💡First, I need to determine the radius of the central fountain by dividing its diameter by 2.Next, I'll add the width of the flower garden ring to the fountain's radius to find the radius up to the inner edge of the walking path.Then, I'll add the width of the walking path to this radius to obtain the total radius of the outer boundary.Finally, I'll calculate the diameter of the outer boundary by multiplying the total radius by 2.

💡Alright, so I have this problem about an Apollonius Circle. Hmm, I remember that an Apollonius Circle is the set of all points in a plane such that the ratio of their distances to two fixed points is constant. In this case, the two fixed points are F₁(-1, 0) and F₂(1, 0), and the ratio is a constant 'a' where a > 1. The question is asking which of the statements A, B, C, or D is correct regarding the symmetry and properties of this curve.First, let me recall the definition. For any point P(x, y) on the curve C, the ratio of its distance to F₁ and F₂ is a constant 'a'. So, mathematically, that would be:√[(x + 1)² + y²] / √[(x - 1)² + y²] = aSince a > 1, this means that the distance from P to F₁ is 'a' times the distance from P to F₂. I think this ratio being greater than 1 implies that the circle will be closer to F₂ than to F₁? Or maybe not necessarily closer, but the shape is determined by this ratio.Anyway, the problem is about symmetry. So, I need to figure out if the curve C is symmetric about the x-axis, y-axis, origin, or passes through the origin.Let me start by analyzing the equation. Let me square both sides to eliminate the square roots:[(x + 1)² + y²] / [(x - 1)² + y²] = a²Multiplying both sides by the denominator:(x + 1)² + y² = a²[(x - 1)² + y²]Expanding both sides:Left side: (x² + 2x + 1) + y²Right side: a²(x² - 2x + 1) + a² y²So, expanding:x² + 2x + 1 + y² = a²x² - 2a²x + a² + a² y²Now, let's bring all terms to one side:x² + 2x + 1 + y² - a²x² + 2a²x - a² - a² y² = 0Combine like terms:(1 - a²)x² + (2 + 2a²)x + (1 - a²)y² + (1 - a²) = 0Wait, let me check that again:From the left side: x² + 2x + 1 + y²From the right side: a²x² - 2a²x + a² + a² y²So subtracting right side from left side:x² - a²x² + 2x + 2a²x + 1 - a² + y² - a² y² = 0So:(1 - a²)x² + (2 + 2a²)x + (1 - a²)y² + (1 - a²) = 0Hmm, that seems correct. Now, I can factor out (1 - a²) from the x², y², and constant terms:(1 - a²)(x² + y² + 1) + (2 + 2a²)x = 0Wait, that might not be the most helpful. Alternatively, perhaps I can rearrange terms:(1 - a²)x² + (2 + 2a²)x + (1 - a²)y² + (1 - a²) = 0Let me factor (1 - a²) from the x², y², and constant terms:(1 - a²)(x² + y² + 1) + (2 + 2a²)x = 0Hmm, maybe not. Alternatively, perhaps I can divide both sides by (1 - a²) to simplify. Since a > 1, (1 - a²) is negative, but that's okay.Dividing both sides by (1 - a²):x² + (2 + 2a²)/(1 - a²) x + y² + 1 = 0Wait, let me compute (2 + 2a²)/(1 - a²):Factor numerator: 2(1 + a²)Denominator: (1 - a²) = -(a² - 1)So, (2 + 2a²)/(1 - a²) = -2(1 + a²)/(a² - 1)Hmm, maybe that's not helpful. Alternatively, perhaps completing the square.Let me try to write the equation in standard circle form.Starting from:(1 - a²)x² + (2 + 2a²)x + (1 - a²)y² + (1 - a²) = 0Let me group x terms and y terms:(1 - a²)x² + (2 + 2a²)x + (1 - a²)y² = -(1 - a²)Factor out (1 - a²) from x² and y² terms:(1 - a²)(x² + y²) + (2 + 2a²)x = -(1 - a²)Hmm, perhaps move the constant term to the other side:(1 - a²)(x² + y²) + (2 + 2a²)x = -(1 - a²)Let me factor out (1 - a²):(1 - a²)(x² + y² + 1) + (2 + 2a²)x = 0Wait, that's similar to what I had before. Maybe I can write it as:(1 - a²)(x² + y² + 1) = -(2 + 2a²)xDivide both sides by (1 - a²):x² + y² + 1 = -(2 + 2a²)/(1 - a²) xLet me compute the coefficient:-(2 + 2a²)/(1 - a²) = (2 + 2a²)/(a² - 1) = 2(1 + a²)/(a² - 1)So, the equation becomes:x² + y² + 1 = [2(1 + a²)/(a² - 1)] xNow, let's rearrange:x² - [2(1 + a²)/(a² - 1)] x + y² + 1 = 0To write this in standard circle form, I need to complete the square for the x terms.Let me denote the coefficient of x as B:B = -2(1 + a²)/(a² - 1)So, the equation is:x² + Bx + y² + 1 = 0Completing the square for x:x² + Bx = (x + B/2)² - (B/2)²So, substituting back:(x + B/2)² - (B/2)² + y² + 1 = 0Which can be written as:(x + B/2)² + y² = (B/2)² - 1So, the center of the circle is at (-B/2, 0) and the radius is sqrt[(B/2)² - 1]Let me compute B:B = -2(1 + a²)/(a² - 1)So, B/2 = - (1 + a²)/(a² - 1)Thus, the center is at (-B/2, 0) = [(1 + a²)/(a² - 1), 0]And the radius squared is (B/2)² - 1:[(1 + a²)/(a² - 1)]² - 1Let me compute that:= (1 + 2a² + a⁴)/(a² - 1)² - 1= [1 + 2a² + a⁴ - (a² - 1)²]/(a² - 1)²Compute (a² - 1)² = a⁴ - 2a² + 1So,= [1 + 2a² + a⁴ - (a⁴ - 2a² + 1)]/(a² - 1)²Simplify numerator:1 + 2a² + a⁴ - a⁴ + 2a² - 1 = (1 - 1) + (2a² + 2a²) + (a⁴ - a⁴) = 4a²So, radius squared is 4a²/(a² - 1)², so radius is 2a/(a² - 1)So, the equation of the circle is:(x + (1 + a²)/(a² - 1))² + y² = [2a/(a² - 1)]²Simplify the center:(1 + a²)/(a² - 1) = (a² + 1)/(a² - 1) = [ (a² - 1) + 2 ]/(a² - 1) = 1 + 2/(a² - 1)But maybe that's not necessary.So, the center is at ( (1 + a²)/(a² - 1), 0 ) and radius is 2a/(a² - 1)Now, knowing that, let's analyze the symmetry.Symmetry about the x-axis: For a curve to be symmetric about the x-axis, if (x, y) is on the curve, then (x, -y) must also be on the curve.Looking at the equation, if we replace y with -y, the equation remains the same because y² is the same as (-y)². So, yes, the curve is symmetric about the x-axis.Symmetry about the y-axis: For symmetry about the y-axis, if (x, y) is on the curve, then (-x, y) must also be on the curve.Looking at the equation, let's see:Original equation: √[(x + 1)² + y²] = a √[(x - 1)² + y²]If we replace x with -x:√[(-x + 1)² + y²] = a √[(-x - 1)² + y²]Which simplifies to:√[(1 - x)² + y²] = a √[(x + 1)² + y²]But the original equation is √[(x + 1)² + y²] = a √[(x - 1)² + y²]These are not the same unless a = 1, which it's not. So, replacing x with -x does not preserve the equation, hence the curve is not symmetric about the y-axis.Symmetry about the origin: For symmetry about the origin, if (x, y) is on the curve, then (-x, -y) must also be on the curve.Let's check:Replace x with -x and y with -y:√[(-x + 1)² + (-y)²] = a √[(-x - 1)² + (-y)²]Which simplifies to:√[(1 - x)² + y²] = a √[(x + 1)² + y²]Again, this is not the same as the original equation unless a = 1, which it's not. So, the curve is not symmetric about the origin.Passing through the origin: Let's check if (0, 0) satisfies the equation.Compute distance from (0,0) to F₁(-1,0): √[(0 + 1)² + 0²] = 1Distance from (0,0) to F₂(1,0): √[(0 - 1)² + 0²] = 1So, the ratio is 1/1 = 1, but the ratio is supposed to be 'a' where a > 1. So, unless a = 1, which it's not, the origin is not on the curve.Therefore, among the options, only A is correct: Curve C is symmetric about the x-axis.

💡Okay, so I have this problem here about a line on a coordinate plane. The equation of the line is given as y = mx + b, where m is the slope and b is the y-intercept. The problem states that the slope m is positive and specifically is 3/4, and the y-intercept b is 6/5. I need to find the correct range for the product of m and b, which is mb.First, I should probably write down what I know. The slope m is 3/4, and the y-intercept b is 6/5. So, m = 3/4 and b = 6/5. The question is asking about the product of these two, which is mb. So, I need to calculate mb and then determine which range it falls into.Let me calculate mb step by step. Multiplying m and b together, I get:mb = (3/4) * (6/5)To multiply these two fractions, I multiply the numerators together and the denominators together. So, 3 times 6 is 18, and 4 times 5 is 20. Therefore, mb equals 18/20.Now, I can simplify 18/20. Both 18 and 20 are divisible by 2. Dividing numerator and denominator by 2, I get 9/10. So, mb simplifies to 9/10.Now, I need to figure out which range this falls into. The options are:A) mb ≤ 0 B) 0 < mb < 1 C) mb = 1 D) mb > 1 E) mb < -1Looking at the value of mb, which is 9/10, I can see that it's a positive number because both m and b are positive. So, it's definitely greater than 0. Next, 9/10 is equal to 0.9, which is less than 1. Therefore, mb is between 0 and 1.Let me double-check my calculations to make sure I didn't make a mistake. Multiplying 3/4 by 6/5:3 times 6 is 18, and 4 times 5 is 20. So, 18/20 simplifies to 9/10. Yes, that seems correct.Is there any chance I could have misread the question? Let me check again. The slope is positive 3/4, and the y-intercept is 6/5. So, both m and b are positive, which means their product must also be positive. That rules out options A and E because they involve negative numbers or zero, which isn't the case here.Option C says mb equals 1, but 9/10 is not equal to 1. Option D says mb is greater than 1, but 9/10 is less than 1. So, the only option that makes sense is B, which states that 0 is less than mb and mb is less than 1.Just to be thorough, let me think about what mb represents. In the equation y = mx + b, m is the slope and b is the y-intercept. The product mb isn't something that has a direct geometric interpretation in the equation, but in this problem, we're just being asked about the numerical value of their product.Since both m and b are positive fractions less than 1, their product will also be a positive fraction less than 1. Specifically, 3/4 is 0.75 and 6/5 is 1.2. Multiplying 0.75 by 1.2 gives 0.9, which is indeed between 0 and 1.I think I've covered all the bases here. I calculated the product, simplified it, checked the options, and even thought about what the product represents. Everything seems to point to option B being the correct answer.

💡Alright, so I have this problem about a store that sells products. The cost price is 30 yuan per item, and they're selling it for 40 yuan each. They can sell 48 items a day. Now, they want to reduce the price to sell more and make a profit of 504 yuan per day. There are two parts to this problem.Starting with part (1): The selling price decreases to 32.4 yuan after two consecutive reductions of the same percentage. I need to find the percentage of each reduction. Hmm, okay, so it's like they reduced the price twice by the same percentage, and after those two reductions, the price is 32.4 yuan.Let me think about how percentage reductions work. If you have a price and you reduce it by a certain percentage, say x%, then the new price is the original price multiplied by (1 - x/100). So, if you do this twice, it's like multiplying by (1 - x/100) each time. So, the final price after two reductions would be the original price times (1 - x/100) squared.Given that, the original price is 40 yuan, and after two reductions, it's 32.4 yuan. So, I can set up the equation:40 * (1 - x/100)^2 = 32.4I need to solve for x. Let me write that down:40 * (1 - x/100)^2 = 32.4First, divide both sides by 40 to simplify:(1 - x/100)^2 = 32.4 / 40Calculating 32.4 divided by 40, that's 0.81.So, (1 - x/100)^2 = 0.81Now, take the square root of both sides to solve for (1 - x/100):sqrt(0.81) = 0.9So, 1 - x/100 = 0.9Now, solving for x:x/100 = 1 - 0.9 = 0.1So, x = 0.1 * 100 = 10%Wait, but when I took the square root, I considered only the positive root. The square root of 0.81 is also -0.9, but since we're talking about percentages, a negative percentage reduction doesn't make sense here. So, x must be 10%.So, the percentage reduction each time is 10%.Okay, that seems straightforward. Let me double-check:Starting price: 40 yuanFirst reduction: 10%, so 40 * 0.9 = 36 yuanSecond reduction: 10%, so 36 * 0.9 = 32.4 yuanYes, that matches the given final price. So, part (1) is 10%.Moving on to part (2): For every 1 yuan reduction in price, the store can sell an additional 8 items per day. They want to make a profit of 504 yuan per day. How much should the price be reduced per item?Alright, so let's break this down. The cost price is 30 yuan, selling price is 40 yuan. If they reduce the price by m yuan, the new selling price is (40 - m) yuan.For each 1 yuan reduction, they sell 8 more items. So, if they reduce by m yuan, they sell 8m more items. The original number of items sold per day is 48, so the new number sold is (48 + 8m).Profit is calculated as (Selling Price - Cost Price) * Number of Items Sold.They want the profit to be 504 yuan. So, setting up the equation:(40 - m - 30) * (48 + 8m) = 504Simplify the terms:(10 - m) * (48 + 8m) = 504Let me expand this:10 * 48 + 10 * 8m - m * 48 - m * 8m = 504Calculating each term:10 * 48 = 48010 * 8m = 80m-m * 48 = -48m-m * 8m = -8m^2So, putting it all together:480 + 80m - 48m - 8m^2 = 504Combine like terms:80m - 48m = 32mSo, the equation becomes:480 + 32m - 8m^2 = 504Now, let's bring all terms to one side to set the equation to zero:-8m^2 + 32m + 480 - 504 = 0Simplify the constants:480 - 504 = -24So, the equation is:-8m^2 + 32m - 24 = 0I can multiply both sides by -1 to make the leading coefficient positive:8m^2 - 32m + 24 = 0Now, let's see if we can simplify this equation. All coefficients are divisible by 8:Divide each term by 8:m^2 - 4m + 3 = 0Now, we have a quadratic equation:m^2 - 4m + 3 = 0Let's factor this:Looking for two numbers that multiply to 3 and add to -4.Those numbers are -1 and -3.So, (m - 1)(m - 3) = 0Thus, m = 1 or m = 3So, the price reduction could be 1 yuan or 3 yuan.But the problem says they want to reduce inventory quickly. So, reducing the price more would likely lead to selling more items, which helps in reducing inventory faster.Therefore, choosing m = 3 yuan reduction.Let me verify:New selling price: 40 - 3 = 37 yuanNumber of items sold: 48 + 8*3 = 48 + 24 = 72 itemsProfit per item: 37 - 30 = 7 yuanTotal profit: 7 * 72 = 504 yuanYes, that matches the desired profit.Alternatively, if m = 1:New selling price: 39 yuanNumber of items sold: 48 + 8 = 56Profit per item: 9 yuanTotal profit: 9 * 56 = 504 yuanSo, both reductions give the desired profit, but since they want to reduce inventory quickly, reducing by 3 yuan is better.So, the answer is 3 yuan reduction.**Final Answer**(1) The percentage of each reduction is boxed{10%}.(2) The price should be reduced by boxed{3} yuan per item.

💡Okay, so I have this problem about dice in a bag, and I need to prove that ( p + (n-1)q = frac{n}{2} ). Let me try to understand what's going on here.First, there are ( n ) dice in the bag. Each die is six-sided, and the faces are colored either white or red. Importantly, the total number of white and red sides across all dice is equal. That means if I count up all the white faces and all the red faces, they should be the same. Since each die has six faces, and there are ( n ) dice, there are a total of ( 6n ) faces. So, half of them, which is ( 3n ), are white, and the other half, ( 3n ), are red.Now, ( p ) is defined as the probability that when I take one die out of the bag and throw it twice, the same color comes up both times. So, I pick a die randomly, roll it, note the color, then roll it again, and check if the color is the same as the first roll.On the other hand, ( q ) is the probability that when I take two dice out of the bag and throw them at the same time, both show the same color. So, I pick two different dice, roll both of them, and check if both show white or both show red.I need to show that ( p + (n-1)q = frac{n}{2} ). Hmm, that seems interesting. Let me try to break it down step by step.First, let's think about ( p ). Since each die has an equal number of white and red faces, the probability of getting white on a single roll is ( frac{1}{2} ), and the same for red. So, when I roll the die twice, the probability that both are white is ( left(frac{1}{2}right)^2 = frac{1}{4} ), and similarly, the probability that both are red is ( frac{1}{4} ). Therefore, the probability that both rolls are the same color is ( frac{1}{4} + frac{1}{4} = frac{1}{2} ). Wait, so is ( p = frac{1}{2} )?But hold on, maybe I'm oversimplifying. The problem says the total number of white and red sides are equal, but it doesn't specify that each die has an equal number of white and red faces. Oh, right, it just says the total number across all dice is equal. So, each die could have a different number of white and red faces, as long as the total across all dice is equal.Hmm, that complicates things. So, each die might have a different number of white and red faces, but overall, across all dice, there are ( 3n ) white faces and ( 3n ) red faces.Therefore, when I pick a die randomly, the probability that it has ( k ) white faces and ( 6 - k ) red faces is something I need to consider. But wait, the problem doesn't specify how the colors are distributed on each die, only that the total number of white and red faces are equal. So, maybe each die has exactly 3 white and 3 red faces? Because if the total is ( 3n ) white and ( 3n ) red, and there are ( n ) dice, each die must have 3 white and 3 red faces. Otherwise, the total wouldn't balance out.Wait, let me think. If each die has 3 white and 3 red faces, then the total number of white faces is ( 3n ) and red faces is ( 3n ), which matches the problem statement. So, maybe each die does have exactly 3 white and 3 red faces. That would make sense. So, each die is balanced with equal numbers of white and red faces.If that's the case, then when I roll a die, the probability of getting white is ( frac{3}{6} = frac{1}{2} ), and the same for red. So, rolling it twice, the probability that both are white is ( frac{1}{2} times frac{1}{2} = frac{1}{4} ), and both red is also ( frac{1}{4} ). So, the total probability ( p ) is ( frac{1}{4} + frac{1}{4} = frac{1}{2} ).Okay, so ( p = frac{1}{2} ).Now, what about ( q )? ( q ) is the probability that when I take two dice out of the bag and throw them at the same time, both show the same color. So, I need to pick two different dice, roll both, and see if both are white or both are red.Since each die is fair and has 3 white and 3 red faces, the probability that the first die shows white is ( frac{1}{2} ), and the same for red. Similarly, for the second die, the probability is ( frac{1}{2} ) for white and ( frac{1}{2} ) for red.But wait, since the two dice are independent, the probability that both show white is ( frac{1}{2} times frac{1}{2} = frac{1}{4} ), and the probability that both show red is also ( frac{1}{4} ). So, the total probability ( q ) is ( frac{1}{4} + frac{1}{4} = frac{1}{2} ).Wait, so both ( p ) and ( q ) are ( frac{1}{2} )? That seems strange because then ( p + (n-1)q = frac{1}{2} + (n-1)times frac{1}{2} = frac{n}{2} ), which matches the equation we need to prove. But that seems too straightforward. Am I missing something?Let me double-check. If each die has exactly 3 white and 3 red faces, then yes, each roll is independent with a 50-50 chance. So, rolling one die twice gives a 50% chance of both being the same color, and rolling two different dice also gives a 50% chance of both being the same color.But wait, is that actually the case? When rolling two different dice, are the outcomes independent? Yes, because the dice are independent. So, the color of one die doesn't affect the color of the other. Therefore, the probability that both are white is ( frac{1}{2} times frac{1}{2} = frac{1}{4} ), and both red is also ( frac{1}{4} ), so total ( frac{1}{2} ).So, both ( p ) and ( q ) are ( frac{1}{2} ). Therefore, ( p + (n-1)q = frac{1}{2} + (n-1)times frac{1}{2} = frac{1}{2} + frac{n-1}{2} = frac{n}{2} ).Wait, that seems to work out. But the problem statement didn't specify that each die has exactly 3 white and 3 red faces, only that the total number of white and red faces across all dice is equal. So, does that necessarily mean each die has 3 white and 3 red faces?Let me think. Suppose we have ( n ) dice, each with 6 faces. The total number of white faces is ( 3n ), and the total number of red faces is ( 3n ). So, if we distribute these white and red faces across the ( n ) dice, each die must have exactly 3 white and 3 red faces. Because if one die had more white faces, another would have to have fewer to maintain the total of ( 3n ) white faces, but since each die has 6 faces, the only way to have the total white faces be ( 3n ) is if each die has exactly 3 white and 3 red faces.Therefore, each die must have exactly 3 white and 3 red faces. So, my initial assumption was correct.Therefore, both ( p ) and ( q ) are ( frac{1}{2} ), and thus ( p + (n-1)q = frac{n}{2} ).But let me try to think differently. Maybe the problem is more general, and the dice don't necessarily have to have exactly 3 white and 3 red faces. Maybe some dice have more white faces and others have more red faces, as long as the total is equal.Wait, but if the total number of white faces is ( 3n ) and red faces is ( 3n ), and there are ( n ) dice, each with 6 faces, then the average number of white faces per die is 3, and the average number of red faces per die is also 3. So, it's possible that some dice have more white faces and others have more red faces, but on average, each die has 3 white and 3 red faces.In that case, the probability ( p ) would still be ( frac{1}{2} ), because when you pick a die at random, the expected number of white faces is 3, so the probability of rolling white is ( frac{3}{6} = frac{1}{2} ), same for red. Therefore, rolling it twice, the probability of both being the same color is ( frac{1}{2} ).Similarly, for ( q ), when you pick two different dice, each has an expected 3 white and 3 red faces, so the probability that both show white is ( frac{1}{2} times frac{1}{2} = frac{1}{4} ), and both red is ( frac{1}{4} ), so total ( frac{1}{2} ).Therefore, regardless of how the white and red faces are distributed among the dice, as long as the total number of white and red faces is equal, the probabilities ( p ) and ( q ) are both ( frac{1}{2} ), leading to ( p + (n-1)q = frac{n}{2} ).But wait, is that always true? Suppose one die has all white faces and another has all red faces, and the rest have a mix. Would that affect the probabilities?Wait, no, because when you pick a die at random, the probability of picking a die with all white faces is ( frac{1}{n} ), and similarly for all red. But since the total number of white and red faces is equal, the number of all-white dice must be balanced by all-red dice, but that's not necessarily the case. Actually, no, because if you have one all-white die, you need another all-red die to balance the total number of white and red faces.Wait, let's say we have one die with 6 white faces and one die with 6 red faces. Then, the total white faces would be 6, and red faces would be 6. But we have ( n ) dice, so if ( n = 2 ), that's fine. But for ( n > 2 ), we would need more dice to balance the total white and red faces.Wait, no, if ( n = 3 ), for example, we could have one die with 6 white, one die with 6 red, and one die with 3 white and 3 red. Then, the total white faces would be 6 + 3 = 9, and red faces would be 6 + 3 = 9, which is equal. So, in that case, the probabilities might change.Wait, let's calculate ( p ) in this case. If I pick a die at random, there's a ( frac{1}{3} ) chance of picking the all-white die, a ( frac{1}{3} ) chance of picking the all-red die, and a ( frac{1}{3} ) chance of picking the balanced die.If I pick the all-white die, the probability of rolling white twice is 1, and red is 0. So, the probability of both being the same color is 1.If I pick the all-red die, similarly, the probability of both being red is 1, so the probability of both being the same color is 1.If I pick the balanced die, the probability of both being the same color is ( frac{1}{2} ).Therefore, the overall probability ( p ) is ( frac{1}{3} times 1 + frac{1}{3} times 1 + frac{1}{3} times frac{1}{2} = frac{1}{3} + frac{1}{3} + frac{1}{6} = frac{2}{3} + frac{1}{6} = frac{5}{6} ).Wait, that's different from ( frac{1}{2} ). So, in this case, ( p = frac{5}{6} ).Similarly, let's calculate ( q ). When picking two dice, there are different combinations:1. Both dice are all-white: probability ( frac{1}{3} times frac{1}{3} = frac{1}{9} ). Both will show white, so same color.2. Both dice are all-red: probability ( frac{1}{3} times frac{1}{3} = frac{1}{9} ). Both will show red, so same color.3. One all-white and one all-red: probability ( 2 times frac{1}{3} times frac{1}{3} = frac{2}{9} ). One will show white, the other red, so different colors.4. One all-white and one balanced: probability ( 2 times frac{1}{3} times frac{1}{3} = frac{2}{9} ). The all-white die will show white, and the balanced die has a ( frac{1}{2} ) chance of white or red. So, the probability that both are white is ( frac{1}{2} ), and different colors is ( frac{1}{2} ).5. One all-red and one balanced: similar to above, probability ( frac{2}{9} ). The all-red die will show red, and the balanced die has a ( frac{1}{2} ) chance of red or white. So, probability both red is ( frac{1}{2} ), different colors is ( frac{1}{2} ).6. Both dice are balanced: probability ( frac{1}{3} times frac{1}{3} = frac{1}{9} ). Both have a ( frac{1}{2} ) chance of white or red. So, probability both white is ( frac{1}{4} ), both red is ( frac{1}{4} ), different colors is ( frac{1}{2} ).Now, let's calculate the total probability ( q ):- From case 1: ( frac{1}{9} times 1 = frac{1}{9} )- From case 2: ( frac{1}{9} times 1 = frac{1}{9} )- From case 3: ( frac{2}{9} times 0 = 0 )- From case 4: ( frac{2}{9} times frac{1}{2} = frac{1}{9} )- From case 5: ( frac{2}{9} times frac{1}{2} = frac{1}{9} )- From case 6: ( frac{1}{9} times frac{1}{2} = frac{1}{18} )Adding these up: ( frac{1}{9} + frac{1}{9} + frac{1}{9} + frac{1}{9} + frac{1}{18} = frac{4}{9} + frac{1}{18} = frac{8}{18} + frac{1}{18} = frac{9}{18} = frac{1}{2} ).Wait, so even in this case where the dice have different distributions of white and red faces, ( q ) still ends up being ( frac{1}{2} ). But ( p ) was ( frac{5}{6} ), which is different from ( frac{1}{2} ).So, in this specific case, ( p = frac{5}{6} ) and ( q = frac{1}{2} ). Let's check if ( p + (n-1)q = frac{n}{2} ).Here, ( n = 3 ). So, ( p + (3-1)q = frac{5}{6} + 2 times frac{1}{2} = frac{5}{6} + 1 = frac{11}{6} ). But ( frac{n}{2} = frac{3}{2} = frac{9}{6} ). So, ( frac{11}{6} neq frac{9}{6} ). That's a problem.Wait, so in this case, the equation doesn't hold. But the problem statement says to prove that ( p + (n-1)q = frac{n}{2} ). So, either my reasoning is wrong, or the initial assumption that each die has exactly 3 white and 3 red faces is necessary.But the problem only states that the total number of white and red faces are equal, not that each die has an equal number. So, in the case where ( n = 3 ), with one all-white, one all-red, and one balanced die, the equation doesn't hold. Therefore, my previous conclusion that ( p + (n-1)q = frac{n}{2} ) is only valid if each die has exactly 3 white and 3 red faces.But the problem doesn't specify that each die has an equal number of white and red faces, only that the total across all dice is equal. So, perhaps there's a different approach.Wait, maybe I need to consider the expected value or something else. Let me think about the overall probability.When you pick one die and roll it twice, the probability of getting the same color twice is ( p ). When you pick two dice and roll them, the probability of both showing the same color is ( q ).But the problem is asking to prove ( p + (n-1)q = frac{n}{2} ). So, perhaps there's a relationship between these probabilities that holds regardless of how the colors are distributed on the dice.Let me consider the total number of possible outcomes when picking one die and rolling it twice, and when picking two dice and rolling them.When picking one die and rolling it twice, there are ( n ) choices for the die, and each die has 6 faces. So, the total number of possible outcomes is ( n times 6 times 6 = 36n ).Similarly, when picking two dice and rolling them, there are ( binom{n}{2} ) ways to choose the two dice, and each die has 6 faces, so the total number of possible outcomes is ( binom{n}{2} times 6 times 6 = frac{n(n-1)}{2} times 36 = 18n(n-1) ).But I'm not sure if counting outcomes is the right approach here. Maybe I should think in terms of expectations or something else.Wait, another approach: consider the probability that two randomly selected faces (with possible repetition) from the entire bag of dice are the same color.Since the total number of white faces is ( 3n ) and red faces is ( 3n ), the probability that two randomly selected faces are the same color is:- Probability both are white: ( frac{3n}{6n} times frac{3n - 1}{6n - 1} )- Probability both are red: ( frac{3n}{6n} times frac{3n - 1}{6n - 1} )So, total probability is ( 2 times frac{3n}{6n} times frac{3n - 1}{6n - 1} = frac{3n - 1}{6n - 1} ).But this seems different from ( frac{1}{2} ). Wait, but when ( n ) is large, this approaches ( frac{1}{2} ), which makes sense because the dependence becomes negligible.But in our problem, we have two different scenarios: picking one die and rolling it twice, and picking two dice and rolling them once each. So, perhaps these two scenarios can be related to the overall probability of picking two faces of the same color.Let me think about it. If I consider all possible pairs of face rolls, some of them involve rolling the same die twice, and others involve rolling two different dice.So, the total probability of getting two faces of the same color can be decomposed into two parts:1. The probability that both rolls are from the same die.2. The probability that both rolls are from different dice.In the first case, the probability is ( p ), and in the second case, the probability is ( q ).But how often do we pick the same die versus different dice?When picking one die and rolling it twice, the probability that both rolls are from the same die is 1, because we're explicitly rolling the same die twice.When picking two dice and rolling them, the probability that both rolls are from different dice is 1, because we're explicitly picking two different dice.Wait, but in the problem statement, ( p ) is defined as the probability when picking one die and rolling it twice, and ( q ) is when picking two dice and rolling them. So, these are two separate experiments.But perhaps there's a way to relate them by considering the overall probability of getting two faces of the same color, regardless of whether they come from the same die or different dice.Wait, let me think about it differently. Suppose I have a large number of trials where I randomly pick either one die and roll it twice or two dice and roll them once each, with some probability. Then, the overall probability of getting two faces of the same color would be a weighted average of ( p ) and ( q ).But the problem is asking to prove ( p + (n-1)q = frac{n}{2} ), which suggests a linear relationship between ( p ) and ( q ).Wait, maybe I can think of it in terms of linearity of expectation. Let me define an indicator variable for each die, but I'm not sure.Alternatively, perhaps I can consider the expected number of same-color pairs.Wait, let me try to model this.Suppose I have ( n ) dice, each with 6 faces, total of ( 6n ) faces, half white and half red.Now, consider all possible pairs of face rolls. There are two types of pairs:1. Pairs where both faces come from the same die.2. Pairs where the two faces come from different dice.For the first type, the number of such pairs is ( n times binom{6}{2} = n times 15 = 15n ).For the second type, the number of such pairs is ( binom{n}{2} times 6 times 6 = frac{n(n-1)}{2} times 36 = 18n(n-1) ).So, total number of pairs is ( 15n + 18n(n-1) = 15n + 18n^2 - 18n = 18n^2 - 3n ).Now, the total number of same-color pairs is:- For same die pairs: each die has ( binom{3}{2} ) white pairs and ( binom{3}{2} ) red pairs, so total same-color pairs per die is ( 3 + 3 = 6 ). Therefore, total same-color pairs from same die is ( n times 6 = 6n ).- For different die pairs: the number of same-color pairs is ( binom{n}{2} times 3 times 3 + binom{n}{2} times 3 times 3 = 2 times binom{n}{2} times 9 = 2 times frac{n(n-1)}{2} times 9 = 9n(n-1) ).Wait, let me explain. For two different dice, the number of white faces on the first die is 3, and on the second die is 3. So, the number of white-white pairs is ( 3 times 3 = 9 ). Similarly, red-red pairs are also ( 9 ). So, total same-color pairs for two different dice is ( 18 ). But since there are ( binom{n}{2} ) pairs of dice, the total same-color pairs from different dice is ( binom{n}{2} times 18 = frac{n(n-1)}{2} times 18 = 9n(n-1) ).Therefore, total same-color pairs is ( 6n + 9n(n-1) = 6n + 9n^2 - 9n = 9n^2 - 3n ).Total number of pairs is ( 18n^2 - 3n ).Therefore, the probability that two randomly selected faces (either from the same die or different dice) are the same color is ( frac{9n^2 - 3n}{18n^2 - 3n} = frac{9n^2 - 3n}{18n^2 - 3n} ).Simplify numerator and denominator:Factor numerator: ( 3n(3n - 1) )Factor denominator: ( 3n(6n - 1) )So, probability is ( frac{3n(3n - 1)}{3n(6n - 1)} = frac{3n - 1}{6n - 1} ).But earlier, I thought this was the probability of two randomly selected faces being the same color, which is different from ( frac{1}{2} ).Wait, but in our problem, we have two specific scenarios: picking one die and rolling it twice, and picking two dice and rolling them once each. So, perhaps the overall probability can be expressed as a combination of these two.Let me think about it. Suppose I have a process where I randomly decide to either pick one die and roll it twice or pick two dice and roll them once each. The probability of choosing the first option is ( frac{1}{n} ) (since there are ( n ) dice, and picking one die is like picking one out of ( n )), and the probability of choosing the second option is ( frac{n-1}{n} ) (since after picking one die, there are ( n-1 ) left).Wait, no, that's not quite right. The process is either:1. Pick one die and roll it twice: probability ( frac{1}{n} ) (since you're picking one specific die out of ( n )).2. Pick two different dice and roll them: probability ( frac{n-1}{n} ) (since after picking the first die, there are ( n-1 ) left).But actually, the probability of picking one die and rolling it twice is ( frac{1}{n} ), and the probability of picking two different dice is ( frac{n-1}{n} ).Therefore, the overall probability of getting two same-color faces is ( frac{1}{n} times p + frac{n-1}{n} times q ).But earlier, we calculated that the overall probability of two same-color faces is ( frac{3n - 1}{6n - 1} ).So, equating these two expressions:( frac{1}{n} p + frac{n-1}{n} q = frac{3n - 1}{6n - 1} )But the problem statement wants us to prove ( p + (n-1) q = frac{n}{2} ). So, if I multiply both sides of the equation by ( n ):( p + (n-1) q = frac{n(3n - 1)}{6n - 1} )But this is not equal to ( frac{n}{2} ) unless ( frac{3n - 1}{6n - 1} = frac{1}{2} ), which would require:( 2(3n - 1) = 6n - 1 )( 6n - 2 = 6n - 1 )Which simplifies to ( -2 = -1 ), which is false. Therefore, my approach must be wrong.Wait, perhaps I made a mistake in calculating the overall probability. Let me double-check.Total number of same-color pairs:- Same die: ( n times 6 = 6n ) (since each die has 6 same-color pairs: 3 white and 3 red, so ( binom{3}{2} times 2 = 6 ) per die).- Different dice: For each pair of dice, the number of same-color pairs is ( 3 times 3 + 3 times 3 = 18 ). There are ( binom{n}{2} ) pairs of dice, so total same-color pairs is ( 18 times binom{n}{2} = 9n(n-1) ).Total same-color pairs: ( 6n + 9n(n-1) = 9n^2 - 3n ).Total number of pairs: same die pairs ( 15n ) plus different die pairs ( 18n(n-1) ), total ( 18n^2 - 3n ).So, probability is ( frac{9n^2 - 3n}{18n^2 - 3n} = frac{3n - 1}{6n - 1} ).But in our problem, the overall probability should be ( frac{1}{2} ), because the total number of white and red faces is equal, so the probability of picking two same-color faces should be ( frac{1}{2} ).Wait, but ( frac{3n - 1}{6n - 1} ) is not equal to ( frac{1}{2} ) unless ( 3n - 1 = 3n - 0.5 ), which is not possible. So, there's a contradiction here.Wait, maybe I made a mistake in counting the same-color pairs. Let me think again.When considering pairs of faces, the total number of same-color pairs is not just the sum of same-die pairs and different-die pairs, because when you pick two faces, they can be from the same die or different dice.But in reality, when you pick two faces, the probability that they are from the same die is ( frac{6}{6n} times frac{5}{6n - 1} ), and the probability they are from different dice is ( 1 - frac{6}{6n} times frac{5}{6n - 1} ).Wait, no, that's not quite right. The probability that two randomly selected faces are from the same die is ( frac{6}{6n} times frac{5}{6n - 1} times n ), because there are ( n ) dice, each contributing ( binom{6}{2} ) pairs.Wait, actually, the number of same-die pairs is ( n times binom{6}{2} = 15n ), and the total number of pairs is ( binom{6n}{2} = frac{6n(6n - 1)}{2} = 18n^2 - 3n ).So, the probability that two randomly selected faces are from the same die is ( frac{15n}{18n^2 - 3n} = frac{15n}{3n(6n - 1)} = frac{5}{6n - 1} ).Similarly, the probability that they are from different dice is ( 1 - frac{5}{6n - 1} = frac{6n - 6}{6n - 1} = frac{6(n - 1)}{6n - 1} ).Now, the probability that two randomly selected faces are the same color is:- If they are from the same die: the probability they are the same color is ( p ).- If they are from different dice: the probability they are the same color is ( q ).Therefore, the total probability is ( frac{5}{6n - 1} times p + frac{6(n - 1)}{6n - 1} times q ).But we also know that the total probability should be ( frac{1}{2} ), because the total number of white and red faces is equal.So, we have:( frac{5}{6n - 1} p + frac{6(n - 1)}{6n - 1} q = frac{1}{2} )Multiplying both sides by ( 6n - 1 ):( 5p + 6(n - 1)q = frac{6n - 1}{2} )But this seems complicated. Let me see if I can relate this to the equation we need to prove, which is ( p + (n - 1)q = frac{n}{2} ).If I can express ( 5p + 6(n - 1)q ) in terms of ( p + (n - 1)q ), maybe I can find a relationship.Let me denote ( S = p + (n - 1)q ). Then, ( 5p + 6(n - 1)q = 5p + 5(n - 1)q + (n - 1)q = 5S + (n - 1)q ).But I'm not sure if this helps.Alternatively, let's try to express ( p ) and ( q ) in terms of ( S ).From ( S = p + (n - 1)q ), we can write ( p = S - (n - 1)q ).Substituting into the equation:( 5(S - (n - 1)q) + 6(n - 1)q = frac{6n - 1}{2} )Simplify:( 5S - 5(n - 1)q + 6(n - 1)q = frac{6n - 1}{2} )( 5S + (n - 1)q = frac{6n - 1}{2} )But we have two variables here, ( S ) and ( q ), so I need another equation.Wait, but ( S = p + (n - 1)q ), and we need to find ( S ). Maybe I can find another relationship.Alternatively, perhaps I can express ( q ) in terms of ( p ).Wait, let me think differently. Since each die has an equal number of white and red faces, the probability that two different dice show the same color is ( q = frac{1}{2} ), as I initially thought. Because each die is fair, so the color of one die doesn't affect the other.But earlier, in the case where one die was all white and another all red, ( q ) was still ( frac{1}{2} ). So, maybe ( q = frac{1}{2} ) regardless of the distribution of colors on the dice.If that's the case, then ( p + (n - 1) times frac{1}{2} = frac{n}{2} ).So, ( p + frac{n - 1}{2} = frac{n}{2} ).Subtracting ( frac{n - 1}{2} ) from both sides:( p = frac{n}{2} - frac{n - 1}{2} = frac{1}{2} ).So, ( p = frac{1}{2} ).Therefore, ( p + (n - 1)q = frac{1}{2} + (n - 1)times frac{1}{2} = frac{n}{2} ).So, this seems to hold if ( q = frac{1}{2} ).But earlier, in the case where one die was all white and another all red, ( q ) was still ( frac{1}{2} ). So, maybe ( q ) is always ( frac{1}{2} ), regardless of the distribution of colors on the dice.Wait, let me test this with another example. Suppose ( n = 2 ), with one die having 4 white and 2 red, and the other having 2 white and 4 red. So, total white faces: 4 + 2 = 6, total red faces: 2 + 4 = 6.Now, let's calculate ( q ). When picking two dice and rolling them, the probability of both being white is ( frac{4}{6} times frac{2}{6} = frac{8}{36} ), and both being red is ( frac{2}{6} times frac{4}{6} = frac{8}{36} ). So, total ( q = frac{16}{36} = frac{4}{9} ), which is not ( frac{1}{2} ).Wait, that's a problem. So, in this case, ( q ) is ( frac{4}{9} ), not ( frac{1}{2} ). Therefore, my previous assumption that ( q = frac{1}{2} ) is not always true.So, in this case, ( n = 2 ), ( q = frac{4}{9} ), and ( p ) would be the probability of rolling the same color twice on a single die.For the first die (4 white, 2 red), ( p_1 = frac{4}{6} times frac{4}{6} + frac{2}{6} times frac{2}{6} = frac{16}{36} + frac{4}{36} = frac{20}{36} = frac{5}{9} ).For the second die (2 white, 4 red), ( p_2 = frac{2}{6} times frac{2}{6} + frac{4}{6} times frac{4}{6} = frac{4}{36} + frac{16}{36} = frac{20}{36} = frac{5}{9} ).Since each die is equally likely to be picked, ( p = frac{5}{9} ).Now, let's check ( p + (n - 1)q = frac{5}{9} + (2 - 1) times frac{4}{9} = frac{5}{9} + frac{4}{9} = frac{9}{9} = 1 ).But ( frac{n}{2} = frac{2}{2} = 1 ). So, in this case, the equation holds.Wait, that's interesting. So, even though ( q ) wasn't ( frac{1}{2} ), the equation ( p + (n - 1)q = frac{n}{2} ) still held.So, perhaps the equation is always true, regardless of how the colors are distributed on the dice, as long as the total number of white and red faces is equal.Therefore, maybe I need to find a general proof that doesn't rely on specific distributions.Let me try to approach it algebraically.Let me denote:- Let ( w_i ) be the number of white faces on die ( i ), so ( r_i = 6 - w_i ) is the number of red faces.Given that the total number of white faces is ( 3n ), so ( sum_{i=1}^{n} w_i = 3n ).Similarly, ( sum_{i=1}^{n} r_i = 3n ).Now, ( p ) is the probability that when picking a die at random and rolling it twice, both rolls are the same color.So, ( p = frac{1}{n} sum_{i=1}^{n} left( frac{w_i}{6} times frac{w_i}{6} + frac{r_i}{6} times frac{r_i}{6} right) ).Simplify:( p = frac{1}{n} sum_{i=1}^{n} left( frac{w_i^2 + r_i^2}{36} right) = frac{1}{36n} sum_{i=1}^{n} (w_i^2 + r_i^2) ).But since ( r_i = 6 - w_i ), we have:( w_i^2 + r_i^2 = w_i^2 + (6 - w_i)^2 = w_i^2 + 36 - 12w_i + w_i^2 = 2w_i^2 - 12w_i + 36 ).Therefore,( p = frac{1}{36n} sum_{i=1}^{n} (2w_i^2 - 12w_i + 36) = frac{1}{36n} left( 2 sum w_i^2 - 12 sum w_i + 36n right) ).We know that ( sum w_i = 3n ), so:( p = frac{1}{36n} left( 2 sum w_i^2 - 12 times 3n + 36n right) = frac{1}{36n} left( 2 sum w_i^2 - 36n + 36n right) = frac{1}{36n} times 2 sum w_i^2 = frac{sum w_i^2}{18n} ).Similarly, let's calculate ( q ). ( q ) is the probability that when picking two different dice and rolling them, both show the same color.So, ( q = frac{sum_{i < j} left( frac{w_i}{6} times frac{w_j}{6} + frac{r_i}{6} times frac{r_j}{6} right)}{binom{n}{2}} ).Simplify:( q = frac{sum_{i < j} left( frac{w_i w_j + r_i r_j}{36} right)}{frac{n(n-1)}{2}} = frac{1}{36 times frac{n(n-1)}{2}} sum_{i < j} (w_i w_j + r_i r_j) ).But ( r_i = 6 - w_i ), so ( r_i r_j = (6 - w_i)(6 - w_j) = 36 - 6w_i - 6w_j + w_i w_j ).Therefore, ( w_i w_j + r_i r_j = w_i w_j + 36 - 6w_i - 6w_j + w_i w_j = 2w_i w_j + 36 - 6(w_i + w_j) ).So,( q = frac{1}{36 times frac{n(n-1)}{2}} sum_{i < j} (2w_i w_j + 36 - 6(w_i + w_j)) ).Let's break this down:First, ( sum_{i < j} 2w_i w_j = 2 times frac{(sum w_i)^2 - sum w_i^2}{2} = (sum w_i)^2 - sum w_i^2 ).Second, ( sum_{i < j} 36 = 36 times binom{n}{2} = 36 times frac{n(n-1)}{2} = 18n(n-1) ).Third, ( sum_{i < j} -6(w_i + w_j) = -6 times sum_{i < j} (w_i + w_j) = -6 times (n - 1) sum w_i ).Because for each ( w_i ), it appears in ( (n - 1) ) terms where ( j neq i ).So, putting it all together:( q = frac{1}{36 times frac{n(n-1)}{2}} left[ (sum w_i)^2 - sum w_i^2 + 18n(n-1) - 6(n - 1) sum w_i right] ).We know that ( sum w_i = 3n ), so:( (sum w_i)^2 = (3n)^2 = 9n^2 ).Also, ( -6(n - 1) sum w_i = -6(n - 1)(3n) = -18n(n - 1) ).Therefore, substituting:( q = frac{1}{36 times frac{n(n-1)}{2}} left[ 9n^2 - sum w_i^2 + 18n(n-1) - 18n(n - 1) right] ).Simplify inside the brackets:( 9n^2 - sum w_i^2 + 18n(n-1) - 18n(n - 1) = 9n^2 - sum w_i^2 ).So,( q = frac{1}{36 times frac{n(n-1)}{2}} (9n^2 - sum w_i^2) = frac{9n^2 - sum w_i^2}{18n(n - 1)} ).Now, recall that ( p = frac{sum w_i^2}{18n} ).So, let's express ( sum w_i^2 = 18n p ).Substituting into the expression for ( q ):( q = frac{9n^2 - 18n p}{18n(n - 1)} = frac{9n^2 - 18n p}{18n(n - 1)} = frac{9n^2}{18n(n - 1)} - frac{18n p}{18n(n - 1)} = frac{n}{2(n - 1)} - frac{p}{n - 1} ).So,( q = frac{n}{2(n - 1)} - frac{p}{n - 1} ).Multiplying both sides by ( n - 1 ):( (n - 1) q = frac{n}{2} - p ).Rearranging:( p + (n - 1) q = frac{n}{2} ).Which is exactly what we needed to prove.So, through algebraic manipulation, we've shown that ( p + (n - 1) q = frac{n}{2} ), regardless of the distribution of white and red faces on the dice, as long as the total number of white and red faces is equal.

💡Alright, so I have this problem here: There's a line passing through two points, P(1 - a, 1 + a) and Q(3, 2a). The angle α that this line makes is obtuse, and I need to find the range of the real number a. Hmm, okay, let's try to figure this out step by step.First, I remember that the slope of a line is given by the formula m = (y2 - y1)/(x2 - x1). So, I need to find the slope of the line passing through points P and Q. Let me write down the coordinates:Point P is (1 - a, 1 + a), and point Q is (3, 2a). So, substituting into the slope formula:m = (2a - (1 + a)) / (3 - (1 - a)).Let me simplify the numerator and the denominator separately. The numerator is 2a - 1 - a, which simplifies to (2a - a) - 1 = a - 1. The denominator is 3 - 1 + a, which is 2 + a. So, the slope m is (a - 1)/(2 + a).Okay, so m = (a - 1)/(a + 2). Now, the problem says that the angle α is obtuse. I need to recall what that means in terms of the slope. I remember that the angle a line makes with the positive direction of the x-axis is called the angle of inclination, and if that angle is obtuse, it means it's greater than 90 degrees but less than 180 degrees.I also recall that the slope of a line is equal to the tangent of the angle it makes with the x-axis. So, if α is obtuse, tan(α) would be negative because tangent is negative in the second quadrant (angles between 90 and 180 degrees). Therefore, the slope m must be negative.So, for the angle α to be obtuse, the slope m must be negative. Therefore, I need to find the values of a for which (a - 1)/(a + 2) < 0.Alright, so now I have the inequality (a - 1)/(a + 2) < 0. To solve this inequality, I need to determine when the numerator and denominator have opposite signs because their ratio will be negative in that case.Let me find the critical points where the numerator or denominator is zero. The numerator a - 1 = 0 when a = 1, and the denominator a + 2 = 0 when a = -2. These points divide the real number line into intervals. I need to test each interval to see where the expression is negative.The critical points are a = -2 and a = 1. So, the intervals are:1. a < -22. -2 < a < 13. a > 1Now, let's test each interval.1. For a < -2, let's pick a = -3. Plugging into (a - 1)/(a + 2):Numerator: -3 - 1 = -4 (negative)Denominator: -3 + 2 = -1 (negative)Negative divided by negative is positive. So, the expression is positive in this interval.2. For -2 < a < 1, let's pick a = 0. Plugging into (a - 1)/(a + 2):Numerator: 0 - 1 = -1 (negative)Denominator: 0 + 2 = 2 (positive)Negative divided by positive is negative. So, the expression is negative in this interval.3. For a > 1, let's pick a = 2. Plugging into (a - 1)/(a + 2):Numerator: 2 - 1 = 1 (positive)Denominator: 2 + 2 = 4 (positive)Positive divided by positive is positive. So, the expression is positive in this interval.Therefore, the expression (a - 1)/(a + 2) is negative only in the interval -2 < a < 1.But wait, I should also check the critical points themselves. At a = -2, the denominator becomes zero, which would make the slope undefined, meaning the line is vertical. A vertical line makes an angle of 90 degrees, which is not obtuse. Similarly, at a = 1, the numerator becomes zero, so the slope is zero, meaning the line is horizontal, making an angle of 0 degrees, which is also not obtuse. So, these points are excluded.Therefore, the range of a is all real numbers between -2 and 1, not including -2 and 1 themselves.Let me just recap to make sure I didn't miss anything. The slope is (a - 1)/(a + 2). For the angle to be obtuse, the slope needs to be negative. Solving the inequality (a - 1)/(a + 2) < 0 gives us the interval (-2, 1). Checking the endpoints shows they don't satisfy the condition, so we don't include them. Therefore, the range of a is (-2, 1).I think that's it. I don't see any mistakes in my reasoning. I considered the slope, related it to the angle, set up the inequality, solved it, and checked the critical points. Everything seems consistent.**Final Answer**The range of the real number ( a ) is boxed{(-2, 1)}.