💡Okay, so I have this probability problem here, and I need to figure out the chance that I draw a white ball from a bag after adding one white ball to it. Let me try to understand the problem step by step.First, there's a bag with 5 balls. I don't know their colors initially—could be all white, all black, or a mix. Then, someone adds one white ball to the bag, making it a total of 6 balls. After that, one ball is drawn at random. I need to find the probability that this drawn ball is white, considering all possible initial color combinations of the original 5 balls, and assuming each combination is equally likely.Hmm, so essentially, I have to consider all possible scenarios of what the original 5 balls could have been in terms of white and black, add one white ball to each scenario, and then find the average probability of drawing a white ball across all these scenarios.Let me break it down:1. **Possible Initial Combinations:** Since there are 5 balls, and each can be either white or black, the number of white balls can range from 0 to 5. So, the possible initial combinations are: - 0 white, 5 black - 1 white, 4 black - 2 white, 3 black - 3 white, 2 black - 4 white, 1 black - 5 white, 0 black2. **After Adding One White Ball:** For each of these initial combinations, we add one white ball. So, the new combinations become: - 1 white, 5 black - 2 white, 4 black - 3 white, 3 black - 4 white, 2 black - 5 white, 1 black - 6 white, 0 black3. **Calculating Probability for Each Case:** Now, for each of these new combinations, I need to calculate the probability of drawing a white ball. Since there are 6 balls in total after adding one, the probability is simply the number of white balls divided by 6. - 1 white, 5 black: Probability = 1/6 - 2 white, 4 black: Probability = 2/6 = 1/3 - 3 white, 3 black: Probability = 3/6 = 1/2 - 4 white, 2 black: Probability = 4/6 = 2/3 - 5 white, 1 black: Probability = 5/6 - 6 white, 0 black: Probability = 6/6 = 14. **Averaging the Probabilities:** Since all initial combinations are equally likely, I need to average these probabilities. There are 6 scenarios, so I'll sum up all the probabilities and divide by 6. Let's compute the sum: - 1/6 + 1/3 + 1/2 + 2/3 + 5/6 + 1 To add these fractions, it's easier if they have a common denominator. The least common denominator here is 6. Converting each term: - 1/6 = 1/6 - 1/3 = 2/6 - 1/2 = 3/6 - 2/3 = 4/6 - 5/6 = 5/6 - 1 = 6/6 Now, adding them up: 1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = (1 + 2 + 3 + 4 + 5 + 6)/6 = 21/6 So, the sum of probabilities is 21/6. Now, divide by 6 to get the average: (21/6) / 6 = 21/36 Simplifying 21/36, both numerator and denominator are divisible by 3: 21 ÷ 3 = 7 36 ÷ 3 = 12 So, the average probability is 7/12.Wait a second, let me double-check my calculations to make sure I didn't make any mistakes.Starting from the sum:1/6 + 1/3 + 1/2 + 2/3 + 5/6 + 1Convert all to sixths:1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6Adding numerators: 1 + 2 + 3 + 4 + 5 + 6 = 21So, total sum is 21/6.Divide by 6: (21/6) / 6 = 21/36 = 7/12.Yes, that seems correct.Alternatively, I could think about it in terms of expected value. The expected number of white balls after adding one is the average of white balls across all scenarios.But since each initial combination is equally likely, and we're averaging the probabilities, which is essentially the same as computing the expected probability.So, I think 7/12 is the correct answer.

💡Alright, so I have this problem with complex numbers, and I need to find the area of triangle OAB where O is the origin, and A and B are points corresponding to complex numbers z₁ and z₂. I know that |z₁| = 4, which means the distance from the origin to point A is 4 units. The equation given is 4z₁² - 2z₁z₂ + z₂² = 0. Hmm, okay, let me try to figure this out step by step.First, I should probably try to simplify the given equation. It looks like a quadratic in terms of z₂. Maybe I can solve for z₂ in terms of z₁. Let me write it down:4z₁² - 2z₁z₂ + z₂² = 0Hmm, this looks like a quadratic equation where z₂ is the variable. Let me rearrange it:z₂² - 2z₁z₂ + 4z₁² = 0Yes, that's a quadratic in z₂. Let me use the quadratic formula to solve for z₂. The quadratic formula is:z₂ = [2z₁ ± √( (2z₁)² - 4 * 1 * 4z₁² ) ] / 2Simplify the discriminant inside the square root:(2z₁)² - 4 * 1 * 4z₁² = 4z₁² - 16z₁² = -12z₁²So, the square root of -12z₁² is √(-12) * z₁, which is 2i√3 * z₁ because √(-12) = 2i√3.So, plugging back into the quadratic formula:z₂ = [2z₁ ± 2i√3 z₁] / 2Simplify by dividing numerator and denominator by 2:z₂ = z₁ ± i√3 z₁So, z₂ = z₁(1 ± i√3)Okay, so z₂ is z₁ multiplied by (1 ± i√3). That means z₂ is a complex number obtained by scaling and rotating z₁. Since z₁ has a magnitude of 4, let's see what the magnitude of z₂ is.The magnitude of z₂ is |z₂| = |z₁| * |1 ± i√3|. The magnitude of 1 ± i√3 is √(1² + (√3)²) = √(1 + 3) = √4 = 2. So, |z₂| = 4 * 2 = 8.Wait, but I need to find the area of triangle OAB. To find the area, I can use the formula involving the cross product of vectors OA and OB. The area is (1/2)|OA × OB|, which in complex numbers can be represented as (1/2)|Im(z₁ * conjugate(z₂))|.But before that, maybe I can find the angle between OA and OB because if I know the angle, I can use the formula (1/2)|OA||OB|sinθ.Since z₂ = z₁(1 ± i√3), let's find the angle between z₁ and z₂. Let me denote θ as the angle between OA and OB.The argument of z₂ is the argument of z₁ plus the argument of (1 ± i√3). The argument of (1 ± i√3) is ±60 degrees because tanθ = √3/1, so θ = 60° or 120°, but since it's ±, it's 60° and -60°, which is equivalent to 300°.Wait, actually, 1 + i√3 has an argument of 60°, and 1 - i√3 has an argument of -60°, which is 300°. So, depending on the sign, the angle between z₁ and z₂ is either 60° or -60°, but the sine of both angles is the same because sin(60°) = sin(300°) = √3/2.So, the angle between OA and OB is 60°, and the lengths of OA and OB are 4 and 8, respectively.Therefore, the area of triangle OAB is (1/2)*|OA|*|OB|*sinθ = (1/2)*4*8*(√3/2) = (1/2)*32*(√3/2) = 16*(√3/2) = 8√3.Wait, but let me double-check. Maybe I should use vectors to compute the area. Let me represent z₁ and z₂ as vectors.Let’s assume z₁ is a vector of length 4, and z₂ is z₁ multiplied by (1 ± i√3). So, if z₁ is 4 units long, then z₂ is 8 units long, as I found earlier.The angle between z₁ and z₂ is 60°, so the area is (1/2)*4*8*sin60° = (1/2)*32*(√3/2) = 8√3.Hmm, that seems consistent.Alternatively, I can use coordinates. Let me assign coordinates to z₁ and z₂. Let’s suppose z₁ is along the real axis for simplicity, so z₁ = 4 + 0i. Then, z₂ = z₁*(1 ± i√3) = 4*(1 ± i√3) = 4 ± 4i√3.So, point A is (4, 0), and point B is (4, ±4√3). The origin is (0,0). So, the triangle OAB has vertices at (0,0), (4,0), and (4, ±4√3).Wait, actually, if z₂ is 4*(1 ± i√3), then the coordinates of B are (4, ±4√3). So, the triangle is formed by (0,0), (4,0), and (4, ±4√3).To find the area, I can use the formula for the area of a triangle given coordinates. The formula is:Area = (1/2)|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|Plugging in the points O(0,0), A(4,0), and B(4, ±4√3):Area = (1/2)|0*(0 - (±4√3)) + 4*((±4√3) - 0) + 4*(0 - 0)|Simplify:= (1/2)|0 + 4*(±4√3) + 0|= (1/2)| ±16√3 |= (1/2)(16√3)= 8√3So, that confirms the area is 8√3.Wait, but I also remember that the area can be found using the determinant method. The area is (1/2)|x₁y₂ - x₂y₁| where vectors OA and OB are (x₁, y₁) and (x₂, y₂). But in this case, OA is (4,0) and OB is (4, ±4√3). So, the determinant is 4*(±4√3) - 0*4 = ±16√3. The absolute value is 16√3, and half of that is 8√3. Yep, same result.Alternatively, thinking in terms of vectors, the area is half the magnitude of the cross product of OA and OB. The cross product in 2D is scalar and equals x₁y₂ - x₂y₁, which is exactly what I just did.So, all methods lead me to the area being 8√3.Wait, but let me think again. The problem says "the area of triangle OAB." I assumed z₁ is along the real axis, but does that affect the generality? Because in the complex plane, we can rotate the coordinate system without changing the area. So, regardless of the initial direction of z₁, the area remains the same because we're dealing with magnitudes and angles, which are invariant under rotation.Therefore, my conclusion is that the area is 8√3, which is option A.But hold on, let me check the quadratic equation step again because sometimes when dealing with complex numbers, especially when taking square roots, there might be multiple solutions or something I might have missed.Starting from 4z₁² - 2z₁z₂ + z₂² = 0.I treated it as a quadratic in z₂ and solved for z₂, which gave me z₂ = z₁(1 ± i√3). That seems correct.Alternatively, maybe I can factor the equation differently. Let me see:4z₁² - 2z₁z₂ + z₂² = 0Hmm, can I factor this as (az₁ + bz₂)²? Let's try:(az₁ + bz₂)² = a²z₁² + 2abz₁z₂ + b²z₂²Comparing to 4z₁² - 2z₁z₂ + z₂², we have:a² = 42ab = -2b² = 1From a² = 4, a = ±2.From b² = 1, b = ±1.Now, 2ab = -2. Let's see:If a = 2, then 2*2*b = -2 => 4b = -2 => b = -0.5. But b² should be 1, but (-0.5)² = 0.25 ≠ 1. So, that doesn't work.If a = -2, then 2*(-2)*b = -2 => -4b = -2 => b = 0.5. Again, b² = 0.25 ≠ 1. So, that doesn't work either.Hmm, so factoring as a perfect square doesn't seem to work here. Maybe I made a mistake earlier?Wait, no, because when I solved the quadratic equation, I got z₂ = z₁(1 ± i√3), which seems correct. So, maybe the equation isn't a perfect square, but the quadratic formula still applies.Alternatively, maybe I can write the equation as:z₂² - 2z₁z₂ + 4z₁² = 0Let me divide both sides by z₁² (assuming z₁ ≠ 0, which it isn't since |z₁| = 4):(z₂/z₁)² - 2(z₂/z₁) + 4 = 0Let me set w = z₂/z₁, then the equation becomes:w² - 2w + 4 = 0Solving for w:w = [2 ± √(4 - 16)] / 2 = [2 ± √(-12)] / 2 = [2 ± 2i√3] / 2 = 1 ± i√3So, w = 1 ± i√3, which means z₂ = z₁(1 ± i√3). So, that's consistent with what I found earlier.Therefore, my previous conclusion still holds. The area is 8√3.Wait, but let me think about the triangle OAB. Points O, A, and B. If z₂ = z₁(1 ± i√3), then vector OB is vector OA multiplied by (1 ± i√3). So, in terms of vectors, OB = OA * (1 ± i√3). So, the transformation from OA to OB is a scaling by |1 ± i√3| and a rotation by the argument of (1 ± i√3), which is 60° or -60°.So, the scaling factor is 2, as |1 ± i√3| = 2, so OB is twice as long as OA, and rotated by 60°. Therefore, the angle between OA and OB is 60°, and the lengths are 4 and 8, respectively.Thus, the area is (1/2)*4*8*sin60° = 8√3.Alternatively, if I think of vectors OA and OB, with OA = (4,0) and OB = (4, ±4√3), then the area is (1/2)|x₁y₂ - x₂y₁| = (1/2)|4*(±4√3) - 0*4| = (1/2)|±16√3| = 8√3.So, all these methods lead me to the same answer, which is 8√3. Therefore, the correct option is A.**Final Answer**boxed{A}

💡Okay, so I have this problem where the Grunters are playing the Screamers five times, and each game the Grunters have a 60% chance of winning. I need to find the probability that the Grunters will win at least 4 out of the 5 games. Hmm, okay, let's break this down step by step.First, I know that when dealing with probabilities of multiple independent events, like each game here, I can use the binomial probability formula. The binomial formula is useful for calculating the probability of having exactly k successes in n trials, where each trial has a success probability of p. The formula is:P(k) = C(n, k) * p^k * (1-p)^(n-k)Where C(n, k) is the combination of n things taken k at a time. So, in this case, n is 5 games, k is the number of wins, and p is 0.6.But the problem asks for the probability of winning at least 4 games. That means I need to consider two scenarios: winning exactly 4 games and winning exactly 5 games. Then, I'll add those probabilities together to get the total probability of winning at least 4 games.Let me start with the probability of winning exactly 4 games. Using the binomial formula:P(4) = C(5, 4) * (0.6)^4 * (0.4)^1First, I need to calculate C(5, 4). That's the number of ways to choose 4 games out of 5. I remember that C(n, k) is calculated as n! / (k!(n-k)!), where "!" denotes factorial. So,C(5, 4) = 5! / (4! * (5-4)!) = (5*4*3*2*1) / ((4*3*2*1)*(1)) = 120 / (24*1) = 5Okay, so C(5, 4) is 5. Now, let's compute (0.6)^4 and (0.4)^1.(0.6)^4 = 0.6 * 0.6 * 0.6 * 0.6 = 0.1296(0.4)^1 = 0.4So, putting it all together:P(4) = 5 * 0.1296 * 0.4 = 5 * 0.05184 = 0.2592Alright, so the probability of winning exactly 4 games is 0.2592.Now, let's calculate the probability of winning all 5 games. Again, using the binomial formula:P(5) = C(5, 5) * (0.6)^5 * (0.4)^0C(5, 5) is the number of ways to choose all 5 games, which is just 1.(0.6)^5 = 0.6 * 0.6 * 0.6 * 0.6 * 0.6 = 0.07776(0.4)^0 = 1, since any number to the power of 0 is 1.So,P(5) = 1 * 0.07776 * 1 = 0.07776Therefore, the probability of winning all 5 games is 0.07776.To find the total probability of winning at least 4 games, I need to add the probabilities of winning exactly 4 games and exactly 5 games:P(at least 4) = P(4) + P(5) = 0.2592 + 0.07776 = 0.33696Hmm, 0.33696 is a decimal, but the problem asks for the probability as a common fraction. So, I need to convert 0.33696 into a fraction.First, let's express 0.33696 as a fraction. Since it's a decimal, I can write it as 33696/100000. But that's a bit messy, so I should simplify it.Let me see if I can find a common factor. Let's try dividing numerator and denominator by 16:33696 ÷ 16 = 2106100000 ÷ 16 = 6250So, now we have 2106/6250. Let's see if this can be simplified further.Divide numerator and denominator by 2:2106 ÷ 2 = 10536250 ÷ 2 = 3125So, now we have 1053/3125.Let me check if 1053 and 3125 have any common factors. 3125 is 5^5, which is 5*5*5*5*5. 1053 divided by 5 is 210.6, which isn't an integer, so 5 isn't a factor. Let's try 3:1053 ÷ 3 = 3513125 ÷ 3 is approximately 1041.666..., which isn't an integer, so 3 isn't a common factor either. Let's check 351:351 ÷ 3 = 117117 ÷ 3 = 3939 ÷ 3 = 13So, 1053 is 3^4 * 13. 3125 is 5^5. There are no common factors between 1053 and 3125, so the fraction is simplified as 1053/3125.Therefore, the probability that the Grunters will win at least 4 out of the 5 games is 1053/3125.Wait a second, let me double-check my calculations to make sure I didn't make any mistakes. So, P(4) was 0.2592 and P(5) was 0.07776. Adding them together gives 0.33696, which is approximately 1053/3125. Let me confirm that 1053 divided by 3125 equals 0.33696.Calculating 1053 ÷ 3125:3125 goes into 1053 zero times. Add a decimal point and a zero: 10530 ÷ 3125.3125 goes into 10530 three times (3*3125=9375). Subtract 9375 from 10530: 1155.Bring down another zero: 11550. 3125 goes into 11550 three times (3*3125=9375). Subtract: 11550 - 9375 = 2175.Bring down another zero: 21750. 3125 goes into 21750 six times (6*3125=18750). Subtract: 21750 - 18750 = 3000.Bring down another zero: 30000. 3125 goes into 30000 nine times (9*3125=28125). Subtract: 30000 - 28125 = 1875.Bring down another zero: 18750. 3125 goes into 18750 exactly six times (6*3125=18750). Subtract: 18750 - 18750 = 0.So, putting it all together: 0.33696. Yep, that matches. So, 1053/3125 is correct.I think that's thorough enough. I considered both cases, calculated each probability, added them, converted to a fraction, and verified the division. I don't see any mistakes in my reasoning.

💡First, I recall that the area of a rhombus can be calculated using the formula:Area = (d₁ * d₂) / 2where d₁ and d₂ are the lengths of the diagonals.In this problem, both diagonals are given as 4 centimeters. So, I substitute the values into the formula:Area = (4 cm * 4 cm) / 2Next, I perform the multiplication:4 cm * 4 cm = 16 cm²Then, I divide by 2 to find the area:16 cm² / 2 = 8 cm²Therefore, the area of the rhombus is 8 square centimeters.

💡Okay, so I have this geometry problem about a regular pentagon and some circles and lines. Let me try to visualize it first. There's a regular pentagon ABCDE, so all sides are equal, and all internal angles are equal. The circle c has its center at point A and passes through points B and E. That means the radius of the circle is equal to the length of AB, which is also equal to AE since it's a regular pentagon.Now, the line BC intersects the circle c again at point F. So, starting from B, going through C, and then intersecting the circle again at F. I need to find where exactly F is located. Since BC is a side of the pentagon, and the circle is centered at A, F must be somewhere beyond C on the line BC.Next, there's a point G on the circle c such that the length of FB is equal to the length of FG, and G is not the same as B. So, starting from F, moving a distance equal to FB along the circle, we reach G. This suggests that G is another point on the circle such that triangle FBG is isosceles with FB = FG.The goal is to prove that the lines AB, EF, and DG all intersect at one point. That means these three lines should meet at a common point, which I'll call P. So, I need to show that P lies on all three lines.Let me start by recalling some properties of regular pentagons. In a regular pentagon, the diagonals intersect each other at the golden ratio. The golden ratio is approximately 1.618, and it often appears in regular pentagons because of their inherent symmetry.Since circle c is centered at A and passes through B and E, the radius is AB. Therefore, points B and E are both on the circle, and so are any other points that are a distance AB away from A. So, F and G are also on this circle.I think it might help to use coordinate geometry here. Let me assign coordinates to the pentagon. Let's place point A at the origin (0,0). Since it's a regular pentagon, I can place the other points on a unit circle for simplicity. The coordinates of the regular pentagon can be given using sine and cosine functions.But wait, the circle c is centered at A, which is the origin, and has radius AB. If I place the pentagon on a unit circle, then AB would be the distance from the origin to point B, which is 1. So, the circle c would have a radius of 1. But in a regular pentagon, the distance from the center to a vertex is the same as the radius, so that's consistent.Let me assign coordinates:- Let’s place point A at (0,0).- Point B can be at (1,0).- Point C can be at (cos(72°), sin(72°)).- Point D can be at (cos(144°), sin(144°)).- Point E can be at (cos(216°), sin(216°)).- Point F is the second intersection of line BC with circle c.Wait, but if I place the pentagon on a unit circle, the center is at the origin, but in the problem, the circle c is centered at A, which is a vertex. So, actually, the regular pentagon is not centered at A, but A is one of its vertices. That complicates things a bit.Maybe it's better to use complex numbers for this problem. Let me consider the regular pentagon inscribed in a unit circle centered at the origin, with point A at (1,0). Then, points B, C, D, E can be represented as complex numbers on the unit circle, each separated by 72 degrees.But in the problem, the circle c is centered at A, which is (1,0), and passes through B and E. So, the radius of circle c is the distance from A to B, which is |B - A|. Since both A and B are on the unit circle, the distance AB is 2*sin(36°), because the angle between them is 72°, and the chord length is 2*R*sin(theta/2), where R is 1.So, the radius of circle c is 2*sin(36°). Therefore, circle c is centered at A (1,0) with radius 2*sin(36°). Points B and E are on this circle, as given.Now, line BC intersects circle c again at F. So, I need to find the coordinates of point F. Let me parameterize line BC.Point B is at (cos(72°), sin(72°)), and point C is at (cos(144°), sin(144°)). The line BC can be parametrized as B + t*(C - B), where t is a real number.I need to find the point F where this line intersects circle c again. Since B is already on circle c, F will be the other intersection point.Let me write the equation of circle c. It's centered at A (1,0) with radius 2*sin(36°). So, the equation is (x - 1)^2 + y^2 = (2*sin(36°))^2.The parametric equation of line BC is:x = cos(72°) + t*(cos(144°) - cos(72°))y = sin(72°) + t*(sin(144°) - sin(72°))I need to substitute these into the circle equation and solve for t. One solution will correspond to point B (t=0), and the other will correspond to point F.This seems a bit involved, but let's try to compute it.First, compute cos(72°), sin(72°), cos(144°), sin(144°):cos(72°) ≈ 0.3090sin(72°) ≈ 0.9511cos(144°) ≈ -0.8090sin(144°) ≈ 0.5878So, the parametric equations become:x = 0.3090 + t*(-0.8090 - 0.3090) = 0.3090 - 1.1180*ty = 0.9511 + t*(0.5878 - 0.9511) = 0.9511 - 0.3633*tNow, substitute into the circle equation:(x - 1)^2 + y^2 = (2*sin(36°))^2 ≈ (2*0.5878)^2 ≈ (1.1756)^2 ≈ 1.38197Compute (x - 1)^2:(0.3090 - 1.1180*t - 1)^2 = (-0.6910 - 1.1180*t)^2 = (0.6910 + 1.1180*t)^2 ≈ (0.6910)^2 + 2*0.6910*1.1180*t + (1.1180*t)^2 ≈ 0.4775 + 1.543*t + 1.25*t^2Compute y^2:(0.9511 - 0.3633*t)^2 ≈ (0.9511)^2 - 2*0.9511*0.3633*t + (0.3633*t)^2 ≈ 0.9046 - 0.688*t + 0.132*t^2Add them together:0.4775 + 1.543*t + 1.25*t^2 + 0.9046 - 0.688*t + 0.132*t^2 ≈ (0.4775 + 0.9046) + (1.543*t - 0.688*t) + (1.25*t^2 + 0.132*t^2) ≈ 1.3821 + 0.855*t + 1.382*t^2Set this equal to 1.38197:1.3821 + 0.855*t + 1.382*t^2 ≈ 1.38197Subtract 1.38197 from both sides:0.00013 + 0.855*t + 1.382*t^2 ≈ 0This is a quadratic equation in t:1.382*t^2 + 0.855*t + 0.00013 ≈ 0Solving for t:t = [-0.855 ± sqrt(0.855^2 - 4*1.382*0.00013)] / (2*1.382)Compute discriminant:0.855^2 ≈ 0.7314*1.382*0.00013 ≈ 0.00072So, sqrt(0.731 - 0.00072) ≈ sqrt(0.7303) ≈ 0.8546Thus,t ≈ [-0.855 ± 0.8546]/(2.764)First solution:t ≈ (-0.855 + 0.8546)/2.764 ≈ (-0.0004)/2.764 ≈ -0.000145Second solution:t ≈ (-0.855 - 0.8546)/2.764 ≈ (-1.7096)/2.764 ≈ -0.618So, t ≈ -0.000145 corresponds to point B (since t=0 is B), and t ≈ -0.618 corresponds to point F.Therefore, point F is at:x ≈ 0.3090 - 1.1180*(-0.618) ≈ 0.3090 + 0.691 ≈ 1.000y ≈ 0.9511 - 0.3633*(-0.618) ≈ 0.9511 + 0.225 ≈ 1.176Wait, that can't be right because point F is supposed to be on circle c, which is centered at (1,0) with radius ≈1.1756. So, the distance from (1,0) to (1.000, 1.176) is sqrt((0)^2 + (1.176)^2) ≈1.176, which is approximately equal to the radius. So, that seems correct.So, point F is approximately at (1, 1.176). Hmm, interesting. So, it's almost directly above point A on the circle.Now, point G is on circle c such that |FB| = |FG| and B ≠ G. So, starting from F, we need to find a point G on the circle such that the distance from F to G is equal to the distance from F to B.First, let's compute |FB|. Point F is at (1, 1.176), and point B is at (0.3090, 0.9511).Distance FB:sqrt[(1 - 0.3090)^2 + (1.176 - 0.9511)^2] ≈ sqrt[(0.691)^2 + (0.225)^2] ≈ sqrt[0.477 + 0.0506] ≈ sqrt[0.5276] ≈ 0.726So, |FB| ≈0.726. Therefore, |FG| should also be ≈0.726.Point G is another point on circle c such that the distance from F to G is ≈0.726. Since F is at (1, 1.176), we need to find another point G on the circle centered at (1,0) with radius ≈1.1756, such that the distance from F to G is ≈0.726.This seems like finding another intersection point of a circle centered at F with radius |FB| and the original circle c.Let me write the equations:Circle c: (x - 1)^2 + y^2 = (1.1756)^2 ≈1.38197Circle centered at F (1,1.176) with radius ≈0.726:(x - 1)^2 + (y - 1.176)^2 ≈0.726^2 ≈0.527Subtracting the two equations to find the radical line:[(x - 1)^2 + y^2] - [(x - 1)^2 + (y - 1.176)^2] ≈1.38197 - 0.527Simplify:y^2 - (y^2 - 2*1.176*y + (1.176)^2) ≈0.85497So,y^2 - y^2 + 2.352*y - 1.382 ≈0.85497Simplify:2.352*y - 1.382 ≈0.854972.352*y ≈0.85497 +1.382 ≈2.23697y ≈2.23697 /2.352 ≈0.951So, the radical line is y ≈0.951. Therefore, the two circles intersect at y ≈0.951.Now, substitute y ≈0.951 into circle c:(x -1)^2 + (0.951)^2 ≈1.38197(x -1)^2 ≈1.38197 -0.904 ≈0.47797So, x -1 ≈±sqrt(0.47797) ≈±0.691Thus, x ≈1 ±0.691, so x ≈1.691 or x≈0.309We already know that point B is at (0.309, 0.951), so the other intersection point is at (1.691, 0.951). Therefore, point G is at approximately (1.691, 0.951).Wait, but point G is supposed to be on circle c. Let me check the distance from A (1,0) to G (1.691, 0.951):sqrt[(1.691 -1)^2 + (0.951 -0)^2] ≈sqrt[(0.691)^2 + (0.951)^2] ≈sqrt[0.477 +0.904] ≈sqrt[1.381] ≈1.175, which is the radius. So, yes, G is on circle c.So, point G is at approximately (1.691, 0.951).Now, I need to find the equations of lines AB, EF, and DG and show that they intersect at one point.First, line AB: Since A is at (0,0) and B is at (0.3090, 0.9511), the slope is (0.9511 -0)/(0.3090 -0) ≈3.077. So, the equation is y ≈3.077x.Second, line EF: Point E is at (cos(216°), sin(216°)) ≈(-0.8090, -0.5878), and point F is at (1,1.176). So, the slope is (1.176 - (-0.5878))/(1 - (-0.8090)) ≈(1.7638)/(1.8090) ≈0.974. So, the equation is y - (-0.5878) ≈0.974(x - (-0.8090)), which simplifies to y +0.5878 ≈0.974x +0.788. So, y ≈0.974x +0.788 -0.5878 ≈0.974x +0.2002.Third, line DG: Point D is at (cos(144°), sin(144°)) ≈(-0.8090, 0.5878), and point G is at (1.691, 0.951). The slope is (0.951 -0.5878)/(1.691 - (-0.8090)) ≈(0.3632)/(2.499) ≈0.145. So, the equation is y -0.5878 ≈0.145(x - (-0.8090)), which simplifies to y -0.5878 ≈0.145x +0.117. So, y ≈0.145x +0.117 +0.5878 ≈0.145x +0.7048.Now, let's find the intersection of AB and EF.AB: y ≈3.077xEF: y ≈0.974x +0.2002Set equal:3.077x ≈0.974x +0.20023.077x -0.974x ≈0.20022.103x ≈0.2002x ≈0.2002 /2.103 ≈0.0952Then, y ≈3.077*0.0952 ≈0.293So, intersection point P is approximately (0.0952, 0.293)Now, check if this point lies on DG.DG: y ≈0.145x +0.7048Plug in x ≈0.0952:y ≈0.145*0.0952 +0.7048 ≈0.0138 +0.7048 ≈0.7186But the y-coordinate from AB and EF is ≈0.293, which is not equal to 0.7186. Hmm, that's a problem. It seems like the lines AB, EF, and DG do not intersect at the same point based on these approximate calculations.Wait, maybe my approximations are too rough. Let me try to do this more accurately.Alternatively, perhaps I made a mistake in the coordinate assignments. Let me think again.Wait, in the problem, the circle c is centered at A, which is a vertex of the pentagon, not the center of the pentagon. So, my initial assumption of placing the pentagon on a unit circle centered at the origin might not be correct because in that case, A is at (1,0), but the center of the pentagon is at the origin. However, in the problem, the circle is centered at A, which is a vertex, not the center.Therefore, perhaps I should model the pentagon such that point A is at (0,0), and the pentagon is not centered at the origin. This complicates things because the coordinates will be different.Alternatively, maybe using complex numbers with A at the origin.Let me try that.Let me place point A at (0,0). Then, since it's a regular pentagon, the other points can be represented as complex numbers rotated around A. However, since A is a vertex, the center of the pentagon is not at A, but somewhere else.Wait, in a regular pentagon, the center is equidistant from all vertices. If I place A at (0,0), the center O would be at some point (d,0), where d is the distance from A to the center.The radius R of the circumscribed circle (distance from center to any vertex) can be related to the side length s. In a regular pentagon, R = s / (2*sin(36°)).But in our case, the circle c is centered at A with radius AB, which is equal to the side length s. So, R = s, but in reality, R = s / (2*sin(36°)). Therefore, s = 2*R*sin(36°). So, if we take R=1, then s=2*sin(36°).But in our problem, the circle c has radius AB = s, so if we take AB = s, then the radius of the circumscribed circle (distance from center to vertex) is R = s / (2*sin(36°)).Therefore, if I place A at (0,0), the center O is at (d,0), where d = R - something? Wait, maybe it's better to use vectors.Alternatively, perhaps using symmetry and properties of the regular pentagon without coordinates.In a regular pentagon, the diagonals intersect at the golden ratio. So, the ratio of the diagonal to the side is the golden ratio φ = (1 + sqrt(5))/2 ≈1.618.Given that, perhaps we can use projective geometry or Ceva's theorem to show that the lines AB, EF, and DG are concurrent.Ceva's theorem states that for concurrent lines from vertices of a triangle, the product of certain ratios equals 1. But in this case, we have a pentagon, not a triangle. Maybe we can consider triangle AEF or something similar.Alternatively, since all points are on circle c, perhaps using power of a point or cyclic quadrilaterals.Wait, let's consider the circle c centered at A, passing through B and E. Points F and G are also on this circle.Given that |FB| = |FG|, triangle FBG is isosceles with FB = FG.Therefore, angle FBG = angle FGB.But since points B, F, G are on circle c, the angles subtended by chords FB and FG at the center A are equal because FB = FG.Therefore, arcs FB and FG are equal, meaning that the central angles for arcs FB and FG are equal.Thus, the angle FAB = angle FAG.But since A is the center, angle FAB is equal to angle FAG.Wait, but F is a point on BC extended, so maybe there's some symmetry here.Alternatively, perhaps using spiral similarity or some rotational symmetry.Given that the pentagon is regular, rotating by 72 degrees around the center maps the pentagon onto itself. But since our circle is centered at A, which is a vertex, the symmetry might be different.Alternatively, perhaps using inversion with respect to circle c.But inversion might be too complicated.Wait, another approach: since AB, EF, and DG are supposed to meet at one point, let's assume that they meet at point P and show that P lies on all three lines.Alternatively, perhaps using Menelaus' theorem.But maybe it's better to use coordinate geometry with more precise calculations.Let me try to set up a coordinate system where point A is at (0,0), and the regular pentagon is such that point B is at (1,0). Then, the other points can be determined based on the regular pentagon structure.In a regular pentagon with point A at (0,0) and point B at (1,0), the center O is at some point (d,0), and the other points C, D, E are located around the center.The coordinates can be determined using the golden ratio.The distance from A to O is d, and the distance from O to any vertex is R.In a regular pentagon, the distance from a vertex to the center is R, and the distance between two adjacent vertices is s = 2*R*sin(36°).Given that AB = s =1, so 1 = 2*R*sin(36°), so R = 1/(2*sin(36°)) ≈1/(2*0.5878)≈0.8507.Therefore, the center O is at (d,0), where d is the distance from A to O.In a regular pentagon, the distance from a vertex to the center is R, so the distance from A to O is R, but since A is at (0,0) and O is at (d,0), then d = R ≈0.8507.Wait, but if A is at (0,0) and O is at (d,0), then the distance from A to O is d, which should be equal to R ≈0.8507. Therefore, d ≈0.8507.But then, the coordinates of the other points can be determined.Point B is at (1,0). Point C is obtained by rotating point B around O by 72 degrees.Wait, actually, in a regular pentagon, each vertex is separated by 72 degrees around the center.But since we have point A at (0,0), which is not the center, this complicates things.Alternatively, perhaps it's better to use vectors.Let me denote vectors from point A.Let me consider point A as the origin.Then, vector AB is vector b, and vector AE is vector e.Since it's a regular pentagon, the angle between AB and AE is 108 degrees (since internal angle at A is 108°).Wait, in a regular pentagon, each internal angle is 108°, so the angle between AB and AE is 108°.But in our case, circle c is centered at A, passing through B and E, so vectors AB and AE have the same length, which is the radius of circle c.Therefore, |b| = |e| = r.Given that, the angle between b and e is 108°, so the dot product b·e = |b||e|cos(108°) = r^2 cos(108°).Now, point F is the second intersection of line BC with circle c.Line BC can be parametrized as B + t*(C - B).But I need to express point C in terms of vectors.In a regular pentagon, point C can be obtained by rotating point B around the center by 72 degrees. But since our center is not at the origin, this is complicated.Alternatively, perhaps using complex numbers.Let me model the regular pentagon in the complex plane with A at 0, B at 1, and the center at some point d on the real axis.Then, the other points can be represented as complex numbers.Let me denote:- A = 0- B = 1- O = d (real number)- The other points C, D, E are located around O.In a regular pentagon, the distance from O to each vertex is R, and the distance between adjacent vertices is s = 2*R*sin(36°).Given that AB = s =1, so R = 1/(2*sin(36°)) ≈0.8507.But the distance from A to O is |O - A| = |d - 0| = d.In a regular pentagon, the distance from a vertex to the center is R, so d = R ≈0.8507.Therefore, O is at d ≈0.8507 on the real axis.Now, the coordinates of point B are (1,0). The coordinates of point C can be found by rotating point B around O by 72 degrees.The rotation of a point z around point O by angle θ is given by:C = (B - O)*e^{iθ} + OSimilarly for other points.Let me compute point C.First, translate point B by -O: B' = B - O = 1 - d ≈1 -0.8507≈0.1493.Now, rotate B' by 72 degrees: B'' = B' * e^{i72°} ≈0.1493*(cos72° + i sin72°) ≈0.1493*(0.3090 + i0.9511) ≈0.0461 + i0.1420.Then, translate back by adding O: C ≈0.0461 + i0.1420 +0.8507 ≈0.8968 + i0.1420.Similarly, point D is obtained by rotating point C around O by 72 degrees.But this is getting complicated. Maybe I can use complex numbers more abstractly.Let me denote the complex plane with A=0, B=1, O=d on the real axis.The other points are:C = (B - O)*e^{i72°} + OD = (C - O)*e^{i72°} + OE = (D - O)*e^{i72°} + OBut this recursive definition might not be the easiest way.Alternatively, since the regular pentagon can be inscribed in a circle of radius R centered at O, with points A, B, C, D, E equally spaced around the circle.But point A is at 0, which is not the center. So, the center O is at d on the real axis, and the other points are located on the circle centered at O with radius R.Therefore, the coordinates of the points can be expressed as:A = 0B = d + R*e^{iθ1}C = d + R*e^{iθ2}D = d + R*e^{iθ3}E = d + R*e^{iθ4}Where θ1, θ2, θ3, θ4 are angles separated by 72 degrees.But since point A is at 0, which is also on the circle centered at O with radius R, we have |A - O| = R, so |0 - d| = R, thus d = R.Therefore, O is at R on the real axis.Given that, the coordinates of the points are:A = 0B = R + R*e^{iθ1}C = R + R*e^{iθ2}D = R + R*e^{iθ3}E = R + R*e^{iθ4}But the angle θ1 is such that point B is at (1,0). So, let's compute θ1.Point B is at 1 on the real axis, so:B = R + R*e^{iθ1} =1Therefore,R*(1 + e^{iθ1}) =1Let me write e^{iθ1} = cosθ1 + i sinθ1So,R*(1 + cosθ1 + i sinθ1) =1Since B is on the real axis, the imaginary part must be zero:R*sinθ1 =0But R ≠0, so sinθ1=0, which implies θ1=0° or 180°. But θ1 cannot be 0° because then point B would coincide with O, which is at R. Therefore, θ1=180°, but that would place point B at R + R*e^{i180°}= R - R=0, which is point A. Contradiction.Wait, that can't be. Maybe my approach is flawed.Alternatively, perhaps point B is not on the real axis. Wait, in my initial setup, I placed point B at (1,0), but if the center O is at (d,0), then point B is not necessarily on the real axis unless θ1=0°, but that leads to a contradiction as above.Therefore, perhaps my initial assumption of placing point A at (0,0) and point B at (1,0) is not compatible with the regular pentagon structure because the center O cannot be on the real axis in that case.This suggests that I need a different coordinate system.Alternatively, perhaps it's better to use a coordinate system where the center of the pentagon is at the origin, and point A is at (R,0), where R is the radius of the circumscribed circle.In that case, point A is at (R,0), and the other points are equally spaced around the circle.Then, the circle c is centered at A (R,0) and passes through B and E.Let me try this approach.Let me denote:- Center of pentagon O at (0,0)- Point A at (R,0)- Points B, C, D, E equally spaced around the circle of radius R.The coordinates of the points are:A = (R,0)B = (R*cos72°, R*sin72°)C = (R*cos144°, R*sin144°)D = (R*cos216°, R*sin216°)E = (R*cos288°, R*sin288°)Now, circle c is centered at A (R,0) and passes through B and E.Compute the radius of circle c: distance from A to B.Distance AB = sqrt[(R*cos72° - R)^2 + (R*sin72°)^2] = R*sqrt[(cos72° -1)^2 + sin^2 72°]Simplify:= R*sqrt[cos^2 72° - 2cos72° +1 + sin^2 72°]= R*sqrt[(cos^2 72° + sin^2 72°) - 2cos72° +1]= R*sqrt[1 - 2cos72° +1]= R*sqrt[2 - 2cos72°]= R*sqrt[2(1 - cos72°)]Using the identity 1 - cosθ = 2sin^2(θ/2):= R*sqrt[2*2sin^2(36°)]= R*sqrt[4sin^2(36°)]= R*2sin36°Therefore, the radius of circle c is 2R sin36°.Now, line BC intersects circle c again at F.First, find the equation of line BC.Points B and C:B = (R*cos72°, R*sin72°)C = (R*cos144°, R*sin144°)Parametrize line BC as:x = R*cos72° + t*(R*cos144° - R*cos72°)y = R*sin72° + t*(R*sin144° - R*sin72°)We need to find t such that (x - R)^2 + y^2 = (2R sin36°)^2Substitute x and y:[R*cos72° + t*(R*cos144° - R*cos72°) - R]^2 + [R*sin72° + t*(R*sin144° - R*sin72°)]^2 = (2R sin36°)^2Simplify:[R*(cos72° -1) + t*R*(cos144° - cos72°)]^2 + [R*sin72° + t*R*(sin144° - sin72°)]^2 = 4R^2 sin^2 36°Factor out R:R^2[(cos72° -1) + t*(cos144° - cos72°)]^2 + R^2[sin72° + t*(sin144° - sin72°)]^2 = 4R^2 sin^2 36°Divide both sides by R^2:[(cos72° -1) + t*(cos144° - cos72°)]^2 + [sin72° + t*(sin144° - sin72°)]^2 = 4 sin^2 36°Let me compute the numerical values:cos72° ≈0.3090, sin72°≈0.9511cos144°≈-0.8090, sin144°≈0.5878sin36°≈0.5878So,First term inside the first square:cos72° -1 ≈0.3090 -1≈-0.6910cos144° - cos72°≈-0.8090 -0.3090≈-1.1180Second term inside the second square:sin72°≈0.9511sin144° - sin72°≈0.5878 -0.9511≈-0.3633So, the equation becomes:[-0.6910 + t*(-1.1180)]^2 + [0.9511 + t*(-0.3633)]^2 ≈4*(0.5878)^2≈4*0.345≈1.38Compute each square:First square:(-0.6910 -1.1180 t)^2 ≈(0.6910 +1.1180 t)^2 ≈0.4775 + 1.543 t +1.25 t^2Second square:(0.9511 -0.3633 t)^2≈0.9046 -0.688 t +0.132 t^2Sum:0.4775 +1.543 t +1.25 t^2 +0.9046 -0.688 t +0.132 t^2≈1.3821 +0.855 t +1.382 t^2Set equal to 1.38:1.3821 +0.855 t +1.382 t^2 ≈1.38Subtract 1.38:0.0021 +0.855 t +1.382 t^2≈0This is a quadratic equation:1.382 t^2 +0.855 t +0.0021≈0Solving for t:t = [-0.855 ± sqrt(0.855^2 -4*1.382*0.0021)]/(2*1.382)Compute discriminant:0.855^2≈0.7314*1.382*0.0021≈0.0114So, sqrt(0.731 -0.0114)=sqrt(0.7196)=≈0.848Thus,t≈[-0.855 ±0.848]/2.764First solution:t≈(-0.855 +0.848)/2.764≈(-0.007)/2.764≈-0.0025Second solution:t≈(-0.855 -0.848)/2.764≈(-1.703)/2.764≈-0.616So, t≈-0.0025 corresponds to point B (t=0), and t≈-0.616 corresponds to point F.Therefore, point F is at:x≈R*cos72° + (-0.616)*(R*cos144° - R*cos72°)≈R*(0.3090) + (-0.616)*(R*(-0.8090) - R*0.3090)≈0.3090 R + (-0.616)*(-0.8090 R -0.3090 R)≈0.3090 R + (-0.616)*(-1.1180 R)≈0.3090 R +0.616*1.1180 R≈0.3090 R +0.690 R≈1.0 RSimilarly,y≈R*sin72° + (-0.616)*(R*sin144° - R*sin72°)≈0.9511 R + (-0.616)*(0.5878 R -0.9511 R)≈0.9511 R + (-0.616)*(-0.3633 R)≈0.9511 R +0.225 R≈1.176 RTherefore, point F is at approximately (R,1.176 R). Since R is the radius of the circumscribed circle, which is also the distance from O to any vertex.Now, point G is on circle c such that |FB| = |FG| and B ≠ G.First, compute |FB|.Point F is at (R,1.176 R), point B is at (R*cos72°, R*sin72°)≈(0.3090 R,0.9511 R).Distance FB:sqrt[(R -0.3090 R)^2 + (1.176 R -0.9511 R)^2]≈sqrt[(0.691 R)^2 + (0.225 R)^2]≈sqrt[0.477 R^2 +0.0506 R^2]≈sqrt[0.5276 R^2]≈0.726 RTherefore, |FG|≈0.726 R.Point G is another point on circle c such that |FG|=0.726 R.Since circle c is centered at A (R,0) with radius≈1.1756 R (since 2R sin36°≈2R*0.5878≈1.1756 R).So, point G lies on circle c and on the circle centered at F with radius≈0.726 R.The intersection points of these two circles will give us point G.The equation of circle c: (x - R)^2 + y^2 = (1.1756 R)^2≈1.38197 R^2The equation of circle centered at F (R,1.176 R) with radius≈0.726 R:(x - R)^2 + (y -1.176 R)^2≈(0.726 R)^2≈0.527 R^2Subtracting the two equations:[(x - R)^2 + y^2] - [(x - R)^2 + (y -1.176 R)^2]≈1.38197 R^2 -0.527 R^2≈0.85497 R^2Simplify:y^2 - (y^2 -2*1.176 R y + (1.176 R)^2)≈0.85497 R^2So,y^2 - y^2 +2.352 R y -1.382 R^2≈0.85497 R^2Simplify:2.352 R y≈0.85497 R^2 +1.382 R^2≈2.23697 R^2Divide both sides by R:2.352 y≈2.23697 RThus,y≈(2.23697 /2.352) R≈0.951 RSo, the radical line is y≈0.951 R.Substitute y≈0.951 R into circle c:(x - R)^2 + (0.951 R)^2≈1.38197 R^2(x - R)^2≈1.38197 R^2 -0.904 R^2≈0.47797 R^2Thus,x - R≈±0.691 RSo,x≈R ±0.691 RTherefore, x≈1.691 R or x≈0.309 RWe already know that point B is at x≈0.309 R, so the other intersection point is at x≈1.691 R.Therefore, point G is at approximately (1.691 R,0.951 R).Now, we have points:- A = (R,0)- B≈(0.309 R,0.951 R)- C≈(-0.809 R,0.5878 R)- D≈(-0.809 R,-0.5878 R)- E≈(0.309 R,-0.951 R)- F≈(R,1.176 R)- G≈(1.691 R,0.951 R)Now, we need to find the equations of lines AB, EF, and DG and show that they intersect at one point.First, line AB: connects A (R,0) and B≈(0.309 R,0.951 R).The slope m_AB≈(0.951 R -0)/(0.309 R - R)≈0.951 R / (-0.691 R)≈-1.376So, equation: y -0 = -1.376(x - R)Simplify: y≈-1.376 x +1.376 RSecond, line EF: connects E≈(0.309 R,-0.951 R) and F≈(R,1.176 R).Slope m_EF≈(1.176 R - (-0.951 R))/(R -0.309 R)≈(2.127 R)/(0.691 R)≈3.077Equation: y - (-0.951 R) =3.077(x -0.309 R)Simplify: y +0.951 R≈3.077 x -0.951 RThus, y≈3.077 x -0.951 R -0.951 R≈3.077 x -1.902 RThird, line DG: connects D≈(-0.809 R,-0.5878 R) and G≈(1.691 R,0.951 R).Slope m_DG≈(0.951 R - (-0.5878 R))/(1.691 R - (-0.809 R))≈(1.5388 R)/(2.5 R)≈0.6155Equation: y - (-0.5878 R)=0.6155(x - (-0.809 R))Simplify: y +0.5878 R≈0.6155 x +0.6155*0.809 R≈0.6155 x +0.5 RThus, y≈0.6155 x +0.5 R -0.5878 R≈0.6155 x -0.0878 RNow, let's find the intersection of AB and EF.AB: y≈-1.376 x +1.376 REF: y≈3.077 x -1.902 RSet equal:-1.376 x +1.376 R≈3.077 x -1.902 RBring all terms to left:-1.376 x -3.077 x +1.376 R +1.902 R≈0-4.453 x +3.278 R≈0Thus,x≈(3.278 R)/4.453≈0.736 RThen, y≈-1.376*(0.736 R)+1.376 R≈-1.012 R +1.376 R≈0.364 RSo, intersection point P≈(0.736 R,0.364 R)Now, check if this point lies on DG.DG: y≈0.6155 x -0.0878 RPlug in x≈0.736 R:y≈0.6155*0.736 R -0.0878 R≈0.453 R -0.0878 R≈0.365 RWhich is approximately equal to the y-coordinate from AB and EF, which is≈0.364 R. The slight difference is due to rounding errors in the calculations.Therefore, point P≈(0.736 R,0.364 R) lies on all three lines AB, EF, and DG, proving that they are concurrent.Thus, the lines AB, EF, and DG intersect at one point.

💡To solve the problem, we start by recognizing that the product ((9 - p)(9 - q)(9 - r)(9 - s)(9 - t) = -120) involves five distinct integers. We need to factorize -120 into five distinct factors. Considering the factors of -120, we can choose (-1, 2, 3, 4, 5) because their product is (-120). These factors correspond to the terms ((9 - p), (9 - q), (9 - r), (9 - s), (9 - t)).Next, we calculate each integer (p, q, r, s, t) by solving the equations:- (9 - p = -1 Rightarrow p = 10)- (9 - q = 2 Rightarrow q = 7)- (9 - r = 3 Rightarrow r = 6)- (9 - s = 4 Rightarrow s = 5)- (9 - t = 5 Rightarrow t = 4)Summing these values, we get:[ p + q + r + s + t = 10 + 7 + 6 + 5 + 4 = 32 ]Thus, the sum of (p, q, r, s, t) is (boxed{mathrm{(D)} 32}).