💡Okay, so I have this problem here about a right-angled triangle with an area of 60 cm². One of its sides is 15 cm, and it touches a circle. I need to find the radius of the circle. Hmm, let me think about how to approach this.First, since it's a right-angled triangle, I know that the area can be calculated using the formula: Area = (1/2) * base * height. They've given the area as 60 cm², and one of the sides is 15 cm. I'm assuming this 15 cm side is either the base or the height. Let me denote the other side as 'h'. So, plugging into the area formula:60 = (1/2) * 15 * hSimplifying that, I get:60 = (15/2) * h 60 = 7.5 * h h = 60 / 7.5 h = 8 cmOkay, so the other side is 8 cm. Now, since it's a right-angled triangle, I can find the hypotenuse using the Pythagorean theorem. Let me denote the hypotenuse as 'c'.c² = 15² + 8² c² = 225 + 64 c² = 289 c = √289 c = 17 cmSo, the sides of the triangle are 8 cm, 15 cm, and 17 cm.Now, the triangle touches a circle. I'm not entirely sure what this means. Does it mean the circle is inscribed in the triangle, or is the triangle inscribed in the circle? The problem says the triangle touches the circle, which makes me think that the circle is tangent to all three sides of the triangle, meaning it's an inscribed circle or incircle.If it's an incircle, the radius can be found using the formula:r = Area / swhere 's' is the semi-perimeter of the triangle.First, let's find the semi-perimeter (s):s = (a + b + c) / 2 s = (8 + 15 + 17) / 2 s = (40) / 2 s = 20 cmNow, plug this into the formula for the radius:r = 60 / 20 r = 3 cmWait, that seems too small. Let me double-check. The area is 60, semi-perimeter is 20, so 60 divided by 20 is indeed 3. But intuitively, if the sides are 8, 15, and 17, the inradius being 3 cm seems a bit small. Maybe I misunderstood the problem.Alternatively, maybe the circle is circumscribed around the triangle, meaning the triangle is inscribed in the circle. In that case, the radius would be the circumradius. For a right-angled triangle, the circumradius is half the hypotenuse.So, hypotenuse is 17 cm, so the circumradius would be 17 / 2 = 8.5 cm.But the problem says the triangle touches the circle. If it's circumscribed, all three vertices lie on the circle, but the sides are tangent to the circle only if it's an incircle. Hmm, this is confusing.Wait, let me clarify. If the circle is tangent to all three sides, it's the incircle, and the radius is 3 cm. If the circle passes through all three vertices, it's the circumcircle, and the radius is 8.5 cm. The problem says "the triangle touches the circle," which could mean that the circle is tangent to the triangle, implying the incircle. But 3 cm seems small.Alternatively, maybe the circle is tangent to one of the sides and passes through the other two vertices? That would be a different scenario. But the problem doesn't specify which sides are tangent or how the triangle touches the circle.Wait, let me read the problem again: "The area of the right-angled triangle is 60 cm². The triangle touches the circle, and one side of the triangle has a length of 15 cm. What is the radius of the circle?"It doesn't specify whether the circle is inscribed or circumscribed. Hmm. Maybe I need to consider both possibilities.If it's the incircle, radius is 3 cm. If it's the circumcircle, radius is 8.5 cm. But 3 cm seems too small for a triangle with sides 8, 15, 17. Let me visualize the triangle. The inradius is the distance from the incenter to each side, which is indeed 3 cm. The circumradius is from the circumcenter to each vertex, which is 8.5 cm.But the problem says "the triangle touches the circle." If it's the incircle, all three sides are tangent to the circle, so the triangle touches the circle at three points. If it's the circumcircle, the triangle touches the circle at three vertices. So, both cases involve the triangle touching the circle, just in different ways.But the problem doesn't specify, so maybe I need to consider both. However, usually, when it's just said that the triangle touches the circle, it might refer to the incircle, as the circumcircle is more about the triangle being inscribed in the circle.But let me think again. The inradius formula is r = Area / s, which gives 3 cm. The circumradius is half the hypotenuse, which is 8.5 cm.Given that one side is 15 cm, and the inradius is 3 cm, that seems consistent. Alternatively, if it's the circumradius, 8.5 cm is also consistent.Wait, maybe the problem is referring to an excircle? An ex-circle is tangent to one side and the extensions of the other two sides. But the problem says the triangle touches the circle, not necessarily all sides.Alternatively, maybe the circle is tangent to one side and the other two sides are chords of the circle. But without more information, it's hard to tell.Wait, let's see. If the circle is tangent to one side and passes through the other two vertices, then we can set up equations to find the radius.Let me denote the triangle as ABC, right-angled at A, with AB = 15 cm, AC = 8 cm, and BC = 17 cm. Suppose the circle is tangent to side AB and passes through points A and C. Then, the center of the circle lies somewhere, and the distance from the center to AB is equal to the radius, and the distance from the center to A and C is also equal to the radius.But this seems complicated. Maybe it's better to stick with the incircle or circumcircle.Given that the problem is likely referring to the incircle, as it's a common problem, and the inradius is 3 cm. But 3 cm seems small, but mathematically, it's correct.Alternatively, maybe the circle is tangent to the hypotenuse and the two legs. That would be the incircle, radius 3 cm.Alternatively, if the circle is tangent to one of the legs and the hypotenuse, but not the other leg, that would be an ex-circle. But the problem doesn't specify.Wait, the problem says "the triangle touches the circle," which is a bit vague. It could mean that the circle is tangent to all three sides, which is the incircle, or it could mean that the circle is tangent to one side and passes through the other two vertices, which would be a different circle.Given the ambiguity, but considering the problem mentions one side is 15 cm, and asks for the radius, and given that the inradius is 3 cm, which is a standard formula, I think the intended answer is 3 cm.But wait, let me check again. The inradius formula is r = (a + b - c)/2 for a right-angled triangle, where c is the hypotenuse. Wait, no, that's not correct. The formula is r = (a + b - c)/2, which for our case:r = (8 + 15 - 17)/2 = (6)/2 = 3 cm. Yes, that's correct.So, the inradius is indeed 3 cm.Alternatively, if it's the circumradius, it's 8.5 cm.But given the problem statement, I think it's referring to the incircle, so the radius is 3 cm.Wait, but 3 cm seems small. Let me think about the triangle. The sides are 8, 15, 17. The inradius is 3 cm, which is the distance from the incenter to each side. The incenter is located at the intersection of the angle bisectors.Alternatively, maybe the circle is tangent to one side and passes through the other two vertices, making it a different circle. Let me try to visualize that.Suppose the circle is tangent to the side of length 15 cm and passes through the other two vertices. Let me denote the triangle as ABC, with AB = 15 cm, AC = 8 cm, and BC = 17 cm, right-angled at A.If the circle is tangent to AB and passes through A and C, then the center of the circle must lie somewhere such that it is equidistant from A and C and also at a distance equal to the radius from AB.Let me set up a coordinate system. Let me place point A at (0,0), point B at (15,0), and point C at (0,8). Then, the circle is tangent to AB, which is the x-axis from (0,0) to (15,0). The circle passes through A (0,0) and C (0,8). Let the center of the circle be at (h, k). Since the circle is tangent to AB (the x-axis), the distance from the center to AB is equal to the radius, which is k.Also, the circle passes through A (0,0) and C (0,8). So, the distance from the center (h, k) to A is equal to the radius, and the distance from the center to C is also equal to the radius.So, distance from center to A:√[(h - 0)² + (k - 0)²] = radius √(h² + k²) = k Squaring both sides: h² + k² = k² So, h² = 0 Thus, h = 0So, the center is at (0, k). Now, the distance from the center to C (0,8) is:√[(0 - 0)² + (8 - k)²] = |8 - k| = radius = kSo, |8 - k| = k Since k is positive, 8 - k = k 8 = 2k k = 4So, the radius is 4 cm.Wait, that's different from the inradius. So, if the circle is tangent to AB and passes through A and C, the radius is 4 cm.Alternatively, if the circle is tangent to AC and passes through A and B, we can do a similar calculation.Let me try that. Suppose the circle is tangent to AC (the y-axis from (0,0) to (0,8)) and passes through A (0,0) and B (15,0). Let the center be at (h, k). Since it's tangent to AC, the distance from the center to AC is equal to the radius, which is h.Also, the distance from the center to A is equal to the radius:√(h² + k²) = h Squaring: h² + k² = h² So, k² = 0 k = 0Thus, the center is at (h, 0). Now, the distance from the center to B (15,0) is:√[(15 - h)² + (0 - 0)²] = |15 - h| = radius = hSo, |15 - h| = h Since h is positive, 15 - h = h 15 = 2h h = 7.5Thus, the radius is 7.5 cm.So, depending on which side the circle is tangent to, the radius can be 4 cm or 7.5 cm.But the problem says "the triangle touches the circle," which is a bit vague. It could mean that the circle is tangent to one side and passes through the other two vertices, which would give us either 4 cm or 7.5 cm as the radius.Alternatively, if the circle is tangent to all three sides, it's the incircle with radius 3 cm.Given that the problem mentions "the triangle touches the circle," without specifying how, it's a bit ambiguous. However, since it's a right-angled triangle, the inradius is a standard calculation, and the problem might be expecting that.But wait, earlier when I considered the circle tangent to AB and passing through A and C, I got a radius of 4 cm. Similarly, tangent to AC and passing through A and B, radius is 7.5 cm.Alternatively, if the circle is tangent to BC (the hypotenuse) and passes through A and B, that would be another case.Let me try that. Suppose the circle is tangent to BC and passes through A and B. Let me denote the center as (h, k). The distance from the center to BC must be equal to the radius, and the distance from the center to A and B must also be equal to the radius.First, let's find the equation of BC. Points B (15,0) and C (0,8). The slope of BC is (8 - 0)/(0 - 15) = -8/15. So, the equation of BC is y = (-8/15)x + 8.The distance from the center (h, k) to the line BC is equal to the radius r. The formula for the distance from a point (h, k) to the line ax + by + c = 0 is |ah + bk + c| / √(a² + b²).Rewriting the equation of BC: (8/15)x + y - 8 = 0. So, a = 8/15, b = 1, c = -8.Thus, the distance is |(8/15)h + k - 8| / √((8/15)² + 1²) = r.Also, the distance from the center to A (0,0) is √(h² + k²) = r.Similarly, the distance from the center to B (15,0) is √((h - 15)² + k²) = r.So, we have three equations:1. |(8/15)h + k - 8| / √((64/225) + 1) = r 2. √(h² + k²) = r 3. √((h - 15)² + k²) = rThis seems complicated, but let's try to solve it.First, let's note that equations 2 and 3 both equal r, so we can set them equal to each other:√(h² + k²) = √((h - 15)² + k²)Squaring both sides:h² + k² = (h - 15)² + k² h² = h² - 30h + 225 0 = -30h + 225 30h = 225 h = 7.5So, h = 7.5 cm.Now, plug h = 7.5 into equation 2:√(7.5² + k²) = r √(56.25 + k²) = rNow, plug h = 7.5 into equation 1:|(8/15)*7.5 + k - 8| / √((64/225) + 1) = rCalculate (8/15)*7.5:(8/15)*7.5 = (8/15)*(15/2) = 8/2 = 4So, the numerator becomes |4 + k - 8| = |k - 4|The denominator is √((64/225) + 1) = √((64 + 225)/225) = √(289/225) = 17/15Thus, equation 1 becomes:|k - 4| / (17/15) = r |k - 4| * (15/17) = rBut from equation 2, r = √(56.25 + k²)So, we have:|k - 4| * (15/17) = √(56.25 + k²)Let me square both sides to eliminate the square root:(|k - 4| * (15/17))² = (56.25 + k²) (k - 4)² * (225/289) = 56.25 + k²Multiply both sides by 289 to eliminate the denominator:(k - 4)² * 225 = (56.25 + k²) * 289Expand (k - 4)²:(k² - 8k + 16) * 225 = 56.25 * 289 + k² * 289Calculate 56.25 * 289:56.25 * 289 = (56 + 0.25) * 289 = 56*289 + 0.25*289 56*289: Let's calculate 56*200 = 11200, 56*89 = 5004, so total 11200 + 5004 = 16204 0.25*289 = 72.25 So, total 16204 + 72.25 = 16276.25Now, expand the left side:225k² - 1800k + 3600 = 16276.25 + 289k²Bring all terms to one side:225k² - 1800k + 3600 - 16276.25 - 289k² = 0 (225k² - 289k²) + (-1800k) + (3600 - 16276.25) = 0 -64k² - 1800k - 12676.25 = 0Multiply both sides by -1:64k² + 1800k + 12676.25 = 0This is a quadratic equation in k. Let's calculate the discriminant:D = 1800² - 4*64*12676.25 D = 3,240,000 - 4*64*12676.25 First, calculate 4*64 = 256 Then, 256*12676.25 = ?Calculate 12676.25 * 256:12676.25 * 200 = 2,535,250 12676.25 * 56 = ?Calculate 12676.25 * 50 = 633,812.5 12676.25 * 6 = 76,057.5 Total = 633,812.5 + 76,057.5 = 709,870 So, total 2,535,250 + 709,870 = 3,245,120Thus, D = 3,240,000 - 3,245,120 = -5,120Wait, the discriminant is negative, which means there are no real solutions. That can't be right. Did I make a mistake in calculations?Let me double-check the earlier steps.We had:(k - 4)² * 225 = 56.25 * 289 + k² * 289 (k² - 8k + 16) * 225 = 16276.25 + 289k² 225k² - 1800k + 3600 = 16276.25 + 289k² 225k² - 1800k + 3600 - 16276.25 - 289k² = 0 (225k² - 289k²) = -64k² -64k² - 1800k + (3600 - 16276.25) = -64k² - 1800k - 12676.25 = 0 Multiply by -1: 64k² + 1800k + 12676.25 = 0Yes, that's correct. Then discriminant D = 1800² - 4*64*12676.25 1800² = 3,240,000 4*64 = 256 256*12676.25 = 3,245,120 So, D = 3,240,000 - 3,245,120 = -5,120Negative discriminant implies no real solution, which suggests that such a circle does not exist. That's confusing because geometrically, I thought it should be possible.Wait, maybe I made a mistake in setting up the equations. Let me check.We assumed the circle is tangent to BC and passes through A and B. But in reality, if the circle is tangent to BC and passes through A and B, the center must lie on the perpendicular bisector of AB. Since AB is horizontal from (0,0) to (15,0), its perpendicular bisector is the vertical line x = 7.5. So, the center must be at (7.5, k). That's why we found h = 7.5 earlier.But then, when we tried to solve for k, we ended up with a negative discriminant, which suggests no solution. That's odd.Alternatively, maybe the circle is tangent to BC and passes through A and C. Let me try that.So, the circle is tangent to BC and passes through A (0,0) and C (0,8). Let the center be at (h, k). The distance from the center to BC is equal to the radius, and the distance from the center to A and C is equal to the radius.Again, the equation of BC is y = (-8/15)x + 8.The distance from (h, k) to BC is |(8/15)h + k - 8| / √((8/15)² + 1) = r.Also, distance from center to A: √(h² + k²) = r Distance from center to C: √(h² + (k - 8)²) = rSo, we have:1. |(8/15)h + k - 8| / √(64/225 + 1) = r 2. √(h² + k²) = r 3. √(h² + (k - 8)²) = rFrom equations 2 and 3:√(h² + k²) = √(h² + (k - 8)²) Squaring both sides: h² + k² = h² + (k - 8)² k² = k² - 16k + 64 0 = -16k + 64 16k = 64 k = 4So, k = 4 cm.Now, plug k = 4 into equation 2:√(h² + 16) = rNow, plug h and k into equation 1:|(8/15)h + 4 - 8| / √(64/225 + 1) = r |(8/15)h - 4| / (17/15) = r |8h/15 - 4| * (15/17) = r |8h - 60| / 17 = rBut from equation 2, r = √(h² + 16)So, |8h - 60| / 17 = √(h² + 16)Square both sides:(8h - 60)² / 289 = h² + 16 (64h² - 960h + 3600) / 289 = h² + 16 Multiply both sides by 289:64h² - 960h + 3600 = 289h² + 4624 64h² - 960h + 3600 - 289h² - 4624 = 0 -225h² - 960h - 1024 = 0 Multiply by -1:225h² + 960h + 1024 = 0Calculate discriminant D:D = 960² - 4*225*1024 D = 921,600 - 4*225*1024 Calculate 4*225 = 900 900*1024 = 921,600 So, D = 921,600 - 921,600 = 0So, D = 0, which means one real solution:h = (-960) / (2*225) = -960 / 450 = -2.133...But h is negative, which would place the center at (-2.133, 4). However, in our coordinate system, the triangle is in the first quadrant, so a negative h would place the center outside the triangle. Is that acceptable?Well, geometrically, the circle could be outside the triangle, but let's check if it makes sense.If h = -2.133, then the center is at (-2.133, 4). The radius r = √(h² + k²) = √(4.55 + 16) = √20.55 ≈ 4.53 cm.But let's check the distance from the center to BC:|(8/15)*(-2.133) + 4 - 8| / (17/15) = |(-1.413) - 4| / (1.133) = | -5.413 | / 1.133 ≈ 5.413 / 1.133 ≈ 4.78But r ≈ 4.53, which is not equal to 4.78. Hmm, that's inconsistent. Did I make a mistake?Wait, let's calculate h more accurately. h = -960 / 450 = -1.666... cm, not -2.133. Wait, 960 / 450 = 2.133, but since it's negative, h = -2.133.Wait, let me recalculate:h = -960 / (2*225) = -960 / 450 = -2.133...So, h ≈ -2.133 cm.Then, r = √(h² + k²) = √(4.55 + 16) = √20.55 ≈ 4.53 cm.Now, distance from center to BC:|(8/15)*(-2.133) + 4 - 8| / (17/15) = |(-1.413) - 4| / (1.133) = | -5.413 | / 1.133 ≈ 5.413 / 1.133 ≈ 4.78 cmBut r ≈ 4.53 cm, which is not equal to 4.78 cm. This inconsistency suggests an error in the calculation.Wait, perhaps I made a mistake in the algebra when squaring both sides. Let me go back.We had:|8h - 60| / 17 = √(h² + 16)Squaring both sides:(8h - 60)² / 289 = h² + 16 (64h² - 960h + 3600) / 289 = h² + 16 Multiply both sides by 289:64h² - 960h + 3600 = 289h² + 4624 64h² - 960h + 3600 - 289h² - 4624 = 0 -225h² - 960h - 1024 = 0 Multiply by -1:225h² + 960h + 1024 = 0Now, discriminant D = 960² - 4*225*1024 = 921,600 - 921,600 = 0So, h = (-960) / (2*225) = -960 / 450 = -2.133...So, h ≈ -2.133 cm.But then, the distance from the center to BC is not equal to r, which suggests an inconsistency. This might be due to the fact that the circle is not only tangent to BC but also passes through A and C, which might not be possible in this configuration.Alternatively, perhaps the circle is tangent to BC and passes through A and C, but the center is outside the triangle, which is acceptable, but the distance from the center to BC must equal the radius.Wait, let's recalculate the distance from the center (-2.133, 4) to BC.The equation of BC is y = (-8/15)x + 8.The distance formula is |(8/15)x + y - 8| / √( (8/15)² + 1 )Plugging in x = -2.133, y = 4:|(8/15)*(-2.133) + 4 - 8| / √(64/225 + 1) = |(-1.413) - 4| / (17/15) = | -5.413 | / (1.133) ≈ 5.413 / 1.133 ≈ 4.78 cmBut the radius r = √(h² + k²) ≈ √(4.55 + 16) ≈ 4.53 cmSo, 4.78 ≈ 4.53? Not quite. There's a discrepancy here. It seems like the calculations are off, possibly due to rounding errors or an incorrect assumption.Alternatively, maybe the circle is tangent to BC and passes through A and B, but as we saw earlier, that leads to a negative discriminant, implying no solution. So, perhaps such a circle does not exist.Given this confusion, I think the problem is more likely referring to the incircle, which has a radius of 3 cm, or the circumcircle, which has a radius of 8.5 cm.But earlier, when considering the circle tangent to AB and passing through A and C, we got a radius of 4 cm, and tangent to AC and passing through A and B, radius of 7.5 cm.Given that the problem mentions "the triangle touches the circle," and without more specifics, it's ambiguous. However, considering the inradius is a standard calculation, and the problem might be expecting that, I think the answer is 3 cm.But wait, let me think again. If the circle is tangent to one side and passes through the other two vertices, the radius could be 4 cm or 7.5 cm, depending on which side it's tangent to. Since the problem mentions one side is 15 cm, which is the longer leg, perhaps the circle is tangent to the hypotenuse, but as we saw, that leads to no solution.Alternatively, if the circle is tangent to the side of 15 cm and passes through the other two vertices, the radius is 4 cm. If it's tangent to the side of 8 cm and passes through the other two vertices, the radius is 7.5 cm.Given that the problem mentions "one side of the triangle has a length of 15 cm," it's possible that the circle is tangent to this side, making the radius 4 cm.But earlier, when I considered the circle tangent to AB (15 cm) and passing through A and C, I got a radius of 4 cm. Let me confirm that.Yes, in that case, the center was at (0,4), and the radius was 4 cm. That seems consistent.So, if the circle is tangent to the side of 15 cm and passes through the other two vertices, the radius is 4 cm.Alternatively, if it's the incircle, radius is 3 cm.Given the ambiguity, but considering the problem mentions "the triangle touches the circle," which could mean that the circle is tangent to one side and passes through the other two vertices, I think the answer is 4 cm.But wait, earlier when I considered the incircle, the radius was 3 cm, which is a standard result. However, the problem didn't specify that the circle is tangent to all three sides, just that the triangle touches the circle. So, it's possible that the circle is tangent to one side and passes through the other two vertices, giving a radius of 4 cm.Alternatively, if the circle is the circumcircle, radius is 8.5 cm.Given the problem's phrasing, I think the intended answer is the inradius, 3 cm. However, considering the possibility of the circle being tangent to one side and passing through the other two vertices, the radius could be 4 cm or 7.5 cm.But since the problem mentions "the triangle touches the circle," without specifying how, and given that the inradius is a standard calculation, I think the answer is 3 cm.Wait, but earlier when I considered the circle tangent to AB and passing through A and C, I got a radius of 4 cm, which is different from the inradius. So, perhaps the problem is referring to that.Given the confusion, I think the problem is more likely referring to the incircle, so the radius is 3 cm.But to be thorough, let me check the inradius formula again.For a right-angled triangle, the inradius is given by r = (a + b - c)/2, where c is the hypotenuse.So, r = (8 + 15 - 17)/2 = (6)/2 = 3 cm.Yes, that's correct.Therefore, the radius of the circle is 3 cm.

💡Alright, so I have this problem where there are n students standing in a line, each in positions 1 to n. When the teacher looks away, some students move to different positions, and when the teacher looks back, the students are in a new line. The goal is to figure out the maximum total number of steps all the students have moved. Each student's movement is measured by the absolute difference between their original position and their new position. So, if a student was in position i and moves to position j, their movement is |i - j| steps. We need to find the maximum sum of these steps for all students.Hmm, okay. So, first, I need to think about permutations because each student is moving to a new position, and no two students can occupy the same position. So, the new arrangement is a permutation of the original positions.Let me denote the original positions as 1, 2, 3, ..., n. After the students move, their new positions are some rearrangement of these numbers. Let's call the new position of the student originally in position i as π(i), where π is a permutation of the numbers 1 to n.So, the total movement S is the sum from i=1 to n of |i - π(i)|. We need to maximize this sum.I remember that in permutations, the maximum sum of |i - π(i)| is related to reversing the order. So, if we reverse the order, the student from position 1 goes to position n, the student from position 2 goes to position n-1, and so on. This seems like it would maximize the distance each student moves.Let me test this idea with a small n, say n=3. The original positions are 1, 2, 3. If we reverse them, we get 3, 2, 1. The movements are |1-3| + |2-2| + |3-1| = 2 + 0 + 2 = 4. Is this the maximum? Let's see other permutations:- 1, 3, 2: movements are |1-1| + |2-3| + |3-2| = 0 + 1 + 1 = 2- 2, 1, 3: movements are |1-2| + |2-1| + |3-3| = 1 + 1 + 0 = 2- 2, 3, 1: movements are |1-2| + |2-3| + |3-1| = 1 + 1 + 2 = 4- 3, 1, 2: movements are |1-3| + |2-1| + |3-2| = 2 + 1 + 1 = 4So, the maximum is indeed 4, which is achieved by reversing the order or some other permutations that also result in the same total movement. So, for n=3, the maximum sum is 4.Wait, let me calculate n=4 as another test. Original positions 1,2,3,4. Reversed, it's 4,3,2,1. Movements are |1-4| + |2-3| + |3-2| + |4-1| = 3 + 1 + 1 + 3 = 8.Let me check another permutation: 4,1,2,3. Movements are |1-4| + |2-1| + |3-2| + |4-3| = 3 + 1 + 1 + 1 = 6. That's less than 8.Another permutation: 3,4,1,2. Movements are |1-3| + |2-4| + |3-1| + |4-2| = 2 + 2 + 2 + 2 = 8. So, same as reversing.So, for n=4, the maximum sum is 8.Hmm, so for n=3, maximum is 4, which is floor(3^2 / 2) = floor(9/2)=4. For n=4, maximum is 8, which is floor(16/2)=8. So, seems like the maximum is floor(n² / 2). Let me test n=2.n=2: original positions 1,2. Reversed: 2,1. Movements: |1-2| + |2-1| =1 +1=2. Which is floor(4/2)=2. Correct.n=1: trivially 0, which is floor(1/2)=0. Correct.So, seems like the formula is floor(n² / 2). So, the maximum sum is floor(n squared over 2).But let me think deeper. Why is this the case? When we reverse the permutation, each student moves as far as possible. For even n, each student moves exactly n/2 positions, but wait, no. For n=4, each student doesn't move 2 positions, but rather, the total is 8, which is 4*2. Wait, n=4, 4² /2=8. So, for n=4, each pair contributes 4 steps? Wait, no, each student contributes their movement.Wait, perhaps another way to think about it: the total movement is the sum over all students of |i - π(i)|. To maximize this, we want to pair the smallest i with the largest π(i), the second smallest with the second largest, etc. So, reversing the order does exactly that.So, for each i from 1 to n, π(i) = n +1 -i. So, the movement for each student is |i - (n +1 -i)| = |2i - (n +1)|. So, the total movement is sum_{i=1}^n |2i - (n +1)|.Let me compute this sum for general n.Case 1: n is even. Let n=2k.Then, for i=1 to 2k, |2i - (2k +1)|.Let me compute this:For i=1: |2 - (2k +1)| = | - (2k -1)| = 2k -1For i=2: |4 - (2k +1)| = | - (2k -3)| = 2k -3...For i=k: |2k - (2k +1)| = | -1| =1For i=k+1: |2(k+1) - (2k +1)| = |2k +2 -2k -1|=1For i=k+2: |2(k+2) - (2k +1)|= |2k +4 -2k -1|=3...For i=2k: |4k - (2k +1)|= |2k -1|So, the sum is (2k -1) + (2k -3) + ... +1 +1 +3 + ... + (2k -1)This is symmetric, so the total sum is 2*(1 +3 +5 +...+(2k -1)).The sum of the first k odd numbers is k². So, 2*(k²) = 2k².But n=2k, so k=n/2. Therefore, 2*(n/2)²=2*(n²/4)=n²/2.So, for even n, the total movement is n²/2.Case 2: n is odd. Let n=2k +1.Similarly, for i=1 to 2k+1, |2i - (2k +2)|.Compute:For i=1: |2 - (2k +2)|= | -2k|=2kFor i=2: |4 - (2k +2)|= | -2k +2|=2k -2...For i=k+1: |2(k+1) - (2k +2)|= |2k +2 -2k -2|=0For i=k+2: |2(k+2) - (2k +2)|= |2k +4 -2k -2|=2...For i=2k+1: |2(2k+1) - (2k +2)|= |4k +2 -2k -2|=2kSo, the sum is 2k + (2k -2) + ... +2 +0 +2 +...+(2k -2)+2k.Wait, let's see:From i=1 to i=k+1, the terms are 2k, 2k-2, ..., 2, 0.From i=k+2 to i=2k+1, the terms are 2, 4, ..., 2k.So, the total sum is (2k + (2k -2) + ... +2 +0) + (2 +4 +...+2k).But notice that the first part is the sum from 0 to 2k with step 2, and the second part is the sum from 2 to 2k with step 2.So, the total sum is (sum from m=0 to m=2k step 2 of m) + (sum from m=2 to m=2k step 2 of m).But the first sum is 0 + 2 + 4 + ... +2k, which is 2*(0 +1 +2 +...+k) =2*(k(k +1)/2)=k(k +1).The second sum is 2 +4 +...+2k, which is 2*(1 +2 +...+k)=2*(k(k +1)/2)=k(k +1).So, total sum is k(k +1) +k(k +1)=2k(k +1).But n=2k +1, so k=(n -1)/2.Substituting, 2*((n -1)/2)*((n -1)/2 +1)=2*((n -1)/2)*((n +1)/2)= (n -1)(n +1)/2=(n² -1)/2.But wait, for n odd, n² is odd, so (n² -1)/2 is an integer, which is equal to floor(n² /2).Because for odd n, n²=2m +1, so floor(n² /2)=m=(n² -1)/2.So, for n odd, the total movement is (n² -1)/2, which is floor(n² /2). For n even, it's n² /2, which is also floor(n² /2) since n² is even.Therefore, regardless of whether n is even or odd, the maximum total movement is floor(n² /2).So, the answer is floor(n squared over 2).**Final Answer**The maximal sum of steps is boxed{leftlfloor dfrac{n^2}{2} rightrfloor}.

💡Okay, so I need to find the cosecant of 315 degrees. Hmm, I remember that cosecant is the reciprocal of sine, right? So, that means csc(315°) is equal to 1 divided by sin(315°). Alright, so first I should figure out what sin(315°) is.I know that 315 degrees is in the fourth quadrant of the unit circle. In the fourth quadrant, sine values are negative because sine corresponds to the y-coordinate on the unit circle, and the y-coordinate is negative there. So, sin(315°) should be negative.Now, 315 degrees is 45 degrees less than 360 degrees. So, I can think of it as 360° - 45°. I remember that sine of an angle in the form of 360° - θ is equal to -sin(θ). So, sin(315°) is equal to -sin(45°). I know that sin(45°) is one of those special angles. It's equal to √2 divided by 2. So, sin(45°) = √2/2. Therefore, sin(315°) is -√2/2.Now, going back to cosecant. Since csc(315°) is 1 divided by sin(315°), that would be 1 divided by (-√2/2). Dividing by a fraction is the same as multiplying by its reciprocal, so 1 divided by (-√2/2) is the same as 1 multiplied by 2/(-√2). Simplifying that, 2 divided by -√2 is equal to -2/√2. But I think it's better to rationalize the denominator. To rationalize -2/√2, I can multiply both the numerator and the denominator by √2. So, (-2/√2) multiplied by (√2/√2) is (-2√2)/(2). The 2 in the numerator and the 2 in the denominator cancel out, leaving me with -√2.Wait, let me double-check that. If I have 2 divided by √2, that's the same as √2 because √2 times √2 is 2. So, 2 divided by √2 is √2. Therefore, -2 divided by √2 is -√2. Yep, that makes sense.So, putting it all together, csc(315°) is equal to -√2. I think that's the answer. Let me just recap to make sure I didn't make any mistakes:1. Start with csc(315°) = 1/sin(315°).2. Determine that 315° is in the fourth quadrant where sine is negative.3. Express 315° as 360° - 45°, so sin(315°) = -sin(45°).4. sin(45°) is √2/2, so sin(315°) is -√2/2.5. Therefore, csc(315°) is 1/(-√2/2) = -2/√2.6. Rationalize the denominator: -2/√2 = -√2.Everything seems to check out. I don't think I made any calculation errors. So, I'm confident that the cosecant of 315 degrees is negative square root of 2.

💡Alright, so I've got this inequality: ( a leqslant x + frac{2}{x - 2} ) and it's supposed to hold true for all ( x > 2 ). I need to find the range of values for ( a ) that makes this inequality true no matter what ( x ) I pick, as long as ( x ) is greater than 2.Hmm, okay. So, essentially, ( a ) has to be less than or equal to this expression ( x + frac{2}{x - 2} ) for every ( x > 2 ). That means ( a ) can't be bigger than the smallest value that ( x + frac{2}{x - 2} ) can take when ( x ) is greater than 2. So, if I can find the minimum value of ( x + frac{2}{x - 2} ) for ( x > 2 ), then ( a ) has to be less than or equal to that minimum value.Alright, so my task is to find the minimum of the function ( f(x) = x + frac{2}{x - 2} ) for ( x > 2 ). Once I find that minimum, that will be the upper bound for ( a ). So, how do I find the minimum of this function?Well, I remember from calculus that to find the extrema of a function, you can take its derivative and set it equal to zero. So, let's try that.First, let's write down the function again: ( f(x) = x + frac{2}{x - 2} ).Now, let's find its derivative with respect to ( x ). The derivative of ( x ) is 1. For the second term, ( frac{2}{x - 2} ), I can rewrite it as ( 2(x - 2)^{-1} ). The derivative of that is ( -2(x - 2)^{-2} ) times the derivative of the inside function, which is 1. So, putting it all together, the derivative ( f'(x) ) is:( f'(x) = 1 - frac{2}{(x - 2)^2} ).Okay, so to find critical points, I set ( f'(x) = 0 ):( 1 - frac{2}{(x - 2)^2} = 0 ).Let's solve for ( x ):( 1 = frac{2}{(x - 2)^2} ).Multiply both sides by ( (x - 2)^2 ):( (x - 2)^2 = 2 ).Take the square root of both sides:( x - 2 = sqrt{2} ) or ( x - 2 = -sqrt{2} ).But since ( x > 2 ), ( x - 2 ) must be positive, so we discard the negative root:( x - 2 = sqrt{2} ).Therefore, ( x = 2 + sqrt{2} ).Alright, so the critical point is at ( x = 2 + sqrt{2} ). Now, I need to check whether this critical point is a minimum or a maximum. Since the function is defined for ( x > 2 ), and we're looking for the minimum, I can use the second derivative test or analyze the behavior of the first derivative around this point.Let me try the second derivative test. First, find the second derivative ( f''(x) ).We already have the first derivative:( f'(x) = 1 - frac{2}{(x - 2)^2} ).Taking the derivative again:( f''(x) = 0 - frac{d}{dx}left(2(x - 2)^{-2}right) ).The derivative of ( 2(x - 2)^{-2} ) is ( -4(x - 2)^{-3} ) times the derivative of the inside function, which is 1. So,( f''(x) = frac{4}{(x - 2)^3} ).Now, evaluate ( f''(x) ) at ( x = 2 + sqrt{2} ):( f''(2 + sqrt{2}) = frac{4}{( (2 + sqrt{2}) - 2 )^3} = frac{4}{(sqrt{2})^3} ).Simplify ( (sqrt{2})^3 ):( (sqrt{2})^3 = (2)^{1.5} = 2 times sqrt{2} ).So,( f''(2 + sqrt{2}) = frac{4}{2 sqrt{2}} = frac{2}{sqrt{2}} = sqrt{2} ).Since ( sqrt{2} ) is positive, the function is concave upward at this point, which means it's a local minimum.Great, so ( x = 2 + sqrt{2} ) is indeed a point of local minimum. Since the function tends to infinity as ( x ) approaches 2 from the right and as ( x ) approaches infinity, this local minimum is actually the global minimum for the function ( f(x) ) on the interval ( x > 2 ).Now, let's find the value of ( f(x) ) at ( x = 2 + sqrt{2} ):( f(2 + sqrt{2}) = (2 + sqrt{2}) + frac{2}{(2 + sqrt{2}) - 2} ).Simplify the denominator in the second term:( (2 + sqrt{2}) - 2 = sqrt{2} ).So,( f(2 + sqrt{2}) = 2 + sqrt{2} + frac{2}{sqrt{2}} ).Simplify ( frac{2}{sqrt{2}} ):( frac{2}{sqrt{2}} = sqrt{2} ).Therefore,( f(2 + sqrt{2}) = 2 + sqrt{2} + sqrt{2} = 2 + 2sqrt{2} ).So, the minimum value of ( f(x) ) is ( 2 + 2sqrt{2} ).Therefore, for the inequality ( a leqslant x + frac{2}{x - 2} ) to hold for all ( x > 2 ), ( a ) must be less than or equal to the minimum value of the function, which is ( 2 + 2sqrt{2} ).Hence, the range of values for ( a ) is all real numbers less than or equal to ( 2 + 2sqrt{2} ).I think that makes sense. I used calculus to find the critical point, checked if it's a minimum, and then evaluated the function there to find the minimum value. Since ( a ) has to be less than or equal to this minimum for all ( x > 2 ), that should be the upper bound for ( a ). I don't think I made any mistakes in the calculations, but let me double-check quickly.Starting from the derivative:( f'(x) = 1 - frac{2}{(x - 2)^2} ).Set to zero:( 1 = frac{2}{(x - 2)^2} ).Multiply both sides:( (x - 2)^2 = 2 ).Square root:( x - 2 = sqrt{2} ).So, ( x = 2 + sqrt{2} ). That seems right.Then, second derivative:( f''(x) = frac{4}{(x - 2)^3} ).At ( x = 2 + sqrt{2} ), ( f''(x) = frac{4}{(sqrt{2})^3} = frac{4}{2 sqrt{2}} = frac{2}{sqrt{2}} = sqrt{2} > 0 ). So, it's a minimum.Then, evaluating ( f(2 + sqrt{2}) ):( 2 + sqrt{2} + frac{2}{sqrt{2}} = 2 + sqrt{2} + sqrt{2} = 2 + 2sqrt{2} ). Yep, that checks out.So, I think my conclusion is correct. ( a ) must be less than or equal to ( 2 + 2sqrt{2} ).**Final Answer**The range of values for ( a ) is boxed{(-infty, 2 + 2sqrt{2}]}.

💡First, determine the number of red roses in each category that Claire plans to sell:- **Small red roses:** 40 total, selling half, so 40 / 2 = 20 small red roses.- **Medium red roses:** 60 total, selling half, so 60 / 2 = 30 medium red roses.- **Large red roses:** Calculate the total number of large red roses. With 400 flowers in total and 120 tulips, there are 280 roses. Subtracting the 80 white roses, there are 200 red roses. After accounting for 40 small and 60 medium red roses, the remaining are large: 200 - (40 + 60) = 100 large red roses. Selling half of them: 100 / 2 = 50 large red roses.Next, calculate the earnings from each group before applying discounts:- **Small red roses:** 20 roses * 0.75 each = 15.00- **Medium red roses:** 30 roses * 1.00 each = 30.00- **Large red roses:** 50 roses * 1.25 each = 62.50Apply the discounts based on the quantity sold:- **Small red roses:** 20 roses qualify for a 10% discount. So, 15.00 - (10% of 15.00) = 15.00 - 1.50 = 13.50- **Medium red roses:** 30 roses qualify for a 15% discount. So, 30.00 - (15% of 30.00) = 30.00 - 4.50 = 25.50- **Large red roses:** 50 roses qualify for a 15% discount. So, 62.50 - (15% of 62.50) = 62.50 - 9.375 = 53.125Finally, sum the earnings from each group after discounts:- **Small red roses:** 13.50- **Medium red roses:** 25.50- **Large red roses:** 53.125Total earnings = 13.50 + 25.50 + 53.125 = 92.125Rounding to the nearest cent, the total earnings are approximately 92.13.

💡Okay, so I have this problem about a quadratic function f(x) = ax² + bx + c. The conditions are that when x is between 0 and 1, the absolute value of f(x) is less than or equal to 1. I need to find the maximum value of (a + b)c. Hmm, okay, let me try to figure this out step by step.First, I know that for a quadratic function, the graph is a parabola. Depending on the coefficient 'a', it can open upwards or downwards. But since we're dealing with an interval [0,1], I should probably evaluate the function at the endpoints and maybe at the vertex if it lies within the interval.Given that |f(x)| ≤ 1 for x in [0,1], that means f(0) and f(1) must each be between -1 and 1. Let me write that down:f(0) = c, so |c| ≤ 1.f(1) = a + b + c, so |a + b + c| ≤ 1.Also, since it's a quadratic, the maximum or minimum occurs at the vertex. The vertex of f(x) is at x = -b/(2a). I need to check if this x-value is within [0,1]. If it is, then the function could attain its maximum or minimum there, which also must satisfy |f(x)| ≤ 1.So, if -b/(2a) is between 0 and 1, then f(-b/(2a)) must also be between -1 and 1. That gives another condition:|f(-b/(2a))| ≤ 1.But this might complicate things because it introduces another variable. Maybe I can express (a + b)c in terms of f(0) and f(1). Let me see:We have f(0) = c and f(1) = a + b + c. So, a + b = f(1) - c. Therefore, (a + b)c = (f(1) - c)c.Hmm, that simplifies to (a + b)c = f(1)c - c². Maybe I can write this as -c² + f(1)c. That looks like a quadratic in terms of c. Maybe I can find its maximum value.Since it's a quadratic in c, the maximum occurs at the vertex. The vertex of a quadratic ax² + bx + c is at x = -b/(2a). In this case, the quadratic is -c² + f(1)c, so the vertex is at c = -f(1)/(2*(-1)) = f(1)/2.So, the maximum value of (a + b)c is when c = f(1)/2. Plugging that back in, we get:-(f(1)/2)² + f(1)*(f(1)/2) = -f(1)²/4 + f(1)²/2 = ( -f(1)² + 2f(1)² ) / 4 = f(1)² / 4.So, the maximum value of (a + b)c is f(1)² / 4. But we know that |f(1)| ≤ 1, so f(1)² ≤ 1. Therefore, the maximum value of (a + b)c is 1/4.Wait, but is this always achievable? I need to make sure that there exists a quadratic function f(x) such that (a + b)c = 1/4 and |f(x)| ≤ 1 on [0,1].Let me try to construct such a function. If I set c = f(1)/2, then from earlier, c = f(1)/2. Also, since f(1) = a + b + c, and a + b = f(1) - c, substituting c gives a + b = f(1) - f(1)/2 = f(1)/2.So, a + b = f(1)/2, and c = f(1)/2. Therefore, (a + b)c = (f(1)/2)*(f(1)/2) = f(1)² / 4. To maximize this, set f(1) = 1, so (a + b)c = 1/4.Now, let's construct f(x). Let me choose f(1) = 1, so c = 1/2. Then, a + b = 1/2. So, f(x) = ax² + bx + 1/2. We need to ensure that |f(x)| ≤ 1 on [0,1].Let me choose a specific function. Let's set a = 1/2 and b = 0. Then, f(x) = (1/2)x² + 1/2. Let's check the values:At x=0: f(0) = 1/2, which is within [-1,1].At x=1: f(1) = (1/2)(1) + 1/2 = 1, which is within [-1,1].Now, let's check the vertex. The vertex is at x = -b/(2a) = 0/(2*(1/2)) = 0. So, the vertex is at x=0, which is already checked. So, the function is increasing on [0,1], since a is positive. Therefore, the maximum is at x=1, which is 1, and the minimum is at x=0, which is 1/2. Both are within [-1,1]. So, this function satisfies the condition, and (a + b)c = (1/2 + 0)*(1/2) = 1/4.Therefore, the maximum value is indeed 1/4.Wait a second, but is this the only function that achieves this maximum? Let me think. Suppose I set f(1) = -1. Then, c = -1/2, and a + b = -1/2. Then, (a + b)c = (-1/2)*(-1/2) = 1/4 as well. So, whether f(1) is 1 or -1, the maximum value of (a + b)c is 1/4.But in this case, if f(1) = -1, then f(x) = ax² + bx -1/2. Let's choose a = -1/2 and b = 0. Then, f(x) = (-1/2)x² -1/2. At x=0, f(0) = -1/2, which is within [-1,1]. At x=1, f(1) = (-1/2)(1) -1/2 = -1, which is within [-1,1]. The vertex is at x = 0, which is already checked. The function is decreasing on [0,1], so the maximum is at x=0, which is -1/2, and the minimum is at x=1, which is -1. Both are within [-1,1]. So, this function also satisfies the condition, and (a + b)c = (-1/2 + 0)*(-1/2) = 1/4.Therefore, regardless of whether f(1) is 1 or -1, we can achieve the maximum value of 1/4 for (a + b)c.But wait, is there a function where f(1) is something else, but (a + b)c is larger than 1/4? Let me test with f(1) = something else.Suppose f(1) = 2/3. Then, c = (2/3)/2 = 1/3. Then, (a + b)c = (2/3)*(1/3) = 2/9 ≈ 0.222, which is less than 1/4.Similarly, if f(1) = 1/2, then c = 1/4, and (a + b)c = (1/2)*(1/4) = 1/8, which is even smaller.So, it seems that the maximum occurs when f(1) is at its extreme value, either 1 or -1, giving (a + b)c = 1/4.Therefore, I think the maximum value of (a + b)c is 1/4.**Final Answer**The maximum value of (a+b)c is boxed{dfrac{1}{4}}.