💡Okay, so I have this quadratic equation: y squared plus 6n times y plus 9n equals zero. I need to find the positive value of n that makes this equation have exactly one real root. Hmm, okay, let me think about how to approach this.First, I remember that for a quadratic equation of the form ay squared plus by plus c equals zero, the discriminant determines the nature of the roots. The discriminant is given by b squared minus 4ac. If the discriminant is zero, the quadratic has exactly one real root, which is what I need here.So, in this equation, the coefficients are: a is 1 because the coefficient of y squared is 1, b is 6n, and c is 9n. Let me write that down:a = 1b = 6nc = 9nNow, I need to calculate the discriminant using these values. The discriminant formula is:Discriminant (Δ) = b² - 4acPlugging in the values I have:Δ = (6n)² - 4 * 1 * 9nLet me compute each part step by step. First, (6n) squared is 36n squared. Then, 4 times 1 times 9n is 36n. So, putting it all together:Δ = 36n² - 36nNow, for the quadratic to have exactly one real root, the discriminant must be zero. So, I set the discriminant equal to zero:36n² - 36n = 0Hmm, okay, so I have this equation to solve for n. Let me factor out the common term. Both terms have a factor of 36n, so factoring that out:36n(n - 1) = 0So, this product equals zero if either factor is zero. That gives me two possible solutions:36n = 0 or n - 1 = 0Solving each equation:If 36n = 0, then n = 0.If n - 1 = 0, then n = 1.Now, the problem asks for the positive value of n. So, n = 0 is not positive, which leaves me with n = 1.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the discriminant:Δ = (6n)² - 4 * 1 * 9n = 36n² - 36nSet to zero:36n² - 36n = 0Factor:36n(n - 1) = 0Solutions: n = 0 or n = 1Yes, that seems correct. So, n = 1 is the positive solution.Just to be thorough, let me plug n = 1 back into the original quadratic equation and see if it indeed has exactly one real root.Plugging n = 1:y² + 6*1*y + 9*1 = y² + 6y + 9This simplifies to y² + 6y + 9. Let me check the discriminant again for this specific case:Δ = (6)² - 4*1*9 = 36 - 36 = 0Yes, the discriminant is zero, so there is exactly one real root. Therefore, n = 1 is indeed the correct answer.I think I've covered all the steps and checked my work. It makes sense, and I don't see any errors in my calculations. So, I'm confident that n = 1 is the positive value that satisfies the condition.

💡Okay, so I have this function f(x) = x³ - x² - x + 2, and I need to do two things: first, find the intervals where the function is increasing or decreasing and identify its extreme values. Second, given that for x in the interval [-1, 2], the inequality -3 ≤ a f(x) + b ≤ 3 holds, I need to find the maximum value of a - b. Hmm, that sounds a bit complex, but let's break it down step by step.Starting with part (I): Finding intervals of monotonicity and extreme values. I remember that to find where a function is increasing or decreasing, I need to look at its first derivative. The first derivative tells me the slope of the function at any point, so if the derivative is positive, the function is increasing, and if it's negative, the function is decreasing. Also, the points where the derivative is zero are potential local maxima or minima.So, let's compute the first derivative of f(x). The function is f(x) = x³ - x² - x + 2. Taking the derivative term by term:- The derivative of x³ is 3x².- The derivative of -x² is -2x.- The derivative of -x is -1.- The derivative of the constant 2 is 0.So, putting it all together, f'(x) = 3x² - 2x - 1.Now, I need to find where f'(x) is zero because those are the critical points where the function could have local maxima or minima. Let's solve the equation 3x² - 2x - 1 = 0.This is a quadratic equation, so I can use the quadratic formula: x = [2 ± sqrt( (-2)² - 4*3*(-1) )]/(2*3). Calculating the discriminant first: (-2)² = 4, 4*3*(-1) = -12, so the discriminant is 4 - (-12) = 16. That's a perfect square, which is good.So, x = [2 ± sqrt(16)]/6 = [2 ± 4]/6. Therefore, two solutions:1. x = (2 + 4)/6 = 6/6 = 12. x = (2 - 4)/6 = (-2)/6 = -1/3So, the critical points are at x = 1 and x = -1/3. These divide the real number line into intervals where the derivative may change sign. Let's test the sign of f'(x) in each interval.The critical points split the number line into three intervals:1. (-∞, -1/3)2. (-1/3, 1)3. (1, ∞)Let's pick test points in each interval to see if f'(x) is positive or negative there.For interval (-∞, -1/3), let's choose x = -1. Plugging into f'(x):f'(-1) = 3*(-1)² - 2*(-1) - 1 = 3*1 + 2 - 1 = 3 + 2 - 1 = 4. That's positive, so f(x) is increasing on (-∞, -1/3).For interval (-1/3, 1), let's choose x = 0. Plugging into f'(x):f'(0) = 3*0² - 2*0 - 1 = 0 - 0 - 1 = -1. That's negative, so f(x) is decreasing on (-1/3, 1).For interval (1, ∞), let's choose x = 2. Plugging into f'(x):f'(2) = 3*(2)² - 2*(2) - 1 = 3*4 - 4 - 1 = 12 - 4 - 1 = 7. That's positive, so f(x) is increasing on (1, ∞).So, summarizing:- f(x) is increasing on (-∞, -1/3)- f(x) is decreasing on (-1/3, 1)- f(x) is increasing on (1, ∞)Now, the critical points at x = -1/3 and x = 1 are potential local maxima or minima. To determine which is which, we can use the second derivative test or analyze the sign changes of the first derivative.Let me try the second derivative test. Compute f''(x):f'(x) = 3x² - 2x - 1, so f''(x) is the derivative of that: 6x - 2.Now, evaluate f''(x) at the critical points:At x = -1/3:f''(-1/3) = 6*(-1/3) - 2 = -2 - 2 = -4. Since f''(-1/3) is negative, the function is concave down at this point, so x = -1/3 is a local maximum.At x = 1:f''(1) = 6*(1) - 2 = 6 - 2 = 4. Since f''(1) is positive, the function is concave up at this point, so x = 1 is a local minimum.Now, let's find the corresponding y-values for these critical points.For x = -1/3:f(-1/3) = (-1/3)³ - (-1/3)² - (-1/3) + 2.Calculating each term:(-1/3)³ = -1/27(-1/3)² = 1/9, so - (1/9) = -1/9- (-1/3) = 1/3So, f(-1/3) = (-1/27) - (1/9) + (1/3) + 2.Let's convert all to 27 denominators:-1/27 - 3/27 + 9/27 + 54/27 = (-1 - 3 + 9 + 54)/27 = (59)/27 ≈ 2.185.So, f(-1/3) = 59/27, which is approximately 2.185.For x = 1:f(1) = (1)³ - (1)² - (1) + 2 = 1 - 1 - 1 + 2 = 1.So, f(1) = 1.Therefore, the function has a local maximum at x = -1/3 with value 59/27 and a local minimum at x = 1 with value 1.So, that completes part (I). Now, moving on to part (II).Part (II) says: If for x ∈ [-1, 2], the inequality -3 ≤ a f(x) + b ≤ 3 holds, find the maximum value of a - b.Hmm, okay. So, we have an inequality involving a linear combination of f(x) with coefficients a and b, and we need to find the maximum value of a - b such that this inequality holds for all x in [-1, 2].First, let's understand what this inequality is saying. For every x in [-1, 2], the expression a f(x) + b must lie between -3 and 3. So, essentially, the function a f(x) + b is bounded between -3 and 3 on the interval [-1, 2].Our goal is to find the maximum value of a - b given this constraint.To approach this, perhaps we can consider the maximum and minimum values of f(x) on the interval [-1, 2]. If we can find the range of f(x) on this interval, then we can set up inequalities for a and b such that a times the maximum of f(x) plus b is ≤ 3, and a times the minimum of f(x) plus b is ≥ -3.So, first, let's find the maximum and minimum values of f(x) on [-1, 2].From part (I), we know that f(x) has critical points at x = -1/3 and x = 1. Since we're looking at the interval [-1, 2], both critical points are within this interval.So, to find the maximum and minimum of f(x) on [-1, 2], we need to evaluate f(x) at the critical points and at the endpoints of the interval.So, let's compute f(-1), f(-1/3), f(1), and f(2).We already have f(-1/3) = 59/27 ≈ 2.185 and f(1) = 1.Compute f(-1):f(-1) = (-1)³ - (-1)² - (-1) + 2 = -1 - 1 + 1 + 2 = (-1 -1) + (1 + 2) = (-2) + 3 = 1.So, f(-1) = 1.Compute f(2):f(2) = (2)³ - (2)² - (2) + 2 = 8 - 4 - 2 + 2 = (8 - 4) + (-2 + 2) = 4 + 0 = 4.So, f(2) = 4.So, summarizing:- f(-1) = 1- f(-1/3) = 59/27 ≈ 2.185- f(1) = 1- f(2) = 4Therefore, on the interval [-1, 2], the maximum value of f(x) is 4, achieved at x = 2, and the minimum value is 1, achieved at x = -1 and x = 1.Wait, hold on. f(-1) is 1, f(1) is 1, and f(-1/3) is approximately 2.185, which is higher than 1, and f(2) is 4, which is the highest.So, the maximum is 4, and the minimum is 1.Therefore, on [-1, 2], f(x) ranges from 1 to 4.So, now, the inequality is -3 ≤ a f(x) + b ≤ 3 for all x in [-1, 2]. Since f(x) ranges from 1 to 4, the expression a f(x) + b must lie between -3 and 3.So, let's denote M = max f(x) = 4 and m = min f(x) = 1.Therefore, for the inequality to hold for all x in [-1, 2], we must have:a*M + b ≤ 3 and a*m + b ≥ -3.So, substituting M and m:a*4 + b ≤ 3a*1 + b ≥ -3So, we have two inequalities:1. 4a + b ≤ 32. a + b ≥ -3Additionally, since f(x) can take any value between 1 and 4, we need to ensure that the entire range of a f(x) + b is within [-3, 3]. So, if we consider a f(x) + b as a linear function in f(x), its maximum and minimum will occur at the endpoints of f(x), which are 1 and 4.Therefore, the constraints are indeed:4a + b ≤ 3a + b ≥ -3But, wait, is that sufficient? Let me think.Suppose a is positive. Then, a f(x) + b will be maximized when f(x) is maximized, i.e., at f(x) = 4, and minimized when f(x) is minimized, i.e., at f(x) = 1.Similarly, if a is negative, then a f(x) + b will be maximized when f(x) is minimized and minimized when f(x) is maximized.Therefore, regardless of the sign of a, the maximum and minimum of a f(x) + b occur at the endpoints of f(x). So, the two inequalities I wrote above are indeed sufficient to ensure that -3 ≤ a f(x) + b ≤ 3 for all x in [-1, 2].Therefore, we have the system of inequalities:1. 4a + b ≤ 32. a + b ≥ -3We need to find the maximum value of a - b.So, we have two inequalities:4a + b ≤ 3a + b ≥ -3We can write this as:4a + b ≤ 3a + b ≥ -3We can represent this system graphically or solve it algebraically. Since we have two inequalities with two variables, we can solve for a and b.But our goal is to maximize a - b. So, perhaps we can express b in terms of a from one inequality and substitute into the other.Let me try that.From the second inequality: a + b ≥ -3, so b ≥ -3 - a.From the first inequality: 4a + b ≤ 3, so b ≤ 3 - 4a.So, combining these two, we have:-3 - a ≤ b ≤ 3 - 4aOur goal is to maximize a - b. Let's express a - b in terms of a.Since b is bounded between -3 - a and 3 - 4a, to maximize a - b, we need to minimize b because a - b = a + (-b). So, to maximize a - b, we need to minimize b.Therefore, the minimal value of b is -3 - a.So, substituting b = -3 - a into a - b, we get:a - (-3 - a) = a + 3 + a = 2a + 3So, now, we need to maximize 2a + 3, but we also have the constraint from the first inequality: b ≤ 3 - 4a.But since we set b = -3 - a, we need to ensure that -3 - a ≤ 3 - 4a.So, let's write that inequality:-3 - a ≤ 3 - 4aLet's solve for a:-3 - a ≤ 3 - 4aAdd 4a to both sides:-3 + 3a ≤ 3Add 3 to both sides:3a ≤ 6Divide by 3:a ≤ 2So, a must be ≤ 2.Therefore, 2a + 3 is maximized when a is as large as possible, which is a = 2.So, substituting a = 2 into b = -3 - a, we get b = -3 - 2 = -5.Therefore, a = 2, b = -5.So, a - b = 2 - (-5) = 2 + 5 = 7.Wait, but let's verify if these values satisfy both inequalities.First inequality: 4a + b = 4*2 + (-5) = 8 - 5 = 3 ≤ 3. Okay, that's good.Second inequality: a + b = 2 + (-5) = -3 ≥ -3. Also good.So, both inequalities are satisfied with equality.Therefore, the maximum value of a - b is 7.But just to be thorough, let's check if this is indeed the case.If a = 2 and b = -5, then the expression a f(x) + b becomes 2 f(x) - 5.We need to ensure that for all x in [-1, 2], -3 ≤ 2 f(x) - 5 ≤ 3.Let's compute 2 f(x) - 5 at the critical points and endpoints.At x = -1: f(-1) = 1, so 2*1 - 5 = 2 - 5 = -3.At x = -1/3: f(-1/3) = 59/27 ≈ 2.185, so 2*(59/27) - 5 ≈ 118/27 - 5 ≈ 4.37 - 5 ≈ -0.63.At x = 1: f(1) = 1, so 2*1 - 5 = -3.At x = 2: f(2) = 4, so 2*4 - 5 = 8 - 5 = 3.So, the expression 2 f(x) - 5 takes values from -3 to 3 on the interval [-1, 2], which satisfies the inequality -3 ≤ 2 f(x) - 5 ≤ 3.Therefore, a = 2 and b = -5 are valid, and a - b = 7 is indeed achievable.Is there a possibility of getting a higher value of a - b? Let's see.Suppose we try a higher a, say a = 3. Then, from the second inequality, b ≥ -3 - 3 = -6.From the first inequality, b ≤ 3 - 4*3 = 3 - 12 = -9.But b must satisfy both b ≥ -6 and b ≤ -9, which is impossible because -6 > -9. So, no solution exists for a = 3.Similarly, trying a = 2.5:From second inequality: b ≥ -3 - 2.5 = -5.5From first inequality: b ≤ 3 - 4*2.5 = 3 - 10 = -7Again, b must be ≥ -5.5 and ≤ -7, which is impossible since -5.5 > -7.So, a cannot be higher than 2.Therefore, the maximum value of a - b is indeed 7.So, summarizing:(I) The function f(x) is increasing on (-∞, -1/3), decreasing on (-1/3, 1), and increasing on (1, ∞). It has a local maximum at x = -1/3 with value 59/27 and a local minimum at x = 1 with value 1.(II) The maximum value of a - b is 7.**Final Answer**The maximum value of (a - b) is (boxed{7}).

💡First, I observe the pattern in the squares of numbers consisting of all nines:- (9^2 = 81) contains 0 zeros.- (99^2 = 9801) contains 1 zero.- (999^2 = 998001) contains 2 zeros.From these examples, it appears that the number of zeros in the square of a number with (n) nines is (n - 1).Applying this pattern to (999,!999,!999), which has 9 nines, the number of zeros in its square should be (9 - 1 = 8).To confirm this, I express (999,!999,!999) as (10^9 - 1) and square it:[(10^9 - 1)^2 = 10^{18} - 2 cdot 10^9 + 1]This results in a number that has 8 zeros in the middle, confirming the pattern.

💡Alright, so I've got this problem here about points on a parabola, and I need to figure out which of the statements A, B, C, or D is correct. Let me try to break this down step by step.First, the problem says that points A, B, and C are all on the parabola given by the equation ( y = -frac{a}{4}x^2 + ax + c ). It also mentions that ( y_2 = a + c ), which is the y-coordinate of point B. I remember that the general form of a parabola is ( y = ax^2 + bx + c ), but this one is a bit different because of the negative coefficient in front of the ( x^2 ) term. That probably means it's a downward-opening parabola if the coefficient is negative, but let me confirm that.Wait, actually, the standard form is ( y = ax^2 + bx + c ), and if the coefficient of ( x^2 ) is negative, it opens downward. In this case, the coefficient is ( -frac{a}{4} ). So, if ( a ) is positive, then ( -frac{a}{4} ) is negative, meaning the parabola opens downward. If ( a ) is negative, then ( -frac{a}{4} ) becomes positive, so the parabola opens upward. Hmm, that's interesting. So the direction the parabola opens depends on the sign of ( a ).But the problem doesn't specify whether ( a ) is positive or negative, so I might have to consider both cases. However, let's see if we can get more information from the given equation.The problem also mentions that ( y_2 = a + c ). Since point B is ( (x_2, y_2) ), plugging that into the equation of the parabola gives:( y_2 = -frac{a}{4}x_2^2 + a x_2 + c )But we know ( y_2 = a + c ), so:( a + c = -frac{a}{4}x_2^2 + a x_2 + c )Subtracting ( c ) from both sides:( a = -frac{a}{4}x_2^2 + a x_2 )Let me solve for ( x_2 ). First, factor out ( a ):( a = a(-frac{1}{4}x_2^2 + x_2) )Assuming ( a neq 0 ), we can divide both sides by ( a ):( 1 = -frac{1}{4}x_2^2 + x_2 )Multiply both sides by 4 to eliminate the fraction:( 4 = -x_2^2 + 4x_2 )Bring all terms to one side:( x_2^2 - 4x_2 + 4 = 0 )This is a quadratic equation in ( x_2 ). Let's factor it:( (x_2 - 2)^2 = 0 )So, ( x_2 = 2 ). That means point B is at ( (2, a + c) ). Wait a second, this looks familiar. The vertex form of a parabola is ( y = a(x - h)^2 + k ), where ( (h, k) ) is the vertex. Let me rewrite the given equation in vertex form to see if I can confirm this.Starting with ( y = -frac{a}{4}x^2 + ax + c ), let's complete the square.Factor out ( -frac{a}{4} ) from the first two terms:( y = -frac{a}{4}(x^2 - 4x) + c )Now, to complete the square inside the parentheses, take half of the coefficient of ( x ), which is ( -4 ), so half of that is ( -2 ), and square it, getting ( 4 ). So, add and subtract 4 inside the parentheses:( y = -frac{a}{4}(x^2 - 4x + 4 - 4) + c )Simplify:( y = -frac{a}{4}((x - 2)^2 - 4) + c )Distribute the ( -frac{a}{4} ):( y = -frac{a}{4}(x - 2)^2 + a + c )So, the vertex form is ( y = -frac{a}{4}(x - 2)^2 + (a + c) ). Therefore, the vertex of the parabola is at ( (2, a + c) ), which is exactly point B. So, point B is the vertex of the parabola.Now, since the vertex is at ( (2, a + c) ), and the coefficient of ( (x - 2)^2 ) is ( -frac{a}{4} ), the parabola opens downward if ( a > 0 ) and upward if ( a < 0 ).Okay, so now that I know point B is the vertex, let's think about the implications.If the parabola opens downward (i.e., ( a > 0 )), then the vertex is the highest point on the parabola. That means ( y_2 = a + c ) is the maximum y-value on the parabola. So, for any other point on the parabola, their y-values will be less than or equal to ( y_2 ).Similarly, if the parabola opens upward (i.e., ( a < 0 )), then the vertex is the lowest point on the parabola, meaning ( y_2 = a + c ) is the minimum y-value, and other points will have y-values greater than or equal to ( y_2 ).But the problem doesn't specify whether ( a ) is positive or negative, so I might need to consider both cases when evaluating the statements.Let's look at the statements one by one.**Statement A: If ( |x_1 - x_2| leq |x_3 - x_2| ), then ( y_2 geq y_3 geq y_1 ).**Okay, so this is saying that if the distance of ( x_1 ) from ( x_2 ) is less than or equal to the distance of ( x_3 ) from ( x_2 ), then ( y_2 ) is greater than or equal to ( y_3 ), which is greater than or equal to ( y_1 ).Since ( x_2 = 2 ), this is comparing the distances of ( x_1 ) and ( x_3 ) from 2. So, if ( x_1 ) is closer to 2 than ( x_3 ) is, then ( y_2 geq y_3 geq y_1 ).But let's think about this in terms of the parabola.If the parabola opens downward (( a > 0 )), then as you move away from the vertex (which is at ( x = 2 )), the y-values decrease. So, the closer a point is to ( x = 2 ), the higher its y-value.Therefore, if ( |x_1 - 2| leq |x_3 - 2| ), then ( y_1 geq y_3 ). But Statement A says ( y_2 geq y_3 geq y_1 ). Wait, that would mean ( y_3 geq y_1 ), but according to the parabola, since ( x_1 ) is closer, ( y_1 ) should be greater than ( y_3 ). So, Statement A seems incorrect in this case.If the parabola opens upward (( a < 0 )), then as you move away from the vertex, the y-values increase. So, the closer a point is to ( x = 2 ), the lower its y-value.In this case, if ( |x_1 - 2| leq |x_3 - 2| ), then ( y_1 leq y_3 ). But Statement A says ( y_2 geq y_3 geq y_1 ). Since ( y_2 ) is the minimum in this case, ( y_2 leq y_3 ) and ( y_2 leq y_1 ). So, ( y_3 geq y_1 ) would depend on the distances. If ( x_1 ) is closer, then ( y_1 ) is smaller, so ( y_3 geq y_1 ) is true, but ( y_2 geq y_3 ) is not necessarily true because ( y_2 ) is the minimum.So, in both cases, Statement A doesn't hold. If the parabola opens downward, ( y_1 geq y_3 ), which contradicts ( y_3 geq y_1 ). If it opens upward, ( y_2 ) is the minimum, so ( y_2 leq y_3 ), which contradicts ( y_2 geq y_3 ). Therefore, Statement A is incorrect.**Statement B: If ( |x_1 - x_2| geq |x_3 - x_2| ), then ( y_2 geq y_3 geq y_1 ).**This is similar to Statement A, but the inequality is reversed. So, if ( x_1 ) is farther from ( x_2 ) than ( x_3 ) is, then ( y_2 geq y_3 geq y_1 ).Again, let's consider the two cases.If the parabola opens downward (( a > 0 )), then moving away from ( x = 2 ) decreases the y-value. So, if ( x_1 ) is farther from 2 than ( x_3 ), then ( y_1 leq y_3 ). But Statement B says ( y_2 geq y_3 geq y_1 ). Since ( y_2 ) is the maximum, ( y_2 geq y_3 ) is true, and ( y_3 geq y_1 ) is also true because ( x_1 ) is farther. So, in this case, Statement B holds.If the parabola opens upward (( a < 0 )), then moving away from ( x = 2 ) increases the y-value. So, if ( x_1 ) is farther from 2 than ( x_3 ), then ( y_1 geq y_3 ). But Statement B says ( y_2 geq y_3 geq y_1 ). However, since ( y_2 ) is the minimum, ( y_2 leq y_3 ) and ( y_2 leq y_1 ). So, ( y_3 geq y_1 ) would be false because ( x_1 ) is farther, so ( y_1 geq y_3 ). Therefore, Statement B is false in this case.Since Statement B is only true when the parabola opens downward, but the problem doesn't specify the sign of ( a ), we can't be certain that Statement B is always correct. Therefore, Statement B is not necessarily correct.**Statement C: If ( y_1 > y_3 geq y_2 ), then ( |x_1 - x_2| < |x_3 - x_2| ).**This is saying that if ( y_1 ) is greater than ( y_3 ), which is greater than or equal to ( y_2 ), then ( x_1 ) is closer to ( x_2 ) than ( x_3 ) is.Let's analyze this.First, ( y_3 geq y_2 ). Since ( y_2 ) is the y-coordinate of the vertex, which is either the maximum or minimum depending on the direction the parabola opens.If the parabola opens downward (( a > 0 )), then ( y_2 ) is the maximum. So, ( y_3 geq y_2 ) would imply ( y_3 = y_2 ), meaning ( x_3 = x_2 ) because the vertex is the only point with that y-value. But if ( x_3 = x_2 ), then ( |x_3 - x_2| = 0 ). Then, ( y_1 > y_3 = y_2 ) would imply that ( x_1 ) is closer to ( x_2 ) than ( x_3 ) is, but since ( x_3 = x_2 ), ( |x_1 - x_2| ) must be greater than 0, which contradicts ( |x_1 - x_2| < |x_3 - x_2| = 0 ). So, this seems impossible. Therefore, in the case of a downward-opening parabola, ( y_3 geq y_2 ) can only happen if ( x_3 = x_2 ), making ( y_3 = y_2 ), but then ( y_1 > y_3 ) would require ( x_1 ) to be closer, which isn't possible because ( x_3 ) is already at the vertex. So, this scenario might not be possible.If the parabola opens upward (( a < 0 )), then ( y_2 ) is the minimum. So, ( y_3 geq y_2 ) is always true because all other points have higher y-values. Now, ( y_1 > y_3 ) implies that ( y_1 ) is greater than ( y_3 ), which is greater than or equal to ( y_2 ). In an upward-opening parabola, as you move away from the vertex, the y-values increase. So, the farther a point is from ( x = 2 ), the higher its y-value. Therefore, if ( y_1 > y_3 ), then ( x_1 ) must be farther from ( x_2 ) than ( x_3 ) is. That would mean ( |x_1 - x_2| > |x_3 - x_2| ), which contradicts Statement C, which says ( |x_1 - x_2| < |x_3 - x_2| ).Therefore, Statement C is incorrect because in the only feasible case (upward-opening parabola), the conclusion is the opposite of what's stated.**Statement D: If ( y_1 > y_3 geq y_2 ), then ( |x_1 - x_2| > |x_3 - x_2| ).**This is similar to Statement C but with the inequality reversed.Again, considering the two cases.If the parabola opens downward (( a > 0 )), as before, ( y_3 geq y_2 ) would imply ( x_3 = x_2 ), making ( y_3 = y_2 ). Then, ( y_1 > y_3 = y_2 ) would require ( x_1 ) to be closer to ( x_2 ), but since ( x_3 = x_2 ), ( |x_1 - x_2| ) would have to be less than 0, which isn't possible. So, this scenario is not possible for a downward-opening parabola.If the parabola opens upward (( a < 0 )), ( y_2 ) is the minimum, so ( y_3 geq y_2 ) is always true. Now, ( y_1 > y_3 ) implies that ( y_1 ) is greater than ( y_3 ), which, in an upward-opening parabola, means ( x_1 ) is farther from ( x_2 ) than ( x_3 ) is. Therefore, ( |x_1 - x_2| > |x_3 - x_2| ), which is exactly what Statement D says.So, in the case of an upward-opening parabola, Statement D holds true. Since the problem doesn't specify the direction, but Statement D is consistent with the behavior when the parabola opens upward, and the other case leads to a contradiction, Statement D is the correct one.

💡Okay, so I have this problem about three unit vectors a, b, and c on a plane. The given conditions are that the absolute value of the dot product of a and b is 1/2, and the absolute value of the dot product of a and c is sqrt(3)/2. I need to find all possible values of the dot product of b and c.First, I remember that the dot product of two unit vectors is equal to the cosine of the angle between them. So, if |a · b| = 1/2, that means the angle between a and b is either 60 degrees or 120 degrees because cos(60°) = 1/2 and cos(120°) = -1/2. Similarly, |a · c| = sqrt(3)/2 implies that the angle between a and c is either 30 degrees or 150 degrees because cos(30°) = sqrt(3)/2 and cos(150°) = -sqrt(3)/2.Since all vectors are on a plane, I can represent them in a 2D coordinate system. Let me assume that vector a is along the x-axis for simplicity. So, vector a can be represented as (1, 0). Then, vectors b and c can be represented in terms of their angles with respect to the x-axis.Let’s denote the angle between a and b as θ, so θ can be 60° or 120°, and the angle between a and c as φ, so φ can be 30° or 150°. Therefore, vector b can be written as (cosθ, sinθ) and vector c as (cosφ, sinφ).Now, the dot product of b and c is given by:b · c = cosθ * cosφ + sinθ * sinφI remember that this expression is equal to cos(θ - φ). So, b · c = cos(θ - φ).So, I need to find all possible values of cos(θ - φ) where θ can be 60°, 120°, and φ can be 30°, 150°. Let me list all possible combinations:1. θ = 60°, φ = 30°: θ - φ = 30°, so cos(30°) = sqrt(3)/22. θ = 60°, φ = 150°: θ - φ = -90°, cos(-90°) = 03. θ = 120°, φ = 30°: θ - φ = 90°, cos(90°) = 04. θ = 120°, φ = 150°: θ - φ = -30°, cos(-30°) = sqrt(3)/2Wait, that's only four combinations, but I think I might have missed some. Let me check again.Actually, since θ can be 60° or 120°, and φ can be 30° or 150°, and both angles can be measured in either direction (clockwise or counterclockwise), but since we're dealing with absolute values, the angles can be positive or negative. So, θ can be ±60°, ±120°, and φ can be ±30°, ±150°. Hmm, that complicates things a bit.But wait, in the plane, the angle between two vectors is always between 0° and 180°, so maybe I don't need to consider negative angles. Let me think.If I fix vector a along the x-axis, vector b can be at 60° above the x-axis or 60° below, which would correspond to 60° and 300°, but since 300° is equivalent to -60°, which is the same as 60° in the opposite direction. Similarly, 120° and 240° are equivalent. So, perhaps I can consider angles in the range of 0° to 180°, and their supplements.But maybe it's simpler to consider all possible angles θ and φ, compute θ - φ, and then take the cosine of that difference.So, let's list all possible θ and φ:θ can be 60°, 120°, 240°, 300° (which are 60°, 120°, -120°, -60°)φ can be 30°, 150°, 210°, 330° (which are 30°, 150°, -150°, -30°)Wait, but in terms of angles between vectors, we usually take the smallest angle between them, so maybe θ and φ are between 0° and 180°, so θ can be 60° or 120°, and φ can be 30° or 150°, as initially thought.So, let's stick with θ = 60°, 120°, and φ = 30°, 150°, and compute θ - φ for all combinations:1. θ = 60°, φ = 30°: θ - φ = 30°, cos(30°) = sqrt(3)/22. θ = 60°, φ = 150°: θ - φ = -90°, cos(-90°) = 03. θ = 120°, φ = 30°: θ - φ = 90°, cos(90°) = 04. θ = 120°, φ = 150°: θ - φ = -30°, cos(-30°) = sqrt(3)/2But wait, what about θ = 60°, φ = 210°? Is that possible? Wait, φ is given as |a · c| = sqrt(3)/2, so φ can be 30° or 150°, but in terms of direction, it can be in either the first or second quadrant, so 30° or 150°, but not 210° or 330°, because those would give the same absolute value for the dot product, but the angle between a and c would be 150° or 30°, respectively.Wait, no, if vector c is at 210°, the angle between a and c would be 210°, but since we take the smallest angle, it would be 150°, because 210° is equivalent to -150°, which is the same as 150° in the opposite direction. Similarly, 330° is equivalent to -30°, which is the same as 30° in the opposite direction.So, in terms of the angle between a and c, it's either 30° or 150°, regardless of the direction. So, φ can be considered as 30° or 150°, but in terms of the actual angle in standard position, it can be 30°, 150°, 210°, or 330°, but the angle between a and c is the smallest one, so 30° or 150°.Similarly, for vector b, the angle between a and b is 60° or 120°, regardless of direction.So, going back, I think the possible differences θ - φ are 30°, -90°, 90°, -30°, but wait, let me check all combinations.Wait, if θ is 60°, φ can be 30°, 150°, 210°, or 330°, but the angle between a and c is 30° or 150°, so φ is either 30° or 150°, but in terms of the actual angle, it's either 30° or 150°, so θ - φ would be:- θ = 60°, φ = 30°: 60 - 30 = 30°- θ = 60°, φ = 150°: 60 - 150 = -90°- θ = 120°, φ = 30°: 120 - 30 = 90°- θ = 120°, φ = 150°: 120 - 150 = -30°So, the differences are 30°, -90°, 90°, -30°, and cos of these angles are sqrt(3)/2, 0, 0, sqrt(3)/2.But wait, cos(-90°) is 0, and cos(-30°) is sqrt(3)/2. So, the possible values of b · c are sqrt(3)/2, 0, and sqrt(3)/2 again, but actually, we have 0 and sqrt(3)/2.Wait, but in the first case, θ = 60°, φ = 30°, so θ - φ = 30°, cos(30°) = sqrt(3)/2.In the second case, θ = 60°, φ = 150°, θ - φ = -90°, cos(-90°) = 0.Third case, θ = 120°, φ = 30°, θ - φ = 90°, cos(90°) = 0.Fourth case, θ = 120°, φ = 150°, θ - φ = -30°, cos(-30°) = sqrt(3)/2.So, the possible values are sqrt(3)/2 and 0.Wait, but that's only two values. But the problem says "the set of all possible values," so maybe there are more possibilities.Wait, perhaps I missed some combinations. Let me think again.If vector b is at 60°, vector c can be at 30°, 150°, 210°, or 330°, but the angle between a and c is 30° or 150°, so the actual angle of c is either 30° or 150°, but in terms of direction, it can be in the first or second quadrant.Similarly, vector b can be at 60° or 120°, but also in the fourth or third quadrant, but since we're dealing with unit vectors on a plane, the angle is measured from the x-axis, so 60°, 120°, 240°, or 300°, but the angle between a and b is 60° or 120°, regardless of direction.Wait, maybe I need to consider all possible positions of b and c, not just in the first and second quadrants.So, if vector b is at 60°, it can also be at 60° + 180° = 240°, and vector c at 30° can also be at 30° + 180° = 210°, and vector c at 150° can be at 150° + 180° = 330°.So, let me consider all possible angles for b and c:Vector b can be at 60°, 120°, 240°, or 300°.Vector c can be at 30°, 150°, 210°, or 330°.Now, let's compute θ - φ for all combinations:1. θ = 60°, φ = 30°: 60 - 30 = 30°, cos(30°) = sqrt(3)/22. θ = 60°, φ = 150°: 60 - 150 = -90°, cos(-90°) = 03. θ = 60°, φ = 210°: 60 - 210 = -150°, cos(-150°) = -sqrt(3)/24. θ = 60°, φ = 330°: 60 - 330 = -270°, cos(-270°) = 05. θ = 120°, φ = 30°: 120 - 30 = 90°, cos(90°) = 06. θ = 120°, φ = 150°: 120 - 150 = -30°, cos(-30°) = sqrt(3)/27. θ = 120°, φ = 210°: 120 - 210 = -90°, cos(-90°) = 08. θ = 120°, φ = 330°: 120 - 330 = -210°, cos(-210°) = -sqrt(3)/29. θ = 240°, φ = 30°: 240 - 30 = 210°, cos(210°) = -sqrt(3)/210. θ = 240°, φ = 150°: 240 - 150 = 90°, cos(90°) = 011. θ = 240°, φ = 210°: 240 - 210 = 30°, cos(30°) = sqrt(3)/212. θ = 240°, φ = 330°: 240 - 330 = -90°, cos(-90°) = 013. θ = 300°, φ = 30°: 300 - 30 = 270°, cos(270°) = 014. θ = 300°, φ = 150°: 300 - 150 = 150°, cos(150°) = -sqrt(3)/215. θ = 300°, φ = 210°: 300 - 210 = 90°, cos(90°) = 016. θ = 300°, φ = 330°: 300 - 330 = -30°, cos(-30°) = sqrt(3)/2Now, let's collect all the results:From the above 16 combinations, the values of cos(θ - φ) are:sqrt(3)/2, 0, -sqrt(3)/2, 0, 0, sqrt(3)/2, 0, -sqrt(3)/2, -sqrt(3)/2, 0, sqrt(3)/2, 0, 0, -sqrt(3)/2, 0, sqrt(3)/2.So, the distinct values are sqrt(3)/2, 0, and -sqrt(3)/2.Therefore, the set of all possible values of b · c is {-sqrt(3)/2, 0, sqrt(3)/2}.Wait, but in the initial four combinations, I only got sqrt(3)/2 and 0, but when considering all possible positions of b and c, I also got -sqrt(3)/2. So, the complete set includes negative sqrt(3)/2 as well.So, the possible values are -sqrt(3)/2, 0, and sqrt(3)/2.

💡Alright, so I have this geometry problem here. Let me read it again: In triangle PQR, PQ is 13 and PR is 20. S is the foot of the perpendicular from P to QR. The ratio of QS to SR is 3:4. I need to find PS. Hmm, okay.First, let me visualize the triangle. Triangle PQR with points P, Q, and R. PQ is 13 units, PR is 20 units. S is the foot of the perpendicular from P to QR, so PS is perpendicular to QR. That means triangle PQS and triangle PRS are both right-angled at S.Given the ratio QS:SR is 3:4. So, if I let QS be 3k, then SR would be 4k for some positive real number k. Therefore, the entire length QR would be QS + SR, which is 3k + 4k = 7k. So, QR is 7k.Now, since PS is the altitude from P to QR, I can use the properties of right triangles here. In triangle PQS, which is right-angled at S, the sides are PQ, QS, and PS. Similarly, in triangle PRS, the sides are PR, SR, and PS.I remember that in a right-angled triangle, the Pythagorean theorem applies: (hypotenuse)^2 = (base)^2 + (height)^2. So, for triangle PQS, we have:PQ² = QS² + PS²Similarly, for triangle PRS:PR² = SR² + PS²Given that PQ is 13 and PR is 20, I can write these equations as:13² = (3k)² + PS²20² = (4k)² + PS²Let me compute these squares:13² is 169, and 20² is 400. So,169 = 9k² + PS² ...(1)400 = 16k² + PS² ...(2)Now, I have two equations with two variables: k² and PS². I can solve for these by subtracting equation (1) from equation (2):400 - 169 = 16k² - 9k² + PS² - PS²Simplifying:231 = 7k²So, 7k² = 231. Then, k² = 231 / 7 = 33. Therefore, k = sqrt(33). But wait, do I need k? Maybe not directly, but let's see.Alternatively, I can express PS² from equation (1):PS² = 169 - 9k²And from equation (2):PS² = 400 - 16k²Since both equal PS², I can set them equal to each other:169 - 9k² = 400 - 16k²Let me solve for k²:169 - 9k² = 400 - 16k²Bring all terms to one side:169 - 9k² - 400 + 16k² = 0Simplify:(169 - 400) + ( -9k² + 16k² ) = 0-231 + 7k² = 0So, 7k² = 231Again, k² = 33, so k = sqrt(33). Hmm, okay.But I need PS, so let's plug k² back into one of the equations for PS². Let's take equation (1):PS² = 169 - 9k² = 169 - 9*33Compute 9*33: 9*30=270, 9*3=27, so 270+27=297So, PS² = 169 - 297 = -128Wait, that can't be right. PS squared can't be negative. Did I make a mistake?Let me check my steps. So, equations:169 = 9k² + PS² ...(1)400 = 16k² + PS² ...(2)Subtracting (1) from (2):400 - 169 = 16k² - 9k² + PS² - PS²231 = 7k²So, k² = 33. Then, plugging back into equation (1):PS² = 169 - 9*33Compute 9*33: 297169 - 297 = -128Hmm, negative. That doesn't make sense. Maybe I messed up the equations.Wait, let's think again. Maybe I set up the equations incorrectly.Wait, in triangle PQS, PQ is 13, QS is 3k, and PS is the height. So, PQ² = QS² + PS², which is 13² = (3k)² + PS². That's correct.Similarly, PR² = SR² + PS², which is 20² = (4k)² + PS². That's also correct.So, equations are correct. Then, why is PS² negative? That must mean I made a mistake in the subtraction.Wait, let me re-examine the subtraction step.Equation (2) minus equation (1):400 - 169 = (16k² + PS²) - (9k² + PS²)So, 231 = 7k²Yes, that's correct. So, 7k² = 231, so k² = 33, so k = sqrt(33). Then, plugging into equation (1):PS² = 169 - 9k² = 169 - 9*33 = 169 - 297 = -128Wait, that's negative. That can't be. So, maybe my initial assumption is wrong.Wait, could it be that the ratio is QS:SR = 3:4, but I assumed QS = 3k and SR = 4k. Maybe I should have considered that QS:SR = 3:4, so QS = (3/7)QR and SR = (4/7)QR. But QR is the length of the base, which we don't know yet.Alternatively, perhaps I should use coordinate geometry to solve this problem.Let me try setting up coordinates. Let me place point P at (0,0). Then, since S is the foot of the perpendicular from P to QR, let me place S at (0,0) as well? Wait, no, because then P would coincide with S, which is not the case.Wait, maybe I should place point S at (0,0), and QR along the x-axis. Then, point Q would be at (-a, 0), and point R at (b, 0), where a and b are positive numbers. Then, point P would be at (0, h), where h is the height PS.Given that QS:SR = 3:4, so the distance from Q to S is 3k, and from S to R is 4k. So, the coordinates would be:Point Q: (-3k, 0)Point S: (0, 0)Point R: (4k, 0)Point P: (0, h)Now, PQ is the distance from P(0,h) to Q(-3k,0). So, PQ² = (0 - (-3k))² + (h - 0)² = (3k)² + h² = 9k² + h². Given PQ = 13, so:9k² + h² = 169 ...(1)Similarly, PR is the distance from P(0,h) to R(4k,0). So, PR² = (0 - 4k)² + (h - 0)² = (4k)² + h² = 16k² + h². Given PR = 20, so:16k² + h² = 400 ...(2)Now, subtract equation (1) from equation (2):(16k² + h²) - (9k² + h²) = 400 - 1697k² = 231So, k² = 33, so k = sqrt(33)Now, plug back k² into equation (1):9*33 + h² = 169297 + h² = 169h² = 169 - 297h² = -128Wait, again, h² is negative. That can't be. What's going on here?Wait, maybe I messed up the coordinate system. If I place S at (0,0), then Q is at (-3k,0) and R at (4k,0). So, the distance from Q to S is 3k, and from S to R is 4k, which gives the ratio 3:4. That seems correct.Then, PQ is the distance from P(0,h) to Q(-3k,0), which is sqrt((3k)^2 + h^2) = sqrt(9k² + h²) = 13. So, 9k² + h² = 169.Similarly, PR is sqrt((4k)^2 + h²) = sqrt(16k² + h²) = 20. So, 16k² + h² = 400.Subtracting the first equation from the second gives 7k² = 231, so k² = 33, which is positive.Then, plugging back into the first equation:9*33 + h² = 169297 + h² = 169h² = 169 - 297 = -128Negative again. That's impossible because h² can't be negative. So, where is the mistake?Wait, maybe the ratio QS:SR is 3:4, but in my coordinate system, QS is 3k and SR is 4k, so QR is 7k. But in reality, QR is the base, and PS is the height. Maybe I need to consider the lengths differently.Alternatively, perhaps I should use the formula for the length of the altitude in terms of the sides and the ratio of the segments.I recall that in a triangle, if a perpendicular is drawn from a vertex to the opposite side, dividing it into segments with a certain ratio, there is a relation involving the squares of the sides.Specifically, the formula is:(PQ² - PR²) = (QS² - SR²)Wait, let me check that.In triangle PQR, with PS perpendicular to QR, then:PQ² - PR² = QS² - SR²Yes, that's a formula I remember. Let me verify it.In triangle PQS, PQ² = QS² + PS²In triangle PRS, PR² = SR² + PS²Subtracting these two equations:PQ² - PR² = QS² - SR²Yes, that's correct.So, given PQ = 13, PR = 20, QS:SR = 3:4.Let me denote QS = 3k, SR = 4k.Then, PQ² - PR² = (3k)² - (4k)²So, 13² - 20² = 9k² - 16k²Compute 13² = 169, 20² = 400So, 169 - 400 = -9k²-231 = -9k²Divide both sides by -9:231 / 9 = k²231 divided by 9: 9*25=225, so 231-225=6, so 25 + 6/9 = 25 + 2/3 = 25.666...Wait, 231 / 9 is 25.666..., which is 77/3.Wait, 231 divided by 9: 9*25=225, so 231-225=6, so 6/9=2/3. So, 25 and 2/3, which is 77/3.So, k² = 77/3Therefore, k = sqrt(77/3)But wait, earlier I had k² = 33, which led to a negative h². So, which one is correct?Wait, in the first approach, I set up the equations as:169 = 9k² + h²400 = 16k² + h²Subtracting, 231 = 7k² => k² = 33But in this formula, PQ² - PR² = QS² - SR², which gives:169 - 400 = 9k² - 16k²-231 = -7k²So, 231 = 7k² => k² = 33Wait, that's the same result as before. So, why is h² negative?Wait, maybe I made a mistake in the formula.Wait, let me rederive the formula.In triangle PQS: PQ² = QS² + PS²In triangle PRS: PR² = SR² + PS²Subtracting: PQ² - PR² = QS² - SR²Yes, that's correct.So, PQ² - PR² = QS² - SR²13² - 20² = (3k)² - (4k)²169 - 400 = 9k² - 16k²-231 = -7k²So, 231 = 7k² => k² = 33So, k² = 33, so k = sqrt(33)Then, plugging back into equation for PS²:From triangle PQS: PS² = PQ² - QS² = 169 - 9k² = 169 - 9*33 = 169 - 297 = -128Negative again. That's impossible.Wait, this suggests that such a triangle cannot exist because PS² is negative. But the problem says it exists. So, maybe I made a wrong assumption.Wait, perhaps the ratio QS:SR is 3:4, but in reality, QS is 3 parts and SR is 4 parts, so the entire QR is 7 parts. But in the coordinate system, I placed Q at (-3k,0) and R at (4k,0). Maybe that's causing the issue.Alternatively, perhaps I should place Q and R on the same side of S, but that would make QR shorter. Wait, no, because S is the foot of the perpendicular, so Q and R must be on opposite sides of S.Wait, maybe I should place S at (0,0), Q at (-3k,0), and R at (4k,0). Then, the distance from Q to S is 3k, and from S to R is 4k, so QR is 7k.Then, point P is at (0,h). So, PQ is the distance from (-3k,0) to (0,h): sqrt((3k)^2 + h^2) = 13Similarly, PR is the distance from (4k,0) to (0,h): sqrt((4k)^2 + h^2) = 20So, equations:9k² + h² = 169 ...(1)16k² + h² = 400 ...(2)Subtract (1) from (2):7k² = 231 => k² = 33Then, h² = 169 - 9*33 = 169 - 297 = -128Again, negative. So, this suggests that with the given side lengths and ratio, such a triangle cannot exist because it leads to an impossible negative height.But the problem states that such a triangle exists, so I must have made a mistake in my approach.Wait, maybe the ratio QS:SR is 3:4, but I assumed QS = 3k and SR = 4k. Maybe it's the other way around? Maybe QS = 4k and SR = 3k?Let me try that.So, if QS:SR = 3:4, then QS = 3k, SR = 4k. But maybe I should consider QS = 4k and SR = 3k? Wait, no, the ratio is QS:SR = 3:4, so QS is 3 parts and SR is 4 parts.Wait, unless the ratio is SR:QS = 3:4, but the problem says QS:SR = 3:4. So, QS is 3 parts, SR is 4 parts.Wait, perhaps I should use Stewart's theorem instead. Stewart's theorem relates the lengths of the sides of a triangle to the length of a cevian.Stewart's theorem states that for triangle PQR with cevian PS, where S divides QR into QS and SR with ratio m:n, then:PQ² * SR + PR² * QS = PS² * QR + QS * SR * QRWait, let me write it properly.Stewart's formula is:PQ² * SR + PR² * QS = PS² * QR + QS * SR * QRGiven that QS:SR = 3:4, let me denote QS = 3k, SR = 4k, so QR = 7k.Then, plugging into Stewart's theorem:PQ² * SR + PR² * QS = PS² * QR + QS * SR * QRSo,13² * 4k + 20² * 3k = PS² * 7k + (3k)(4k)(7k)Compute each term:13² = 169, 20² = 400So,169 * 4k + 400 * 3k = PS² * 7k + 12k² * 7kSimplify:(676k) + (1200k) = 7k PS² + 84k³So,1876k = 7k PS² + 84k³Divide both sides by k (assuming k ≠ 0, which it isn't):1876 = 7 PS² + 84k²Now, from earlier, we have from the Pythagorean theorem:From triangle PQS: 169 = 9k² + PS² => PS² = 169 - 9k²Similarly, from triangle PRS: 400 = 16k² + PS² => PS² = 400 - 16k²So, both equal to PS², so 169 - 9k² = 400 - 16k²Which gives:7k² = 231 => k² = 33So, k² = 33, so k = sqrt(33)Now, plug k² = 33 into the equation from Stewart's theorem:1876 = 7 PS² + 84k²Compute 84k² = 84*33 = 2772So,1876 = 7 PS² + 2772Subtract 2772 from both sides:1876 - 2772 = 7 PS²-896 = 7 PS²So, PS² = -896 / 7 = -128Again, negative. So, PS² is negative, which is impossible.Wait, this is the same result as before. So, regardless of the method, I end up with PS² being negative, which suggests that such a triangle cannot exist with the given side lengths and ratio.But the problem states that it does exist, so I must have made a wrong assumption somewhere.Wait, maybe the ratio QS:SR is 3:4, but in reality, QS is longer than SR. Wait, no, the ratio is 3:4, so QS is 3 parts, SR is 4 parts, so SR is longer.Wait, maybe I should consider that the foot S is not between Q and R, but beyond R or Q. But in that case, the ratio would be different.Wait, in the problem, S is the foot of the perpendicular from P to QR, so S must lie on QR. So, it's between Q and R.Wait, unless QR is not a straight line, but that's not possible in a triangle.Wait, maybe I should consider that the triangle is obtuse, so the foot S lies outside the segment QR. But the problem says S is the foot, so it must lie on QR.Wait, but if S is the foot, it can lie inside or outside QR depending on the type of triangle. If the triangle is acute, S is inside; if it's obtuse, S is outside.But in this case, since we have PS perpendicular to QR, and S is the foot, it can be inside or outside.Wait, but if S is outside QR, then QS and SR would have different signs. But the ratio is given as 3:4, which are positive.Wait, maybe I should consider that S is beyond R, so QR is extended beyond R to S, making SR negative. But the ratio QS:SR is 3:4, so if S is beyond R, then QS would be QR + RS, which would complicate things.Alternatively, maybe I should use coordinate geometry again, but this time place S somewhere else.Wait, let me try a different approach. Let me denote QR as a line segment, and S divides QR into QS:SR = 3:4. So, QR = QS + SR = 3k + 4k = 7k.Now, in triangle PQR, with PS perpendicular to QR, we can use the formula for the length of the altitude:PS = (2 * area of triangle PQR) / QRBut to find the area, we can use Heron's formula, but we don't know QR yet.Alternatively, since we have two sides and the ratio of the segments, maybe we can express QR in terms of k and then find the area.Wait, but without knowing QR, it's difficult.Alternatively, maybe use the formula for the area in terms of the sides and the altitude.Wait, area = (1/2)*QR*PSBut also, area can be found using Heron's formula if we know all three sides, but we don't know QR.Wait, but we can express QR in terms of k, which is 7k, and then express the area in two ways.Wait, let me try that.Let QR = 7k.Then, area = (1/2)*7k*PSAlso, using Heron's formula, we need all three sides: PQ=13, PR=20, QR=7k.Compute semi-perimeter, s = (13 + 20 + 7k)/2 = (33 + 7k)/2Then, area = sqrt[s(s - PQ)(s - PR)(s - QR)]So,Area = sqrt[ ( (33 + 7k)/2 ) ( (33 + 7k)/2 - 13 ) ( (33 + 7k)/2 - 20 ) ( (33 + 7k)/2 - 7k ) ]Simplify each term:s = (33 + 7k)/2s - PQ = (33 + 7k)/2 - 13 = (33 + 7k - 26)/2 = (7 + 7k)/2 = 7(k + 1)/2s - PR = (33 + 7k)/2 - 20 = (33 + 7k - 40)/2 = (7k - 7)/2 = 7(k - 1)/2s - QR = (33 + 7k)/2 - 7k = (33 + 7k - 14k)/2 = (33 - 7k)/2So, area = sqrt[ ( (33 + 7k)/2 ) * (7(k + 1)/2 ) * (7(k - 1)/2 ) * ( (33 - 7k)/2 ) ]Simplify:= sqrt[ ( (33 + 7k)(33 - 7k) * 7(k + 1) * 7(k - 1) ) / (2^4) ]= sqrt[ ( (33² - (7k)²) * 49(k² - 1) ) / 16 ]= sqrt[ ( (1089 - 49k²) * 49(k² - 1) ) / 16 ]= (7/4) * sqrt[ (1089 - 49k²)(k² - 1) ]Now, equate this to the other expression for area:(1/2)*7k*PS = (7/4) * sqrt[ (1089 - 49k²)(k² - 1) ]Simplify:(7k/2)*PS = (7/4) * sqrt[ (1089 - 49k²)(k² - 1) ]Divide both sides by 7:(k/2)*PS = (1/4) * sqrt[ (1089 - 49k²)(k² - 1) ]Multiply both sides by 4:2k*PS = sqrt[ (1089 - 49k²)(k² - 1) ]Now, square both sides:(2k*PS)^2 = (1089 - 49k²)(k² - 1)4k²*PS² = (1089 - 49k²)(k² - 1)Now, from earlier, we have from the Pythagorean theorem:PS² = 169 - 9k²So, plug that into the equation:4k²*(169 - 9k²) = (1089 - 49k²)(k² - 1)Expand both sides:Left side: 4k²*169 - 4k²*9k² = 676k² - 36k⁴Right side: 1089*(k² - 1) - 49k²*(k² - 1) = 1089k² - 1089 - 49k⁴ + 49k² = (1089k² + 49k²) - 49k⁴ - 1089 = 1138k² - 49k⁴ - 1089So, equation becomes:676k² - 36k⁴ = 1138k² - 49k⁴ - 1089Bring all terms to left side:676k² - 36k⁴ - 1138k² + 49k⁴ + 1089 = 0Combine like terms:(676k² - 1138k²) + (-36k⁴ + 49k⁴) + 1089 = 0(-462k²) + (13k⁴) + 1089 = 0Rearrange:13k⁴ - 462k² + 1089 = 0Let me write this as:13k⁴ - 462k² + 1089 = 0Let me let u = k², so equation becomes:13u² - 462u + 1089 = 0Now, solve for u using quadratic formula:u = [462 ± sqrt(462² - 4*13*1089)] / (2*13)Compute discriminant:D = 462² - 4*13*1089Compute 462²:462 * 462: Let's compute 400² = 160000, 60²=3600, 2²=4, and cross terms.Wait, 462² = (400 + 62)² = 400² + 2*400*62 + 62² = 160000 + 49600 + 3844 = 160000 + 49600 = 209600 + 3844 = 213,444Now, 4*13*1089 = 52*1089Compute 52*1000=52,000, 52*89=4,628, so total 52,000 + 4,628 = 56,628So, D = 213,444 - 56,628 = 156,816Now, sqrt(156,816). Let's see:400² = 160,000, which is larger. 396² = (400 - 4)² = 160,000 - 2*400*4 + 16 = 160,000 - 3,200 + 16 = 156,816So, sqrt(156,816) = 396So, u = [462 ± 396] / 26Compute both possibilities:First, u = (462 + 396)/26 = 858/26 = 33Second, u = (462 - 396)/26 = 66/26 = 33/13 ≈ 2.538So, u = 33 or u = 33/13But u = k², so k² = 33 or k² = 33/13Now, let's check k² = 33:From earlier, PS² = 169 - 9k² = 169 - 297 = -128, which is negative. So, invalid.Now, k² = 33/13Then, PS² = 169 - 9*(33/13) = 169 - 297/13Compute 297/13: 13*22=286, so 297-286=11, so 22 + 11/13 = 22.846So, 169 - 22.846 ≈ 146.154But let's compute exactly:169 = 169/1 = (169*13)/13 = 2197/13So, PS² = 2197/13 - 297/13 = (2197 - 297)/13 = 1900/13So, PS² = 1900/13Therefore, PS = sqrt(1900/13) = sqrt(1900)/sqrt(13) = (sqrt(100*19))/sqrt(13) = (10*sqrt(19))/sqrt(13)Rationalize the denominator:= 10*sqrt(19)*sqrt(13)/13 = 10*sqrt(247)/13But 247 factors: 247 = 13*19, so sqrt(247) = sqrt(13*19) = sqrt(13)*sqrt(19)So, PS = 10*sqrt(13)*sqrt(19)/13 = 10*sqrt(19)/sqrt(13)Wait, that seems complicated. Maybe I made a mistake in the calculation.Wait, let me recompute PS² when k² = 33/13.From triangle PQS: PS² = 169 - 9k² = 169 - 9*(33/13) = 169 - (297/13)Convert 169 to 13ths: 169 = 13*13 = 169/1 = 2197/13So, PS² = 2197/13 - 297/13 = (2197 - 297)/13 = 1900/13So, PS = sqrt(1900/13) = sqrt(1900)/sqrt(13) = (10*sqrt(19))/sqrt(13)Rationalize:= (10*sqrt(19)*sqrt(13))/13 = 10*sqrt(247)/13But 247 = 13*19, so sqrt(247) = sqrt(13*19) = sqrt(13)*sqrt(19)So, PS = 10*sqrt(13)*sqrt(19)/13 = (10/13)*sqrt(247)But 247 is 13*19, so sqrt(247) = sqrt(13*19). So, it's already simplified.Alternatively, sqrt(1900/13) can be written as sqrt(1900)/sqrt(13) = (10*sqrt(19))/sqrt(13) = (10*sqrt(247))/13But perhaps the answer is expected in a simpler form. Let me check if 1900/13 can be simplified.1900 divided by 13: 13*146=1898, so 1900-1898=2, so 1900/13=146 + 2/13=146.1538...But that's not helpful.Wait, maybe I made a mistake in the earlier steps. Let me check.We had the equation:13k⁴ - 462k² + 1089 = 0Let u = k², so 13u² - 462u + 1089 = 0Solutions:u = [462 ± sqrt(462² - 4*13*1089)] / (2*13)We computed discriminant as 156,816, which is 396², correct.So, u = (462 ± 396)/26So, u1 = (462 + 396)/26 = 858/26 = 33u2 = (462 - 396)/26 = 66/26 = 33/13So, k² = 33 or 33/13But when k²=33, PS² is negative, so invalid.When k²=33/13, PS²=1900/13, which is positive.So, PS = sqrt(1900/13) = sqrt(1900)/sqrt(13) = (10*sqrt(19))/sqrt(13) = (10*sqrt(247))/13But 247=13*19, so sqrt(247)=sqrt(13*19). So, it's as simplified as it gets.But the problem asks for PS, so maybe we can rationalize or write it differently.Alternatively, maybe I made a mistake in the earlier step when setting up Stewart's theorem.Wait, let me check Stewart's theorem again.Stewart's theorem states that for triangle PQR with cevian PS, where S divides QR into QS and SR with ratio m:n, then:PQ² * SR + PR² * QS = PS² * QR + QS * SR * QRIn our case, QS:SR = 3:4, so QS=3k, SR=4k, QR=7kSo,13² * 4k + 20² * 3k = PS² * 7k + (3k)(4k)(7k)Which is:169*4k + 400*3k = 7k PS² + 84k³So,676k + 1200k = 7k PS² + 84k³1876k = 7k PS² + 84k³Divide both sides by k:1876 = 7 PS² + 84k²From earlier, we have PS² = 169 - 9k²So, plug into equation:1876 = 7*(169 - 9k²) + 84k²Compute 7*169 = 1183, 7*(-9k²)= -63k²So,1876 = 1183 - 63k² + 84k²Simplify:1876 = 1183 + 21k²Subtract 1183:1876 - 1183 = 21k²693 = 21k²So, k² = 693 / 21 = 33So, k²=33, which leads to PS²=169 - 9*33= -128, which is negative.Wait, but earlier with Heron's formula, we got k²=33/13. So, which one is correct?Wait, in Stewart's theorem, I think I made a mistake in the formula.Wait, let me double-check Stewart's theorem.Stewart's theorem is: b²m + c²n = a(d² + mn)Where, for triangle ABC with cevian AD, where D divides BC into BD=m and DC=n, and AD=d, BC=a.So, in our case, triangle PQR, cevian PS, with S dividing QR into QS=3k, SR=4k, so QR=7k.So, in Stewart's terms:PQ² * SR + PR² * QS = QR*(PS² + QS*SR)So,13² * 4k + 20² * 3k = 7k*(PS² + 3k*4k)So,169*4k + 400*3k = 7k*(PS² + 12k²)Compute:676k + 1200k = 7k*(PS² + 12k²)1876k = 7k*PS² + 84k³Divide both sides by k:1876 = 7 PS² + 84k²Which is the same as before.From Pythagorean theorem, PS² = 169 - 9k²So,1876 = 7*(169 - 9k²) + 84k²Compute:7*169=1183, 7*(-9k²)= -63k²So,1876 = 1183 - 63k² + 84k²Simplify:1876 = 1183 + 21k²Subtract 1183:693 = 21k²So, k²=693/21=33Thus, k²=33, leading to PS²=169 - 9*33= -128, which is impossible.This suggests that there is no such triangle with the given conditions, which contradicts the problem statement.But the problem says "In triangle PQR, PQ=13, PR=20. Let S be the foot of the perpendicular from P to QR. If QS:SR=3:4, then find PS."So, the problem assumes that such a triangle exists. Therefore, my approach must be wrong.Wait, maybe I should consider that the foot S is not between Q and R, but beyond R, making SR negative. So, QS:SR=3:4, but SR is negative.Let me try that.So, let me denote QS=3k, SR=-4k, so QR=QS + SR=3k -4k= -k. But QR is a length, so it can't be negative. So, perhaps QR=|QS + SR|=|3k -4k|=k.Wait, but then the ratio QS:SR=3:4, but SR is negative, so the ratio would be 3:(-4), which is not the same as 3:4.Alternatively, maybe I should consider the directed segments.In that case, QS=3k, SR=4k, but S is beyond R, so QR=QS - SR=3k -4k= -k, but again, QR is a length, so it's | -k | =k.But then, the ratio QS:SR=3:4, but in directed segments, it's 3k:4k, but since S is beyond R, SR is negative, so ratio is 3k:(-4k)= -3:4, which is not the same as 3:4.So, perhaps this approach is not correct.Alternatively, maybe the triangle is such that PS is longer than both PQ and PR, but that's not possible because PS is the altitude, which is shorter than the sides.Wait, but in reality, in a right-angled triangle, the altitude can be shorter or longer depending on the sides.Wait, but in this case, PS is the altitude from P to QR, so it must be shorter than both PQ and PR if the triangle is acute, but if the triangle is obtuse, PS can be longer.Wait, but in our case, PQ=13, PR=20, so PR is longer. If the triangle is obtuse at P, then the foot S would lie outside QR, making QS:SR=3:4, but with S beyond R.Wait, let me try that.Assume that S is beyond R, so QR is extended beyond R to S, making SR negative.So, QS=QR + RS, but since S is beyond R, RS is negative.Given QS:SR=3:4, but SR is negative, so QS=3k, SR=-4k.Thus, QR=QS - SR=3k - (-4k)=7kWait, that doesn't make sense because QR is a fixed length.Wait, perhaps QR=QS - SR=3k -4k= -k, but QR is positive, so QR=k.But then, QS=3k, SR= -4k, so QR=QS + SR=3k -4k= -k, which is negative, but QR is positive, so QR=k.So, QR=k, QS=3k, SR= -4k, so QR=QS + SR=3k -4k= -k, which contradicts QR=k.This is getting too confusing.Alternatively, maybe I should consider that the ratio QS:SR=3:4, but S is beyond R, so QS=QR + RS, with RS negative.Let me denote QR=x, then QS=x + RS, but RS is negative.Given QS:SR=3:4, so QS=3k, SR=4k, but SR is negative, so SR=-4k.Thus, QS=3k, SR=-4k, so QR=QS - SR=3k - (-4k)=7kWait, that makes QR=7k, which is positive.So, QR=7k, QS=3k, SR=-4k.Thus, in this case, S is beyond R, so QR=7k, QS=3k, SR=-4k.Now, let's apply Stewart's theorem again.Stewart's theorem: PQ² * SR + PR² * QS = QR*(PS² + QS*SR)So,13²*(-4k) + 20²*(3k) = 7k*(PS² + (3k)*(-4k))Compute:169*(-4k) + 400*(3k) = 7k*(PS² -12k²)Simplify:-676k + 1200k = 7k*(PS² -12k²)524k = 7k*(PS² -12k²)Divide both sides by k:524 = 7*(PS² -12k²)So,524 = 7 PS² -84k²From triangle PQS, which is right-angled at S, we have:PQ² = QS² + PS²13² = (3k)² + PS²169 = 9k² + PS²So, PS² = 169 -9k²Similarly, in triangle PRS, which is right-angled at S, but since S is beyond R, PR² = SR² + PS², but SR is negative, so:20² = (-4k)² + PS²400 = 16k² + PS²So, PS² = 400 -16k²Thus, from both equations:169 -9k² = 400 -16k²So,7k² = 231k²=33Thus, k=√33Now, plug k²=33 into PS²=169 -9k²=169 -297= -128Again, negative. So, impossible.Wait, but in this case, S is beyond R, so maybe the formula for the area is different.Alternatively, maybe the area is negative, but that doesn't make sense.Wait, perhaps the problem is that when S is beyond R, the formula for the area using Heron's formula would involve negative lengths, which complicates things.Alternatively, maybe the problem is designed such that the foot S is inside QR, but the given side lengths and ratio make it impossible, leading to a negative PS², which suggests that such a triangle cannot exist.But the problem states that it does exist, so I must have made a wrong assumption.Wait, maybe the ratio QS:SR is 3:4, but in reality, QS is 4 and SR is 3, making QR=7, but that contradicts the ratio.Wait, no, the ratio is given as QS:SR=3:4, so QS=3k, SR=4k.Wait, perhaps the sides PQ and PR are not the ones adjacent to the right angle, but that's not the case because PS is the altitude from P to QR.Wait, maybe I should use trigonometry. Let me denote angle at P as θ.Then, in triangle PQS, cosθ = QS/PQ = 3k/13In triangle PRS, cosθ = SR/PR = 4k/20= k/5So, 3k/13 = k/5Cross-multiply: 15k =13k => 2k=0 => k=0Which is impossible because k>0.Thus, this suggests that such a triangle cannot exist because the angles would not match.Therefore, the problem as stated has no solution because it leads to a contradiction.But the problem says "In triangle PQR...", implying that such a triangle exists. So, perhaps I made a mistake in the ratio.Wait, maybe the ratio is SR:QS=3:4 instead of QS:SR=3:4. Let me try that.If SR:QS=3:4, then SR=3k, QS=4k, QR=7k.Then, from triangle PQS: 13²= (4k)² + PS² => 169=16k² + PS²From triangle PRS:20²=(3k)² + PS² =>400=9k² + PS²Subtract:400 -169=9k² -16k²231= -7k²Which is impossible because k² cannot be negative.So, that doesn't work either.Thus, it seems that regardless of the approach, the problem leads to an impossible solution, suggesting that such a triangle cannot exist with the given conditions.But since the problem is given, perhaps I made a mistake in the initial setup.Wait, maybe the ratio QS:SR=3:4 is given, but in reality, QS is 3 units and SR is 4 units, making QR=7 units. Then, using the Pythagorean theorem:From PQ=13, QS=3, so PS²=13² -3²=169-9=160From PR=20, SR=4, so PS²=20² -4²=400-16=384But 160≠384, so PS cannot be both sqrt(160) and sqrt(384). Contradiction.Thus, the problem is impossible as stated.But the problem says "In triangle PQR...", so perhaps I misread the problem.Wait, let me read the problem again:"In triangle PQR, PQ = 13 and PR = 20. Let S be the foot of the perpendicular from P to QR. If QS:SR = 3:4, then find PS."So, perhaps the problem is correct, and I need to find PS=8√2, as in the initial solution, but that led to a negative PS².Wait, but in the initial solution, the user wrote:"Let h = PS. By applying Pythagoras' theorem on right triangle PQS and PRS, we have:QS² = 13² - h² = 169 - h²,SR² = 20² - h² = 400 - h².Given the ratio QS:SR = 3:4, so QS²:SR² = 9:16. Therefore, we set up the equation:(169 - h²)/(400 - h²) = 9/16.To solve for h, cross-multiply and simplify:16(169 - h²) = 9(400 - h²),2704 - 16h² = 3600 - 9h²,7h² = 896,h² = 128,h = sqrt(128) = 8√2.Thus, PS = 8√2."But in this solution, they assume that QS²:SR²=9:16, which is correct because QS:SR=3:4.But when I tried this approach, I ended up with h²=128, which is positive, so PS=8√2.Wait, but earlier when I tried this, I got negative h². Wait, no, in the initial approach, I set up the equations correctly and got h²=128, which is positive.Wait, but in my later approaches, I tried to use Stewart's theorem and Heron's formula, which led to contradictions. So, perhaps the initial approach is correct, and the later approaches have errors.Let me re-examine the initial approach.Let h=PS.Then, QS²=13² - h²=169 - h²SR²=20² - h²=400 - h²Given QS:SR=3:4, so QS²:SR²=9:16Thus,(169 - h²)/(400 - h²)=9/16Cross-multiply:16(169 - h²)=9(400 - h²)2704 -16h²=3600 -9h²Bring all terms to left:2704 -16h² -3600 +9h²=0-896 -7h²=0-7h²=896h²= -128Wait, that's negative again. So, the initial solution is wrong.Wait, but the user's initial solution says:16(169 - h²)=9(400 - h²)2704 -16h²=3600 -9h²Then,2704 -3600=16h² -9h²-896=7h²h²= -128So, h² negative.But the user wrote:7h²=896, h²=128Which is incorrect because -896=7h² leads to h²= -128.So, the initial solution is wrong.Thus, the correct conclusion is that such a triangle cannot exist because it leads to a negative PS².But the problem states that it does exist, so perhaps there is a mistake in the problem statement.Alternatively, maybe the ratio is SR:QS=3:4 instead of QS:SR=3:4.Let me try that.If SR:QS=3:4, then SR=3k, QS=4k, QR=7k.Then,From triangle PQS: QS²=16k²=13² -h²=169 -h²From triangle PRS: SR²=9k²=20² -h²=400 -h²So,16k²=169 -h² ...(1)9k²=400 -h² ...(2)Subtract (2) from (1):7k²= -231Which is impossible.Thus, the problem is impossible as stated.Therefore, the answer is that such a triangle cannot exist.But since the problem asks to find PS, perhaps the answer is 8√2, as in the initial incorrect solution, but that leads to a contradiction.Alternatively, the problem may have a typo, and the ratio should be 4:3 instead of 3:4.If QS:SR=4:3, then QS=4k, SR=3k, QR=7k.Then,From triangle PQS: QS²=16k²=13² -h²=169 -h²From triangle PRS: SR²=9k²=20² -h²=400 -h²So,16k²=169 -h² ...(1)9k²=400 -h² ...(2)Subtract (1)-(2):7k²= -231Again, impossible.Thus, regardless of the ratio, it's impossible.Therefore, the conclusion is that such a triangle cannot exist with the given conditions.