💡Okay, so I have this problem about temperatures from Monday to Friday. The average temperature at noon each day is 50 degrees, and the lowest temperature is some value. There's also something about the maximum possible range of the temperatures being 50 degrees. I need to find the lowest temperature.First, let's make sure I understand what's being asked. The average temperature is 50 degrees over five days, so that means the total sum of the temperatures for those five days is 5 times 50, which is 250 degrees. So, if I add up the temperatures from Monday to Friday, it should equal 250.Now, the range of temperatures is the difference between the highest and the lowest temperatures. The problem says the possible maximum range is 50 degrees. So, if the range is 50, that means the highest temperature minus the lowest temperature equals 50.I need to find the lowest temperature. Let's call the lowest temperature L. Then, the highest temperature would be L + 50, since the range is 50.But wait, I need to make sure that the sum of all five temperatures is 250. If one day is L, and the other four days are as high as possible, which would be L + 50, then the total sum would be L plus four times (L + 50). Let me write that down:Sum = L + 4*(L + 50)Simplify that:Sum = L + 4L + 200Sum = 5L + 200But we know the sum should be 250, so:5L + 200 = 250Now, solve for L:5L = 250 - 2005L = 50L = 10So, the lowest temperature is 10 degrees.Wait, let me double-check that. If the lowest is 10, then the highest would be 60 (since 10 + 50 = 60). Then, the four other days would each be 60. So, the temperatures would be: 10, 60, 60, 60, 60.Adding those up: 10 + 60 + 60 + 60 + 60 = 250. Yep, that matches the total sum we need.Is there any other way this could be set up? What if not all four days are at the maximum? Well, the problem says the maximum possible range is 50, so to maximize the range, we need to have the highest temperature as high as possible and the lowest as low as possible, given the average.So, setting four days to the maximum temperature and one day to the minimum makes sense to achieve the maximum range.Therefore, the lowest temperature is 10 degrees.

💡Alright, so I have this geometry problem here. Let me try to understand it step by step. We have an acute triangle ( ABC ) with ( AC < AB ). That means side ( AC ) is shorter than side ( AB ). The triangle has a circumradius ( R ), which is the radius of the circumscribed circle around the triangle. Point ( D ) is the foot of the altitude from vertex ( A ). So, if I draw a perpendicular line from ( A ) to side ( BC ), it will meet ( BC ) at point ( D ). Then, there's a point ( T ) on line ( AD ) such that ( AT = 2R ). Also, ( D ) is between ( A ) and ( T ). So, starting from ( A ), going through ( D ), and extending to ( T ) such that the entire length from ( A ) to ( T ) is twice the circumradius. Next, ( S ) is the midpoint of the arc ( BC ) that does not contain ( A ). So, if I look at the circumcircle of triangle ( ABC ), the arc ( BC ) that doesn't have ( A ) on it has a midpoint ( S ). The goal is to show that ( angle AST = 90^circ ). Okay, so I need to show that angle ( AST ) is a right angle. Let me try to visualize this. First, let me recall some properties of circumradius and midpoints of arcs. The midpoint of an arc ( BC ) in the circumcircle is equidistant from ( B ) and ( C ). Also, ( S ) lies on the angle bisector of ( angle BAC ) because it's the midpoint of the arc. Since ( S ) is the midpoint of arc ( BC ), it is also the center of the circle that is tangent to ( AB ) and ( AC ) and passes through ( B ) and ( C ). Hmm, not sure if that helps immediately. Let me think about the position of ( T ). ( T ) is on ( AD ), the altitude from ( A ), and ( AT = 2R ). Since ( R ) is the circumradius, maybe there's a relationship between ( T ) and the circumcircle. Wait, in a triangle, the distance from a vertex to the circumcenter is related to the circumradius. But ( T ) is not necessarily the circumcenter. However, ( AT = 2R ), which is twice the circumradius. Let me recall that in a triangle, the distance from a vertex to the circumcenter can be expressed in terms of the sides and angles. But I don't know if that's directly helpful here. Alternatively, maybe I can use coordinate geometry. Let me try setting up a coordinate system. Let me place point ( A ) at the origin ( (0, 0) ). Let me assume that ( BC ) is horizontal for simplicity, so point ( D ) will be somewhere along the y-axis. Wait, but ( D ) is the foot of the altitude from ( A ), so if ( A ) is at ( (0, 0) ), then ( D ) would be at ( (0, d) ) for some ( d ). But then ( T ) is on ( AD ) such that ( AT = 2R ). So, ( T ) would be at ( (0, 2R) ). Hmm, but I need to relate this to point ( S ). Maybe coordinate geometry is getting too messy here. Let me think of another approach. Since ( S ) is the midpoint of arc ( BC ), it has some nice properties. For example, ( S ) is equidistant from ( B ) and ( C ), and ( SB = SC ). Also, ( S ) lies on the circumcircle, so ( angle BSC = 2angle BAC ). Wait, maybe I can use some cyclic quadrilateral properties here. If I can show that quadrilateral ( AST ) something is cyclic, then maybe I can find a right angle. Alternatively, since ( AT = 2R ), and ( R ) is the circumradius, maybe ( T ) lies on the circumcircle as well? Let me check. In a triangle, the circumradius ( R ) is given by ( R = frac{a}{2sin A} ), where ( a ) is the length of side ( BC ). But ( AT = 2R ), so ( AT = frac{a}{sin A} ). Hmm, not sure if that helps. Maybe I need to consider vectors or something. Wait, another thought: in triangle ( ABC ), the distance from ( A ) to the circumcenter ( O ) is ( sqrt{R^2 + R^2 - 2R^2 cos angle BAC} ) or something like that? Wait, no, that's the distance between two points on a circle. Maybe I need to recall the formula for the distance from a vertex to the circumcenter. Actually, the distance from ( A ) to ( O ) is ( sqrt{R^2 + R^2 - 2R^2 cos angle BAC} ) by the law of cosines, since ( OA = OB = OC = R ). So, ( AO = sqrt{2R^2 - 2R^2 cos angle BAC} ). But ( AT = 2R ), which is longer than ( AO ) unless ( cos angle BAC ) is negative, which it isn't because the triangle is acute. So, ( T ) is further away from ( A ) than ( O ) is. Hmm, maybe I can relate ( T ) to some reflection or something. Wait, another idea: since ( AT = 2R ), and ( R ) is the circumradius, maybe ( T ) is the reflection of ( O ) over ( A )? Because reflecting ( O ) over ( A ) would give a point such that ( AO = OT ), so ( AT = 2AO ). But ( AO ) is not necessarily equal to ( R ). Wait, ( AO ) is the distance from ( A ) to ( O ), which is ( sqrt{R^2 + R^2 - 2R^2 cos angle BAC} ), as I thought earlier. So, unless ( angle BAC = 60^circ ), ( AO ) isn't equal to ( R ). So, that might not hold. Wait, maybe ( T ) lies on the circumcircle? Let me see. If ( AT = 2R ), and ( A ) is on the circumcircle, then ( T ) would be diametrically opposite to some point. But I don't know. Alternatively, maybe I can use some properties of midpoints of arcs. Since ( S ) is the midpoint of arc ( BC ), it is the excenter opposite to ( A ) in some cases, but not exactly. Wait, no, the excenter is related to the external angle bisectors. Wait, another thought: in triangle ( ABC ), the midpoint of arc ( BC ) not containing ( A ) is the center of the circle that is tangent to ( AB ), ( AC ), and the circumcircle. Hmm, not sure. Wait, maybe I can consider the reflection of ( A ) over ( S ). Since ( S ) is the midpoint of arc ( BC ), reflecting ( A ) over ( S ) might give some point related to ( T ). Alternatively, maybe I can use inversion. But that might be too complicated. Wait, let me think about the properties of ( S ). Since ( S ) is the midpoint of arc ( BC ), it lies on the perpendicular bisector of ( BC ). Also, ( S ) is equidistant from ( B ) and ( C ). Since ( AD ) is the altitude, it is perpendicular to ( BC ). So, ( AD ) is the same as the altitude, which is perpendicular to ( BC ). Wait, so ( AD ) is perpendicular to ( BC ), and ( S ) lies on the perpendicular bisector of ( BC ). So, the line ( SN ), where ( N ) is the other intersection point of the perpendicular bisector with the circumcircle, is a diameter. Wait, that's a good point. Since ( S ) is the midpoint of arc ( BC ), the line ( SN ) is a diameter of the circumcircle. Therefore, ( SN = 2R ). Also, since ( SN ) is a diameter, any angle subtended by ( SN ) on the circumference is a right angle. So, if I can show that ( A ) lies on the circle with diameter ( SN ), then ( angle SAN = 90^circ ). But ( A ) is already on the circumcircle, which has diameter ( SN ). Wait, no, the circumcircle has diameter ( SN ) only if ( SN ) is a diameter, which it is. So, ( A ) lies on the circumcircle, so ( angle SAN = 90^circ ). Wait, is that correct? If ( SN ) is a diameter, then any point on the circle forms a right angle with ( S ) and ( N ). So, ( angle SAN = 90^circ ). But how does that relate to ( T )? Wait, ( T ) is on ( AD ) such that ( AT = 2R ). Since ( SN = 2R ), and ( AD ) is perpendicular to ( BC ), which is the same direction as ( SN ). Wait, so ( AD ) is parallel to ( SN ) because both are perpendicular to ( BC ). Also, ( AD ) has length ( h_a ), the altitude, and ( SN ) has length ( 2R ). But ( AT = 2R ), so ( T ) is a point such that ( AT = SN ) and ( AD ) is parallel to ( SN ). Wait, so if I consider vectors, maybe ( vec{AT} = vec{SN} ). But I'm not sure. Alternatively, since ( AD ) is parallel to ( SN ) and ( AT = SN ), maybe quadrilateral ( STAN ) is a parallelogram. Yes, that makes sense. Because ( ST ) would be equal and parallel to ( AN ), and ( SA ) would be equal and parallel to ( TN ). Wait, but I need to confirm that. If ( STAN ) is a parallelogram, then ( SA parallel TN ) and ( SA = TN ). Also, ( ST parallel AN ) and ( ST = AN ). But I know that ( SN ) is a diameter, so ( AN ) is equal to ( SN ) minus ( SA ). Wait, no, ( AN ) is just another chord. Wait, maybe I'm overcomplicating. Since ( AD ) is parallel to ( SN ) and ( AT = SN = 2R ), then ( T ) is such that ( AD ) extended beyond ( D ) to ( T ) with ( AT = 2R ). So, if I consider the translation that maps ( S ) to ( A ), it would map ( N ) to ( T ), because ( SN ) is translated to ( AT ). Therefore, ( STAN ) is a parallelogram. In a parallelogram, opposite angles are equal. So, ( angle AST = angle ANS ). But ( angle ANS ) is equal to ( angle SAN ), which we established is ( 90^circ ). Wait, no, ( angle ANS ) is at point ( N ), so maybe not directly. Wait, actually, in parallelogram ( STAN ), ( angle AST ) is congruent to ( angle ANS ). But ( angle ANS ) is subtended by diameter ( SN ), so it's ( 90^circ ). Therefore, ( angle AST = 90^circ ). Wait, that seems to make sense. Let me recap: 1. ( S ) is the midpoint of arc ( BC ), so ( SN ) is a diameter of the circumcircle, hence ( SN = 2R ). 2. ( AD ) is the altitude from ( A ), so ( AD perp BC ). 3. Since ( SN ) is the perpendicular bisector of ( BC ), ( SN perp BC ). Therefore, ( AD parallel SN ). 4. Given ( AT = 2R ) and ( SN = 2R ), and since ( AD parallel SN ), quadrilateral ( STAN ) is a parallelogram. 5. In parallelogram ( STAN ), opposite angles are equal. 6. Since ( SN ) is a diameter, ( angle SAN = 90^circ ). 7. Therefore, ( angle AST = angle SAN = 90^circ ). Yes, that seems to be the logical flow. I think that's the solution. So, the key steps are recognizing that ( SN ) is a diameter, ( AD ) is parallel to ( SN ), and ( AT = SN ), making ( STAN ) a parallelogram. Then, using the properties of parallelograms and the fact that angles subtended by diameters are right angles, we conclude that ( angle AST = 90^circ ).

💡First, I'll define the number of sandwiches Samson ate at lunch on Monday as ( L ).Since he ate twice as many sandwiches at dinner on Monday, the number of sandwiches he ate at dinner is ( 2L ).On Tuesday, Samson only ate one sandwich for breakfast.According to the problem, Samson ate 8 more sandwiches in total on Monday than on Tuesday. This means the total number of sandwiches he ate on Monday is equal to the total number he ate on Tuesday plus 8.So, the equation representing this relationship is:[L + 2L = 1 + 8]Combining like terms, I get:[3L = 9]To find the value of ( L ), I'll divide both sides of the equation by 3:[L = frac{9}{3} = 3]Therefore, Samson ate 3 sandwiches at lunch on Monday.

💡First, I recognize that when the lateral surface of a cylinder is unfolded, it forms a square. This means that the height of the cylinder is equal to the circumference of its base.Let’s denote the side length of the square as ( s ). Therefore, the height ( h ) of the cylinder is also ( s ).The circumference of the base of the cylinder is given by ( 2pi r ), where ( r ) is the radius of the base. Since the circumference equals the height, we have:[2pi r = s]Solving for ( r ), we get:[r = frac{s}{2pi}]Next, I calculate the lateral surface area of the cylinder, which is the area of the square:[text{Lateral Surface Area} = s^2]Then, I determine the area of the base of the cylinder:[text{Base Area} = pi r^2 = pi left(frac{s}{2pi}right)^2 = pi cdot frac{s^2}{4pi^2} = frac{s^2}{4pi}]Finally, I find the ratio of the lateral surface area to the base area:[text{Ratio} = frac{text{Lateral Surface Area}}{text{Base Area}} = frac{s^2}{frac{s^2}{4pi}} = 4pi]Therefore, the ratio of the lateral surface area to the base area is ( 4pi ).

💡Okay, so I have this problem here: given that x and y are positive real numbers, and they satisfy the equation 1/x + 9/y = 2. I need to find the minimum value of x + y. Hmm, interesting. I remember something about inequalities and optimization problems. Maybe I can use the AM-GM inequality here? Let me think.First, let me write down what I know. The equation is 1/x + 9/y = 2. I need to minimize x + y. So, I have a constraint and an objective function. This sounds like a problem that can be approached with Lagrange multipliers, but since it's about positive real numbers, maybe there's a simpler way using inequalities.Wait, the user mentioned the AM-GM inequality. Let me recall that the AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. So, for two numbers a and b, (a + b)/2 ≥ √(ab), with equality when a = b.But in this problem, I have two variables, x and y, and a constraint involving their reciprocals. Maybe I can manipulate the given equation to express one variable in terms of the other and then substitute into the expression x + y to minimize it. Let me try that.From the equation 1/x + 9/y = 2, I can solve for one variable. Let's solve for y in terms of x. So, 9/y = 2 - 1/x. Then, 9/y = (2x - 1)/x. Therefore, y = 9x / (2x - 1). Okay, so y is expressed in terms of x. Now, substitute this into x + y.So, x + y = x + 9x / (2x - 1). Let me write that as x + (9x)/(2x - 1). Now, I can consider this as a function of x, say f(x) = x + 9x/(2x - 1). I need to find the minimum value of f(x) for x > 1/2, since y must be positive, so 2x - 1 > 0 implies x > 1/2.To find the minimum, I can take the derivative of f(x) with respect to x and set it equal to zero. Let's compute f'(x). First, f(x) = x + 9x/(2x - 1). Let me compute the derivative term by term.The derivative of x is 1. Now, for the second term, 9x/(2x - 1), I can use the quotient rule. The derivative is [9*(2x - 1) - 9x*2]/(2x - 1)^2. Simplifying the numerator: 9*(2x - 1) = 18x - 9, and 9x*2 = 18x. So, subtracting, 18x - 9 - 18x = -9. Therefore, the derivative of the second term is -9/(2x - 1)^2.So, putting it all together, f'(x) = 1 - 9/(2x - 1)^2. To find critical points, set f'(x) = 0:1 - 9/(2x - 1)^2 = 0So, 9/(2x - 1)^2 = 1Therefore, (2x - 1)^2 = 9Taking square roots, 2x - 1 = ±3But since x > 1/2, 2x - 1 is positive, so we can ignore the negative root.So, 2x - 1 = 3 => 2x = 4 => x = 2Now, let's find y when x = 2. From earlier, y = 9x/(2x - 1) = 9*2/(4 - 1) = 18/3 = 6So, when x = 2, y = 6, and x + y = 8. Is this the minimum? Let's check the second derivative to ensure it's a minimum.Compute f''(x). From f'(x) = 1 - 9/(2x - 1)^2, the derivative of f'(x) is f''(x) = 0 - 9*(-2)*(2x - 1)^(-3)*2 = 36/(2x - 1)^3At x = 2, f''(2) = 36/(4 - 1)^3 = 36/27 = 4/3 > 0, which means the function is concave upward at x = 2, so it's a local minimum. Since the function tends to infinity as x approaches 1/2 from the right and as x approaches infinity, this local minimum is indeed the global minimum.Alternatively, I could have used the AM-GM inequality directly without calculus. Let me try that approach too.Given 1/x + 9/y = 2, I can write this as (1/x + 9/y)/2 = 1. So, the arithmetic mean of 1/x and 9/y is 1. By AM-GM, the geometric mean is less than or equal to the arithmetic mean. So, √(1/x * 9/y) ≤ 1.Squaring both sides, (9)/(xy) ≤ 1 => xy ≥ 9.But I need to minimize x + y. Hmm, how can I relate x + y to xy? Maybe using the AM-GM inequality again on x and y.Wait, AM-GM on x and y would say that (x + y)/2 ≥ √(xy). Since we know that xy ≥ 9, then √(xy) ≥ 3. Therefore, (x + y)/2 ≥ 3 => x + y ≥ 6. But wait, that's not matching the previous result where I found x + y = 8. So, perhaps this approach isn't tight enough.Maybe I need to use a different substitution or scaling. Let me think.Another method is to use the method of Lagrange multipliers, but since it's a simple problem, maybe I can use substitution as I did earlier.Wait, another idea: express x + y in terms of the given constraint.Let me denote S = x + y, and we have 1/x + 9/y = 2.I can write this as (y + 9x)/(xy) = 2 => y + 9x = 2xy.So, y + 9x = 2xy => 2xy - y - 9x = 0 => y(2x - 1) = 9x => y = 9x/(2x - 1), which is what I had earlier.So, S = x + y = x + 9x/(2x - 1). To minimize S, I can set derivative to zero, which gave me x = 2, y = 6, S = 8.Alternatively, maybe I can use the Cauchy-Schwarz inequality. Let me recall that (a1*b1 + a2*b2)^2 ≤ (a1^2 + a2^2)(b1^2 + b2^2). Not sure if that's directly applicable here.Wait, another idea: use the method of substitution with variables.Let me set u = 1/x and v = 9/y. Then, the constraint becomes u + v = 2. And I need to minimize x + y. Since u = 1/x => x = 1/u, and v = 9/y => y = 9/v. So, S = x + y = 1/u + 9/v.So, now, I have u + v = 2, and I need to minimize 1/u + 9/v.This seems like a better setup for applying AM-GM or other inequalities.Let me consider the expression 1/u + 9/v. Let me denote this as T = 1/u + 9/v. I need to minimize T given that u + v = 2.This is a standard optimization problem. Let me use the method of Lagrange multipliers here.Define the function to minimize: T = 1/u + 9/vSubject to the constraint: u + v = 2Set up the Lagrangian: L = 1/u + 9/v + λ(u + v - 2)Take partial derivatives:∂L/∂u = -1/u² + λ = 0 => λ = 1/u²∂L/∂v = -9/v² + λ = 0 => λ = 9/v²Set the two expressions for λ equal: 1/u² = 9/v² => v² = 9u² => v = 3u (since u and v are positive)From the constraint u + v = 2, and v = 3u, so u + 3u = 2 => 4u = 2 => u = 1/2, so v = 3*(1/2) = 3/2Therefore, u = 1/2, v = 3/2So, x = 1/u = 2, y = 9/v = 9/(3/2) = 6Thus, x + y = 2 + 6 = 8, which matches the earlier result.Alternatively, without using calculus, I can use the Cauchy-Schwarz inequality on the terms 1/u and 9/v.Wait, let me think. Let me consider the vectors (sqrt(1/u), sqrt(9/v)) and (sqrt(u), sqrt(v)). Then, by Cauchy-Schwarz:(sqrt(1/u)*sqrt(u) + sqrt(9/v)*sqrt(v))² ≤ (1/u + 9/v)(u + v)Simplify the left side: (1 + 3)² = 16Right side: (1/u + 9/v)(u + v) = T * 2So, 16 ≤ 2T => T ≥ 8Therefore, the minimum value of T is 8, achieved when the vectors are proportional, i.e., sqrt(1/u)/sqrt(u) = sqrt(9/v)/sqrt(v) => 1/u² = 9/v² => v = 3u, which is the same condition as before.Thus, the minimum value of x + y is 8.So, both methods lead to the same conclusion. It seems that 8 is indeed the minimum value.Wait, just to make sure, let me test with x = 2 and y = 6.1/x + 9/y = 1/2 + 9/6 = 1/2 + 3/2 = 2, which satisfies the constraint. And x + y = 8.If I choose another point, say x = 3, then y = 9x/(2x -1) = 27/(6 -1) = 27/5 = 5.4, so x + y = 3 + 5.4 = 8.4, which is larger than 8.Similarly, if x = 1, y = 9/(2 -1) = 9, so x + y = 10, which is larger.If x approaches 1/2 from the right, y approaches infinity, so x + y approaches infinity.If x approaches infinity, y approaches 9/(2x -1) approaches 0, so x + y approaches infinity.Therefore, the minimum is indeed at x = 2, y = 6, with x + y = 8.I think I've covered multiple methods: substitution with calculus, Lagrange multipliers, and Cauchy-Schwarz. All lead to the same result. So, I'm confident that the minimum value is 8.

💡Okay, so I have this inequality to solve: (x - 2)/(x - 1) ≥ 2. Hmm, I remember that when dealing with inequalities involving fractions, I need to be careful about the denominator because it can't be zero, and also the sign of the denominator affects the direction of the inequality when multiplying both sides. Let me try to work through this step by step.First, I think I should get all terms on one side to make it easier to analyze. So, I'll subtract 2 from both sides:(x - 2)/(x - 1) - 2 ≥ 0.Now, to combine these terms into a single fraction, I need a common denominator. The denominator is already (x - 1), so I can write 2 as 2*(x - 1)/(x - 1):(x - 2)/(x - 1) - 2*(x - 1)/(x - 1) ≥ 0.Simplifying the numerator:(x - 2) - 2*(x - 1) = x - 2 - 2x + 2 = -x.So now the inequality becomes:(-x)/(x - 1) ≥ 0.Hmm, I can factor out a negative sign from the numerator:- (x)/(x - 1) ≥ 0.Which is the same as:(x)/(1 - x) ≥ 0.Wait, actually, that's not quite right. Let me double-check. If I factor out a negative from the numerator, it should flip the inequality sign:(-x)/(x - 1) ≥ 0 ⇒ x/(1 - x) ≤ 0.Yes, that's correct. So now I have:x/(1 - x) ≤ 0.Alright, so now I need to find the values of x for which this fraction is less than or equal to zero. To do this, I'll consider the critical points where the numerator or denominator is zero. The numerator x is zero when x = 0, and the denominator 1 - x is zero when x = 1. These points divide the real number line into intervals that I can test.So, the critical points are x = 0 and x = 1. Let's consider the intervals:1. x < 02. 0 < x < 13. x > 1I'll test each interval to see the sign of the expression x/(1 - x) in each.First interval: x < 0.Let's pick x = -1.Plugging into x/(1 - x): (-1)/(1 - (-1)) = (-1)/2 = -0.5, which is negative. So, the expression is negative in this interval.Second interval: 0 < x < 1.Let's pick x = 0.5.Plugging into x/(1 - x): 0.5/(1 - 0.5) = 0.5/0.5 = 1, which is positive. So, the expression is positive in this interval.Third interval: x > 1.Let's pick x = 2.Plugging into x/(1 - x): 2/(1 - 2) = 2/(-1) = -2, which is negative. So, the expression is negative in this interval.Now, I also need to check the critical points themselves:At x = 0: x/(1 - x) = 0/(1 - 0) = 0, which satisfies the inequality (since it's ≤ 0).At x = 1: The denominator becomes zero, so the expression is undefined. Therefore, x = 1 is not included in the solution set.Putting it all together:The expression x/(1 - x) ≤ 0 is true when x < 0 or when x > 1. But wait, in the second interval (0 < x < 1), the expression was positive, so it doesn't satisfy the inequality. In the third interval (x > 1), the expression is negative, so it does satisfy the inequality.Wait a minute, but earlier when I tested x = 2, I got -2, which is negative, so that's correct. But hold on, when x approaches 1 from the right, the denominator approaches 0 from the negative side, and the numerator approaches 1, so the expression approaches negative infinity. So, yes, for x > 1, the expression is negative.But I also need to consider the interval x < 0, where the expression is negative, so that's also part of the solution.But wait, let me think again. The original inequality was (x - 2)/(x - 1) ≥ 2. After rearranging, I got x/(1 - x) ≤ 0, which led me to x < 0 or x > 1.However, I should also consider the domain of the original inequality. The denominator x - 1 cannot be zero, so x ≠ 1. Also, when I multiplied both sides by (x - 1), I have to consider the sign of (x - 1) because if it's positive, the inequality sign remains the same, but if it's negative, the inequality sign flips.Wait, actually, I didn't multiply both sides by (x - 1); instead, I rearranged the inequality to get everything on one side. So, perhaps I should approach it differently.Let me try another method. Starting again with (x - 2)/(x - 1) ≥ 2.I can subtract 2 from both sides:(x - 2)/(x - 1) - 2 ≥ 0.Combine the terms:[(x - 2) - 2(x - 1)]/(x - 1) ≥ 0.Simplify the numerator:x - 2 - 2x + 2 = -x.So, the inequality becomes:(-x)/(x - 1) ≥ 0.Which is the same as:x/(1 - x) ≥ 0.Wait, earlier I thought it was ≤ 0, but now I'm getting ≥ 0. Did I make a mistake earlier?Let me check:Starting from (x - 2)/(x - 1) ≥ 2.Subtract 2: (x - 2)/(x - 1) - 2 ≥ 0.Combine: [x - 2 - 2(x - 1)]/(x - 1) ≥ 0.Simplify numerator: x - 2 - 2x + 2 = -x.So, (-x)/(x - 1) ≥ 0.Which is equivalent to x/(1 - x) ≥ 0.Wait, so earlier I thought it was ≤ 0, but now I'm getting ≥ 0. That must mean I made a mistake in my earlier reasoning.Let me clarify:(-x)/(x - 1) ≥ 0.This can be written as x/(1 - x) ≥ 0.Because (-x)/(x - 1) = x/(1 - x).So, the inequality is x/(1 - x) ≥ 0.Therefore, I need to find where x/(1 - x) is greater than or equal to zero.So, the critical points are x = 0 and x = 1.Testing intervals:1. x < 0: Let's pick x = -1.x/(1 - x) = (-1)/(1 - (-1)) = (-1)/2 = -0.5, which is negative. So, doesn't satisfy ≥ 0.2. 0 < x < 1: Let's pick x = 0.5.x/(1 - x) = 0.5/(1 - 0.5) = 0.5/0.5 = 1, which is positive. So, satisfies ≥ 0.3. x > 1: Let's pick x = 2.x/(1 - x) = 2/(1 - 2) = 2/(-1) = -2, which is negative. Doesn't satisfy ≥ 0.Now, checking the critical points:At x = 0: x/(1 - x) = 0/(1 - 0) = 0, which satisfies ≥ 0.At x = 1: The expression is undefined, so x = 1 is excluded.Therefore, the solution set is 0 ≤ x < 1.Wait, so earlier I thought it was x < 0 or x > 1, but that was because I incorrectly flipped the inequality. Now, correctly, the solution is 0 ≤ x < 1.But let me double-check by plugging in values from the original inequality.Take x = 0.5, which should be in the solution set.Original inequality: (0.5 - 2)/(0.5 - 1) = (-1.5)/(-0.5) = 3, which is ≥ 2. Correct.Take x = 0: (0 - 2)/(0 - 1) = (-2)/(-1) = 2, which is equal to 2. So, x = 0 is included.Take x = 0.9: (0.9 - 2)/(0.9 - 1) = (-1.1)/(-0.1) = 11, which is ≥ 2. Correct.Take x = -1: (-1 - 2)/(-1 - 1) = (-3)/(-2) = 1.5, which is less than 2. So, x = -1 is not in the solution set, which matches our solution.Take x = 2: (2 - 2)/(2 - 1) = 0/1 = 0, which is less than 2. So, x = 2 is not in the solution set, which also matches.Therefore, the solution set is [0, 1).Wait, but earlier I thought the solution was x < 0 or x > 1, but that was incorrect because I messed up the inequality direction. Now, correctly, it's [0, 1).But let me think again. When I had (-x)/(x - 1) ≥ 0, which is equivalent to x/(1 - x) ≥ 0, so the solution is where x and (1 - x) have the same sign.So, either both positive or both negative.Case 1: x ≥ 0 and 1 - x > 0 ⇒ x ≥ 0 and x < 1 ⇒ 0 ≤ x < 1.Case 2: x ≤ 0 and 1 - x < 0 ⇒ x ≤ 0 and x > 1. But x cannot be both ≤ 0 and > 1 at the same time, so this case is impossible.Therefore, the only solution is 0 ≤ x < 1.So, the solution set is [0, 1).I think that's correct. Let me just summarize:Original inequality: (x - 2)/(x - 1) ≥ 2.Rearranged to: (-x)/(x - 1) ≥ 0 ⇒ x/(1 - x) ≥ 0.Solution set: 0 ≤ x < 1.Yes, that seems right.