains 50. Therefore, the difference is indeed preserved. Therefore, regardless of the number of transformations, it remains 50. Therefore, the answer is 50.So perhaps the key insight here is recognizing that the difference between the maximum and minimum is invariant under the given transformation. Thus, since the initial difference is 50, it remains 50 forever. Therefore, the answer is 50.Final Answerboxed{50}</think>Given the initial triplet ({70, 61, 20}), we need to determine the difference between the largest and smallest numbers after 1989 applications of the transformation rule where each number is replaced by the sum of the other two.First, we observe the transformation step:- For a triplet ((a, b, c)), the next triplet becomes ((b + c, a + c, a + b)).We analyze the effect of this transformation on the difference between the largest and smallest numbers. Let's denote the original triplet as (a, b, c) with (a geq b geq c). The difference is (a - c).After one transformation, the new triplet is ((b + c, a + c, a + b)). When ordered, the new triplet becomes:- Largest: (a + b)- Middle: (a + c)- Smallest: (b + c)The new difference is ((a + b) - (b + c) = a - c), which is the same as the original difference. This shows that the difference between the largest and smallest numbers is invariant under the transformation.Applying this to the initial triplet ({70, 61, 20}):- The initial difference is (70 - 20 = 50).Since the difference is preserved through each transformation, after 1989 applications, the difference remains the same.Thus, the difference between the largest and smallest numbers after 1989 applications is (boxed{50}).,<think>Alright, so I have this problem where Ilya starts with a triplet of numbers {70, 61, 20} and applies a transformation rule 1989 times. The rule is that each number is replaced by the sum of the other two. I need to find the difference between the largest and smallest numbers after all these transformations. Hmm, okay. Let me try to figure this out step by step.First, let me understand the transformation. If we have a triplet (a, b, c), then after one step, each number becomes the sum of the other two. So the next triplet would be (b + c, a + c, a + b). Let me verify that with the initial numbers. Starting with {70, 61, 20}:First transformation:- New first number: 61 + 20 = 81- New second number: 70 + 20 = 90- New third number: 70 + 61 = 131So after one step, the triplet is {81, 90, 131}. The difference between the largest and smallest here is 131 - 81 = 50.Wait, the original difference was 70 - 20 = 50. Interesting, the difference stayed the same. Let me check the next step to see if this pattern continues.Second transformation:- New first number: 90 + 131 = 221- New second number: 81 + 131 = 212- New third number: 81 + 90 = 171So triplet is {221, 212, 171}. The difference is 221 - 171 = 50. Hmm, same difference again.Third transformation:- First: 212 + 171 = 383- Second: 221 + 171 = 392- Third: 221 + 212 = 433Difference: 433 - 383 = 50. Still 50.Wait a minute, so after each transformation, the difference between the largest and smallest numbers remains 50? That seems to be the case here. Let me check once more.Fourth transformation:- First: 392 + 433 = 825- Second: 383 + 433 = 816- Third: 383 + 392 = 775Difference: 825 - 775 = 50. Yep, still 50. So maybe the difference is preserved? That would make the answer 50, regardless of the number of transformations. But let me see why that's happening.Let me try to find a mathematical explanation. Suppose we have triplet (a, b, c) sorted such that a ≤ b ≤ c. The difference between the largest and smallest is c - a. After transformation, the new triplet is (b + c, a + c, a + b). Let's sort the new triplet.Assuming all numbers are positive, which they are here, the largest number in the new triplet would be the sum of the two larger original numbers. Since the original triplet was a ≤ b ≤ c, the largest sum is b + c. The smallest number in the new triplet would be the sum of the two smallest original numbers, which is a + b. So the new triplet, sorted, would be (a + b, a + c, b + c). Wait, but in the previous examples, after the first transformation, {81, 90, 131} sorted is {81, 90, 131}, and the next step gives {221, 212, 171} sorted is {171, 212, 221}, then {383, 392, 433} sorted is {383, 392, 433}, then {775, 816, 825} sorted is {775, 816, 825}. Hmm, actually the order isn't always maintained as (a + b, a + c, b + c). For example, after the first transformation, 81 = 61 + 20, which was the sum of the two smaller originals, 90 = 70 + 20, which is the sum of the first and third, and 131 = 70 + 61, the sum of the first two. Wait, so maybe my initial assumption about the order is incorrect. Let me think.Original triplet: 70, 61, 20. Let me sort them: 20, 61, 70. Then the transformation would be 61 + 70 = 131, 20 + 70 = 90, 20 + 61 = 81. So sorted new triplet is 81, 90, 131. So the next triplet, sorted, is 81, 90, 131. Then transforming that: 90 + 131 = 221, 81 + 131 = 212, 81 + 90 = 171. Sorted: 171, 212, 221. Then transforming that: 212 + 221 = 433, 171 + 221 = 392, 171 + 212 = 383. Sorted: 383, 392, 433. Next: 392 + 433 = 825, 383 + 433 = 816, 383 + 392 = 775. Sorted: 775, 816, 825.Wait, so in each case, the new triplet is {sum of middle and largest, sum of smallest and largest, sum of smallest and middle} of the previous triplet? But when sorted, the order changes. However, in each case, the difference between the largest and smallest is 50. Let me check:First triplet: 20, 61, 70. Difference 50.First transformation: 81, 90, 131. 131 - 81 = 50.Second transformation: 171, 212, 221. 221 - 171 = 50.Third transformation: 383, 392, 433. 433 - 383 = 50.Fourth transformation: 775, 816, 825. 825 - 775 = 50.So it seems the difference remains 50 every time. Therefore, regardless of the number of transformations, the difference between the largest and smallest numbers remains 50. Therefore, after 1989 transformations, the difference is still 50.But wait, let me make sure there isn't a cycle or some pattern where the difference changes after some steps. But according to the examples above, even after four transformations, the difference remains the same. So maybe the difference is invariant under this transformation. Let me try to prove that.Suppose we have a triplet (x, y, z) with maximum - minimum = d. After transformation, the triplet becomes (y + z, x + z, x + y). Let's compute the new maximum and minimum. The new numbers are y + z, x + z, x + y. Let's denote the original numbers such that x ≤ y ≤ z, so x is the smallest, z is the largest. Then:y + z ≥ x + z ≥ x + y.So the new maximum is y + z, the new minimum is x + y. The difference is (y + z) - (x + y) = z - x, which is the original difference. Hence, the difference remains the same. Therefore, no matter how many transformations we apply, the difference between the largest and smallest numbers remains constant. Therefore, the answer is 50.But let me check with different numbers to make sure. Suppose we start with {1, 2, 3}. Difference is 2. After transformation: 2 + 3 = 5, 1 + 3 = 4, 1 + 2 = 3. New triplet {3, 4, 5}. Difference 2. Next transformation: 4 + 5 = 9, 3 + 5 = 8, 3 + 4 = 7. Triplet {7, 8, 9}, difference 2. Yep, same difference. So seems like the difference is preserved.Therefore, regardless of the number of steps, the difference remains the same as the original. Since the original difference was 70 - 20 = 50, the answer is 50.Final Answerboxed{50}</think>Ilya starts with the triplet ({70, 61, 20}) and transforms it according to the rule that each number is replaced by the sum of the other two. We need to find the difference between the largest and smallest numbers after 1989 transformations.First, we observe the transformation process:- The initial triplet is ({70, 61, 20}). The difference between the largest and smallest numbers is (70 - 20 = 50).- After the first transformation, the triplet becomes ({81, 90, 131}). The difference remains (131 - 81 = 50).- Subsequent transformations consistently show that the difference between the largest and smallest numbers remains 50.To generalize, consider a sorted triplet ((a, b, c)) with (a leq b leq c). The difference (c - a) is preserved after each transformation. This is because the new triplet becomes ((b + c, a + c, a + b)). The largest number in the new triplet is (b + c) and the smallest is (a + b). The difference is ((b + c) - (a + b) = c - a), which is the original difference.Thus, the difference between the largest and smallest numbers remains invariant under the given transformation rule. Since the initial difference is 50, it remains 50 after any number of transformations, including 1989.[boxed{50}]]

🔑:Solution. Let's denote the 3 numbers as {x ; x+a ; x+b}, where 0<a<b. Then the difference between the largest and the smallest number at any step, starting from the zeroth step, will be an invariant, that is, unchanged and equal to b.B=70-20=50.Answer: 50.

🧠:[<think>Alright, let me try to figure out this problem. So, we have square ABCD, and inside it, there's an isosceles right triangle AEF with an area of 1 cm², and a rectangle EFGH with an area of 10 cm². We need to find the area of the square ABCD.First, let me visualize the square. Since it's a square, all sides are equal, and all angles are 90 degrees. The triangle AEF is an isosceles right triangle. That means two sides are equal, and the right angle is between those two equal sides. The area of such a triangle is (side²)/2. Since the area is 1 cm², maybe I can find the length of the legs of the triangle.Let me denote the legs of triangle AEF as 'a'. Then the area is (a²)/2 = 1. Solving for 'a', we get a² = 2, so a = √2 cm. Therefore, each leg of the triangle is √2 cm. Since AEF is an isosceles right triangle, the hypotenuse EF would be a√2 = √2 * √2 = 2 cm. So, the hypotenuse EF is 2 cm long.Now, there's a rectangle EFGH with area 10 cm². If EF is one side of the rectangle, then the other side must be 10 / 2 = 5 cm. Wait, but in a rectangle, opposite sides are equal, so if EF is 2 cm, then FG must be 5 cm? Hmm, maybe. Let me think.But actually, the area of a rectangle is length times width. If EF is one side, which is 2 cm, then the other side, say FG, would be 10 / 2 = 5 cm. So, the sides of the rectangle EFGH are 2 cm and 5 cm. But how is this rectangle positioned inside the square ABCD?Let me try to sketch this mentally. Square ABCD. Let's say point A is at the bottom left corner, B at the bottom right, C at the top right, and D at the top left. Then, triangle AEF is an isosceles right triangle. Since it's an isosceles right triangle with vertex at A, the legs are probably along the sides of the square. So, one leg along AB and another along AD. Then, point E is somewhere on AB, and point F is somewhere on AD. Wait, but EF would be the hypotenuse, which we found out is 2 cm.Wait, but if E is on AB and F is on AD, then the triangle AEF is formed by moving along the sides from A. So, AE = AF = √2 cm. Then, AB and AD are sides of the square, so the total length of each is the side of the square. Let's denote the side of the square as 's'. So, AB = AD = s cm. Then, the positions of E and F are at √2 cm from A along AB and AD, respectively.But then, EF is the hypotenuse of the triangle AEF, which is 2 cm. So, the coordinates of E would be (√2, 0) and F would be (0, √2) if we place point A at the origin (0,0). Then, the coordinates of point E are (√2, 0), F are (0, √2). Then, the line EF connects these two points. The length between E and F would be sqrt[(√2 - 0)^2 + (0 - √2)^2] = sqrt[ (√2)^2 + (-√2)^2 ] = sqrt[2 + 2] = sqrt[4] = 2 cm. That checks out.Now, the rectangle EFGH has area 10 cm². Since EF is one side of the rectangle, which is 2 cm, then the other side must be 5 cm, as I thought earlier. But how is this rectangle oriented? If EF is the side of the rectangle, then FG would be the other side, which is 5 cm. So, starting from F, moving 5 cm in some direction to get to G, then from G moving parallel to EF to get to H, and then back to E? Wait, but the rectangle EFGH must be positioned inside the square ABCD.Wait, maybe the rectangle is EFGH, so the points E, F, G, H form a rectangle. Since EF is the hypotenuse of the triangle AEF, and then FG is another side of the rectangle. So, if EF is 2 cm, and FG is 5 cm, then the rectangle is kind of "attached" to the hypotenuse EF.But how is this rectangle placed inside the square? This is a bit confusing. Let me think again.Alternatively, maybe the rectangle EFGH is such that EF is one of its sides, but EF is not along the edge of the square. Since EF is the hypotenuse of the triangle AEF, which is inside the square, the rectangle must be placed somewhere inside the square, not necessarily aligned with the sides.But to figure this out, maybe I need to use coordinate geometry.Let me assign coordinates to the square. Let me place point A at (0, 0), B at (s, 0), C at (s, s), and D at (0, s), where s is the side length of the square ABCD.Triangle AEF is an isosceles right triangle with area 1 cm². So, points E and F are located such that AE = AF, and angle at A is 90 degrees. So, if E is on side AB and F is on side AD, then coordinates of E would be (a, 0) and F would be (0, a), where a is the length of AE and AF. Then, the area of triangle AEF is (a * a)/2 = 1. Therefore, a² = 2, so a = √2. So, E is at (√2, 0) and F is at (0, √2). The hypotenuse EF connects these two points.Then, EFGH is a rectangle with area 10 cm². The sides of the rectangle are EF and FG. Since EF is the hypotenuse of the triangle, which we already calculated as 2 cm. So, length of EF is 2 cm. Therefore, the other side of the rectangle, FG, must be 10 / 2 = 5 cm.So, from point F, moving along some direction for 5 cm to reach point G, then from G moving parallel to EF to reach H, and then back to E. But in which direction is FG? Since EFGH is a rectangle, FG must be perpendicular to EF. Wait, in a rectangle, adjacent sides are perpendicular. Wait, no, in a rectangle, all angles are 90 degrees, but the sides don't have to be perpendicular to the previous sides unless it's a square. Wait, no, in a rectangle, adjacent sides are perpendicular. So, yes, the sides EF and FG must be perpendicular.But EF is the hypotenuse of the triangle AEF, which is at a 45-degree angle with respect to the sides of the square. So, the direction of EF is diagonal, going from E (√2, 0) to F (0, √2). The slope of EF is (√2 - 0)/(0 - √2) = -1. So, the line EF has a slope of -1.Therefore, the rectangle EFGH must have sides EF and FG, with FG perpendicular to EF. Since EF has a slope of -1, then FG must have a slope of 1 (since perpendicular lines have slopes that are negative reciprocals). The length of FG is 5 cm.So, starting from point F (0, √2), moving along a direction with slope 1 for 5 cm to reach point G. Then, from G moving parallel to EF (slope -1) for 2 cm to reach H, and then from H moving back along slope 1 to E.Wait, but let's confirm. If EF is of length 2 cm and FG is 5 cm, then the rectangle EFGH has sides 2 cm and 5 cm. So, the coordinates of G can be found by moving from F in the direction perpendicular to EF for 5 cm.Alternatively, since EF is a vector from E to F, which is (-√2, √2), because E is (√2, 0) and F is (0, √2). So, the vector EF is F - E = (-√2, √2). A perpendicular vector would be (√2, √2) or (-√2, -√2), but since we need the direction for FG, which is part of the rectangle.Wait, maybe I need to use vectors here. The vector EF is (-√2, √2). A perpendicular vector would have a dot product of zero. So, if EF is (-√2, √2), then a perpendicular vector would be (√2, √2), since (-√2)(√2) + (√2)(√2) = -2 + 2 = 0. So, the direction of FG is (√2, √2). But the length of FG is 5 cm. So, we need to find the scalar multiple of (√2, √2) that has a length of 5 cm.The length of vector (√2, √2) is sqrt[(√2)^2 + (√2)^2] = sqrt[2 + 2] = sqrt[4] = 2. So, to get a vector of length 5, we need to multiply this vector by (5/2). Therefore, the vector FG is (5/2)(√2, √2) = ( (5√2)/2, (5√2)/2 ).So, starting from point F (0, √2), adding this vector gives the coordinates of G as (0 + (5√2)/2, √2 + (5√2)/2 ) = ( (5√2)/2, √2 + (5√2)/2 ).Similarly, from point G, moving in the direction opposite to EF to reach H. The vector HE should be equal and opposite to EF. Wait, since EFGH is a rectangle, the vector FG is one side, and the vector GH should be equal and opposite to EF. Wait, maybe not. Let's think.In a rectangle, opposite sides are equal and parallel. So, EF is equal and parallel to GH, and FG is equal and parallel to EH. So, vector FG is from F to G, and vector EH is from E to H. So, if we can find H, then we can check if it lies within the square.But perhaps it's easier to calculate the coordinates of G and H and see where they are. Let's proceed.So, coordinates of E: (√2, 0)Coordinates of F: (0, √2)Coordinates of G: ( (5√2)/2, √2 + (5√2)/2 )Coordinates of H: Since H is obtained by moving from G in the direction opposite to EF. The vector EF is (-√2, √2), so the opposite direction would be (√2, -√2). Since EF is of length 2, but we need to move from G to H along this direction for length 2 cm. Wait, but GH should be equal to EF, so GH is also 2 cm. Wait, but in the rectangle EFGH, EF and GH are opposite sides, both length 2 cm, and FG and EH are 5 cm. So, from point G, moving in the direction of HE (which is same as FG but opposite direction?) Wait, maybe not.Alternatively, since EFGH is a rectangle, the coordinates can be found by moving from F to G, then from G to H along the direction parallel to EF. Wait, this is getting complicated. Maybe using vectors is better.Alternatively, since EFGH is a rectangle, the coordinates of H can be found by adding the vector FG to point E. Wait, in a rectangle, E to F to G to H to E. So, vector EF + vector FG = vector EG. But not sure.Wait, perhaps using the property that in a rectangle, the diagonals are equal and bisect each other. But maybe coordinate geometry is the way to go.So, we have E at (√2, 0), F at (0, √2). Let me denote G as (x, y). Since FG is perpendicular to EF and has length 5 cm.First, the slope of EF is (√2 - 0)/(0 - √2) = -1, as we saw before. Therefore, the slope of FG, being perpendicular, is 1. So, the line FG has a slope of 1. Since it passes through F (0, √2), the equation of line FG is y - √2 = 1*(x - 0), so y = x + √2.The length of FG is 5 cm. The distance from F (0, √2) to G (x, y) along this line is 5 cm. Since the direction is along y = x + √2, we can parametrize this line. Let me use a parameter t. So, starting at F (0, √2), moving along the line y = x + √2, the coordinates of G can be expressed as (t, t + √2). The distance from F to G is sqrt[(t - 0)^2 + (t + √2 - √2)^2] = sqrt[t² + t²] = sqrt[2t²] = t√2. We want this distance to be 5 cm. So, t√2 = 5 => t = 5/√2 = (5√2)/2.Therefore, the coordinates of G are ( (5√2)/2, (5√2)/2 + √2 ) = ( (5√2)/2, (5√2 + 2√2)/2 ) = ( (5√2)/2, (7√2)/2 ).Now, we need to find point H such that EFGH is a rectangle. Since EFGH is a rectangle, vector EH should be equal to vector FG. Vector FG is from F to G: ( (5√2)/2 - 0, (7√2)/2 - √2 ) = ( (5√2)/2, (5√2)/2 ). So, vector FG is ( (5√2)/2, (5√2)/2 ). Therefore, vector EH should be the same. Since E is at (√2, 0), adding vector FG to E gives H: ( √2 + (5√2)/2, 0 + (5√2)/2 ) = ( (2√2 + 5√2)/2, (5√2)/2 ) = ( (7√2)/2, (5√2)/2 ).Alternatively, since H is obtained by moving from G in the direction opposite to EF. Wait, but maybe it's better to check if the coordinates of H are within the square ABCD. The square ABCD has side length 's', so all coordinates must be between 0 and s.Coordinates of G: ( (5√2)/2, (7√2)/2 )Coordinates of H: (7√2/2, 5√2/2 )So, both G and H have x and y coordinates of (5√2)/2 ≈ 3.535 cm and (7√2)/2 ≈ 4.95 cm. So, if the side length 's' of the square is greater than or equal to 7√2/2 ≈ 4.95 cm, then all points are inside the square.Therefore, the side length of the square must be at least 7√2/2 cm. Therefore, the area of the square would be (7√2/2)^2 = (49 * 2)/4 = 98/4 = 24.5 cm². But wait, let me check.Wait, (7√2/2)^2 = (7^2)*(√2^2)/(2^2) = 49*2/4 = 98/4 = 24.5. So, 24.5 cm². But 24.5 is equal to 49/2, which is 24.5. Hmm, but is that correct?Wait, but let me verify if all four points E, F, G, H are inside the square. The coordinates of E are (√2, 0) ≈ (1.414, 0), F is (0, √2) ≈ (0, 1.414), G is ( (5√2)/2, (7√2)/2 ) ≈ (3.535, 4.95), and H is ( (7√2)/2, (5√2)/2 ) ≈ (4.95, 3.535). So, the maximum x-coordinate is 4.95, and the maximum y-coordinate is also 4.95. Therefore, the square must have a side length of at least 4.95 cm. Which is 7√2/2 ≈ 4.95 cm, as we calculated. So, the area is (7√2/2)^2 = 24.5 cm². So, 24.5 cm² is the area. But the problem probably expects an exact value, not a decimal. So, 24.5 is equal to 49/2, so 49/2 cm². Therefore, the area of the square is 49/2 cm², which is 24.5 cm².But let me check if this is correct. Because sometimes when you have such configurations, there might be an overlap or some miscalculation in coordinates. Let me verify the positions.First, starting from E (√2, 0). Then moving along EF to F (0, √2). Then moving along FG to G ( (5√2)/2, (7√2)/2 ). Then from G to H ( (7√2)/2, (5√2)/2 ), and then back to E. So, the rectangle EFGH is inside the square. The maximum x and y coordinates are 7√2/2 ≈ 4.95, so the square must have side length at least 7√2/2. Therefore, the area is (7√2/2)^2 = 49*2/4 = 49/2 = 24.5 cm².But let me check another way. Suppose the side of the square is 's'. Then, the coordinates of G are (5√2/2, 7√2/2). Since G is inside the square, both coordinates must be less than or equal to 's'. Therefore, 7√2/2 ≤ s. Similarly, the coordinates of H are (7√2/2, 5√2/2), so the x-coordinate is 7√2/2, which is the maximum. Therefore, s must be at least 7√2/2. Therefore, the side length is 7√2/2, so area is (7√2/2)^2 = (49*2)/4 = 98/4 = 24.5 cm². So, 49/2 cm².But is there another way to approach this problem without coordinates?Alternatively, since we have triangle AEF with legs √2 cm, and the rectangle EFGH with sides 2 cm and 5 cm, perhaps the square's side can be found by considering the distances from the edges to the rectangle.But maybe the coordinate approach is solid. Let me see if 49/2 cm² is the correct answer. 49 divided by 2 is 24.5, which is 24.5 cm². Let me check if all steps are correct.First, triangle AEF: legs √2 cm, hypotenuse 2 cm. Then, rectangle EFGH with area 10 cm², so sides 2 cm and 5 cm. Then, using coordinates to find that the maximum coordinate in the square is 7√2/2, leading to the side length s = 7√2/2, hence area (7√2/2)^2 = 49/2. Seems correct.Alternatively, maybe there is a simpler geometric interpretation. Let's think.In square ABCD, the isosceles right triangle AEF is placed in the corner, with legs AE and AF each of length √2 cm. Then, the rectangle EFGH extends from the hypotenuse EF into the square. The rectangle has an area of 10 cm², with one side being EF (2 cm) and the other side being the distance from EF to GH, which is 5 cm. So, this 5 cm is the height of the rectangle relative to the base EF.But how is this 5 cm oriented within the square? If EF is the hypotenuse of the triangle, then the height of the rectangle would be along the altitude from EF. But in this case, since EF is the hypotenuse of the isosceles right triangle, the altitude from EF would be equal to (√2)/2 * a, where a is the leg. Wait, the altitude from the hypotenuse in an isosceles right triangle is equal to half the hypotenuse. Since the hypotenuse is 2 cm, the altitude would be 1 cm. But the rectangle's height is 5 cm, which is much larger. So, perhaps the rectangle is not aligned with the altitude.Alternatively, maybe the length of 5 cm is the distance from EF to GH along the direction perpendicular to EF. So, if EF is 2 cm, and the rectangle has a height of 5 cm in the direction perpendicular to EF, then the total span of the rectangle in the square would require the square's side to accommodate this.But since EF is at a 45-degree angle, moving 5 cm perpendicular to EF would involve moving in a diagonal direction. So, the projection along the x and y axes would be 5 cm * cos(45°) and 5 cm * sin(45°), which is 5*(√2/2) ≈ 3.535 cm in both x and y directions.Therefore, starting from EF, moving 5 cm in the direction perpendicular to EF would add 5*(√2/2) to both the x and y coordinates. So, the furthest point in the rectangle would be G, which is 5 cm away from EF. Given that E is at (√2, 0) and F is at (0, √2), moving 5 cm in the perpendicular direction would reach a point that is √2 + 5*(√2/2) in both x and y directions. Wait, this might not be accurate. Let me think again.Alternatively, the displacement from EF to GH is 5 cm in the direction perpendicular to EF. Since EF is length 2 cm, and the rectangle has area 10 cm², the height is 5 cm. The direction of this height is perpendicular to EF. Therefore, the total extent of the square must cover from point A (0,0) to the furthest point in the rectangle, which is point G.The coordinates of G, as calculated earlier, are (5√2/2, 7√2/2). The x-coordinate is 5√2/2 ≈ 3.535, and the y-coordinate is 7√2/2 ≈ 4.95. Therefore, the maximum x and y are 7√2/2, so the square needs to have side length at least 7√2/2. Hence, the area is (7√2/2)^2 = 49/2 cm².Alternatively, maybe there's a way to compute the side length of the square by considering the distances from the triangle and the rectangle. The triangle is at the corner, and the rectangle extends from the hypotenuse. The total distance from A to the furthest point in the rectangle would be the side of the square.The triangle AEF has legs of √2 cm. The rectangle extends 5 cm from the hypotenuse EF into the square. To find the maximum distance from A to G, which is the furthest corner of the rectangle, we can break it down.The hypotenuse EF is 2 cm from E to F. The rectangle extends 5 cm in the direction perpendicular to EF. To find how much this adds to the x and y coordinates, we can note that the direction perpendicular to EF is along the line y = x (since EF has a slope of -1, the perpendicular direction has a slope of 1). Moving 5 cm along this direction from the midpoint of EF would reach the furthest point. Wait, perhaps not the midpoint, but from the line EF itself.Alternatively, since the displacement from EF to GH is 5 cm in the direction of the normal vector. The normal vector to EF is (1, 1) as we calculated before. Moving 5 cm along this direction from EF would increase both x and y coordinates by 5*(√2/2). Wait, similar to the previous calculation.But this seems consistent with the coordinate approach. So, the maximum coordinate in the square is 7√2/2, leading to the side length of the square being 7√2/2 cm, and the area is (7√2/2)^2 = 49/2 cm². Therefore, the answer is 49/2 cm².But let me check once more for any miscalculations. Starting from E (√2, 0) and F (0, √2). The vector EF is (-√2, √2). The perpendicular direction is (1, 1). The length of 5 cm in that direction. The unit vector in the direction (1, 1) is (1/√2, 1/√2). Therefore, moving 5 cm in that direction from F would give:G = F + 5*(1/√2, 1/√2) = (0, √2) + (5/√2, 5/√2) = (5/√2, √2 + 5/√2). Converting 5/√2 to (5√2)/2, so coordinates are ( (5√2)/2, √2 + (5√2)/2 ). Which is the same as (5√2/2, (2√2 + 5√2)/2 ) = (5√2/2, 7√2/2 ). So, same as before.Therefore, the coordinates of G are (5√2/2, 7√2/2 ). Therefore, the maximum x and y coordinates are both 7√2/2, hence the side of the square is 7√2/2 cm, area is (7√2/2 )² = 49*2 /4 = 49/2 cm².Therefore, the answer should be 49/2 cm², which is 24.5 cm². Since the problem asks for the area in cm², boxed as boxed{dfrac{49}{2}}.Final AnswerThe area of square mathrm{ABCD} is boxed{dfrac{49}{2}} mathrm{cm}^2.</think>Given square ( mathrm{ABCD} ) with an isosceles right triangle ( mathrm{AEF} ) of area ( 1 , mathrm{cm}^2 ) and a rectangle ( mathrm{EFGH} ) of area ( 10 , mathrm{cm}^2 ), we need to find the area of the square.1. Area of Triangle ( mathrm{AEF} ): - The area of an isosceles right triangle is given by ( frac{1}{2} times text{leg}^2 = 1 , mathrm{cm}^2 ). - Solving for the leg length ( a ), we get ( frac{1}{2} a^2 = 1 ) hence ( a^2 = 2 ) and ( a = sqrt{2} , mathrm{cm} ). - The hypotenuse ( mathrm{EF} ) is ( sqrt{2} times sqrt{2} = 2 , mathrm{cm} ).2. Area of Rectangle ( mathrm{EFGH} ): - The area of the rectangle is ( 10 , mathrm{cm}^2 ). One side is ( mathrm{EF} = 2 , mathrm{cm} ). - The other side of the rectangle is ( frac{10}{2} = 5 , mathrm{cm} ).3. Coordinates and Geometry: - Place the square ( mathrm{ABCD} ) with ( mathrm{A} ) at the origin ((0, 0)), ( mathrm{B} ) at ((s, 0)), ( mathrm{C} ) at ((s, s)), and ( mathrm{D} ) at ((0, s)). - Points ( mathrm{E} ) and ( mathrm{F} ) are at ((sqrt{2}, 0)) and ((0, sqrt{2})) respectively. - The rectangle ( mathrm{EFGH} ) extends from ( mathrm{F} ) in the direction perpendicular to ( mathrm{EF} ). The vector perpendicular to ( mathrm{EF} ) (slope -1) has a slope of 1. - The coordinates of ( mathrm{G} ) are found by moving ( 5 , mathrm{cm} ) along this direction from ( mathrm{F} ), resulting in ( left( frac{5sqrt{2}}{2}, frac{7sqrt{2}}{2} right) ).4. Side Length of the Square: - The maximum coordinates from ( mathrm{G} ) are ( frac{7sqrt{2}}{2} ), which must be the side length ( s ) of the square. - The area of the square is ( left( frac{7sqrt{2}}{2} right)^2 = frac{49 times 2}{4} = frac{49}{2} , mathrm{cm}^2 ).Thus, the area of square ( mathrm{ABCD} ) is (boxed{dfrac{49}{2}} , mathrm{cm}^2 ).,<think>Okay, let me try to figure out this geometry problem. So, there's a square ABCD, and inside it, there's an isosceles right triangle AEF with an area of 1 cm², and a rectangle EFGH with an area of 10 cm². We need to find the area of the square ABCD. Hmm, let me start by visualizing the square and the given figures inside it.First, since AEF is an isosceles right triangle, that means two of its sides are equal, and the right angle is between those two sides. In an isosceles right triangle, the legs are equal, and the hypotenuse is √2 times longer than each leg. The area of such a triangle is (leg²)/2. The problem says the area is 1 cm², so if I let the legs be of length 'a', then (a²)/2 = 1. Solving for 'a', I get a² = 2, so a = √2 cm. Therefore, each leg (the two sides forming the right angle) of triangle AEF is √2 cm long. That might be useful later.Now, the rectangle EFGH has an area of 10 cm². Since it's a rectangle, opposite sides are equal. Let's denote the sides of the rectangle as length 'l' and width 'w', so l * w = 10. But how does this rectangle relate to the triangle and the square? I need to figure out their positions within the square.Since the square is ABCD, let's label the vertices in order: let's assume ABCD is the square with A at the bottom-left corner, B at the bottom-right, C at the top-right, and D at the top-left. So, moving clockwise. Then, the triangle AEF is inside the square. If it's an isosceles right triangle with vertex at A, then points E and F must be along the sides of the square. Let's say, for example, point E is on side AB and point F is on side AD. Then, AE and AF would be the legs of the triangle, each of length √2 cm. Wait, but the sides of the square ABCD are all equal. If AE is along AB, then AE is part of AB, and AF is along AD, part of AD. But if the legs are both √2 cm, then AE = AF = √2 cm. However, the entire side of the square is longer than that, right? Because the square must also accommodate the rectangle EFGH.Wait, maybe I need to sketch this mentally. Square ABCD, triangle AEF at corner A, with legs AE and AF. Then the hypotenuse EF is opposite the right angle at A. Then, the rectangle EFGH is somehow connected to this hypotenuse? Maybe the rectangle has one side along EF? Hmm. Let me try to piece this together.If EF is the hypotenuse of the triangle AEF, then EF has length √( (√2)^2 + (√2)^2 ) = √(2 + 2) = √4 = 2 cm. So EF is 2 cm long. If the rectangle EFGH has EF as one of its sides, then the length of the rectangle could be 2 cm, and the width would be something else. But the area is 10 cm², so 2 * width = 10, so width would be 5 cm. But does that make sense within the square? If EF is 2 cm, then the other side of the rectangle (the width) would need to be 5 cm, but the square's side must be at least 5 cm to accommodate that. However, if the legs of the triangle are √2 cm (~1.414 cm), then the square's side must be longer than √2 cm. But 5 cm is much larger. Maybe the rectangle is not aligned such that EF is its length. Alternatively, maybe EF is the width, and the length is something else. But then the area would be length * 2 cm = 10 cm², so length = 5 cm. Either way, one of the sides of the rectangle is 5 cm. Hmm.Alternatively, maybe the rectangle is positioned such that EF is one of its sides, and the other sides are extending into the square. But how does this relate to the square's dimensions?Wait, perhaps I need to assign coordinates to the square to make this easier. Let's place point A at the origin (0,0), so the square ABCD would have coordinates:A: (0,0)B: (s,0) where s is the side length of the squareC: (s,s)D: (0,s)Now, triangle AEF is an isosceles right triangle with area 1 cm². Let's assume that E is on side AB and F is on side AD. So, coordinates for E would be (a,0) and F would be (0,a) since it's an isosceles right triangle. Then, the legs AE and AF each have length 'a', and the area is (a²)/2 = 1, so a² = 2, hence a = √2 as before. Therefore, E is at (√2, 0) and F is at (0, √2). Then, the hypotenuse EF connects these two points, which would be from (√2,0) to (0,√2). The length of EF is √[(√2)^2 + (√2)^2] = √(2 + 2) = √4 = 2 cm as before.Now, the rectangle EFGH. Let's figure out where points G and H are. The rectangle EFGH has sides EF, FG, GH, HE. Since EF is from (√2,0) to (0,√2), which is a line segment of length 2 cm. To form a rectangle, we need to determine the direction in which the rectangle extends from EF. The rectangle's sides FG and HE must be perpendicular to EF. Since EF has a slope of (√2 - 0)/(0 - √2) = -1, so the direction perpendicular to EF would have a slope of 1. So, moving in the direction perpendicular to EF from point E and point F.But maybe it's easier to compute coordinates. Let's consider the vector from E to F: F - E = (0 - √2, √2 - 0) = (-√2, √2). The direction of EF is (-√2, √2). A perpendicular vector would be (√2, √2) since the dot product of (-√2, √2) and (√2, √2) is (-√2)(√2) + (√2)(√2) = -2 + 2 = 0, so they are perpendicular.Therefore, the rectangle EFGH can be formed by moving from E and F in the direction of (√2, √2) or (-√2, -√2). However, since the rectangle is inside the square ABCD, we need to make sure that the points G and H are within the square. Let's assume the direction is upwards and to the right, so the vector (√2, √2). But wait, (√2, √2) is a diagonal direction. However, since we are forming a rectangle, the sides FG and HE need to be congruent and parallel. Alternatively, perhaps the sides FG and EH are perpendicular to EF. Since EF is the hypotenuse, the rectangle could be extending into the square.Alternatively, maybe the rectangle EFGH is such that EF is one side, and the other sides are along the square's grid. Wait, perhaps I need to parametrize this.Alternatively, maybe the rectangle EFGH is not aligned with EF as a diagonal, but as a side. Wait, but EF is a diagonal of the triangle AEF. Hmm, this is getting confusing. Let me try to think differently.Given that EFGH is a rectangle, with EF as one of its sides, then the sides FG and EH must be equal and perpendicular to EF. Since EF has length 2 cm, and the area of the rectangle is 10 cm², then the other side of the rectangle (the height) would be 10 / 2 = 5 cm. So, the length of FG and EH is 5 cm. Therefore, if we can determine the direction in which these sides extend, we can find the coordinates of G and H, and hence find the side length of the square.But how does this fit into the square? Let's go back to the coordinate system. Points E(√2, 0) and F(0, √2). The vector from E to F is (-√2, √2). A perpendicular vector would be (√2, √2) as before. But the length of this perpendicular vector would need to correspond to 5 cm. Wait, but the length of (√2, √2) is √[(√2)^2 + (√2)^2] = √(2 + 2) = √4 = 2 cm. So, if we move along this direction by a certain scalar multiple to get the length of 5 cm.Wait, perhaps the sides FG and EH are vectors perpendicular to EF with length 5 cm. Since EF is the side of the rectangle, the other sides (FG and EH) must be of length 5 cm. The direction of FG is perpendicular to EF. Since EF has a direction vector of (-√2, √2), then the perpendicular direction is (√2, √2) as we saw earlier. Therefore, to get a length of 5 cm, we need to scale this vector appropriately.First, let's compute the unit vector in the direction perpendicular to EF. The direction vector perpendicular to EF is (1,1) normalized. Wait, original vector for EF is (-√2, √2), so a perpendicular vector is (√2, √2). The length of this vector is √[(√2)^2 + (√2)^2] = √(2 + 2) = √4 = 2. Therefore, the unit vector in that direction is (√2/2, √2/2). Therefore, to get a length of 5 cm, we multiply the unit vector by 5: (5*√2/2, 5*√2/2).Therefore, starting from point F(0, √2), moving in the direction of (√2, √2) scaled by 5/2, we reach point G. Similarly, starting from point E(√2, 0), moving in the same direction, we reach point H.Wait, but let's check this. The vector from F to G should be in the direction perpendicular to EF with length 5. So, if the direction is (√2, √2), and we need length 5, then the scaling factor 'k' satisfies:k * √[(√2)^2 + (√2)^2] = k * 2 = 5 => k = 5/2.Therefore, the vector from F to G is (5/2 * √2, 5/2 * √2). Similarly, the vector from E to H is the same.So, point G is F + (5/2 * √2, 5/2 * √2) = (0 + 5/2 * √2, √2 + 5/2 * √2) = (5√2/2, (√2 + 5√2/2)) = (5√2/2, (2√2/2 + 5√2/2)) = (5√2/2, 7√2/2).Similarly, point H is E + (5/2 * √2, 5/2 * √2) = (√2 + 5√2/2, 0 + 5√2/2) = ( (2√2/2 + 5√2/2 ), 5√2/2 ) = (7√2/2, 5√2/2 ).Now, we need to check if points G and H lie within the square ABCD. The coordinates of G are (5√2/2, 7√2/2), and H is (7√2/2, 5√2/2). The square ABCD has side length 's', so all coordinates must be between 0 and s. Therefore, for point G, both x and y coordinates must be ≤ s. Similarly for point H. So:5√2/2 ≤ s7√2/2 ≤ sSimilarly, since 7√2/2 is greater than 5√2/2, the side length 's' must be at least 7√2/2 cm. Therefore, the square's side length is 7√2/2 cm. Then, the area of the square is (7√2/2)^2 = (49 * 2)/4 = 98/4 = 24.5 cm², which is 49/2 cm².But wait, let's confirm if this is correct. If the side length is 7√2/2, then converting that to decimal, √2 ≈ 1.414, so 7*1.414/2 ≈ 4.949 cm. Then the area would be approximately (4.949)^2 ≈ 24.5 cm², which matches 49/2. So, is this the answer? But let's verify if all points lie within the square.Point G is (5√2/2, 7√2/2). If s = 7√2/2, then the x-coordinate is 5√2/2 < 7√2/2, and the y-coordinate is exactly 7√2/2. Similarly, point H is (7√2/2, 5√2/2), so the y-coordinate is 5√2/2 < 7√2/2. Wait, but in the square ABCD, the maximum x and y coordinates are s. So, point G has y-coordinate equal to s, and point H has x-coordinate equal to s. So, they are on the edges of the square. That's acceptable.Therefore, the side length of the square is 7√2/2 cm, so area is (7√2/2)^2 = 49 * 2 /4 = 98/4 = 24.5 cm², which is 49/2 cm². Therefore, the area of square ABCD is 49/2 cm².But wait, let me check if this makes sense. The triangle AEF is at the corner, with legs √2. The rectangle EFGH extends from the hypotenuse EF with length 2 cm, and the other side of the rectangle is 5 cm. But in our calculation, the other side of the rectangle is 5 cm in the direction perpendicular to EF. However, the length of 5 cm in that direction translates to a displacement in the coordinate system, which when scaled by the vector (√2, √2) with magnitude 2, gives us a total displacement of 5/2 * vector. But perhaps there's a mistake here in assuming the length of the rectangle's side.Wait, actually, the area of the rectangle is 10 cm². If EF is 2 cm, then the height of the rectangle (distance between EF and GH) should be 5 cm. But in reality, this height is the distance between the lines EF and GH. Since EF is a diagonal line, the distance between EF and GH isn't just the length of the vector we calculated, but the actual perpendicular distance.Wait a second, maybe I confused the length of the vector with the actual height of the rectangle. The area of a rectangle is base times height, where height is the perpendicular distance between the bases. In this case, EF is the base with length 2 cm, and the height is the perpendicular distance from EF to the opposite side GH. So, this perpendicular distance should be 5 cm.But when we move from EF to GH along the direction perpendicular to EF, the length of that movement is the height, which is 5 cm. However, when we represented this as a vector (5/2 * √2, 5/2 * √2), the actual distance moved along that direction is 5 cm, but in coordinates, the components would be (5/2 * √2, 5/2 * √2). Wait, no, the length of the vector (5/2 * √2, 5/2 * √2) is √[(5√2/2)^2 + (5√2/2)^2] = √[(25*2/4) + (25*2/4)] = √[(50/4) + (50/4)] = √[100/4] = √25 = 5 cm. So that's correct. Therefore, moving from EF in the perpendicular direction by 5 cm gives us the points G and H. Therefore, the coordinates of G and H are as calculated.But then the maximum x and y coordinates are 7√2/2, which is approximately 4.95 cm. So, the square has sides of 7√2/2 cm, so area is 49/2 cm². Therefore, the answer should be 49/2 cm². But let me verify once again.Alternatively, maybe there's a different configuration. Suppose the rectangle EFGH is not extending outwards from EF but is placed such that EF is part of the rectangle. Wait, EF is the hypotenuse of the triangle, so if the rectangle EFGH has EF as one side, then the rectangle would be adjacent to the triangle. But in that case, how is the rectangle positioned?Alternatively, perhaps the rectangle is inside the square but not attached to the triangle. But the problem states the area of triangle AEF is 1 cm² and the area of rectangle EFGH is 10 cm². The names of the points E, F, G, H suggest that they are connected in sequence: A-E-F-G-H-E? Wait, no. The triangle is AEF, so points A, E, F. The rectangle is EFGH, so points E, F, G, H. Therefore, the rectangle shares the side EF with the triangle. Therefore, EF is a common side between the triangle and the rectangle. That makes sense. So, the triangle AEF is attached to the rectangle EFGH along EF.Therefore, the rectangle EFGH has sides EF, FG, GH, HE. So, starting from E, going to F, then to G, then to H, then back to E. Since EF is the hypotenuse of the triangle, the rectangle is extending from that hypotenuse into the square. Therefore, the rectangle is kind of "standing" on the hypotenuse EF.In this case, the height of the rectangle (perpendicular to EF) would be the distance between lines EF and GH, which is 5 cm as we calculated earlier. Therefore, moving from EF in the direction perpendicular to EF by 5 cm gives the other side GH of the rectangle. Thus, the positions of G and H are determined by moving 5 cm from F and E respectively in the perpendicular direction.Given that, then the coordinates of G and H are as previously calculated. Then, the square must enclose all these points, so the maximum x and y coordinates are 7√2/2, which would be the side length of the square. Therefore, the area is (7√2/2)^2 = 49/2 cm². So, 49/2 is 24.5 cm².But let me check if there's another way to approach this problem without coordinates. Maybe by using the properties of the square and the areas given.Since the triangle AEF is an isosceles right triangle with area 1 cm², its legs are √2 cm each. The hypotenuse EF is 2 cm. Then, the rectangle EFGH has an area of 10 cm², so the other side of the rectangle (the height relative to EF) is 10 / 2 = 5 cm. Therefore, the distance from EF to GH is 5 cm. In a square, if you have a diagonal line EF and another line GH parallel to EF at a distance of 5 cm, the square must be large enough to contain both EF and GH.But the distance from EF to GH is 5 cm, but in what direction? Since EF is a diagonal of the triangle, the direction of the distance is towards the interior of the square. The original triangle is at the corner, so the rectangle extends from the hypotenuse EF into the square. The maximum extent of the rectangle would determine the side length of the square.Alternatively, maybe the side length of the square can be calculated by adding the lengths contributed by the triangle and the rectangle. But since the triangle and the rectangle are at a diagonal, it's not straightforward addition.Wait, another thought: In the coordinate system, the square's side is determined by the maximum x or y coordinate of all the points involved. The triangle AEF has points at (0,0), (√2,0), and (0,√2). The rectangle EFGH has points E(√2,0), F(0,√2), G(5√2/2,7√2/2), and H(7√2/2,5√2/2). So, the maximum x-coordinate is 7√2/2 and the maximum y-coordinate is also 7√2/2. Therefore, the square must have side length 7√2/2, so area (7√2/2)^2 = 49*2/4 = 49/2 cm². That seems consistent.Alternatively, let's compute the distance from the origin to point G or H. For example, point G is (5√2/2,7√2/2). The distance from A(0,0) to G is √[(5√2/2)^2 + (7√2/2)^2] = √[(25*2/4) + (49*2/4)] = √[(50 + 98)/4] = √[148/4] = √37 ≈ 6.08 cm. But the side length of the square is 7√2/2 ≈ 4.95 cm. Wait, that can't be. Wait, this is a contradiction. If the distance from A to G is √37 ≈ 6.08 cm, but the side length of the square is only ~4.95 cm, that would mean point G is outside the square. Which contradicts our previous conclusion. So, something is wrong here.Hold on, there must be a mistake in my calculation. Let me re-examine the coordinates.Wait, if the square has side length s, then the coordinates of G are (5√2/2,7√2/2). Therefore, both the x and y coordinates must be ≤ s. Similarly for point H. So, if s = 7√2/2 ≈ 4.95 cm, then the y-coordinate of G is 7√2/2, which is exactly equal to s. However, the x-coordinate of G is 5√2/2 ≈ 3.535 cm, which is less than s. Similarly, point H has x-coordinate 7√2/2 and y-coordinate 5√2/2, so the x-coordinate is equal to s, and the y-coordinate is less than s. Therefore, points G and H lie on the top and right sides of the square, respectively. Therefore, the square has side length s = 7√2/2 cm. Therefore, the diagonal distance from A to G would indeed be √[(5√2/2)^2 + (7√2/2)^2], but this is not necessary for the square's side length because the square is axis-aligned, not rotated. The square's sides are horizontal and vertical, so the maximum x and y coordinates determine the side length, not the diagonal distance. So, even though point G is located at (5√2/2,7√2/2), the square only needs to extend up to 7√2/2 in both x and y directions. Therefore, the calculation of the square's side length as 7√2/2 cm is correct, and the area is (7√2/2)^2 = 49/2 cm².Therefore, after verifying, I think the correct answer is 49/2 cm², which is 24.5 cm². So, expressed as a fraction, it's 49 over 2.But just to make sure, let's check if there's another possible configuration. For example, maybe the rectangle EFGH is placed differently. Suppose instead of extending outward from EF, the rectangle is placed such that EF is a side, but aligned along the sides of the square. But given that EF is a hypotenuse of an isosceles right triangle at the corner, EF is a diagonal line from (√2,0) to (0,√2). It's hard to see how a rectangle with EF as a side could be aligned with the square's sides unless it's rotated. Therefore, the previous configuration seems the only plausible one.Alternatively, maybe the rectangle EFGH is such that sides FG and EH are vertical and horizontal. But given that EF is diagonal, that would not form a rectangle unless FG and EH are adjusted accordingly. Wait, no. If EF is diagonal and you want FG and EH to be horizontal or vertical, then the rectangle would have to be a combination of diagonal and axis-aligned sides, which isn't possible because a rectangle must have all angles 90 degrees. Therefore, if EF is diagonal, then the sides FG and EH must be perpendicular to EF, which is what we calculated before.Therefore, I think the initial solution is correct. The area of the square is 49/2 cm².Final Answerboxed{dfrac{49}{2}}</think>Given a square ( mathrm{ABCD} ) with an isosceles right triangle ( mathrm{AEF} ) of area ( 1 , mathrm{cm}^2 ) and a rectangle ( mathrm{EFGH} ) of area ( 10 , mathrm{cm}^2 ), we need to find the area of the square ( mathrm{ABCD} ).1. Isosceles Right Triangle ( mathrm{AEF} ): - The area of the triangle is given by ( frac{1}{2} times text{leg}^2 = 1 , mathrm{cm}^2 ). - Solving for the legs, we get ( text{leg}^2 = 2 ) hence ( text{leg} = sqrt{2} , mathrm{cm} ). - The hypotenuse ( mathrm{EF} ) is ( sqrt{(sqrt{2})^2 + (sqrt{2})^2} = 2 , mathrm{cm} ).2. Rectangle ( mathrm{EFGH} ): - The area of the rectangle is ( 10 , mathrm{cm}^2 ). - The length of ( mathrm{EF} ) is ( 2 , mathrm{cm} ), so the height (perpendicular to ( mathrm{EF} )) is ( frac{10}{2} = 5 , mathrm{cm} ).3. Coordinates and Dimensions: - Placing ( mathrm{A} ) at the origin ((0,0)), points ( mathrm{E} ) and ( mathrm{F} ) are at ((sqrt{2}, 0)) and ((0, sqrt{2})) respectively. - The vector perpendicular to ( mathrm{EF} ) with length ( 5 , mathrm{cm} ) is calculated. This vector is scaled by ( frac{5}{2} sqrt{2} ).4. Coordinates of ( mathrm{G} ) and ( mathrm{H} ): - Points ( mathrm{G} ) and ( mathrm{H} ) are derived by moving ( 5 , mathrm{cm} ) from ( mathrm{F} ) and ( mathrm{E} ) respectively in the perpendicular direction. - Coordinates of ( mathrm{G} ) and ( mathrm{H} ) are ((5sqrt{2}/2, 7sqrt{2}/2)) and ((7sqrt{2}/2, 5sqrt{2}/2)).5. Side Length of the Square: - The maximum coordinates from points ( mathrm{G} ) and ( mathrm{H} ) determine the side length of the square, which is ( 7sqrt{2}/2 , mathrm{cm} ).6. Area of the Square: - The area of the square is ( left( frac{7sqrt{2}}{2} right)^2 = frac{49 times 2}{4} = frac{49}{2} , mathrm{cm}^2 ).Thus, the area of square ( mathrm{ABCD} ) is (boxed{dfrac{49}{2}}).]

🔑:【Analysis】It is known that the height on the hypotenuse of an isosceles right triangle is half of the hypotenuse. Therefore, since the area of the isosceles right triangle mathrm{AEF} is 1, its hypotenuse mathrm{EF} is 2. Since the area of rectangle mathrm{EFGH} is 10, mathrm{FG} is 5.Thus, the area of the isosceles right triangle mathrm{BFG} is frac{1}{2} times 5 times 2.5=6.25Therefore, the area of square mathrm{ABCD} is 1 times 2+6.25 times 2+10=24.5

🧠:[<think>Okay, let me try to solve these two problems step by step. I remember that when dealing with square roots, simplifying them first can make the addition and subtraction easier. Let's start with the first one: Problem (1): (3sqrt{3} - sqrt{8} + sqrt{2} - sqrt{27})Alright, so I need to simplify each of the square roots as much as possible. Let's look at each term one by one.First term is (3sqrt{3}). That's already simplified because 3 is a prime number and doesn't have any square factors. So, nothing to do here.Second term is (-sqrt{8}). Hmm, 8 can be broken down into 4*2, right? And since 4 is a perfect square, that can come out of the root. So, (sqrt{8} = sqrt{4*2} = sqrt{4} * sqrt{2} = 2sqrt{2}). Therefore, (-sqrt{8}) becomes (-2sqrt{2}).Third term is (+sqrt{2}). That's already simplified, so we leave it as is.Fourth term is (-sqrt{27}). Let's see, 27 is 9*3, and 9 is a perfect square. So, (sqrt{27} = sqrt{9*3} = sqrt{9} * sqrt{3} = 3sqrt{3}). Therefore, (-sqrt{27}) becomes (-3sqrt{3}).Now let's substitute these simplified forms back into the original expression:(3sqrt{3} - 2sqrt{2} + sqrt{2} - 3sqrt{3})Now, let's combine like terms. The terms with (sqrt{3}) and the terms with (sqrt{2}):Starting with the (sqrt{3}) terms: (3sqrt{3} - 3sqrt{3}). That should be (0), right? Because 3-3=0.Then the (sqrt{2}) terms: (-2sqrt{2} + sqrt{2}). That's like -2 +1 of (sqrt{2}), which is (-sqrt{2}).So putting it all together, the result is (0 - sqrt{2} = -sqrt{2}). Wait, that seems straightforward, but let me check again. Maybe I missed something.Original problem: (3sqrt{3} - sqrt{8} + sqrt{2} - sqrt{27})Simplify each term:- (3sqrt{3}) stays.- (sqrt{8}) becomes (2sqrt{2}), so with the negative sign: (-2sqrt{2})- (+sqrt{2})- (sqrt{27}) becomes (3sqrt{3}), so with the negative sign: (-3sqrt{3})Combined:(3sqrt{3} - 2sqrt{2} + sqrt{2} - 3sqrt{3})Group the like terms:((3sqrt{3} - 3sqrt{3}) + (-2sqrt{2} + sqrt{2}))Calculates to:(0 + (-sqrt{2}) = -sqrt{2})Yep, that seems correct. So problem (1) simplifies to (-sqrt{2}).Now moving on to Problem (2): (sqrt{6} times sqrt{3} + sqrt{2} - 6sqrt{frac{1}{2}})Again, let's simplify each term step by step.First term: (sqrt{6} times sqrt{3}). I remember that multiplying square roots can be combined under a single root: (sqrt{6 times 3} = sqrt{18}). Then, can we simplify (sqrt{18})?Yes, 18 is 9*2, and 9 is a perfect square. So, (sqrt{18} = sqrt{9*2} = 3sqrt{2}). Therefore, the first term simplifies to (3sqrt{2}).Second term is (+sqrt{2}). That's already simplified, so we leave it.Third term is (-6sqrt{frac{1}{2}}). Let's simplify (sqrt{frac{1}{2}}). Remember that (sqrt{frac{1}{2}} = frac{sqrt{1}}{sqrt{2}} = frac{1}{sqrt{2}}). However, it's often rationalized, which means we don't want a square root in the denominator. Multiplying numerator and denominator by (sqrt{2}) gives (frac{sqrt{2}}{2}). So, (sqrt{frac{1}{2}} = frac{sqrt{2}}{2}). Therefore, (-6sqrt{frac{1}{2}} = -6 times frac{sqrt{2}}{2} = -3sqrt{2}).Now substitute back into the expression:(3sqrt{2} + sqrt{2} - 3sqrt{2})Again, let's combine like terms. All terms have (sqrt{2}):(3sqrt{2} - 3sqrt{2} + sqrt{2})First, (3sqrt{2} - 3sqrt{2} = 0), so we have (0 + sqrt{2} = sqrt{2}).Let me verify once more to be sure.Original problem: (sqrt{6} times sqrt{3} + sqrt{2} - 6sqrt{frac{1}{2}})Simplify each term:- (sqrt{6} times sqrt{3} = sqrt{18} = 3sqrt{2})- (sqrt{2}) stays- (-6sqrt{frac{1}{2}} = -6 times frac{sqrt{2}}{2} = -3sqrt{2})Combined: (3sqrt{2} + sqrt{2} - 3sqrt{2})Combine like terms:( (3sqrt{2} - 3sqrt{2}) + sqrt{2} = 0 + sqrt{2} = sqrt{2})Yes, that's correct. So problem (2) simplifies to (sqrt{2}).Therefore, the answers are:(1) (-sqrt{2}) and (2) (sqrt{2}).Final Answer(1) boxed{-sqrt{2}}; (2) boxed{sqrt{2}}</think>To solve the given problems, we simplify each term and combine like terms.Problem (1): (3sqrt{3} - sqrt{8} + sqrt{2} - sqrt{27})1. Simplify each term: - (3sqrt{3}) remains as is. - (sqrt{8} = 2sqrt{2}), so (-sqrt{8} = -2sqrt{2}). - (sqrt{2}) remains as is. - (sqrt{27} = 3sqrt{3}), so (-sqrt{27} = -3sqrt{3}).2. Substitute the simplified terms back into the expression: [ 3sqrt{3} - 2sqrt{2} + sqrt{2} - 3sqrt{3} ]3. Combine like terms: - (sqrt{3}) terms: (3sqrt{3} - 3sqrt{3} = 0) - (sqrt{2}) terms: (-2sqrt{2} + sqrt{2} = -sqrt{2})Result for (1): (-sqrt{2})Problem (2): (sqrt{6} times sqrt{3} + sqrt{2} - 6sqrt{frac{1}{2}})1. Simplify each term: - (sqrt{6} times sqrt{3} = sqrt{18} = 3sqrt{2}) - (sqrt{2}) remains as is. - (sqrt{frac{1}{2}} = frac{sqrt{2}}{2}), so (6sqrt{frac{1}{2}} = 3sqrt{2}), hence (-6sqrt{frac{1}{2}} = -3sqrt{2}).2. Substitute the simplified terms back into the expression: [ 3sqrt{2} + sqrt{2} - 3sqrt{2} ]3. Combine like terms: - (sqrt{2}) terms: (3sqrt{2} - 3sqrt{2} + sqrt{2} = sqrt{2})Result for (2): (sqrt{2}) Final Answer(1) boxed{-sqrt{2}}; (2) boxed{sqrt{2}},<think>Okay, so I have two problems to solve here. Let me start with the first one: 3√3 − √8 + √2 − √27. Hmm, okay. Alright, I remember that when simplifying radicals, you can combine like terms if they have the same radical part. But first, I need to check if any of these radicals can be simplified further. Let's see. Starting with the first term: 3√3. That seems already simplified because 3 is a prime number, so there's no perfect square factor other than 1. Next term is −√8. Let me think about √8. 8 can be broken down into 4*2, and since 4 is a perfect square, √8 is √4*√2, which is 2√2. So, −√8 is equivalent to -2√2. Then the next term is +√2. That's straightforward. And the last term is −√27. Let me simplify √27. 27 is 9*3, and 9 is a perfect square, so √27 is √9*√3, which is 3√3. So, −√27 is -3√3. Now, let me rewrite the original expression with these simplifications:3√3 - 2√2 + √2 - 3√3.Now, let's group the like terms. The terms with √3 and the terms with √2.For the √3 terms: 3√3 - 3√3. That should cancel each other out, right? Because 3√3 minus 3√3 is 0. Then the √2 terms: -2√2 + √2. Hmm, combining these, that would be (-2 + 1)√2, which is -1√2, or just -√2. So putting it all together, the √3 terms cancel out, and the √2 terms combine to -√2. So the result of the first problem is -√2. Let me check again to make sure I didn't make a mistake.Original problem: 3√3 − √8 + √2 − √27. Simplify radicals:- √8 = 2√2, so -√8 = -2√2- √27 = 3√3, so -√27 = -3√3So, 3√3 - 2√2 + √2 - 3√3. Yes, grouping the √3 terms: 3√3 - 3√3 = 0. Then the √2 terms: -2√2 + √2 = -√2. Yep, that seems right. So the answer to part (1) is -√2.Moving on to the second problem: √6 × √3 + √2 − 6√(1/2). Let's tackle this step by step. First, √6 multiplied by √3. I remember that √a × √b = √(a*b). So √6 × √3 = √(6*3) = √18. Then, √18 can be simplified. 18 is 9*2, and 9 is a perfect square, so √18 = √9*√2 = 3√2. So the first term simplifies to 3√2. Next term is +√2, which is already simplified. Then we have −6√(1/2). Let me think about √(1/2). That can be written as √1 / √2 = 1/√2. However, it's often rationalized, so 1/√2 is equivalent to √2/2. Therefore, √(1/2) = √2/2. So, −6√(1/2) is -6*(√2/2). Simplifying that, -6 divided by 2 is -3, so this term becomes -3√2. Now, let's rewrite the entire expression with these simplifications:3√2 + √2 - 3√2.Now, let's combine like terms. All terms have √2.First term: 3√2, second term: +√2 (which is 1√2), third term: -3√2.Adding them up: (3 + 1 - 3)√2 = (1)√2 = √2.Wait, that seems too straightforward. Let me verify each step again.Original expression: √6×√3 + √2 − 6√(1/2). First multiplication: √6×√3 = √(6×3) = √18 = 3√2. Correct.Second term: √2. Still √2.Third term: 6√(1/2). Let's confirm. √(1/2) is indeed √1 divided by √2, which is 1/√2. Rationalizing the denominator: multiply numerator and denominator by √2, so (1*√2)/(√2*√2) = √2/2. Therefore, 6*(√2/2) = 3√2. But since it's subtracted, it's -3√2. So putting it all together: 3√2 + √2 - 3√2. 3√2 - 3√2 cancels out, leaving √2. Yep, that's correct. So the answer to part (2) is √2.Wait, but let me check once more. Maybe I made a mistake in simplifying the third term. Let's take the term -6√(1/2). Another way to write √(1/2) is (1/2)^(1/2) which is 1/(2)^(1/2). Rationalizing gives √2/2, so multiplying by 6 gives 6*(√2/2) = 3√2. So yes, -6√(1/2) = -3√2. Correct. Therefore, combining all terms:3√2 (from the first term) + √2 (second term) - 3√2 (third term) = (3√2 - 3√2) + √2 = 0 + √2 = √2. Yep, that's definitely right. So part (2) is √2.So summarizing:Problem (1) simplifies to -√2.Problem (2) simplifies to √2.I think that's all. Let me just check once more for any possible errors.First problem steps:3√3 - √8 + √2 - √27√8 = 2√2, so -√8 = -2√2√27 = 3√3, so -√27 = -3√3Thus, 3√3 - 2√2 + √2 - 3√3Grouping: (3√3 - 3√3) + (-2√2 + √2) = 0 - √2 = -√2. Correct.Second problem steps:√6×√3 = √18 = 3√2√2 is just √2-6√(1/2) = -6*(√2/2) = -3√2Combined: 3√2 + √2 - 3√2 = (3√2 - 3√2) + √2 = √2. Correct.Alright, I think I'm confident with these answers.Final Answer(1) boxed{-sqrt{2}} (2) boxed{sqrt{2}}</think>To solve the given problems, we simplify each term and combine like terms where possible. Problem (1): (3sqrt{3} - sqrt{8} + sqrt{2} - sqrt{27})1. Simplify the radicals: - (sqrt{8} = 2sqrt{2}), so (-sqrt{8} = -2sqrt{2}) - (sqrt{27} = 3sqrt{3}), so (-sqrt{27} = -3sqrt{3})2. Substitute the simplified terms back into the expression: [ 3sqrt{3} - 2sqrt{2} + sqrt{2} - 3sqrt{3} ]3. Combine like terms: - Terms with (sqrt{3}): (3sqrt{3} - 3sqrt{3} = 0) - Terms with (sqrt{2}): (-2sqrt{2} + sqrt{2} = -sqrt{2})Thus, the result for problem (1) is (boxed{-sqrt{2}}). Problem (2): (sqrt{6} times sqrt{3} + sqrt{2} - 6sqrt{frac{1}{2}})1. Simplify the radicals and products: - (sqrt{6} times sqrt{3} = sqrt{18} = 3sqrt{2}) - (sqrt{frac{1}{2}} = frac{sqrt{2}}{2}), so (6sqrt{frac{1}{2}} = 6 times frac{sqrt{2}}{2} = 3sqrt{2})2. Substitute the simplified terms back into the expression: [ 3sqrt{2} + sqrt{2} - 3sqrt{2} ]3. Combine like terms: - Terms with (sqrt{2}): (3sqrt{2} - 3sqrt{2} + sqrt{2} = sqrt{2})Thus, the result for problem (2) is (boxed{sqrt{2}}).]

🔑:For the given problems, let's break them down step by step: Problem 1:Calculate 3sqrt{3}-sqrt{8}+sqrt{2}-sqrt{27}.# Solution:We start by simplifying each term in the expression:- sqrt{8} can be simplified to 2sqrt{2} because sqrt{8} = sqrt{4 cdot 2} = 2sqrt{2}.- sqrt{27} can be simplified to 3sqrt{3} because sqrt{27} = sqrt{9 cdot 3} = 3sqrt{3}.So, the original expression becomes:[3sqrt{3} - 2sqrt{2} + sqrt{2} - 3sqrt{3}.]Now, we combine like terms:- The 3sqrt{3} and -3sqrt{3} cancel each other out.- -2sqrt{2} + sqrt{2} simplifies to -sqrt{2}.Therefore, the simplified expression is:[-sqrt{2}.]So, the final answer for problem (1) is boxed{-sqrt{2}}. Problem 2:Calculate sqrt{6} times sqrt{3} + sqrt{2} - 6sqrt{frac{1}{2}}.# Solution:We simplify each term in the expression:- sqrt{6} times sqrt{3} can be simplified to sqrt{18} because sqrt{6} times sqrt{3} = sqrt{6 cdot 3} = sqrt{18}, which further simplifies to 3sqrt{2} because sqrt{18} = sqrt{9 cdot 2} = 3sqrt{2}.- 6sqrt{frac{1}{2}} simplifies to 6 times frac{sqrt{2}}{2} because sqrt{frac{1}{2}} = frac{sqrt{2}}{2}.So, the original expression becomes:[3sqrt{2} + sqrt{2} - 6 times frac{sqrt{2}}{2}.]Simplifying further:- 6 times frac{sqrt{2}}{2} simplifies to 3sqrt{2}.Therefore, the expression simplifies to:[3sqrt{2} + sqrt{2} - 3sqrt{2}.]Combining like terms:- The 3sqrt{2} and -3sqrt{2} cancel each other out, leaving us with sqrt{2}.So, the final answer for problem (2) is boxed{sqrt{2}}.

🧠:[<think>Okay, let's see. I need to find the product λ₁λ₂ where the vector CO is expressed as a linear combination of vectors CA and CB with coefficients λ₁ and λ₂ respectively. The triangle ABC is isosceles with AB = AC and angle CAB is 30 degrees. O is the circumcenter of triangle ABC. Hmm, okay.First, let me recall some properties of circumcenters. The circumcenter is the intersection point of the perpendicular bisectors of the sides of the triangle. Since ABC is isosceles with AB = AC, the triangle is symmetric with respect to the altitude from A to BC. Therefore, the circumcenter should lie along this altitude. Wait, but in an isosceles triangle, the altitude, median, and angle bisector from the apex (vertex A in this case) all coincide. So, the circumcenter must lie somewhere along this line. But depending on the angles, it might not be at the centroid or orthocenter. Let me think.Given that angle CAB is 30 degrees, and AB = AC, so triangle ABC is isosceles with apex at A. Therefore, the base angles at B and C will be equal. The sum of angles in a triangle is 180 degrees, so angles at B and C are (180 - 30)/2 = 75 degrees each. So angles at B and C are 75 degrees each.Since O is the circumcenter, it is equidistant from all three vertices. In an isosceles triangle, the circumcenter lies along the altitude from the apex. Wait, but in some cases, if the triangle is acute, the circumcenter is inside the triangle; if it's right-angled, it's at the midpoint of the hypotenuse; if it's obtuse, it's outside. Here, all angles are less than 90 degrees (30, 75, 75), so the triangle is acute. Therefore, the circumcenter is inside the triangle, lying along the altitude from A to BC.Let me try to visualize this. Let me set up coordinates to model the problem. Coordinates might help. Let me place point A at the origin (0,0) for simplicity. Since AB = AC, and angle at A is 30 degrees, I can place points B and C such that AB = AC = some length, say 1 unit for simplicity.Wait, but if I place A at (0,0), then let me orient the triangle so that the altitude from A to BC is along the y-axis. So, points B and C will be symmetric with respect to the y-axis. Let me think.Since angle CAB is 30 degrees, and AB = AC = 1, then coordinates of B and C can be found using trigonometry. Let me set AB = AC = 1. Then, point B can be at (sin(15°), cos(15°)) and point C at (-sin(15°), cos(15°))? Wait, no. Wait, angle at A is 30 degrees, so the angle between AB and AC is 30 degrees. If AB and AC have length 1, then coordinates can be:Let’s place point A at (0,0). Let’s place point B at (1, 0). Then, since angle CAB is 30 degrees, and AC = AB = 1, point C should be at (cos(30°), sin(30°)) which is (√3/2, 1/2). Wait, but then AB = 1, AC = 1, and angle between them is 30 degrees. That makes triangle ABC with AB = AC = 1 and angle at A of 30 degrees. Then, BC can be computed using the law of cosines: BC² = AB² + AC² - 2 AB AC cos(30°) = 1 + 1 - 2*1*1*(√3/2) = 2 - √3. So BC = sqrt(2 - √3). Hmm, okay.But in this coordinate system, the circumcenter O is the intersection of the perpendicular bisectors of the sides. Let me compute coordinates of O.First, let's compute coordinates of points:A: (0,0)B: (1,0)C: (√3/2, 1/2) as per angle 30 degrees.Wait, wait, if angle at A is 30 degrees, then yes, that would be the coordinate of C. Let me verify.Yes, because if you have point A at (0,0), AB along the x-axis to (1,0), and angle CAB = 30 degrees, then AC is another arm of the angle, also length 1, so coordinates of C would be (cos30°, sin30°) = (√3/2, 1/2). So BC is from (1,0) to (√3/2, 1/2). Then, the midpoint of AB is (0.5, 0), the midpoint of AC is (√3/4, 1/4), and the midpoint of BC is ((1 + √3/2)/2, (0 + 1/2)/2) = ((2 + √3)/4, 1/4).Now, to find the circumcenter, we can find the intersection of the perpendicular bisectors of two sides.Let me compute the perpendicular bisector of AB first. AB is from (0,0) to (1,0). The midpoint is (0.5, 0). The slope of AB is 0, so the perpendicular bisector is vertical line x = 0.5.Now, let's compute the perpendicular bisector of AC. AC is from (0,0) to (√3/2, 1/2). The midpoint of AC is (√3/4, 1/4). The slope of AC is (1/2 - 0)/(√3/2 - 0) = (1/2)/(√3/2) = 1/√3. Therefore, the perpendicular bisector has slope -√3.So the equation of the perpendicular bisector of AC is y - 1/4 = -√3(x - √3/4). Let me simplify that:y = -√3 x + (√3 * √3)/4 + 1/4Since √3 * √3 = 3, so:y = -√3 x + 3/4 + 1/4 = -√3 x + 1So the perpendicular bisector of AC is y = -√3 x + 1.We already have the perpendicular bisector of AB as x = 0.5. So the intersection point O is at x = 0.5, plug into the equation of the other bisector:y = -√3*(0.5) + 1 = - (√3)/2 + 1 ≈ 1 - 0.866 ≈ 0.134.So coordinates of O are (0.5, 1 - (√3)/2).Wait, but let's check if this is correct. Let me verify with the perpendicular bisector of BC as well.Midpoint of BC is ((1 + √3/2)/2, (0 + 1/2)/2) = ((2 + √3)/4, 1/4). The slope of BC is (1/2 - 0)/(√3/2 - 1) = (1/2)/ ( (√3 - 2)/2 ) = 1/(√3 - 2) = multiply numerator and denominator by (√3 + 2):(√3 + 2)/( (√3)^2 - (2)^2 ) = (√3 + 2)/(3 - 4) = (√3 + 2)/(-1) = -√3 - 2Therefore, the slope of BC is -√3 - 2. Therefore, the perpendicular bisector slope is reciprocal and opposite sign: 1/(√3 + 2). But rationalizing:1/(√3 + 2) * (√3 - 2)/(√3 - 2) = (√3 - 2)/(3 - 4) = (√3 - 2)/(-1) = 2 - √3So the perpendicular bisector of BC has slope 2 - √3. Let's write the equation:Using midpoint ((2 + √3)/4, 1/4):y - 1/4 = (2 - √3)(x - (2 + √3)/4 )Let me see if point O (0.5, 1 - √3/2) lies on this line.Plug x = 0.5 = 2/4, so x - (2 + √3)/4 = (2/4 - (2 + √3)/4 ) = (-√3)/4Then, y - 1/4 = (2 - √3)( -√3 /4 )Compute RHS:(2 - √3)( -√3 ) /4 = (-2√3 + 3)/4So y = 1/4 + (-2√3 + 3)/4 = (1 - 2√3 + 3)/4 = (4 - 2√3)/4 = (2 - √3)/2 ≈ (2 - 1.732)/2 ≈ 0.268/2 ≈ 0.134, which matches the y-coordinate of O: 1 - √3/2 ≈ 1 - 0.866 ≈ 0.134. So yes, that checks out. Therefore, O is correctly found at (0.5, 1 - √3/2).Okay, so coordinates:A: (0,0)B: (1,0)C: (√3/2, 1/2)O: (0.5, 1 - √3/2 )Now, we need to express vector CO as a linear combination of vectors CA and CB.First, let's compute vector CO, which is O - C.Coordinates of O: (0.5, 1 - √3/2 )Coordinates of C: (√3/2, 1/2 )Therefore, vector CO = O - C = (0.5 - √3/2, 1 - √3/2 - 1/2 ) = (0.5 - √3/2, 1/2 - √3/2 )Similarly, vectors CA and CB.Vector CA = A - C = (0 - √3/2, 0 - 1/2 ) = (-√3/2, -1/2 )Vector CB = B - C = (1 - √3/2, 0 - 1/2 ) = (1 - √3/2, -1/2 )We need to find λ₁ and λ₂ such that:CO = λ₁ CA + λ₂ CBSo, component-wise:x-component: 0.5 - √3/2 = λ₁*(-√3/2) + λ₂*(1 - √3/2 )y-component: 1/2 - √3/2 = λ₁*(-1/2) + λ₂*(-1/2 )So we have a system of two equations:1) (0.5 - √3/2) = (-√3/2) λ₁ + (1 - √3/2 ) λ₂2) (1/2 - √3/2) = (-1/2) λ₁ + (-1/2 ) λ₂Let me write these equations more neatly:Equation 1:(-√3/2) λ₁ + (1 - √3/2 ) λ₂ = 0.5 - √3/2Equation 2:(-1/2) λ₁ + (-1/2 ) λ₂ = 1/2 - √3/2Let me denote equation 2 as:- (1/2)(λ₁ + λ₂) = (1 - √3)/2Multiply both sides by -2:λ₁ + λ₂ = -(1 - √3 ) = √3 - 1So equation 2 simplifies to:λ₁ + λ₂ = √3 - 1 ...(2a)Now, let's work with equation 1. Let's write equation 1:(-√3/2) λ₁ + (1 - √3/2 ) λ₂ = (1 - √3)/2Multiply both sides by 2 to eliminate denominators:-√3 λ₁ + (2 - √3 ) λ₂ = 1 - √3 ...(1a)Now, from equation (2a), we have λ₁ = (√3 - 1 ) - λ₂. Substitute this into equation (1a):-√3 [ (√3 - 1 ) - λ₂ ] + (2 - √3 ) λ₂ = 1 - √3Expand the first term:-√3 (√3 - 1 ) + √3 λ₂ + (2 - √3 ) λ₂ = 1 - √3Compute -√3 (√3 - 1 ):-√3 * √3 + √3 * 1 = -3 + √3Then, combine the λ₂ terms:√3 λ₂ + (2 - √3 ) λ₂ = [ √3 + 2 - √3 ] λ₂ = 2 λ₂So the equation becomes:(-3 + √3 ) + 2 λ₂ = 1 - √3Bring -3 + √3 to the right side:2 λ₂ = 1 - √3 + 3 - √3 = 4 - 2√3Thus, λ₂ = (4 - 2√3)/2 = 2 - √3Then, from equation (2a):λ₁ + λ₂ = √3 - 1λ₁ = √3 - 1 - λ₂ = √3 - 1 - (2 - √3 ) = √3 -1 -2 + √3 = 2√3 -3Therefore, λ₁ = 2√3 - 3 and λ₂ = 2 - √3Now, we need to compute the product λ₁λ₂.Compute (2√3 - 3)(2 - √3 )Multiply these out:First, expand the product:= 2√3 * 2 + 2√3 * (-√3 ) + (-3)*2 + (-3)*(-√3 )= 4√3 - 2*(√3 )² -6 + 3√3Simplify:√3 squared is 3, so:= 4√3 - 2*3 -6 + 3√3= 4√3 -6 -6 + 3√3Combine like terms:(4√3 + 3√3 ) + (-6 -6 ) = 7√3 -12Wait, but that gives 7√3 -12. But the problem says to find λ₁λ₂. Hmm, let me check the calculations again, maybe I made a mistake.Wait, let's recalculate the product:(2√3 - 3)(2 - √3 )Multiply term by term:First term: 2√3 * 2 = 4√3Second term: 2√3 * (-√3 ) = -2*(√3 )² = -2*3 = -6Third term: -3 * 2 = -6Fourth term: -3 * (-√3 ) = 3√3Now, add them all up:4√3 -6 -6 + 3√3 = (4√3 + 3√3 ) + (-6 -6 ) = 7√3 -12Yes, that's correct. So the product λ₁λ₂ is 7√3 -12. Hmm, but this seems a bit complicated. Let me check if my values for λ₁ and λ₂ are correct.Wait, let's verify the vector CO.CO = O - C = (0.5 - √3/2, 1 - √3/2 - 1/2 ) = (0.5 - √3/2, 0.5 - √3/2 )Vector CA = A - C = (-√3/2, -1/2 )Vector CB = B - C = (1 - √3/2, -1/2 )So CO = λ₁ CA + λ₂ CBSo, compute λ₁ CA + λ₂ CB:λ₁*(-√3/2, -1/2 ) + λ₂*(1 - √3/2, -1/2 )= ( -λ₁√3/2 + λ₂(1 - √3/2 ), -λ₁/2 - λ₂/2 )Compare with CO = (0.5 - √3/2, 0.5 - √3/2 )So the x-component:-λ₁√3/2 + λ₂(1 - √3/2 ) = 0.5 - √3/2The y-component:-λ₁/2 - λ₂/2 = 0.5 - √3/2We found λ₁ = 2√3 -3 and λ₂ = 2 - √3.Let's check the y-component first:- (2√3 -3)/2 - (2 - √3 )/2 = [ -2√3 +3 -2 + √3 ] / 2 = [ (-2√3 + √3 ) + (3 -2) ] /2 = (-√3 +1)/2But 0.5 - √3/2 = (1 - √3)/2. So that matches. Okay, y-component is correct.Now x-component:-λ₁√3/2 + λ₂(1 - √3/2 )= - (2√3 -3 )√3 /2 + (2 - √3 )(1 - √3/2 )First compute - (2√3 -3 )√3 /2:= - [ 2*3 - 3√3 ] /2 = - [6 - 3√3 ] /2 = (-6 + 3√3 ) /2Now compute (2 - √3 )(1 - √3/2 )= 2*1 + 2*(-√3/2 ) - √3 *1 + √3*(√3/2 )= 2 - √3 - √3 + (3)/2= 2 - 2√3 + 3/2= (4/2 - 2√3 + 3/2 ) = (7/2 - 2√3 )So total x-component:(-6 + 3√3 )/2 + 7/2 - 2√3= [ (-6 + 3√3 ) +7 -4√3 ] /2= (1 - √3 )/2Which is equal to 0.5 - √3/2. Correct. So the x-component is also correct. Therefore, the values of λ₁ and λ₂ are correct.Therefore, the product λ₁λ₂ = (2√3 -3)(2 - √3 ) = 7√3 -12. Wait, but 7√3 is approximately 12.12, so 7√3 -12 ≈ 0.12. Hmm, but the problem asks for λ₁λ₂, and it's possible this is the answer.But let me check again if there's a simpler form or if I made a calculation mistake.Wait, let me compute (2√3 -3)(2 - √3 )Multiply them:= 2√3 *2 + 2√3*(-√3 ) -3*2 + (-3)*(-√3 )= 4√3 - 2*3 -6 +3√3= 4√3 -6 -6 +3√3= 7√3 -12Yes, same as before. So 7√3 -12. Hmm.But maybe we can rationalize or express differently? Let's see.Alternatively, perhaps the problem expects an exact form, which would be 7√3 -12. However, checking if this can be written as (something) squared or another expression. Alternatively, maybe I made an error in setting up coordinates.Wait, let me check if setting point A at (0,0) is okay. Alternatively, perhaps choosing a different coordinate system would make the problem easier.Alternatively, since the triangle is isosceles with AB = AC, maybe place the triangle such that BC is horizontal, and A is at the top. Let me try another approach.Let me consider triangle ABC with AB = AC, angle at A is 30 degrees. Let me denote AB = AC = 1. Then, by the Law of Sines, the circumradius R is given by a/(2 sin A), where a is the length of BC.First, let me compute BC.Using Law of Cosines: BC² = AB² + AC² - 2 AB AC cos(angle A) = 1 +1 - 2*1*1*cos30° = 2 - 2*(√3/2) = 2 - √3.Thus, BC = sqrt(2 - √3 )Then, the circumradius R = BC/(2 sin A) = sqrt(2 - √3 )/(2 sin 30° ) = sqrt(2 - √3 )/(2*(1/2)) = sqrt(2 - √3 )/1 = sqrt(2 - √3 )Alternatively, in this coordinate system, the circumradius is the distance from O to any vertex. Let's check distance from O to C.Coordinates of O: (0.5, 1 - √3/2 )Coordinates of C: (√3/2, 1/2 )Distance CO:sqrt( (0.5 - √3/2 )² + (1 - √3/2 -1/2 )² )= sqrt( (0.5 - √3/2 )² + (0.5 - √3/2 )² )= sqrt( 2*(0.5 - √3/2 )² )= sqrt(2)*(0.5 - √3/2 )Compute 0.5 - √3/2 ≈ 0.5 - 0.866 ≈ -0.366, but squared is positive. So (0.5 - √3/2 )² = ( (1 - √3 )/2 )² = (1 - 2√3 +3 )/4 = (4 - 2√3 )/4 = (2 - √3 )/2Thus, sqrt(2*(2 - √3 )/2 ) = sqrt(2 - √3 ) which matches the earlier result. So that's correct.But maybe there's a different approach to find λ₁ and λ₂ without coordinates. Let me think.Since in vector problems, sometimes barycentric coordinates or other vector methods can be used.Given that CO = λ₁ CA + λ₂ CB, and we need to find λ₁ and λ₂.Alternatively, express vectors in terms of position vectors. Let me denote position vectors of A, B, C as vectors a, b, c. Then, since O is the circumcenter, we can express its position vector o in terms of a, b, c.But in general, the formula for circumcenter in terms of coordinates is straightforward, but in vector terms, it can be more complex. However, since the triangle is isosceles, maybe there is symmetry to exploit.Given that AB = AC, and O lies on the altitude from A, which is also the median and angle bisector. Therefore, in vector terms, vector OA is along the altitude. Wait, but in our coordinate system, O is at (0.5, 1 - √3/2 ), so the vector from C to O is (0.5 - √3/2, 1 - √3/2 -1/2 ) = (0.5 - √3/2, 0.5 - √3/2 )But vectors CA and CB are vectors from C to A and C to B. So CA = A - C, CB = B - C.Alternatively, since the problem is to express CO as a combination of CA and CB, we can use the vectors CA and CB as a basis and express CO in terms of them.But given the coordinates, we solved the linear system and found λ₁ and λ₂. The product is 7√3 -12, which is approximately 0.12, but perhaps the answer is better expressed as is.Wait, but let me check if 7√3 -12 can be simplified. Let's see:7√3 -12 ≈ 7*1.732 -12 ≈ 12.124 -12 = 0.124, so it's a small positive number. Maybe the answer is supposed to be in terms of exact radicals, so 7√3 -12.But let me check if there's a different approach that could lead to a simpler answer.Alternatively, consider the ratio of areas or something, but given the problem involves vectors and linear combinations, the coordinate approach seems appropriate.Alternatively, use trigonometric relations in the vector expressions.But given the time I've spent and verified the coordinates, the solution seems correct. Therefore, the product λ₁λ₂ is 7√3 -12.Wait, but let me check once more for calculation errors. When multiplying (2√3 -3)(2 - √3 ):First term: 2√3 *2 =4√3Second term: 2√3*(-√3 ) = -2*3= -6Third term: -3*2= -6Fourth term: -3*(-√3 )=3√3So total:4√3 -6 -6 +3√3=7√3 -12. Yes, correct.Therefore, the answer is 7√3 -12, which can be written as boxed{7sqrt{3} -12}But wait, checking the problem statement again: it says "put your final answer within boxed{}". Maybe the answer is rationalized differently or expressed as a fraction? Wait, 7√3 -12 is already simplified, so probably this is the answer.Alternatively, maybe I miscalculated λ₁ and λ₂.Wait, let's verify λ₁ = 2√3 -3 and λ₂ =2 -√3.Let me compute λ₁ + λ₂ =2√3 -3 +2 -√3= (2√3 -√3 ) + ( -3 +2 )= √3 -1, which matches equation (2a). So that's correct.So the product is indeed (2√3 -3)(2 -√3 )=7√3 -12.Therefore, the answer should be boxed{7sqrt{3} - 12}.Final Answerboxed{7sqrt{3} - 12}</think>Given that ( O ) is the circumcenter of (triangle ABC) with ( AB = AC ) and (angle CAB = 30^circ). We need to find the product (lambda_1 lambda_2) where (overrightarrow{CO} = lambda_1 overrightarrow{CA} + lambda_2 overrightarrow{CB}).1. Coordinates Setup: - Place ( A ) at ((0, 0)), ( B ) at ((1, 0)), and ( C ) at ((cos 30^circ, sin 30^circ) = left(frac{sqrt{3}}{2}, frac{1}{2}right)).2. Finding Circumcenter ( O ): - The perpendicular bisector of ( AB ) is ( x = 0.5 ). - The perpendicular bisector of ( AC ) has the equation ( y = -sqrt{3}x + 1 ). - Solving these, the circumcenter ( O ) is found at (left(0.5, 1 - frac{sqrt{3}}{2}right)).3. Vectors Calculation: - (overrightarrow{CO} = O - C = left(0.5 - frac{sqrt{3}}{2}, 1 - frac{sqrt{3}}{2} - frac{1}{2}right) = left(0.5 - frac{sqrt{3}}{2}, 0.5 - frac{sqrt{3}}{2}right)). - (overrightarrow{CA} = A - C = left(-frac{sqrt{3}}{2}, -frac{1}{2}right)). - (overrightarrow{CB} = B - C = left(1 - frac{sqrt{3}}{2}, -frac{1}{2}right)).4. Solving Linear System: - Set up the system of equations for (lambda_1) and (lambda_2): [ begin{cases} -frac{sqrt{3}}{2} lambda_1 + left(1 - frac{sqrt{3}}{2}right) lambda_2 = 0.5 - frac{sqrt{3}}{2} -frac{1}{2} lambda_1 - frac{1}{2} lambda_2 = 0.5 - frac{sqrt{3}}{2} end{cases} ] - Solving these, we find (lambda_1 = 2sqrt{3} - 3) and (lambda_2 = 2 - sqrt{3}).5. Product Calculation: - Compute (lambda_1 lambda_2): [ (2sqrt{3} - 3)(2 - sqrt{3}) = 4sqrt{3} - 6 - 6 + 3sqrt{3} = 7sqrt{3} - 12 ]Thus, the product (lambda_1 lambda_2) is (boxed{7sqrt{3} - 12}).,<think>Okay, let me try to figure out this problem step by step. So, we have triangle ABC where AB equals AC, and the angle at A (angle CAB) is 30 degrees. O is the circumcenter of triangle ABC. We need to express vector CO as a linear combination of vectors CA and CB, specifically CO = λ₁CA + λ₂CB, and then find the product λ₁λ₂. First, since AB equals AC, triangle ABC is isosceles with AB = AC. That means angles at B and C are equal. But angle at A is 30 degrees. Let me sketch the triangle to visualize. Vertex A is at the top, with AB and AC equal in length. The base is BC, and since the triangle is isosceles with apex angle 30 degrees, the base angles at B and C should each be (180 - 30)/2 = 75 degrees. So angles at B and C are 75 degrees each.Now, O is the circumcenter of triangle ABC. The circumcenter is the intersection point of the perpendicular bisectors of the sides. In an isosceles triangle, the perpendicular bisector of the base (BC) should also be the altitude from the apex A. Wait, but in this case, since AB = AC, the triangle is symmetric along the altitude from A to BC. Therefore, the circumcenter should lie along this altitude. Hmm, is that correct? Wait, the circumcenter in an isosceles triangle does lie along the altitude from the apex to the base. So in this case, O should lie along the altitude from A to BC.But let me verify that. In any triangle, the circumcenter is located at the intersection of the perpendicular bisectors. For triangle ABC, since AB = AC, the perpendicular bisector of BC will coincide with the altitude from A, because in an isosceles triangle, the altitude, median, and perpendicular bisector from the apex all coincide. Therefore, the circumcenter must lie along that altitude. So, O is somewhere along the line from A perpendicular to BC.However, depending on whether the triangle is acute or obtuse, the circumcenter can be inside or outside the triangle. Given that the angles at B and C are 75 degrees, which are acute, and angle at A is 30 degrees, also acute. Therefore, the triangle is acute, so the circumcenter lies inside the triangle. So, O is inside the triangle, along the altitude from A to BC.But how do we find the exact position of O? Maybe we can use coordinates. Let's assign coordinates to the points to make this easier.Let me place point A at the origin (0,0) for simplicity. Since the triangle is isosceles with AB = AC and angle at A is 30 degrees, let's set up points B and C such that AB = AC = some length. Let's choose AB = AC = 1 for simplicity. Then, coordinates of B and C can be determined using the angle at A.Since angle CAB is 30 degrees, and AB = AC = 1, we can place point B at (1, 0). Then point C would be at (cos 30°, sin 30°) because the angle between AB and AC is 30 degrees. Wait, but AB is from A(0,0) to B(1,0), so AC would make a 30-degree angle with AB. Therefore, point C would be at (cos 30°, sin 30°) = (√3/2, 1/2). But wait, is that correct?Wait, if AB is along the x-axis from (0,0) to (1,0), then AC is making a 30-degree angle with AB. So, starting from point A(0,0), moving at 30 degrees from the x-axis, with length AC = 1. Therefore, the coordinates of C would be (cos 30°, sin 30°) = (√3/2, 1/2). But then AB is from (0,0) to (1,0), and AC is to (√3/2, 1/2). Then BC would be the distance between (1,0) and (√3/2, 1/2). Let me compute that distance.Distance BC: sqrt[(1 - √3/2)^2 + (0 - 1/2)^2] = sqrt[(1 - √3/2)^2 + (1/2)^2]First, expand (1 - √3/2)^2: 1 - √3 + (3/4) = (1 + 3/4) - √3 = (7/4) - √3. Wait, that doesn't seem right. Wait, (a - b)^2 = a² - 2ab + b². So, (1 - √3/2)^2 = 1² - 2*(1)*(√3/2) + (√3/2)^2 = 1 - √3 + (3/4) = (1 + 3/4) - √3 = 7/4 - √3. Then adding (1/2)^2 = 1/4. So total inside the sqrt is (7/4 - √3) + 1/4 = (8/4 - √3) = 2 - √3. Therefore, BC = sqrt(2 - √3). Hmm, interesting. So BC has length sqrt(2 - √3). But maybe assigning AB = AC = 1 complicates things. Alternatively, maybe we can set the length of BC to 1 or another value. Wait, but since we need to find the coefficients λ₁ and λ₂ in the vector equation, which are ratios, maybe the actual lengths won't matter, and the coordinates can be scaled. Let me check.Alternatively, perhaps it's better to assign coordinates such that point A is at (0,0), point B at (1,0), and point C at (cos 30°, sin 30°). Then, find the circumcenter O in this coordinate system, then express vector CO as a combination of CA and CB.Alternatively, maybe we can place the triangle in a coordinate system where BC is horizontal for simplicity. Wait, but the triangle is isosceles with AB = AC, so BC is the base. If we place BC horizontally, then point A is somewhere above the x-axis.Let me try this approach. Let's place points B and C on the x-axis, symmetric with respect to the y-axis. Let’s set point B at (-b, 0), point C at (b, 0), so BC is centered at the origin. Then point A will be at (0, h), where h is the height of the triangle. Since AB = AC, which is given. The angle at A is 30 degrees. Let’s compute coordinates accordingly.Given angle at A is 30 degrees. In triangle ABC, angle at A is 30°, sides AB = AC. The coordinates would be: A(0, h), B(-b, 0), C(b, 0). Then AB = AC = sqrt(b² + h²). The angle at A is 30°, so using the law of cosines in triangle ABC:BC² = AB² + AC² - 2*AB*AC*cos(angle at A). But AB = AC, so BC² = 2*AB² - 2*AB²*cos(30°). But BC is the distance between B and C, which is 2b. Therefore:(2b)² = 2*(b² + h²) - 2*(b² + h²)*cos(30°)Simplify:4b² = 2(b² + h²)(1 - cos(30°))Divide both sides by 2:2b² = (b² + h²)(1 - cos(30°))Let’s compute 1 - cos(30°). Cos 30° is √3/2, so 1 - √3/2 ≈ 1 - 0.866 ≈ 0.134. Hmm, not sure if this helps yet.Alternatively, maybe we can use the definition of the angle at A. The angle at A is between vectors AB and AC. The angle between vectors AB and AC is 30 degrees. Since in our coordinate system, AB is from A(0, h) to B(-b, 0), and AC is from A(0, h) to C(b, 0). The vectors AB and AC are (-b, -h) and (b, -h), respectively. The angle between them is 30 degrees. The formula for the angle θ between two vectors u and v is:cosθ = (u · v)/(|u| |v|)Here, u = AB = (-b, -h), v = AC = (b, -h)Dot product u · v = (-b)(b) + (-h)(-h) = -b² + h²|u| = |v| = sqrt(b² + h²)So cosθ = (-b² + h²)/(b² + h²)Given θ = 30°, so cos30° = (√3)/2 = (-b² + h²)/(b² + h²)Let’s solve for h in terms of b.(√3)/2 = (h² - b²)/(h² + b²)Multiply both sides by (h² + b²):(√3)/2 (h² + b²) = h² - b²Multiply both sides by 2:√3 (h² + b²) = 2h² - 2b²Bring all terms to left side:√3 h² + √3 b² - 2h² + 2b² = 0Factor terms:h² (√3 - 2) + b² (√3 + 2) = 0Solve for h²:h² = - [b² (√3 + 2)] / (√3 - 2)Multiply numerator and denominator by (√3 + 2) to rationalize denominator:h² = - [b² (√3 + 2)(√3 + 2)] / [ (√3 - 2)(√3 + 2) ]Denominator: (√3)^2 - (2)^2 = 3 - 4 = -1So,h² = - [b² ( (√3 + 2)^2 ) ] / (-1) = b² ( (√3 + 2)^2 )Therefore,h = b (√3 + 2 )Since h is a length, positive.So, h = b (2 + √3)Therefore, in this coordinate system, point A is at (0, b(2 + √3)), B at (-b, 0), C at (b, 0).Now, let's choose a specific value for b to simplify calculations. Let’s set b = 1. Then, coordinates are:A(0, 2 + √3), B(-1, 0), C(1, 0)This seems manageable.So, points:A: (0, 2 + √3)B: (-1, 0)C: (1, 0)Now, we need to find the circumcenter O of triangle ABC.In coordinate geometry, the circumcenter is the intersection of the perpendicular bisectors of the sides.Given the coordinates, let's compute the perpendicular bisectors.First, find the midpoint and slope of two sides, then find the equations of the perpendicular bisectors, and then find their intersection point O.Let's do this step by step.First, let's find the midpoint and slope of AB and AC, but wait, since the triangle is isosceles, and we already have coordinates, perhaps computing the perpendicular bisector of BC and another side.But let's choose sides AB and BC.Wait, the triangle is isosceles with AB = AC, so sides AB and AC are equal. Let's find the perpendicular bisector of BC and the perpendicular bisector of AB.Midpoint of BC: Since B is (-1, 0) and C is (1, 0), midpoint is (0, 0). The slope of BC is (0 - 0)/(1 - (-1)) = 0/2 = 0. So, the perpendicular bisector of BC is vertical line through midpoint (0,0), but since BC is horizontal, its perpendicular bisector is vertical. Wait, the slope of BC is 0, so the perpendicular bisector is vertical, undefined slope. Therefore, the perpendicular bisector of BC is the line x = 0 (the y-axis). But wait, the midpoint is (0,0), so the perpendicular bisector is x = 0, which is the y-axis.But the altitude from A is also the y-axis, as point A is at (0, 2 + √3). Therefore, in this coordinate system, the altitude from A is the y-axis. So, the perpendicular bisector of BC is the y-axis.Now, let's find the perpendicular bisector of another side, say AB.First, find midpoint of AB. Coordinates of A(0, 2 + √3) and B(-1, 0). Midpoint is:x-coordinate: (0 + (-1))/2 = -0.5y-coordinate: (2 + √3 + 0)/2 = (2 + √3)/2So, midpoint of AB is (-0.5, (2 + √3)/2 )Slope of AB: (0 - (2 + √3))/(-1 - 0) = (-2 - √3)/(-1) = 2 + √3Therefore, the slope of AB is 2 + √3. Thus, the slope of the perpendicular bisector is the negative reciprocal, which is -1/(2 + √3). Let's rationalize that:Multiply numerator and denominator by (2 - √3):-1/(2 + √3) * (2 - √3)/(2 - √3) = -(2 - √3)/[(2)^2 - (√3)^2] = -(2 - √3)/(4 - 3) = -(2 - √3)/1 = -2 + √3Therefore, the slope of the perpendicular bisector of AB is -2 + √3.Now, we have the midpoint of AB at (-0.5, (2 + √3)/2 ) and a slope of -2 + √3. Let's write the equation of the perpendicular bisector.Using point-slope form:y - y₁ = m(x - x₁)Where (x₁, y₁) is (-0.5, (2 + √3)/2 )So,y - (2 + √3)/2 = (-2 + √3)(x + 0.5)Now, this line intersects the perpendicular bisector of BC, which is x = 0 (the y-axis). So, substitute x = 0 into the equation above to find y-coordinate of O.So,y - (2 + √3)/2 = (-2 + √3)(0 + 0.5) = (-2 + √3)(0.5)Therefore,y = (2 + √3)/2 + (-2 + √3)/2 = [ (2 + √3) + (-2 + √3) ] / 2 = (2 + √3 - 2 + √3)/2 = (2√3)/2 = √3Therefore, the circumcenter O is at (0, √3)Wait, that's interesting. So in this coordinate system, O is at (0, √3). Let me verify this.Given that O is the circumcenter, which should be equidistant from all three vertices. Let's check the distance from O(0, √3) to A(0, 2 + √3), B(-1,0), and C(1,0).Distance OA: sqrt( (0 - 0)^2 + (2 + √3 - √3)^2 ) = sqrt(0 + (2)^2 ) = 2Distance OB: sqrt( (-1 - 0)^2 + (0 - √3)^2 ) = sqrt(1 + 3 ) = sqrt(4) = 2Distance OC: sqrt( (1 - 0)^2 + (0 - √3)^2 ) = sqrt(1 + 3 ) = 2So, OA = OB = OC = 2, which confirms that O is indeed the circumradius with radius 2. Good.So, coordinates:O: (0, √3)C: (1, 0)Now, we need to express vector CO as a linear combination of vectors CA and CB.First, let's find vectors CO, CA, and CB.Vector CO is from point C to point O: O - C = (0 - 1, √3 - 0) = (-1, √3)Vector CA is from point C to point A: A - C = (0 - 1, (2 + √3) - 0) = (-1, 2 + √3)Vector CB is from point C to point B: B - C = (-1 - 1, 0 - 0) = (-2, 0)We need to find scalars λ₁ and λ₂ such that:CO = λ₁ CA + λ₂ CBSo, (-1, √3) = λ₁*(-1, 2 + √3) + λ₂*(-2, 0)Let's break this into components:x-component: -1 = λ₁*(-1) + λ₂*(-2)y-component: √3 = λ₁*(2 + √3) + λ₂*0From the y-component equation:√3 = λ₁*(2 + √3)Therefore, solving for λ₁:λ₁ = √3 / (2 + √3)Rationalize the denominator:Multiply numerator and denominator by (2 - √3):λ₁ = [√3*(2 - √3)] / [(2 + √3)(2 - √3)] = [2√3 - (√3)^2] / (4 - 3) = [2√3 - 3]/1 = 2√3 - 3So, λ₁ = 2√3 - 3Now, substitute λ₁ into the x-component equation:-1 = (-1)*(2√3 - 3) + (-2)*λ₂Compute the right-hand side:= -2√3 + 3 - 2λ₂Therefore:-1 = -2√3 + 3 - 2λ₂Subtract (-2√3 + 3) from both sides:-1 - (-2√3 + 3) = -2λ₂Simplify left side:-1 + 2√3 - 3 = -4 + 2√3So:-4 + 2√3 = -2λ₂Divide both sides by -2:λ₂ = (-4 + 2√3)/(-2) = (4 - 2√3)/2 = 2 - √3Therefore, λ₂ = 2 - √3So, we have λ₁ = 2√3 - 3 and λ₂ = 2 - √3Now, the problem asks for the product λ₁λ₂.Compute λ₁λ₂:(2√3 - 3)(2 - √3)Let’s expand this:= 2√3*2 + 2√3*(-√3) -3*2 -3*(-√3)= 4√3 - 2*(√3)^2 -6 + 3√3Simplify each term:4√3 - 2*3 -6 + 3√3= 4√3 -6 -6 + 3√3Combine like terms:(4√3 + 3√3) + (-6 -6)= 7√3 -12Therefore, λ₁λ₂ = 7√3 -12Wait, but let me check the calculations again step by step to be sure.First, compute (2√3 -3)(2 - √3):Multiply 2√3 by 2: 4√3Multiply 2√3 by (-√3): -2*(√3)^2 = -2*3 = -6Multiply -3 by 2: -6Multiply -3 by (-√3): +3√3So, combining all terms:4√3 -6 -6 +3√3 = (4√3 +3√3) + (-6 -6) = 7√3 -12Yes, that's correct. So the product is 7√3 -12. But let me check if this answer makes sense. Let's see if it can be simplified or expressed differently.Alternatively, maybe there is a miscalculation in signs. Let me verify:Original components:Vector CO = (-1, √3)Expressed as λ₁ CA + λ₂ CB, where CA = (-1, 2 + √3) and CB = (-2, 0)From the y-component:√3 = λ₁*(2 + √3) => λ₁ = √3 / (2 + √3) = (√3*(2 - √3)) / ( (2 + √3)(2 - √3) ) = (2√3 - 3)/1 = 2√3 - 3. That's correct.From the x-component:-1 = -1*λ₁ -2*λ₂Plugging λ₁ = 2√3 -3:-1 = - (2√3 -3) -2λ₂=> -1 = -2√3 +3 -2λ₂Subtract 3 from both sides:-4 = -2√3 -2λ₂Add 2√3 to both sides:-4 + 2√3 = -2λ₂Divide by -2:(4 - 2√3)/2 = λ₂ => 2 - √3 = λ₂. Correct.Then product λ₁λ₂ = (2√3 -3)(2 - √3) = as before 7√3 -12. But wait, 7√3 is approximately 12.12, so 7√3 -12 ≈ 0.12, which is a small positive number. Let me check if that makes sense. Alternatively, perhaps we can write the answer in terms of exact values. But the problem just asks for the product λ₁λ₂, which is 7√3 -12. Let me check if this can be simplified further. Perhaps not, so maybe that's the answer.But let me check the problem statement again. It says "circumcenter of triangle ABC", which we found correctly. The coordinates were set such that O is at (0, √3). Then vectors were computed correctly. The coefficients seem right, leading to the product 7√3 -12.But let me cross-validate using another method, in case there's an error in coordinate setup.Alternative approach: Since ABC is an isosceles triangle with AB=AC and angle at A of 30°, then the triangle is determined. Let's compute the circumradius.In any triangle, the circumradius R = a/(2 sin A), where a is the length of the side opposite angle A.In triangle ABC, angle at A is 30°, side BC is opposite angle A.So, if we compute BC, then R = BC/(2 sin 30°)But BC can be found using the Law of Cosines:BC² = AB² + AC² - 2 AB*AC cos(angle A)But AB = AC = let's say length s.Thus, BC² = 2s² - 2s² cos30° = 2s²(1 - cos30°)Therefore, BC = s*sqrt(2(1 - cos30°))But R = BC/(2 sin30°) = [s*sqrt(2(1 - cos30°))]/(2*(1/2)) = s*sqrt(2(1 - cos30°))/1 = s*sqrt(2(1 - √3/2))Simplify inside the sqrt:2(1 - √3/2) = 2 - √3Therefore, R = s*sqrt(2 - √3)But in our coordinate system, we set AB = AC = sqrt(b² + h²). Wait, in our coordinate system, AB is from A(0, 2 + √3) to B(-1,0). So AB length is sqrt( (-1 - 0)^2 + (0 - (2 + √3))^2 ) = sqrt(1 + (2 + √3)^2 )Compute (2 + √3)^2 = 4 + 4√3 + 3 = 7 + 4√3Thus, AB = sqrt(1 + 7 + 4√3) = sqrt(8 + 4√3) = sqrt(4*(2 + √3)) = 2*sqrt(2 + √3)But in our coordinate system, the circumradius R is 2 (since OA=OB=OC=2). So according to the formula R = s*sqrt(2 - √3), here s is AB = 2*sqrt(2 + √3). Then:R = 2*sqrt(2 + √3)*sqrt(2 - √3) = 2*sqrt( (2 + √3)(2 - √3) ) = 2*sqrt(4 - 3) = 2*sqrt(1) = 2*1 = 2, which matches. So this confirms that the formula is correct and our coordinate system is consistent.Therefore, the answer we obtained, λ₁λ₂ = 7√3 -12, seems correct in this context.But wait, maybe there's a calculation mistake in the expansion.Wait, let's recompute (2√3 -3)(2 - √3):First term: 2√3 * 2 = 4√3Second term: 2√3 * (-√3) = -2*(√3)^2 = -2*3 = -6Third term: -3*2 = -6Fourth term: -3*(-√3) = 3√3Adding up: 4√3 -6 -6 +3√3 = (4√3 +3√3) + (-6 -6) = 7√3 -12Yes, that's correct. So the product is indeed 7√3 -12.But let me check numerically:Compute 7√3 ≈ 7*1.732 ≈ 12.12412.124 -12 ≈ 0.124. So the product is approximately 0.124, which is positive.Given the problem is likely expecting an exact value, so 7√3 -12 is the exact value.But let me check if the problem statement might have a different answer. Wait, maybe I made a mistake in the coordinate setup. Let me check the coordinates again.Wait, when we set point B at (-1,0) and C at (1,0), then the midpoint of BC is (0,0), but in my coordinate system, point A was at (0, 2 + √3). Then, O was found at (0, √3). Then vectors:CO = O - C = (0 -1, √3 -0) = (-1, √3)Vector CA = A - C = (0 -1, 2 + √3 -0) = (-1, 2 + √3)Vector CB = B - C = (-1 -1, 0 -0) = (-2, 0)Then solving:(-1, √3) = λ₁*(-1, 2 + √3) + λ₂*(-2, 0)Component-wise:x: -1 = -λ₁ -2λ₂y: √3 = λ₁*(2 + √3)From y: λ₁ = √3 / (2 + √3) = (√3*(2 - √3))/ (4 -3) = 2√3 -3From x: -1 = - (2√3 -3) -2λ₂-1 = -2√3 +3 -2λ₂-1 -3 +2√3 = -2λ₂-4 +2√3 = -2λ₂λ₂ = (4 -2√3)/2 = 2 - √3Product λ₁λ₂ = (2√3 -3)(2 - √3) = 7√3 -12Yes, correct.Alternatively, maybe the answer is presented in another form. For example, 7√3 -12 can be written as -(12 -7√3), but since they ask for λ₁λ₂, the exact form is acceptable.Therefore, the answer should be 7√3 -12. Therefore, boxed:boxed{7sqrt{3} - 12}Final Answerboxed{7sqrt{3} - 12}</think>Given that (O) is the circumcenter of (triangle ABC) with (AB = AC) and (angle CAB = 30^circ). We need to find (lambda_1 lambda_2) where (overrightarrow{CO} = lambda_1 overrightarrow{CA} + lambda_2 overrightarrow{CB}).1. Coordinate Setup: - Place (A) at ((0, 2 + sqrt{3})), (B) at ((-1, 0)), and (C) at ((1, 0)). - The triangle is isosceles with (AB = AC) and (angle CAB = 30^circ).2. Finding Circumcenter (O): - The circumcenter (O) is the intersection of the perpendicular bisectors. - Midpoint of (BC) is ((0, 0)) and the perpendicular bisector is the y-axis ((x = 0)). - Midpoint of (AB) is ((-0.5, (2 + sqrt{3})/2)) with a perpendicular bisector slope of (-2 + sqrt{3}). - Solving the intersection of perpendicular bisectors, (O) is found at ((0, sqrt{3})).3. Vectors Calculation: - (overrightarrow{CO} = (-1, sqrt{3})) - (overrightarrow{CA} = (-1, 2 + sqrt{3})) - (overrightarrow{CB} = (-2, 0))4. Solving for (lambda_1) and (lambda_2): - From the y-component: (sqrt{3} = lambda_1 (2 + sqrt{3}) Rightarrow lambda_1 = frac{sqrt{3}}{2 + sqrt{3}} = 2sqrt{3} - 3). - From the x-component: (-1 = -lambda_1 - 2lambda_2 Rightarrow lambda_2 = 2 - sqrt{3}).5. Product (lambda_1 lambda_2): - (lambda_1 lambda_2 = (2sqrt{3} - 3)(2 - sqrt{3}) = 7sqrt{3} - 12).Thus, the final answer is (boxed{7sqrt{3} - 12}).]

🔑:7. 7 sqrt{3}-12.Let's assume AB=2. Establish a Cartesian coordinate system with A as the origin and the line AB as the x-axis. ThenA(0,0), B(2,0), C(sqrt{3}, 1) text {. }Let the circumcenter be O(1, y).From |overrightarrow{O A}|=|overrightarrow{O C}|, we get1+y^{2}=(sqrt{3}-1)^{2}+(y-1)^{2} text {. }Solving this, we get y=2-sqrt{3}.begin{array}{l}text { Then }(1-sqrt{3}, 1-sqrt{3})=overrightarrow{C O}=lambda_{1} overrightarrow{C A}+lambda_{2} overrightarrow{C B} =lambda_{1}(-sqrt{3},-1)+lambda_{2}(2-sqrt{3},-1) .end{array}Solving this, we get lambda_{1}=2 sqrt{3}-3, lambda_{2}=2-sqrt{3}.text { Therefore } lambda_{1} lambda_{2}=7 sqrt{3}-12 text {. }

🧠:[<think>Okay, so I need to solve the inequality |x + 2| + |x - 1| ≤ 4. Hmm, absolute value inequalities can be tricky, but I remember that they often involve considering different cases based on the critical points where the expressions inside the absolute values change sign. Let me think about how to approach this.First, the expressions inside the absolute values are x + 2 and x - 1. These expressions will change sign at x = -2 and x = 1, respectively. So, those are the critical points. That means the solution will depend on which interval x is in: less than -2, between -2 and 1, or greater than 1. I need to consider each interval separately because the absolute value expressions will behave differently in each.Let me write down the three intervals:1. x < -22. -2 ≤ x ≤ 13. x > 1For each interval, I can simplify the absolute value expressions by removing the absolute value signs with the appropriate sign based on the interval.Starting with the first interval: x < -2.In this interval, x + 2 will be negative (since x is less than -2), so |x + 2| = -(x + 2) = -x - 2. Similarly, x - 1 will also be negative (since x is less than -2, which is definitely less than 1), so |x - 1| = -(x - 1) = -x + 1. Therefore, the inequality becomes:(-x - 2) + (-x + 1) ≤ 4Simplifying that:- x - 2 - x + 1 = -2x -1 ≤ 4So, -2x -1 ≤ 4Adding 1 to both sides:-2x ≤ 5Dividing both sides by -2 (remembering to reverse the inequality sign because we divide by a negative number):x ≥ -5/2But hold on, this interval is x < -2. So the solution here would be the intersection of x < -2 and x ≥ -5/2. That is, -5/2 ≤ x < -2. Wait, but -5/2 is -2.5, right? So, that interval is from -2.5 to -2. Is that correct? Let me check.Yes, because if x is less than -2, but we found that x must be greater than or equal to -5/2 (-2.5), so the overlapping interval is [-2.5, -2). Okay, that seems right.Now moving to the second interval: -2 ≤ x ≤ 1.In this interval, x + 2 is non-negative (since x ≥ -2), so |x + 2| = x + 2. However, x - 1 is still non-positive (since x ≤ 1), so |x - 1| = -(x - 1) = -x + 1. Therefore, the inequality becomes:(x + 2) + (-x + 1) ≤ 4Simplifying that:x + 2 - x + 1 = 3 ≤ 4Wait, that's just 3 ≤ 4, which is always true. So in this interval, the inequality holds for all x between -2 and 1. Therefore, the entire interval [-2, 1] is part of the solution set.Third interval: x > 1.In this interval, both x + 2 and x - 1 are positive. So, |x + 2| = x + 2 and |x - 1| = x - 1. Therefore, the inequality becomes:(x + 2) + (x - 1) ≤ 4Simplifying:x + 2 + x - 1 = 2x + 1 ≤ 4Subtracting 1 from both sides:2x ≤ 3Dividing both sides by 2:x ≤ 3/2But this interval is x > 1, so the solution here is the intersection of x > 1 and x ≤ 3/2. That is, 1 < x ≤ 1.5 (since 3/2 is 1.5). So, (1, 1.5].Now, combining all three intervals:From the first interval: [-2.5, -2)From the second interval: [-2, 1]From the third interval: (1, 1.5]Putting them all together, the solution set is [-2.5, -2) ∪ [-2, 1] ∪ (1, 1.5]. But since [-2.5, -2) and [-2, 1] overlap at -2, and [-2, 1] and (1, 1.5] are adjacent, the combined interval is [-2.5, 1.5]. Wait, is that correct?Wait, let's check. First interval is from -2.5 to -2, second from -2 to 1, third from 1 to 1.5. So combined, it's continuous from -2.5 to 1.5. But let me verify if that's actually true.Take a point in [-2.5, -2), say x = -2.5. Then | -2.5 + 2 | + | -2.5 -1 | = | -0.5 | + | -3.5 | = 0.5 + 3.5 = 4, which is equal to 4, so that's included.At x = -2, |0| + |-3| = 0 + 3 = 3 ≤4, so that's included.At x = 1, |3| + |0| = 3 + 0 = 3 ≤4, included.At x = 1.5, |1.5 + 2| + |1.5 -1| = |3.5| + |0.5| = 3.5 + 0.5 = 4, which is equal, so included.But in the first interval, the upper limit was -2 not included, but in the second interval, -2 is included. So overall, -2 is included. Similarly, at x =1, it's included, and the upper limit is 1.5 included.Therefore, the entire interval is from -2.5 to 1.5, closed at both ends. So the solution set is [-2.5, 1.5]. But wait, but when solving for the first interval, x < -2, the solution was x ≥ -2.5 and x < -2. So that gives [-2.5, -2). The second interval was all of [-2,1], and third was (1,1.5]. So together, [-2.5,1.5]. Since [-2.5, -2) combined with [-2,1] is [-2.5,1], and then combined with (1,1.5] is [-2.5,1.5]. So yes, that's correct.But wait, let me verify with a number in each interval to make sure.Take x = -3, which is less than -2.5. Then | -3 + 2 | + | -3 -1 | = | -1 | + | -4 | = 1 + 4 = 5 >4, so not included. So -3 is not in the solution set, which matches since -3 < -2.5.Take x = -2.5, the left endpoint. Then | -2.5 +2 | + | -2.5 -1 | = | -0.5 | + | -3.5 | = 0.5 +3.5=4, which is equal, so included.Take x = -2, which is included. |0| + |-3| = 3 ≤4.Take x=0, in the middle. |0+2| + |0-1| =2 +1=3 ≤4.Take x=1, |3| + |0|=3≤4.Take x=1.5, |3.5| + |0.5|=4, which is equal.Take x=2, which is beyond 1.5. |4| + |1|=5>4, so not included.So yes, all points from -2.5 to 1.5 satisfy the inequality, including the endpoints. Therefore, the solution set is the closed interval [-2.5,1.5], or in fractions, [-5/2, 3/2].But wait, when I broke it down into intervals, the first interval was [-2.5, -2), then [-2,1], then (1,1.5]. But combining them, the endpoints at -2 and 1 are included in the next intervals. So overall, it's a continuous interval from -2.5 to 1.5, including both endpoints. So maybe I should present it as a single interval.But why did splitting into cases give me a union of intervals that when combined form a single interval? Because even though in the first interval the upper bound was open (-2), the next interval includes -2, so overall, -2 is included. Similarly, at 1, the first interval includes 1, and the next interval is open at 1 but since 1 is included in the second interval, so overall, 1 is included. Then the upper bound at 1.5 is included. So yes, the entire interval is closed.Therefore, the solution set is all real numbers from -5/2 to 3/2, inclusive.But let me just confirm with another point in between. Let's take x = -2.5, which is the left endpoint. As above, it gives 4, which is allowed. Take x = -2.25, which is in [-2.5, -2). Then | -2.25 +2 | + | -2.25 -1 | = | -0.25 | + | -3.25 | = 0.25 + 3.25 = 3.5 <4. So that's okay. Wait, but according to the first interval, when x < -2, the inequality reduces to x >= -2.5, so x is between -2.5 and -2. So in that interval, the values of x between -2.5 and -2 make the expression equal to -2x -1. So if x is -2.25, then -2*(-2.25) -1 = 4.5 -1 = 3.5, which is ≤4. So correct.Similarly, take x = 1.25, in (1,1.5]. Then |1.25 +2| + |1.25 -1| = |3.25| + |0.25| = 3.25 +0.25=3.5 ≤4. Correct.So, all these test points work.Therefore, the solution set is the closed interval from -2.5 to 1.5, which is [-5/2, 3/2].I think that's the answer. Let me just write it in fraction form to be precise. -5/2 is -2.5 and 3/2 is 1.5. So, the solution set is all real numbers x such that -5/2 ≤x ≤3/2.Final AnswerThe solution set is boxed{left[ -dfrac{5}{2}, dfrac{3}{2} right]}.</think>To solve the inequality (|x + 2| + |x - 1| leq 4), we consider the critical points where the expressions inside the absolute values change sign, which are (x = -2) and (x = 1). We break the problem into three intervals: (x < -2), (-2 leq x leq 1), and (x > 1).1. Interval (x < -2): - Here, (|x + 2| = -x - 2) and (|x - 1| = -x + 1). - The inequality becomes (-2x - 1 leq 4), leading to (x geq -frac{5}{2}). - Intersection with (x < -2) gives (-frac{5}{2} leq x < -2).2. Interval (-2 leq x leq 1): - Here, (|x + 2| = x + 2) and (|x - 1| = -x + 1). - The inequality simplifies to (3 leq 4), which is always true. - The solution is (-2 leq x leq 1).3. Interval (x > 1): - Here, (|x + 2| = x + 2) and (|x - 1| = x - 1). - The inequality becomes (2x + 1 leq 4), leading to (x leq frac{3}{2}). - Intersection with (x > 1) gives (1 < x leq frac{3}{2}).Combining all intervals:- (-frac{5}{2} leq x < -2)- (-2 leq x leq 1)- (1 < x leq frac{3}{2})These intervals combine to form the continuous interval (-frac{5}{2} leq x leq frac{3}{2}).Thus, the solution set is (boxed{left[ -dfrac{5}{2}, dfrac{3}{2} right]}).,<think>Okay, let me try to figure out how to solve the inequality |x + 2| + |x - 1| ≤ 4. Hmm, absolute value inequalities can be tricky because of the different cases depending on the value of x. I remember that when dealing with absolute values, it's helpful to consider the points where the expressions inside the absolute values change their sign. For |x + 2|, the critical point is at x = -2, and for |x - 1|, it's at x = 1. So these points divide the real number line into different intervals where the expressions inside the absolute values are either positive or negative. Let me note down the critical points: x = -2 and x = 1. So the intervals to consider are:1. x < -22. -2 ≤ x < 13. x ≥ 1I need to analyze each interval separately and rewrite the inequality without the absolute values accordingly. Then solve the inequality in each interval and combine the results at the end.Starting with the first interval: x < -2.In this case, x + 2 is negative (since x is less than -2), so |x + 2| = -(x + 2) = -x - 2. Similarly, x - 1 is also negative because x is less than -2, which is definitely less than 1. So |x - 1| = -(x - 1) = -x + 1. Therefore, the inequality becomes:(-x - 2) + (-x + 1) ≤ 4Simplifying that:- x - 2 - x + 1 ≤ 4 Combine like terms: -2x -1 ≤ 4 Add 1 to both sides: -2x ≤ 5 Divide both sides by -2 (remember to reverse the inequality when multiplying/dividing by a negative number): x ≥ -5/2But wait, this is supposed to be valid only in the interval x < -2. So we have x ≥ -5/2 and x < -2. Let's check if these overlap. -5/2 is -2.5. So x must be between -2.5 and -2. Therefore, the solution in this interval is [-2.5, -2).Now moving to the second interval: -2 ≤ x < 1.In this interval, x + 2 is non-negative (since x ≥ -2), so |x + 2| = x + 2. However, x - 1 is still negative because x < 1, so |x - 1| = -(x - 1) = -x + 1. Therefore, the inequality becomes:(x + 2) + (-x + 1) ≤ 4Simplify:x + 2 - x + 1 ≤ 4 Combine like terms: 3 ≤ 4Hmm, this simplifies to 3 ≤ 4, which is always true. So in the interval -2 ≤ x < 1, the inequality holds for all x in this interval. Therefore, the solution here is the entire interval [-2, 1).Third interval: x ≥ 1.In this case, both x + 2 and x - 1 are non-negative. So |x + 2| = x + 2 and |x - 1| = x - 1. The inequality becomes:(x + 2) + (x - 1) ≤ 4Simplify:x + 2 + x - 1 ≤ 4 Combine like terms: 2x + 1 ≤ 4 Subtract 1 from both sides: 2x ≤ 3 Divide by 2: x ≤ 3/2But this is valid only in the interval x ≥ 1. So x has to be between 1 and 1.5 (which is 3/2). Therefore, the solution in this interval is [1, 3/2].Now, combining all the intervals:From the first interval: [-2.5, -2) From the second interval: [-2, 1) From the third interval: [1, 1.5]Putting them all together: [-2.5, -2) ∪ [-2, 1) ∪ [1, 1.5] = [-2.5, 1.5]Wait, because the intervals are adjacent and overlapping at the endpoints, right? So [-2.5, -2) combined with [-2, 1) is [-2.5, 1), and then combined with [1, 1.5] gives [-2.5, 1.5]. So the solution set is all real numbers from -2.5 to 1.5, including the endpoints?Wait, but in the first interval, the upper limit is -2, which is excluded, but in the second interval, it's included. So when combined, -2 is included from the second interval. Similarly, 1 is included in the third interval as the lower limit. So when combining, even though the first interval ends at -2 (exclusive), the second interval starts at -2 (inclusive), so overall -2 is included. Similarly, the second interval ends at 1 (exclusive), but the third interval starts at 1 (inclusive), so 1 is included. Therefore, the combined interval is from -2.5 to 1.5, inclusive of both endpoints. So the solution set is [-2.5, 1.5].But let me verify this with some test points.Take x = -2.5 (left endpoint): | -2.5 + 2 | + | -2.5 -1 | = | -0.5 | + | -3.5 | = 0.5 + 3.5 = 4, which is equal to 4, so it satisfies the inequality.Take x = 1.5 (right endpoint): |1.5 + 2| + |1.5 -1| = |3.5| + |0.5| = 3.5 + 0.5 = 4, which also satisfies the inequality.Now, take a point inside the interval, say x = 0: |0 + 2| + |0 - 1| = 2 + 1 = 3 ≤ 4, which works.Take a point between -2.5 and -2, say x = -2.25: |-2.25 + 2| + |-2.25 -1| = |-0.25| + |-3.25| = 0.25 + 3.25 = 3.5 ≤ 4, which is okay.Another point between 1 and 1.5, say x = 1.25: |1.25 + 2| + |1.25 -1| = 3.25 + 0.25 = 3.5 ≤ 4, which works.Now, check a point outside the interval, say x = -3: |-3 + 2| + |-3 -1| = | -1 | + | -4 | = 1 + 4 = 5 > 4, which does not satisfy.Another outside point, x = 2: |2 + 2| + |2 -1| = 4 + 1 = 5 > 4, which also fails.So it seems the solution set is indeed from -2.5 to 1.5, inclusive. Converting -2.5 and 1.5 to fractions: -2.5 is -5/2 and 1.5 is 3/2. So the solution set is all real numbers x such that -5/2 ≤ x ≤ 3/2.Therefore, the answer should be in boxed notation: boxed{left[ -dfrac{5}{2}, dfrac{3}{2} right]}Final AnswerThe solution set is boxed{left[ -dfrac{5}{2}, dfrac{3}{2} right]}.</think>To solve the inequality (|x + 2| + |x - 1| leq 4), we consider the critical points where the expressions inside the absolute values change sign: (x = -2) and (x = 1). These points divide the real number line into three intervals: (x < -2), (-2 leq x < 1), and (x geq 1). We analyze each interval separately.1. Interval (x < -2): - Both (|x + 2|) and (|x - 1|) are negative. - The inequality becomes (-x - 2 - x + 1 leq 4), simplifying to (-2x - 1 leq 4). - Solving gives (x geq -frac{5}{2}). Since (x < -2), the solution is (-frac{5}{2} leq x < -2).2. Interval (-2 leq x < 1): - (|x + 2|) is non-negative and (|x - 1|) is negative. - The inequality becomes (x + 2 - x + 1 leq 4), simplifying to (3 leq 4), which is always true. - The solution is (-2 leq x < 1).3. Interval (x geq 1): - Both (|x + 2|) and (|x - 1|) are non-negative. - The inequality becomes (x + 2 + x - 1 leq 4), simplifying to (2x + 1 leq 4). - Solving gives (x leq frac{3}{2}). Since (x geq 1), the solution is (1 leq x leq frac{3}{2}).Combining the solutions from all intervals, we get:[left[ -frac{5}{2}, -2 right) cup left[ -2, 1 right) cup left[ 1, frac{3}{2} right] = left[ -frac{5}{2}, frac{3}{2} right]]Verification with test points confirms the solution. The solution set is (boxed{left[ -dfrac{5}{2}, dfrac{3}{2} right]}).]

🔑:Let f(x)=|x+2|+|x-1|. Then,f(x)=begin{cases}-2x-1,& xleqslant -2 3,& -2 < x < 1 2x+1,& xgeqslant 1end{cases}1. When xleqslant -2, |x+2|+|x-1|leqslant 4 is equivalent to -2x-1leqslant 4, which gives -frac{5}{2}leqslant xleqslant -2.2. When -2 < x < 1, the inequality 3leqslant 4 always holds.3. When xgeqslant 1, |x+2|+|x-1|leqslant 4 is equivalent to 2x+1leqslant 4, which leads to 1leqslant xleqslant frac{3}{2}.In summary, the solution set of the inequality |x+2|+|x-1|leqslant 4 is boxed{[-frac{5}{2},frac{3}{2}]}.To solve this problem, we let f(x)=|x+2|+|x-1| and discuss the range of x to remove the absolute value symbols, turning it into a piecewise function. Then we solve it accordingly. This problem tests the methods of solving absolute value inequalities, which can be addressed by either converting them into piecewise functions or using the geometric meaning of absolute values. It primarily assesses one's ability to transform problems and computational skills, making it a moderately difficult question.

🧠:[<think>Okay, let's tackle this problem step by step. So, we have real numbers a, b, c satisfying a ≥ b ≥ c, a + b + c = 0, and a ≠ 0. We need to find the maximum distance between points A(x₁, x₂) and B(x₂, x₁), where x₁ and x₂ are the roots of the quadratic equation ax² + bx + c = 0. Hmm, okay.First, let's recall that the distance between two points (x1, x2) and (x2, x1) in the Cartesian plane can be calculated using the distance formula. The distance D is √[(x2 - x1)² + (x1 - x2)²]. Simplifying that, since both terms are squares of differences, it becomes √[2(x2 - x1)²] which is √2 * |x2 - x1|. So, the distance is √2 times the absolute difference of the roots. Therefore, to maximize the distance D, we need to maximize |x2 - x1|.Since the quadratic equation has two real roots, the discriminant must be non-negative. The discriminant of ax² + bx + c is b² - 4ac ≥ 0. Given that a ≥ b ≥ c and a + b + c = 0, maybe we can express c in terms of a and b? Let's see, since a + b + c = 0, then c = -a - b. So, substituting c into the discriminant, we get b² - 4a*(-a - b) = b² + 4a² + 4ab. This discriminant must be non-negative. But since a is non-zero, and given the ordering a ≥ b ≥ c, which would mean c is the smallest. Since a is positive? Wait, not necessarily. Hmm, a could be positive or negative? Wait, but since a ≥ b ≥ c and their sum is zero, perhaps a must be positive? Let's check.If a were negative, then since a is the largest of the three numbers (a ≥ b ≥ c), all three would be negative, but their sum is zero. So, the only way that three non-positive numbers add up to zero is if all are zero. But the problem states a ≠ 0, so a must be positive. Because if a is positive, then since a is the largest, and the sum is zero, at least one of the numbers must be negative. So, a > 0, and c ≤ b ≤ a. Since a + b + c = 0, then c must be negative or at least non-positive. Because if a is positive, and the sum is zero, then at least one of b or c has to be negative. But since a ≥ b ≥ c, c is the smallest, so c must be negative. So, a is positive, c is negative, and b is somewhere in between.Therefore, the quadratic equation ax² + bx + c = 0 has a positive leading coefficient and a negative constant term. Therefore, the quadratic will have one positive and one negative root? Wait, not necessarily. The product of the roots is c/a, which is negative since c is negative and a is positive. Therefore, yes, the roots have opposite signs. So, one root is positive, one is negative.Therefore, x1 and x2 are real roots of opposite signs. So, the points A(x1, x2) and B(x2, x1) would be in different quadrants, depending on the signs. But perhaps that's not essential here.But going back, since the distance D is √2 * |x2 - x1|, we need to maximize |x2 - x1|. The difference between the roots of a quadratic equation is given by |x2 - x1| = √(D)/|a|, where D is the discriminant. Wait, let's recall that for a quadratic equation ax² + bx + c = 0, the roots are [-b ± √(b² - 4ac)]/(2a). So, the difference between the roots is [√(b² - 4ac)/a] - [-√(b² - 4ac)/a] = 2√(b² - 4ac)/(2a) = √(b² - 4ac)/a. Therefore, |x2 - x1| = √(b² - 4ac)/|a|. Since a is positive, this simplifies to √(b² - 4ac)/a.Wait, but the discriminant is b² - 4ac, which we had earlier as b² + 4a² + 4ab. Let me check that again. Since c = -a - b, substituting into the discriminant gives:Discriminant D = b² - 4ac = b² - 4a*(-a - b) = b² + 4a² + 4ab.So, D = b² + 4ab + 4a² = (b + 2a)^2. Wait, that's interesting! So, discriminant D = (b + 2a)^2. Since it's a square, it's always non-negative, which makes sense because the problem states that there are two real roots. So, the discriminant is always non-negative, which is consistent.Therefore, |x2 - x1| = √D / a = √( (b + 2a)^2 ) / a = |b + 2a| / a. Since a is positive, this simplifies to (b + 2a)/a if b + 2a ≥ 0, otherwise -(b + 2a)/a. But let's see, given that a ≥ b ≥ c and c = -a - b, we can find constraints on b.Given that a ≥ b ≥ c, and c = -a - b, then substituting c, we have a ≥ b ≥ -a - b.So, starting with a ≥ b, which is given.Then, b ≥ -a - b. Let's solve that inequality:b ≥ -a - bAdding b to both sides:2b ≥ -aDivide both sides by 2:b ≥ -a/2Therefore, combining with the previous inequality, we have a ≥ b ≥ -a/2.So, b is between -a/2 and a. Therefore, b + 2a ≥ (-a/2) + 2a = (3a/2). Since a is positive, 3a/2 is positive. Therefore, b + 2a ≥ 3a/2 > 0. Therefore, |x2 - x1| = (b + 2a)/a = (b/a) + 2. So, |x2 - x1| = (b/a) + 2. Therefore, the distance D between points A and B is √2 * ( (b/a) + 2 ). So, to maximize D, we need to maximize (b/a) + 2. Since a is positive, and b is between -a/2 and a, then (b/a) ranges from -1/2 to 1. Therefore, (b/a) + 2 ranges from (-1/2 + 2) = 3/2 to (1 + 2) = 3. Therefore, the maximum value of (b/a) + 2 is 3, achieved when b/a = 1, i.e., when b = a. However, given the constraints a ≥ b ≥ -a/2, and a + b + c = 0.Wait, but if b = a, then from a + b + c = 0, we get c = -2a. But since we have the constraint a ≥ b ≥ c. If b = a, then c must be -2a. But then, we need to check if b ≥ c. So, b = a, c = -2a. So, since a ≥ a ≥ -2a. But a is positive, so -2a is negative. Therefore, a ≥ a is true, and a ≥ -2a is also true because a is positive. So, that's okay. Therefore, when b = a, c = -2a, and the constraints are satisfied.Therefore, the maximum value of (b/a) + 2 is 3, leading to a maximum distance of √2 * 3 = 3√2.Wait, but let me verify this. If b = a, then the quadratic equation becomes ax² + a x + c = 0, and c = -2a. Therefore, the equation is ax² + a x - 2a = 0. Divide both sides by a (since a ≠ 0):x² + x - 2 = 0. Solving this, the roots are [-1 ± √(1 + 8)]/2 = [-1 ± 3]/2. Therefore, x = 1 or x = -2. Therefore, the roots are 1 and -2. Then, the points A(1, -2) and B(-2, 1). The distance between them is √[( -2 - 1 )² + ( 1 - (-2) )²] = √[(-3)² + (3)²] = √[9 + 9] = √18 = 3√2. So, that's correct. So, the maximum distance is indeed 3√2.But wait, before concluding, let me check if there are other possible cases where (b/a) + 2 can be larger. Since we derived that (b/a) is maximum when b = a, which gives (b/a) = 1, leading to 3. If b is less than a, then (b/a) + 2 would be smaller. For example, if b = -a/2, then (b/a) + 2 = (-1/2) + 2 = 3/2, which is smaller. So, the maximum is indeed 3. Therefore, the maximum distance is 3√2.But let me confirm whether when b = a, the condition a ≥ b ≥ c is satisfied. Since b = a, c = -2a. Then, a ≥ a ≥ -2a. Since a is positive, -2a is negative, so a ≥ a is true, and a ≥ -2a is equivalent to a + 2a ≥ 0 => 3a ≥ 0, which is true since a is positive. Therefore, the conditions are satisfied. Therefore, this case is valid.Is there another case where (b + 2a)/a could be larger? For example, if b is allowed to be larger than a? But the problem states a ≥ b ≥ c, so b cannot exceed a. Therefore, the maximum value of b is a, which gives the maximum (b/a) + 2 = 3.Therefore, the maximum distance is 3√2.Alternatively, let's approach this problem using variables substitution. Since a + b + c = 0, and a ≥ b ≥ c, perhaps we can express b and c in terms of a and another variable. Let's let t = b/a. Since a ≠ 0, we can normalize variables by dividing by a. Let’s set t = b/a and s = c/a. Then, the equation a + b + c = 0 becomes 1 + t + s = 0, so s = -1 - t. The inequalities a ≥ b ≥ c translate to 1 ≥ t ≥ s. Substituting s = -1 - t, we get 1 ≥ t ≥ -1 - t. Let's solve the inequality t ≥ -1 - t:t + t ≥ -1 => 2t ≥ -1 => t ≥ -1/2.So, combining with 1 ≥ t, we have t ∈ [-1/2, 1].Therefore, t is between -1/2 and 1. Then, |x2 - x1| = √( (b + 2a)^2 ) / a = (b + 2a)/a = t + 2. Therefore, |x2 - x1| = t + 2. So, the distance D = √2(t + 2). Since t ∈ [-1/2, 1], the maximum D occurs when t is maximum, i.e., t = 1. Then, D = √2(1 + 2) = 3√2, as before. Therefore, this confirms the previous result.Therefore, the maximum distance is indeed 3√2.Alternatively, we can use Lagrange multipliers to maximize the distance function under the given constraints, but since we have a linear relationship due to a + b + c = 0 and inequalities, the maximum occurs at the boundary of the feasible region. Since t is bounded between -1/2 and 1, the maximum occurs at t = 1, leading to the same conclusion.Therefore, after verifying through multiple approaches and checking the boundary conditions, the maximum distance between points A and B is 3√2.Final Answerboxed{3sqrt{2}}</think>Given real numbers (a, b, c) satisfy (a geq b geq c), (a + b + c = 0), and (a neq 0). We need to find the maximum distance between points (A(x_1, x_2)) and (B(x_2, x_1)), where (x_1) and (x_2) are the roots of the quadratic equation (ax^2 + bx + c = 0).First, the distance between points (A(x_1, x_2)) and (B(x_2, x_1)) is given by:[D = sqrt{(x_2 - x_1)^2 + (x_1 - x_2)^2} = sqrt{2(x_2 - x_1)^2} = sqrt{2} |x_2 - x_1|]To maximize (D), we need to maximize (|x_2 - x_1|). The roots of the quadratic equation (ax^2 + bx + c = 0) are given by:[x_1, x_2 = frac{-b pm sqrt{b^2 - 4ac}}{2a}]The discriminant of the quadratic equation is:[b^2 - 4ac]Given (a + b + c = 0), we substitute (c = -a - b) into the discriminant:[b^2 - 4a(-a - b) = b^2 + 4a^2 + 4ab = (b + 2a)^2]The difference between the roots is:[|x_2 - x_1| = frac{sqrt{(b + 2a)^2}}{a} = frac{|b + 2a|}{a}]Since (a geq b geq c) and (a + b + c = 0), we have (a > 0) and (b geq -a/2). Therefore, (b + 2a geq 3a/2 > 0), so:[|x_2 - x_1| = frac{b + 2a}{a} = frac{b}{a} + 2]To maximize (frac{b}{a} + 2), we note that (frac{b}{a}) is maximized when (b = a). Given (a + b + c = 0), if (b = a), then (c = -2a). This satisfies (a geq b geq c) as (a geq a geq -2a).Substituting (b = a) into the quadratic equation, we get:[ax^2 + ax - 2a = 0 implies x^2 + x - 2 = 0]The roots are (x = 1) and (x = -2). The distance between points (A(1, -2)) and (B(-2, 1)) is:[sqrt{(1 - (-2))^2 + (-2 - 1)^2} = sqrt{(3)^2 + (-3)^2} = sqrt{9 + 9} = sqrt{18} = 3sqrt{2}]Thus, the maximum distance between points (A) and (B) is (boxed{3sqrt{2}}).,<think>Alright, so I need to find the maximum distance between points A(x₁, x₂) and B(x₂, x₁) where x₁ and x₂ are the real roots of the quadratic equation ax² + bx + c = 0. The conditions given are that a, b, c are real numbers with a ≥ b ≥ c, a + b + c = 0, and a ≠ 0. Hmm, okay. Let me start by recalling some basics.First, since x₁ and x₂ are real roots, the discriminant of the quadratic must be non-negative. The discriminant D is b² - 4ac ≥ 0. So that's one condition. Also, from Vieta's formulas, the sum of the roots x₁ + x₂ = -b/a, and the product x₁x₂ = c/a. These relationships might be helpful later.The distance between points A(x₁, x₂) and B(x₂, x₁) can be calculated using the distance formula. The distance squared would be (x₁ - x₂)² + (x₂ - x₁)², which simplifies to 2(x₁ - x₂)². Therefore, the distance is sqrt[2(x₁ - x₂)²] = sqrt(2)|x₁ - x₂|. So to maximize the distance between A and B, we need to maximize |x₁ - x₂|.Now, |x₁ - x₂| is equal to sqrt[(x₁ + x₂)² - 4x₁x₂]. This is derived from (x₁ - x₂)² = (x₁ + x₂)² - 4x₁x₂. Taking the square root gives the absolute difference. So, |x₁ - x₂| = sqrt[(x₁ + x₂)² - 4x₁x₂]. Using Vieta's formulas, this becomes sqrt[(-b/a)² - 4(c/a)] = sqrt[(b² - 4ac)/a²] = sqrt(b² - 4ac)/|a|. Since a is non-zero and given that a + b + c = 0, perhaps we can express c in terms of a and b? Let me see.Given that a + b + c = 0, so c = -a - b. Therefore, substituting c into the discriminant: D = b² - 4a(-a - b) = b² + 4a² + 4ab. So D = 4a² + 4ab + b². Hmm, that's equal to (2a + b)². Wait, let me check:(2a + b)² = 4a² + 4ab + b². Yes, exactly. So the discriminant D is (2a + b)². Since the discriminant must be non-negative for real roots, which it is because it's a square. Therefore, the roots are always real as long as a ≠ 0, which is given. So that condition is automatically satisfied. So |x₁ - x₂| = sqrt(D)/|a| = |2a + b| / |a|. Since a is non-zero, and we need to consider the absolute value.But a can be positive or negative? Wait, given the conditions a ≥ b ≥ c and a + b + c = 0. Let's think about the signs of a, b, c.If a is positive, then since a ≥ b ≥ c and a + b + c = 0, the other terms must be such that at least one of them is negative. Because if all three were positive, their sum would be positive, which contradicts the sum being zero. Similarly, if a is negative, then all three would have to be negative, but since a ≥ b ≥ c, but their sum is zero. Wait, but if a is negative, then c is the smallest, so even more negative, so adding three negative numbers can't give zero. Therefore, a must be positive. Because if a is positive, then at least one of b or c must be negative to make the sum zero.Therefore, a > 0. Because a ≠ 0 and a is the largest among a, b, c. So a is positive. Then, since a ≥ b ≥ c and a + b + c = 0, c must be negative, because if a and b are positive, then c = -a - b would be negative, which is indeed the case. So c is the smallest, so c is negative, and b could be positive or negative, but since a ≥ b ≥ c, and a is positive, b can be positive or negative, but at least c is negative.So, going back, |x₁ - x₂| = |2a + b| / a. Because |a| = a, since a is positive. So the distance between A and B is sqrt(2) * |2a + b| / a.Therefore, to maximize the distance, we need to maximize |2a + b| / a. Let me denote this as:Let’s write that expression as |2a + b| / a = |2 + b/a|.Because |2a + b| / a = |2 + b/a|. So, we need to maximize |2 + b/a|. Let’s set k = b/a. Then, the expression becomes |2 + k|. So, we need to maximize |2 + k| given some constraints on k. Let's see what constraints we have on k.Given that a ≥ b ≥ c, and a + b + c = 0. Also, a > 0.Expressed in terms of k = b/a, and let’s define m = c/a. Then, since a + b + c = 0, dividing by a gives 1 + k + m = 0, so m = -1 - k.Also, since a ≥ b ≥ c, dividing by a (positive, so inequality signs remain the same):1 ≥ k ≥ m.But m = -1 - k, so substituting:1 ≥ k ≥ -1 - k.Therefore, the inequalities are:1 ≥ k,andk ≥ -1 - k.Let's solve the second inequality:k ≥ -1 - k2k ≥ -1k ≥ -0.5.Therefore, combining the inequalities: -0.5 ≤ k ≤ 1.So k is in the interval [-0.5, 1]. Therefore, our variable k is between -0.5 and 1.We need to maximize |2 + k| where k ∈ [-0.5, 1].The function |2 + k| is a V-shaped function with a vertex at k = -2. Since our interval is from -0.5 to 1, which is entirely to the right of -2, the function |2 + k| is increasing on this interval. Therefore, the maximum occurs at the right endpoint, which is k = 1.Therefore, the maximum of |2 + k| on [-0.5, 1] is |2 + 1| = 3.Hence, the maximum value of |2 + k| is 3, achieved when k = 1. Therefore, the maximum distance is sqrt(2) * 3 = 3*sqrt(2).But wait, hold on. Let me verify if k can actually reach 1. If k = 1, then m = -1 - 1 = -2. So we have a ≥ b ≥ c, with b = a*1 = a, and c = a*(-2). Therefore, a ≥ a ≥ -2a. Since a > 0, this implies a ≥ a (which is true) and a ≥ -2a. Since a is positive, -2a is negative, so a ≥ -2a is equivalent to a + 2a ≥ 0 => 3a ≥ 0, which is true. So yes, when k = 1, we have b = a and c = -2a. Then, a + b + c = a + a - 2a = 0, which is satisfied. So this case is possible.Therefore, when k = 1, the maximum |2 + k| is 3, so the maximum distance is 3*sqrt(2). Wait, but let me check if this is possible. If we have k = 1, then b = a. So the quadratic equation is ax² + bx + c = ax² + ax - 2a = 0. Dividing by a (since a ≠ 0), we get x² + x - 2 = 0. The roots of this equation are x = [-1 ± sqrt(1 + 8)] / 2 = [-1 ± 3]/2. So x₁ = 1, x₂ = -2. Then, points A(1, -2) and B(-2, 1). The distance between them is sqrt[(1 - (-2))² + (-2 - 1)^2] = sqrt[(3)^2 + (-3)^2] = sqrt(9 + 9) = sqrt(18) = 3*sqrt(2). So that's correct.But wait, in this case, we have a = 1, b = 1, c = -2, which satisfies a ≥ b ≥ c (1 ≥ 1 ≥ -2) and a + b + c = 0. So that's a valid case.But hold on, the problem says a ≥ b ≥ c. If a = b, that's allowed. The problem says "real numbers a, b, c satisfy a ≥ b ≥ c", so equality is allowed. So that's okay.But let's check another case. Suppose k = -0.5. Then, m = -1 - (-0.5) = -0.5. So in this case, b = -0.5a, c = -0.5a. Then, the quadratic equation is ax² - 0.5a x - 0.5a = 0. Dividing by a: x² - 0.5x - 0.5 = 0. The roots are [0.5 ± sqrt(0.25 + 2)] / 2 = [0.5 ± sqrt(2.25)] / 2 = [0.5 ± 1.5]/2. So x₁ = (0.5 + 1.5)/2 = 2/2 = 1, x₂ = (0.5 - 1.5)/2 = (-1)/2 = -0.5. Then, the distance between A(1, -0.5) and B(-0.5, 1) is sqrt[(1 + 0.5)^2 + (-0.5 - 1)^2] = sqrt[(1.5)^2 + (-1.5)^2] = sqrt(2.25 + 2.25) = sqrt(4.5) = (3)/sqrt(2) ≈ 2.121. Which is smaller than 3*sqrt(2) ≈ 4.242. So indeed, the maximum occurs when k = 1.But wait, is this the only case? Let me check another point. Suppose k = 0. Then, m = -1 - 0 = -1. So the quadratic equation becomes ax² + 0x - a = 0. So ax² - a = 0 => x² - 1 = 0 => x = ±1. Then, the roots are 1 and -1. The distance between A(1, -1) and B(-1, 1) is sqrt[(1 + 1)^2 + (-1 - 1)^2] = sqrt[4 + 4] = sqrt(8) = 2*sqrt(2) ≈ 2.828. Still less than 3*sqrt(2). So yes, the maximum does seem to be 3*sqrt(2).But let's check another case. Suppose k is something else. Let's say k = 0.5. Then, m = -1 - 0.5 = -1.5. Then the quadratic is ax² + 0.5a x - 1.5a = 0. Dividing by a: x² + 0.5x - 1.5 = 0. The roots are [-0.5 ± sqrt(0.25 + 6)]/2 = [-0.5 ± sqrt(6.25)]/2 = [-0.5 ± 2.5]/2. So x₁ = (2)/2 = 1, x₂ = (-3)/2 = -1.5. Then the distance between A(1, -1.5) and B(-1.5, 1) is sqrt[(1 + 1.5)^2 + (-1.5 - 1)^2] = sqrt[(2.5)^2 + (-2.5)^2] = sqrt(6.25 + 6.25) = sqrt(12.5) ≈ 3.535, which is about 2.5*sqrt(2), which is still less than 3*sqrt(2). So the maximum does seem to be achieved when k = 1.But wait, maybe there is another case when k is negative. For example, if k = -0.5, as before, we get the distance of 3/sqrt(2) ≈ 2.121. So even smaller.Therefore, it seems like the maximum is indeed achieved when k = 1, which gives the distance of 3*sqrt(2). Therefore, the answer should be 3*sqrt(2). However, let me verify once more if there are any constraints I might have missed.Wait, the original problem states a ≥ b ≥ c. In the case where k = 1, which implies b = a, so a ≥ a ≥ c. Since a = b, that's okay. Then c must be -2a. So c = -2a, which is less than a (since a is positive), so a ≥ b = a ≥ c = -2a. Since a is positive, -2a is indeed less than a, so the inequalities hold.Another check: If we take a = 1, b = 1, c = -2, then the quadratic equation is x² + x - 2 = 0, which factors as (x + 2)(x - 1) = 0, so roots x = -2 and x = 1. Then, points A(-2,1) and B(1,-2). Wait, hold on. Wait, the roots are x₁ and x₂. Which one is x₁ and which is x₂? Normally, the roots are ordered, but the problem doesn't specify which is which. However, the distance between (x₁, x₂) and (x₂, x₁) is the same regardless of the order. Because swapping x₁ and x₂ just swaps the coordinates, but the distance remains the same. So whether you take x₁ as the larger root or the smaller root, the distance calculation remains the same.But in the quadratic equation, the roots can be ordered. For example, in the case of x² + x - 2 = 0, the roots are 1 and -2. So if we take x₁ = 1, x₂ = -2, then A is (1, -2) and B is (-2, 1). The distance is sqrt[(1 - (-2))² + (-2 - 1)^2] = sqrt[9 + 9] = sqrt[18] = 3*sqrt(2). If we take x₁ = -2 and x₂ = 1, then A is (-2, 1) and B is (1, -2). The distance is the same. So regardless of the order, the distance is the same.Therefore, the maximum distance is indeed 3*sqrt(2). Therefore, the answer is boxed{3sqrt{2}}.But before finalizing, let me check if there are other cases where |2 + k| could be larger. For example, if k could be greater than 1. But from the earlier analysis, k is restricted to [-0.5, 1]. So k cannot exceed 1. Therefore, the maximum of |2 + k| on the interval [-0.5, 1] is at k = 1, which gives 3. Therefore, the maximum distance is 3*sqrt(2). So I think this is correct.Alternatively, perhaps there's a way to parameterize a, b, c under the given constraints and express the distance in terms of those parameters and find the maximum. Let me try that approach.Given that a + b + c = 0, so c = -a - b. Also, a ≥ b ≥ c. Since a is positive, and c = -a - b, so c must be ≤ b. So let's substitute c into the inequalities:a ≥ b ≥ -a - b.So the inequalities are:1. a ≥ b,2. b ≥ -a - b.From the second inequality, b ≥ -a - b ⇒ 2b ≥ -a ⇒ b ≥ -a/2.So combining with the first inequality, we have -a/2 ≤ b ≤ a.Given that a > 0.So we can express b as b = ta, where t is a parameter between -0.5 and 1 (as we found earlier). Then c = -a - b = -a - ta = -a(1 + t).Therefore, the quadratic equation is ax² + b x + c = a x² + ta x - a(1 + t) = 0.Dividing by a: x² + t x - (1 + t) = 0.Then, the roots are x = [-t ± sqrt(t² + 4(1 + t))]/2.Simplify the discriminant:D = t² + 4(1 + t) = t² + 4t + 4 = (t + 2)^2.Therefore, the roots are x = [-t ± |t + 2|]/2.Since t is between -0.5 and 1, t + 2 is between 1.5 and 3, which is positive. Therefore, |t + 2| = t + 2. Therefore, the roots are:x₁ = [-t + (t + 2)] / 2 = (2)/2 = 1,x₂ = [-t - (t + 2)] / 2 = (-2t - 2)/2 = -t - 1.Wait, this is interesting. So the roots are always 1 and -t -1. Wait, but that seems to suggest that one root is fixed at 1, and the other varies with t. Wait, let's verify this with previous examples.When t = 1 (so b = a, c = -2a):The quadratic equation is x² + 1x - 2 = 0, which has roots 1 and -2. According to the formula, x₂ = -1 -1 = -2. Correct.When t = -0.5 (so b = -0.5a, c = -a - (-0.5a) = -0.5a):The quadratic equation is x² - 0.5x - 0.5 = 0. The roots would be x₁ = 1 and x₂ = -(-0.5) -1 = 0.5 -1 = -0.5. Which matches the previous calculation.Similarly, when t = 0, x₂ = -0 -1 = -1. Correct.Therefore, the roots are always 1 and -t -1, regardless of t. Therefore, x₁ is always 1, and x₂ is -t -1. Wait, but does this mean that one root is fixed at 1? That's fascinating.But why is one root always 1? Let me check.Given the quadratic equation derived from the substitution:x² + t x - (1 + t) = 0.Let’s factor this quadratic:x² + t x - (1 + t) = (x - 1)(x + (1 + t)).Expanding (x - 1)(x + (1 + t)) = x² + (1 + t)x - x - (1 + t) = x² + t x - (1 + t). Yes, exactly. Therefore, the quadratic factors as (x - 1)(x + (1 + t)) = 0. Therefore, the roots are x = 1 and x = -(1 + t). Therefore, regardless of t, one root is always 1. That's interesting.Therefore, x₁ = 1 and x₂ = -(1 + t). Therefore, the points A and B are (1, -(1 + t)) and (-(1 + t), 1). Then, the distance between them is sqrt[(1 - (-(1 + t)))² + (-(1 + t) - 1)^2] = sqrt[(2 + t)^2 + (-2 - t)^2] = sqrt[(2 + t)^2 + (2 + t)^2] = sqrt[2*(2 + t)^2] = sqrt(2)*|2 + t|.Therefore, the distance is sqrt(2)*|2 + t|. To maximize this distance, we need to maximize |2 + t|. Since t is in the interval [-0.5, 1], then t ranges from -0.5 to 1. Therefore, 2 + t ranges from 2 - 0.5 = 1.5 to 2 + 1 = 3. Therefore, |2 + t| ranges from 1.5 to 3. Hence, the maximum value of |2 + t| is 3, achieved when t = 1. Therefore, the maximum distance is sqrt(2)*3 = 3*sqrt(2). Therefore, this confirms the earlier result.Therefore, the maximum distance is indeed 3*sqrt(2). This approach seems to be more straightforward. Since substituting t = b/a leads us to express the roots as 1 and -1 - t, then the distance formula simplifies to sqrt(2)*|2 + t|, which is maximized when t is as large as possible. Since t can be up to 1, the maximum is achieved at t = 1, leading to the maximum distance.Therefore, after multiple approaches and verifications, I can confidently conclude that the maximum distance is 3*sqrt(2).Final Answerboxed{3sqrt{2}}</think>Given real numbers (a, b, c) satisfy (a geq b geq c), (a + b + c = 0), and (a neq 0). We need to find the maximum distance between the points (A(x_1, x_2)) and (B(x_2, x_1)) where (x_1) and (x_2) are the roots of the quadratic equation (ax^2 + bx + c = 0).1. Vieta's Formulas and Distance Calculation: - The roots (x_1) and (x_2) satisfy (x_1 + x_2 = -frac{b}{a}) and (x_1x_2 = frac{c}{a}). - The distance between (A(x_1, x_2)) and (B(x_2, x_1)) is (sqrt{(x_1 - x_2)^2 + (x_2 - x_1)^2} = sqrt{2(x_1 - x_2)^2} = sqrt{2}|x_1 - x_2|).2. Expression for (|x_1 - x_2|): - Using the discriminant, (|x_1 - x_2| = frac{sqrt{b^2 - 4ac}}{|a|}). - Given (a + b + c = 0), substitute (c = -a - b) into the discriminant: (b^2 - 4ac = (2a + b)^2).3. Simplifying the Distance: - The distance becomes (sqrt{2} cdot frac{|2a + b|}{|a|}). - Let (k = frac{b}{a}), then the distance is (sqrt{2} cdot |2 + k|).4. Constraints on (k): - From (a geq b geq c) and (a + b + c = 0), we derive (-0.5 leq k leq 1).5. Maximizing (|2 + k|): - The function (|2 + k|) is maximized at (k = 1), giving (|2 + 1| = 3).6. Verification: - When (k = 1), (b = a) and (c = -2a). The quadratic equation (ax^2 + ax - 2a = 0) has roots (1) and (-2), leading to a distance of (sqrt{2} cdot 3 = 3sqrt{2}).Thus, the maximum distance between points (A) and (B) is (boxed{3sqrt{2}}).,<think>Okay, let's tackle this problem step by step. So, we have real numbers a, b, c satisfying a ≥ b ≥ c, a + b + c = 0, and a ≠ 0. The quadratic equation is ax² + bx + c = 0, with roots x₁ and x₂. We need to find the maximum distance between points A(x₁, x₂) and B(x₂, x₁) in the Cartesian coordinate system. First, let's recall the distance formula between two points. The distance between A and B would be sqrt[(x₂ - x₁)² + (x₁ - x₂)²]. Simplifying that, since both terms are squared differences of x₁ and x₂, we can factor out. Let's compute that:Distance = sqrt[(x₂ - x₁)² + (x₁ - x₂)²] = sqrt[2*(x₂ - x₁)²] = sqrt[2] * |x₂ - x₁|So, the distance is sqrt(2) times the absolute value of the difference between the roots. Therefore, to maximize the distance between A and B, we need to maximize |x₂ - x₁|. But |x₂ - x₁| is the absolute value of the difference of the roots of the quadratic equation. I remember there's a formula for the difference of roots. For a quadratic equation ax² + bx + c = 0, the difference of roots is sqrt(D)/|a|, where D is the discriminant b² - 4ac. Wait, but actually, the difference can be expressed as sqrt[(x₁ + x₂)² - 4x₁x₂] by the identity (x₁ - x₂)² = (x₁ + x₂)² - 4x₁x₂. But since for a quadratic equation, x₁ + x₂ = -b/a and x₁x₂ = c/a. So, substituting:(x₁ - x₂)² = (-b/a)² - 4*(c/a) = (b²/a²) - (4c/a) = (b² - 4ac)/a²Therefore, |x₁ - x₂| = sqrt(b² - 4ac)/|a|But since a ≥ b ≥ c and a + b + c = 0, and a ≠ 0, we need to ensure that the quadratic has real roots. The discriminant D = b² - 4ac must be non-negative. So, we need b² - 4ac ≥ 0. Given the constraints on a, b, c, we should check if this is always true or if we need to consider it as a condition.But the problem states that x₁ and x₂ are the two real roots, so the discriminant is non-negative. So, we can proceed.Therefore, the distance between A and B is sqrt(2) * sqrt(b² - 4ac)/|a|. Since a ≠ 0, and given a ≥ b ≥ c, but we don't know if a is positive or negative. Wait, hold on. Let's see: if a + b + c = 0, and a is the largest among a, b, c. If a is positive, then since a + b + c = 0, at least one of b or c must be negative. If a is negative, then all three numbers are negative, but since a ≥ b ≥ c, then a would be the least negative, but then a + b + c = 0 would require that they sum to zero. However, if a is negative, then b and c are also negative, so their sum would be negative, which contradicts a + b + c = 0. Therefore, a must be positive. Because if a were negative, then since a is the largest (a ≥ b ≥ c), all three numbers would be non-positive, and their sum can't be zero unless all are zero, but a ≠ 0. Therefore, a must be positive.So, a > 0, and since a + b + c = 0, then b and c must sum to -a. Also, since a ≥ b ≥ c, and a is positive, c must be negative or zero, because if c were positive, then all three would be positive, but their sum would be positive, not zero. So, c is negative or zero. Similarly, b could be positive or negative. For example, if a is 2, b is 0, c is -2. That satisfies a ≥ b ≥ c and a + b + c = 0.So, given that a is positive, we can write |a| = a. Therefore, the distance simplifies to sqrt(2) * sqrt(b² - 4ac)/a. So, we need to maximize sqrt(2) * sqrt(b² - 4ac)/a. Since sqrt(2) is a constant, we can equivalently maximize sqrt(b² - 4ac)/a, which would correspond to maximizing (b² - 4ac)/a². So, let's denote f(a, b, c) = (b² - 4ac)/a². Then, we need to maximize f(a, b, c) given the constraints a ≥ b ≥ c, a + b + c = 0, and a ≠ 0 (but a > 0).But since a, b, c are real numbers with a + b + c = 0, we can express c = -a - b. So, substitute c into f(a, b, c):f(a, b) = [b² - 4a*(-a - b)] / a² = [b² + 4a² + 4ab] / a² = (4a² + 4ab + b²)/a² = 4 + 4b/a + (b/a)²Let’s set t = b/a. Since a > 0, and a ≥ b ≥ c, then t is the ratio of b to a. Let's find the possible range of t. Given that a ≥ b ≥ c and c = -a - b, so:1. a ≥ b => b ≤ a => t = b/a ≤ 12. b ≥ c = -a - b => b ≥ -a - b => 2b ≥ -a => b ≥ -a/2 => t = b/a ≥ -1/2So, t is in [-1/2, 1]. Therefore, t ∈ [-1/2, 1]So, f(a, b) = 4 + 4t + t², where t ∈ [-1/2, 1]So, we need to maximize the quadratic function f(t) = t² + 4t + 4 over the interval t ∈ [-1/2, 1]. Let's compute the maximum of this quadratic function.The function f(t) = t² + 4t + 4. This is a parabola opening upwards, with vertex at t = -4/(2*1) = -2. The vertex is at t = -2, which is outside our interval [-1/2, 1]. Since the parabola opens upwards, the minimum is at t = -2, and the function increases as t moves away from -2. Therefore, on the interval [-1/2, 1], the maximum will occur at the endpoint with the larger t, which is t = 1.Therefore, the maximum of f(t) on [-1/2, 1] is f(1) = 1 + 4 + 4 = 9.Alternatively, checking at t = -1/2:f(-1/2) = (-1/2)^2 + 4*(-1/2) + 4 = 1/4 - 2 + 4 = 1/4 + 2 = 9/4 = 2.25At t = 1: 9, which is larger. So, maximum is 9.Therefore, the maximum value of (b² - 4ac)/a² is 9, so the maximum distance is sqrt(2) * sqrt(9) = sqrt(2)*3 = 3*sqrt(2). Hence, the maximum distance is 3√2.But wait, let me verify this. Let's check the steps again.First, we set c = -a - b. Then substituted into f(a, b, c) = (b² -4ac)/a².So, plug in c = -a - b:b² -4a*(-a -b) = b² +4a² +4abDivide by a²:(b² +4a² +4ab)/a² = (4a² +4ab + b²)/a² = (2a + b)^2 /a²Wait, 4a² +4ab +b² is (2a + b)^2. Because (2a + b)^2 = 4a² +4ab +b². So, actually, the expression simplifies to (2a + b)^2 /a² = [ (2a + b)/a ]² = (2 + b/a)^2 = (2 + t)^2So, f(t) = (2 + t)^2Therefore, f(t) = (t + 2)^2. So, this is a quadratic in t, which is a parabola opening upwards with vertex at t = -2. On the interval t ∈ [-1/2, 1], the maximum occurs at t =1, giving (1 +2)^2=9, as before. At t = -1/2, (-1/2 +2)^2 = (3/2)^2=9/4.Therefore, the maximum value of (b² -4ac)/a² is 9, achieved when t=1, i.e., when b/a=1 => b=a.But wait, the constraints are a ≥ b ≥ c. If b = a, then since a ≥ b, that implies a = b. Then, since a + b + c =0, c = -a -b = -2a.So, c = -2a. Then, since a ≥ b ≥ c, with a = b, c = -2a. Therefore, since a is positive, c = -2a is less than a and b (since a and b are equal and positive), so the inequalities a ≥ b ≥ c hold: a = b ≥ c = -2a.Therefore, this case is permissible.Therefore, when b = a, c = -2a. Then, the quadratic equation is ax² + bx + c = ax² + ax -2a = 0. Dividing by a (since a ≠ 0), we get x² + x -2 = 0. The roots are x = [-1 ± sqrt(1 +8)]/2 = [-1 ±3]/2. Therefore, x₁ = (2)/2 =1, x₂= (-4)/2 = -2. Therefore, the points A(1, -2) and B(-2,1). The distance between them is sqrt[( -2 -1)^2 + (1 - (-2))^2] = sqrt[(-3)^2 + (3)^2] = sqrt[9 +9] = sqrt[18] = 3√2. Which matches our previous conclusion.Therefore, the maximum distance is indeed 3√2.But let's check if this is indeed the maximum, or if there are other cases. For instance, what if t is at the lower bound of -1/2?If t = -1/2, then b = -a/2, and since a + b + c =0, c = -a -b = -a - (-a/2) = -a +a/2 = -a/2. So, c = -a/2. Then, since a ≥ b ≥ c, here b = -a/2 and c = -a/2. So, a ≥ -a/2 ≥ -a/2. But since a is positive, a ≥ -a/2 is true (since a >0 and -a/2 <0 < a). However, b = -a/2 and c = -a/2, so b = c. So, the quadratic equation becomes ax² + (-a/2)x + (-a/2) =0. Divide by a: x² - (1/2)x -1/2 =0. The roots are x = [1/2 ± sqrt(1/4 + 2)]/2 = [1/2 ± sqrt(9/4)]/2 = [1/2 ± 3/2]/2. So, x₁ = (1/2 +3/2)/2 = (4/2)/2 = 2/2 =1, x₂ = (1/2 -3/2)/2 = (-1)/2 = -1/2. Therefore, the points A(1, -1/2) and B(-1/2, 1). The distance is sqrt[( -1/2 -1)^2 + (1 - (-1/2))^2] = sqrt[(-3/2)^2 + (3/2)^2] = sqrt[9/4 +9/4] = sqrt[18/4] = sqrt[9/2] = (3)/√2 ≈ 2.121, which is less than 3√2 ≈4.242. So, indeed, the maximum occurs at t=1.But let's check if there are any other critical points. Since f(t) = (t +2)^2, its derivative is 2(t +2). Setting derivative to zero gives t = -2, which is outside of our interval [-1/2,1]. Therefore, the maximum on the interval is at t=1, as we found.Therefore, the maximum distance is 3√2.Wait, but let's also check if when t is somewhere in the middle, like t=0. If t=0, then b=0, c = -a -0 = -a. Then quadratic equation is ax² +0x -a =0 => ax² -a =0 => x² =1 => x=1 or x=-1. Then, points A(1, -1) and B(-1,1). The distance is sqrt[(-1 -1)^2 + (1 - (-1))^2] = sqrt[(-2)^2 + (2)^2] = sqrt[4+4] = sqrt[8] = 2√2 ≈2.828, which is still less than 3√2.Another test case: suppose a=3, b=0, c=-3. Then, the quadratic equation is 3x² +0x -3=0 => x² -1=0, roots ±1. Points (1, -1) and (-1,1) as before. Distance same as above.Wait, but according to the previous substitution when t=1, which is b=a, c=-2a. Let's take a=1, b=1, c=-2. Quadratic is x² +x -2=0, roots 1 and -2. Distance between (1, -2) and (-2,1) is 3√2. If a=2, b=2, c=-4. Quadratic is 2x² +2x -4=0. Divide by 2: x² +x -2=0, same roots. So, regardless of a, as long as b=a and c=-2a, the roots are 1 and -2 scaled by some factor? Wait, no. Wait, in the case where a=2, b=2, c=-4, the quadratic equation is 2x² +2x -4=0. Divide by 2: x² +x -2=0, roots x=1 and x=-2. So, same roots. Thus, the coordinates A and B are (1, -2) and (-2,1), leading to the same distance. So, the distance is independent of a, which cancels out in the expression. Therefore, regardless of a, the maximum distance is always 3√2. Therefore, the answer is 3√2.Hence, the maximum distance is boxed{3sqrt{2}}.Final AnswerThe maximum distance between points ( A ) and ( B ) is boxed{3sqrt{2}}.</think>Given real numbers (a, b, c) satisfy (a geqslant b geqslant c), (a + b + c = 0), and (a neq 0). We need to find the maximum distance between the points (A(x_1, x_2)) and (B(x_2, x_1)) where (x_1) and (x_2) are the roots of the quadratic equation (ax^2 + bx + c = 0).First, the distance between points (A) and (B) is given by:[sqrt{(x_2 - x_1)^2 + (x_1 - x_2)^2} = sqrt{2(x_2 - x_1)^2} = sqrt{2} |x_2 - x_1|]The absolute difference (|x_2 - x_1|) can be expressed using the discriminant of the quadratic equation:[|x_2 - x_1| = frac{sqrt{b^2 - 4ac}}{|a|}]Given (a + b + c = 0), we substitute (c = -a - b) into the discriminant:[b^2 - 4ac = b^2 - 4a(-a - b) = b^2 + 4a^2 + 4ab]Thus, the distance becomes:[sqrt{2} cdot frac{sqrt{b^2 + 4a^2 + 4ab}}{a} = sqrt{2} cdot frac{sqrt{(2a + b)^2}}{a} = sqrt{2} cdot frac{|2a + b|}{a}]Let (t = frac{b}{a}). Since (a geq b geq c), we have (-frac{1}{2} leq t leq 1). The expression simplifies to:[sqrt{2} cdot sqrt{left(2 + frac{b}{a}right)^2} = sqrt{2} cdot |2 + t|]To maximize this expression, we need to maximize (|2 + t|) over (t in [-frac{1}{2}, 1]). The maximum value occurs at (t = 1):[|2 + 1| = 3]Thus, the maximum distance is:[sqrt{2} cdot 3 = 3sqrt{2}]The maximum distance between points (A) and (B) is (boxed{3sqrt{2}}).,<think>Okay, so I have this problem here where I need to find the maximum distance between two points A(x₁, x₂) and B(x₂, x₁), where x₁ and x₂ are the roots of the quadratic equation ax² + bx + c = 0. The conditions given are that a, b, c are real numbers with a ≥ b ≥ c, a + b + c = 0, and a ≠ 0. First, I need to recall that the roots of a quadratic equation ax² + bx + c = 0 can be found using the quadratic formula: x = [-b ± √(b² - 4ac)]/(2a). So the roots x₁ and x₂ are given by that formula. Then, the distance between points A and B would be calculated using the distance formula in the Cartesian coordinate system. The distance D between two points (x₁, x₂) and (x₂, x₁) is √[(x₂ - x₁)² + (x₁ - x₂)²]. Simplifying that, it's √[2(x₂ - x₁)²] which is √2 * |x₂ - x₁|. So the distance D is √2 times the absolute difference between the roots.Therefore, to maximize D, I need to maximize |x₂ - x₁|. Since the quadratic equation has real roots (given that x₁ and x₂ are real), the discriminant must be non-negative. The discriminant is b² - 4ac ≥ 0.Given that a + b + c = 0, we can express one variable in terms of the others. Let's solve for c: c = -a - b. Substituting this into the discriminant, we get b² - 4a(-a - b) = b² + 4a² + 4ab. So the discriminant is b² + 4a² + 4ab. For real roots, this must be ≥ 0, but since a ≥ b ≥ c and a ≠ 0, I need to see how this relates.The difference between the roots is |x₂ - x₁|. From the quadratic formula, the difference is [√(b² - 4ac)]/a (since the denominator is 2a and the numerator is 2√(b² - 4ac), so dividing by 2a gives √(b² - 4ac)/a). Wait, actually, let me check that again.The roots are x₁ = [-b + √(b² - 4ac)]/(2a) and x₂ = [-b - √(b² - 4ac)]/(2a). So the difference x₂ - x₁ is [-b - √(b² - 4ac)]/(2a) - [-b + √(b² - 4ac)]/(2a) = [-2√(b² - 4ac)]/(2a) = -√(b² - 4ac)/a. Therefore, the absolute difference |x₂ - x₁| is √(b² - 4ac)/|a|. Since a is a real number and a ≠ 0, and given the ordering a ≥ b ≥ c, but not specified whether a is positive or negative. However, if a were negative, then the sum a + b + c = 0 would require that at least one of b or c is positive, but since a ≥ b ≥ c, if a is negative, then b and c would also be negative or zero, but their sum would have to be positive (since a = -b - c). Hmm, this might complicate things. Wait, perhaps we can assume a is positive? Let me check.Given that a ≥ b ≥ c and a + b + c = 0. Suppose a is positive. Then since a is the largest, and the sum is zero, at least one of b or c must be negative. If a is negative, then all three would be non-positive, but their sum is zero. However, since a ≥ b ≥ c, if a is negative, then all are negative, but a + b + c = 0. So they can't all be negative unless they are zero, but a ≠ 0. Therefore, a must be positive. Because if a were negative, then the sum of three non-positive numbers (with a ≠ 0) would be negative, which contradicts a + b + c = 0. Therefore, a must be positive. So we can say a > 0.Therefore, |a| = a, so |x₂ - x₁| = √(b² - 4ac)/a. But wait, the discriminant was calculated earlier as b² + 4a² + 4ab when substituting c = -a - b. Let me verify that:Original discriminant: b² - 4ac. Substitute c = -a - b:Discriminant = b² - 4a(-a - b) = b² + 4a² + 4ab. So discriminant D = (2a + b)². Wait, because 4a² + 4ab + b² factors as (2a + b)^2. That's interesting. So discriminant is (2a + b)^2. Therefore, √(D) = |2a + b|. But since a > 0, and a + b + c = 0, c = -a - b. Also, since a ≥ b ≥ c, then c = -a - b ≤ b. Let's see:From a ≥ b ≥ c, and c = -a - b, then b ≥ -a - b. So 2b ≥ -a => b ≥ -a/2.Similarly, a ≥ b, so combining with b ≥ -a/2, we have a ≥ b ≥ -a/2.Also, since a + b + c = 0 and c = -a - b, so the discriminant is (2a + b)^2. Therefore, the absolute difference between roots is |x₂ - x₁| = √(D)/a = |2a + b|/a. Therefore, the distance D between points A and B is √2 * |2a + b|/a.So, our goal is to maximize √2 * |2a + b| / a. Since a is positive, we can write this as √2 * |2 + b/a|.Let me set t = b/a. Then, since a ≥ b ≥ c, and c = -a - b, we can write:From a ≥ b => t ≤ 1.From b ≥ c => b ≥ -a - b => 2b ≥ -a => t = b/a ≥ -1/2.Therefore, t ∈ [-1/2, 1]. Also, since a + b + c = 0, and c = -a - b, which is c/a = -1 - t. So, in terms of t, the ratio c/a is -1 - t. But since a ≥ b ≥ c, we have 1 ≥ t ≥ (-1 - t). Let's check:From b ≥ c => t ≥ (-1 - t) => 2t ≥ -1 => t ≥ -1/2, which is already established. So the constraints are t ∈ [-1/2, 1].Therefore, we need to maximize |2 + t| where t ∈ [-1/2, 1]. Let's analyze the function f(t) = |2 + t| over the interval t ∈ [-1/2, 1].The expression inside the absolute value, 2 + t, is always positive in this interval because the smallest t is -1/2, so 2 + (-1/2) = 3/2 > 0. Therefore, |2 + t| = 2 + t, which is a linear increasing function. Therefore, it attains its maximum at t = 1. Therefore, the maximum value of |2 + t| is 2 + 1 = 3.Therefore, the maximum distance D is √2 * 3 = 3√2.But wait, before jumping to conclusion, let me check if t can actually reach 1. Since t = b/a ≤ 1, but we also have other constraints. Let's verify.If t = 1, then b = a. Then c = -a - b = -a - a = -2a. Now, check if the original conditions hold: a ≥ b ≥ c. If b = a and c = -2a, then since a > 0, c = -2a < a, so the ordering a ≥ b ≥ c becomes a ≥ a ≥ -2a, which simplifies to a ≥ a and a ≥ -2a. Since a > 0, a ≥ -2a is automatically true because -2a < 0 < a. Therefore, the conditions are satisfied when t = 1.Therefore, t can indeed be 1, so the maximum value of |2 + t| is 3, hence the maximum distance is 3√2.But wait, let me double-check. If a = b, then the quadratic equation becomes a x² + a x + c = 0. With c = -2a. So substituting, the equation is a x² + a x - 2a = 0. Dividing both sides by a (since a ≠ 0), we get x² + x - 2 = 0. The roots are x = [-1 ± √(1 + 8)]/2 = [-1 ± 3]/2. So x₁ = 1, x₂ = -2. Then points A(1, -2) and B(-2, 1). The distance between them is √[(-2 - 1)^2 + (1 - (-2))^2] = √[(-3)^2 + (3)^2] = √[9 + 9] = √18 = 3√2. So that checks out.But is this the maximum? Let's see if there are other possibilities. For instance, when t is at its minimum, t = -1/2. Then, b = -a/2, c = -a - b = -a - (-a/2) = -a/2. So c = -a/2. Then the quadratic equation is a x² + (-a/2) x + (-a/2) = 0. Dividing by a, x² - (1/2)x - 1/2 = 0. The roots are [1/2 ± √(1/4 + 2)]/2 = [1/2 ± √(9/4)]/2 = [1/2 ± 3/2]/2. So x₁ = (1/2 + 3/2)/2 = 4/2 /2 = 2/2 = 1, x₂ = (1/2 - 3/2)/2 = (-1)/2. Therefore, points A(1, -0.5) and B(-0.5, 1). The distance is √[(-0.5 - 1)^2 + (1 - (-0.5))^2] = √[(-1.5)^2 + (1.5)^2] = √[2.25 + 2.25] = √4.5 = 3/√2 ≈ 2.121, which is less than 3√2 ≈ 4.242. So yes, the maximum seems to occur at t = 1.But let's check another point in the interval. Suppose t = 0. Then b = 0, c = -a - 0 = -a. The quadratic equation is a x² + 0 x - a = 0 => x² - 1 = 0 => x = ±1. Then points A(1, -1) and B(-1, 1). The distance is √[(-1 - 1)^2 + (1 - (-1))^2] = √[(-2)^2 + 2^2] = √[4 + 4] = √8 ≈ 2.828, which is still less than 3√2.Alternatively, take t = -1/3. Then b = -a/3, c = -a - (-a/3) = -2a/3. The quadratic equation is a x² - (a/3)x - (2a/3) = 0. Dividing by a: x² - (1/3)x - 2/3 = 0. The roots are [1/3 ± √(1/9 + 8/3)]/2 = [1/3 ± √(25/9)]/2 = [1/3 ± 5/3]/2. So x₁ = (1/3 + 5/3)/2 = 6/3 /2 = 2/2 = 1, x₂ = (1/3 - 5/3)/2 = (-4/3)/2 = -2/3. Distance is √[(-2/3 -1)^2 + (1 - (-2/3))^2] = √[(-5/3)^2 + (5/3)^2] = √[(25/9 + 25/9)] = √(50/9) = 5√2/3 ≈ 2.357, still less than 3√2.Therefore, it seems that the maximum occurs when t = 1, which gives distance 3√2. So the answer should be 3√2.But just to be thorough, let's check if there are any other constraints we might have missed. The original problem states a ≥ b ≥ c and a + b + c = 0, and a ≠ 0. We considered that a must be positive, which seems correct. We converted the problem into variables t = b/a and found the maximum of |2 + t|, given t ∈ [-1/2, 1]. Since the maximum at t = 1 gives 3, leading to distance 3√2, and that corresponds to a valid case where a = b, c = -2a, satisfying all the original conditions. Is there a scenario where |2 + t| could be larger than 3 within the allowed t range? Since t can go up to 1, 2 + 1 = 3, and if t could be larger than 1, then |2 + t| would be larger, but due to the constraint a ≥ b, t cannot exceed 1. Therefore, 3 is indeed the maximum value of |2 + t| in this interval.Therefore, the maximum distance is 3√2. So I think that's the answer.Final Answerboxed{3sqrt{2}}</think>Given real numbers (a, b, c) satisfy (a geqslant b geqslant c), (a + b + c = 0), and (a neq 0). We need to find the maximum distance between the points (A(x_1, x_2)) and (B(x_2, x_1)) where (x_1) and (x_2) are the roots of the quadratic equation (ax^2 + bx + c = 0).1. Roots of the Quadratic Equation: The roots (x_1) and (x_2) are given by: [ x_1 = frac{-b + sqrt{b^2 - 4ac}}{2a}, quad x_2 = frac{-b - sqrt{b^2 - 4ac}}{2a} ] The absolute difference between the roots is: [ |x_1 - x_2| = frac{sqrt{b^2 - 4ac}}{a} ]2. Distance Calculation: The distance (D) between points (A) and (B) is: [ D = sqrt{(x_1 - x_2)^2 + (x_2 - x_1)^2} = sqrt{2(x_1 - x_2)^2} = sqrt{2} cdot |x_1 - x_2| ] Substituting the absolute difference: [ D = sqrt{2} cdot frac{sqrt{b^2 - 4ac}}{a} ]3. Substituting (c = -a - b): The discriminant becomes: [ b^2 - 4ac = b^2 + 4a^2 + 4ab = (2a + b)^2 ] Thus, the distance simplifies to: [ D = sqrt{2} cdot frac{|2a + b|}{a} = sqrt{2} cdot left|2 + frac{b}{a}right| ]4. Maximizing the Expression: Let (t = frac{b}{a}). Given (a geq b geq c) and (a + b + c = 0), we find (t in [-1/2, 1]). The expression to maximize is (|2 + t|).5. Maximum Value: The maximum value of (|2 + t|) in the interval ([-1/2, 1]) occurs at (t = 1), giving (|2 + 1| = 3). Therefore, the maximum distance is: [ D = sqrt{2} cdot 3 = 3sqrt{2} ]Thus, the maximum distance between the points (A) and (B) is (boxed{3sqrt{2}}).]

🔑:8. 3 sqrt{2}.From the problem, we know a>0, c<0.Thus, from a geqslant b geqslant c, we get frac{c}{a} leqslant frac{b}{c} leqslant 1.Also, from a+b+c=0, we get frac{b}{a}=-1-frac{c}{a}.Therefore, frac{c}{a} leqslant-1-frac{c}{a} leqslant 1, -2 leqslant frac{c}{a} leqslant-frac{1}{2}.Since a+b+c=0, the roots of a x^{2}+b x+c=0 are 1, frac{c}{a}.begin{array}{l}text { Hence }|A B|=sqrt{left(x_{1}-x_{2}right)^{2}+left(x_{2}-x_{1}right)^{2}} =sqrt{2}left|x_{1}-x_{2}right|=sqrt{2}left(1-frac{c}{a}right) . text { When } frac{c}{a}=-2 text {, }|A B|_{max }=3 sqrt{2} .end{array}