💡Okay, so I have this problem here where I need to find the sum of the coordinates of point X. The given information is that the ratios of the distances XZ/XY and ZY/XY are both 1/2. Points Y and Z are given as Y = (1, 7) and Z = (-1, -7). Hmm, let me try to visualize this.First, I know that ratios of distances often relate to points dividing a line segment in a particular ratio. Since both XZ/XY and ZY/XY are 1/2, that suggests that Z is somehow between X and Y, or maybe X is between Z and Y? Wait, let's think about it.If XZ/XY = 1/2, that means the distance from X to Z is half the distance from X to Y. Similarly, ZY/XY is also 1/2, so the distance from Z to Y is also half the distance from X to Y. That seems interesting. So, both XZ and ZY are half of XY. That would imply that Z is the midpoint between X and Y, right? Because if both segments from X to Z and Z to Y are equal and each is half of XY, then Z must be exactly halfway between X and Y.Let me confirm that. If Z is the midpoint of XY, then the coordinates of Z should be the average of the coordinates of X and Y. So, if I let X = (x, y), then the midpoint formula tells me that Z = ((x + 1)/2, (y + 7)/2). But wait, Z is given as (-1, -7). So, I can set up equations based on that.So, for the x-coordinate: (x + 1)/2 = -1. Multiplying both sides by 2 gives x + 1 = -2, so x = -3. Similarly, for the y-coordinate: (y + 7)/2 = -7. Multiplying both sides by 2 gives y + 7 = -14, so y = -21. Therefore, X must be at (-3, -21). Wait, let me double-check that. If X is (-3, -21) and Y is (1, 7), then the midpoint should be ((-3 + 1)/2, (-21 + 7)/2) = (-2/2, -14/2) = (-1, -7), which is exactly point Z. So that checks out.But just to make sure I didn't make a mistake, let me think about the distances. The distance from X to Z should be half the distance from X to Y. Let's calculate the distance between X and Z. Using the distance formula: sqrt[(-1 - (-3))^2 + (-7 - (-21))^2] = sqrt[(2)^2 + (14)^2] = sqrt[4 + 196] = sqrt[200] = 10*sqrt(2).Now, the distance from X to Y: sqrt[(1 - (-3))^2 + (7 - (-21))^2] = sqrt[(4)^2 + (28)^2] = sqrt[16 + 784] = sqrt[800] = 20*sqrt(2). So, indeed, the distance from X to Z is half of the distance from X to Y, which is consistent with the given ratio.Similarly, the distance from Z to Y: sqrt[(1 - (-1))^2 + (7 - (-7))^2] = sqrt[(2)^2 + (14)^2] = sqrt[4 + 196] = sqrt[200] = 10*sqrt(2), which is again half of XY. So everything seems to line up.Therefore, the coordinates of X are (-3, -21). The sum of the coordinates is -3 + (-21) = -24. I think that's solid. I considered the midpoint, checked the distances, and everything matches up with the given ratios. So I'm confident that the sum is -24.**Final Answer**The sum of the coordinates of ( X ) is boxed{-24}.

💡Okay, so I've got this problem here about a function and some propositions. Let me try to understand it step by step. The function is given as f(x) = x(x - a)(x - 2), where a is a real number. There are two propositions: p is "0 < a < 2" and q is "f'(a) < 0". The question is asking what kind of condition p is for q. The options are A, B, C, D, which correspond to different types of conditions.First, I think I need to figure out what f'(a) is. To do that, I should find the derivative of f(x). The function f(x) is a cubic polynomial, so its derivative should be a quadratic function. Let me expand f(x) first to make it easier to differentiate.So, f(x) = x(x - a)(x - 2). Let me multiply these terms out. First, multiply x and (x - a):x*(x - a) = x^2 - a x.Now, multiply that result by (x - 2):(x^2 - a x)*(x - 2) = x^3 - 2x^2 - a x^2 + 2a x.Combine like terms:x^3 - (2 + a)x^2 + 2a x.So, f(x) = x^3 - (a + 2)x^2 + 2a x.Now, to find f'(x), I'll differentiate term by term:The derivative of x^3 is 3x^2.The derivative of -(a + 2)x^2 is -2(a + 2)x.The derivative of 2a x is 2a.So, putting it all together:f'(x) = 3x^2 - 2(a + 2)x + 2a.Now, I need to evaluate this derivative at x = a, so f'(a):f'(a) = 3a^2 - 2(a + 2)a + 2a.Let me simplify this expression step by step.First, expand the second term:-2(a + 2)a = -2a^2 - 4a.So now, f'(a) becomes:3a^2 - 2a^2 - 4a + 2a.Combine like terms:3a^2 - 2a^2 = a^2.-4a + 2a = -2a.So, f'(a) = a^2 - 2a.I can factor this expression:a(a - 2).So, f'(a) = a(a - 2).Now, the proposition q is f'(a) < 0, which translates to a(a - 2) < 0.I need to find when this inequality holds. Let's analyze the expression a(a - 2).This is a quadratic inequality. The roots are at a = 0 and a = 2. The quadratic opens upwards because the coefficient of a^2 is positive.So, the expression a(a - 2) will be negative between the roots, i.e., when 0 < a < 2.That means a(a - 2) < 0 if and only if 0 < a < 2.Wait a minute, that's exactly the proposition p: 0 < a < 2.So, if p is true, then q is true, and if q is true, then p must be true.Therefore, p is both a necessary and sufficient condition for q.So, looking back at the options, C is "Necessary and sufficient condition", which should be the correct answer.But let me double-check to make sure I didn't make any mistakes.Starting from f(x) = x(x - a)(x - 2), expanding it correctly gives f(x) = x^3 - (a + 2)x^2 + 2a x.Differentiating term by term, f'(x) = 3x^2 - 2(a + 2)x + 2a. That seems correct.Evaluating at x = a:f'(a) = 3a^2 - 2(a + 2)a + 2a.Expanding the second term: -2a^2 - 4a.So, f'(a) = 3a^2 - 2a^2 - 4a + 2a = a^2 - 2a.Factoring gives a(a - 2), which is correct.The inequality a(a - 2) < 0 holds when 0 < a < 2, which is exactly proposition p.So, p is equivalent to q in this case, meaning p is a necessary and sufficient condition for q.Therefore, the answer should be C.**Final Answer**boxed{C}

💡To determine the resulting sum when Jeremy, Joan, and Jack add their counts of edges, vertices, and faces of a regular tetrahedron, we can follow these steps:1. **Count the Edges**: A regular tetrahedron has 6 edges.2. **Count the Vertices**: A regular tetrahedron has 4 vertices.3. **Count the Faces**: A regular tetrahedron has 4 faces.Adding these quantities together:[6 text{ (edges)} + 4 text{ (vertices)} + 4 text{ (faces)} = 14]Thus, the resulting sum is (14).

💡Alright, so I've got this problem about a circle and some tangents, and I need to find the equation of the circle and then the minimum area of a triangle formed by points A, B, and C, where C is the intersection of two tangents from A and B to the circle. Let me try to break this down step by step.First, part (1) is about finding the equation of circle M. The problem says that the center of the circle lies on the y-axis, so its coordinates must be (0, b) for some b. The radius is given as 1, so the equation of the circle will be x² + (y - b)² = 1.Next, it mentions that the chord intercepted by the line l: y = 2x + 2 on the circle has a length of 4√5 / 5. Also, the center M is below the line l, so b must be less than the y-intercept of line l, which is 2. Therefore, b < 2.To find b, I remember that the length of a chord in a circle can be related to the distance from the center to the chord. The formula for the length of a chord is 2√(r² - d²), where r is the radius and d is the distance from the center to the chord.Given the chord length is 4√5 / 5, so:2√(1² - d²) = 4√5 / 5Divide both sides by 2:√(1 - d²) = 2√5 / 5Square both sides:1 - d² = (4 * 5) / 25 = 20 / 25 = 4/5So,d² = 1 - 4/5 = 1/5Therefore, d = √(1/5) = √5 / 5But d is the distance from the center M(0, b) to the line l: y = 2x + 2.The distance from a point (x0, y0) to the line ax + by + c = 0 is |ax0 + by0 + c| / √(a² + b²).First, let me write line l in standard form:y = 2x + 2 => 2x - y + 2 = 0So, a = 2, b = -1, c = 2.The distance from M(0, b) to this line is |2*0 + (-1)*b + 2| / √(2² + (-1)²) = | -b + 2 | / √5We found earlier that this distance d is √5 / 5, so:| -b + 2 | / √5 = √5 / 5Multiply both sides by √5:| -b + 2 | = (√5 * √5) / 5 = 5 / 5 = 1So,| -b + 2 | = 1This gives two possibilities:1. -b + 2 = 1 => -b = -1 => b = 12. -b + 2 = -1 => -b = -3 => b = 3But the problem states that the center M is below the line l, which has a y-intercept at 2. So, b must be less than 2. Therefore, b = 1 is the valid solution.So, the equation of circle M is x² + (y - 1)² = 1.Okay, that seems solid. I think I got part (1) down.Now, moving on to part (2). This seems more involved. Let me read it again.We have points A(t, 0) and B(t + 5, 0), where t is between -4 and -1. So, A and B are points on the x-axis, 5 units apart, and their positions vary as t changes from -4 to -1. So, when t = -4, A is at (-4, 0) and B is at (1, 0). When t = -1, A is at (-1, 0) and B is at (4, 0).We are told that AC and BC are tangent lines to circle M. So, point C is the intersection of the two tangents from A and B to the circle M.We need to find the minimum value of the area of triangle ABC.Hmm, okay. So, first, I need to find the coordinates of point C in terms of t, then express the area of triangle ABC in terms of t, and then find its minimum value over the interval t ∈ [-4, -1].Let me recall that the area of triangle ABC can be found using the formula:Area = (1/2) * base * heightHere, AB is the base, which is fixed at 5 units since the distance between A(t, 0) and B(t + 5, 0) is always 5. So, the base is 5. Therefore, the area will be (1/2)*5*height, where height is the distance from point C to the line AB, which is the x-axis. So, the y-coordinate of point C is the height.Therefore, to minimize the area, we need to minimize the y-coordinate of point C.Alternatively, since the base is fixed, minimizing the area is equivalent to minimizing the height, which is the y-coordinate of C.So, if I can find the y-coordinate of C in terms of t, then I can find its minimum.But how do I find the coordinates of point C?Point C is the intersection of the two tangents from A and B to the circle M. So, I need to find the equations of these two tangents and then find their intersection point C.To find the equations of the tangents from a point to a circle, I can use the formula for the tangent lines from an external point to a circle.Given a circle with center (h, k) and radius r, the equation of the tangent from an external point (x1, y1) is:(y - k) = m(x - h) ± r√(1 + m²)But this might not be the easiest way. Alternatively, I can use the condition that the distance from the point to the tangent line is equal to the radius.Wait, maybe a better approach is to use the fact that the tangent from a point to a circle satisfies the condition that the power of the point with respect to the circle is equal to the square of the length of the tangent.The power of point A with respect to circle M is equal to the square of the length of the tangent from A to M.The power of a point (x1, y1) with respect to the circle (x - h)² + (y - k)² = r² is (x1 - h)² + (y1 - k)² - r².So, for point A(t, 0), the power is (t - 0)² + (0 - 1)² - 1² = t² + 1 - 1 = t².Similarly, for point B(t + 5, 0), the power is (t + 5 - 0)² + (0 - 1)² - 1² = (t + 5)² + 1 - 1 = (t + 5)².So, the lengths of the tangents from A and B to the circle are √(t²) = |t| and √((t + 5)²) = |t + 5|, respectively.But since t is between -4 and -1, t is negative, so |t| = -t, and t + 5 is between 1 and 4, so |t + 5| = t + 5.Okay, so the lengths are -t and t + 5.But how does this help me find the coordinates of C?Alternatively, I can parametrize the tangent lines from A and B to the circle.Let me recall that the equation of a tangent to a circle can be written in the form:For circle x² + (y - 1)² = 1, the tangent lines from point A(t, 0) can be found by solving for the condition that the distance from A to the tangent line is equal to the radius.Wait, perhaps another approach is to use the parametric equations of the tangents.Alternatively, I can use the fact that the tangent lines from A to the circle will satisfy the condition that the line from A to the point of tangency is perpendicular to the radius at the point of tangency.So, if I denote the point of tangency as (x1, y1) on the circle, then the vector from M(0,1) to (x1, y1) is perpendicular to the vector from A(t, 0) to (x1, y1).So, the dot product of these two vectors is zero.So, (x1 - 0, y1 - 1) · (x1 - t, y1 - 0) = 0Which simplifies to:x1(x1 - t) + (y1 - 1)y1 = 0Also, since (x1, y1) lies on the circle, x1² + (y1 - 1)² = 1So, we have two equations:1. x1² - t x1 + y1² - y1 = 02. x1² + y1² - 2 y1 + 1 = 1 => x1² + y1² - 2 y1 = 0Subtracting equation 2 from equation 1:(x1² - t x1 + y1² - y1) - (x1² + y1² - 2 y1) = 0 - 0Simplify:- t x1 + y1² - y1 - y1² + 2 y1 = 0Simplify further:- t x1 + y1 = 0 => y1 = t x1So, from this, y1 = t x1.Now, substitute y1 = t x1 into equation 2:x1² + (t x1)² - 2 (t x1) = 0Simplify:x1² + t² x1² - 2 t x1 = 0Factor x1:x1 (1 + t²) x1 - 2 t x1 = 0Wait, that's:x1² (1 + t²) - 2 t x1 = 0Factor x1:x1 (x1 (1 + t²) - 2 t) = 0So, either x1 = 0 or x1 (1 + t²) - 2 t = 0If x1 = 0, then from y1 = t x1, y1 = 0. But (0, 0) is not on the circle x² + (y - 1)² = 1, because 0 + (0 - 1)² = 1, which is true. Wait, actually, (0, 0) is on the circle? Let me check:0² + (0 - 1)² = 1, yes, that's correct. So, (0, 0) is on the circle.But wait, if x1 = 0, then the tangent from A(t, 0) would be the line connecting A(t, 0) to (0, 0). But is that a tangent?Wait, (0,0) is on the circle, and the line from A(t, 0) to (0,0) is just the x-axis. But the x-axis is not a tangent to the circle x² + (y - 1)² = 1 because the distance from the center (0,1) to the x-axis is 1, which is equal to the radius. So, the x-axis is actually a tangent to the circle at (0,0).So, that's one tangent line: the x-axis itself. But in our case, point A is (t, 0), which is on the x-axis, so the tangent from A to the circle is the x-axis itself.But wait, if A is on the x-axis, which is a tangent to the circle, then the tangent from A is the x-axis. But then, the other tangent from A would be another line.Wait, but when I solved the equations, I got x1 = 0 or x1 = 2 t / (1 + t²). So, if x1 = 0, we get the x-axis as a tangent, but there's another tangent with x1 = 2 t / (1 + t²).So, let's compute that.x1 = 2 t / (1 + t²)Then, y1 = t x1 = t * (2 t / (1 + t²)) = 2 t² / (1 + t²)So, the other point of tangency is (2 t / (1 + t²), 2 t² / (1 + t²))Therefore, the equation of the tangent line from A(t, 0) to this point can be found using the two-point form.But actually, since we know the point of tangency, we can write the equation of the tangent line.Alternatively, since we know the slope of the tangent line at point (x1, y1) on the circle is perpendicular to the radius.The slope of the radius from M(0,1) to (x1, y1) is (y1 - 1)/x1.Therefore, the slope of the tangent line is the negative reciprocal, which is -x1 / (y1 - 1)So, the slope of the tangent line at (x1, y1) is m = -x1 / (y1 - 1)Therefore, the equation of the tangent line is:y - y1 = m (x - x1)Plugging in m:y - y1 = (-x1 / (y1 - 1)) (x - x1)But since y1 = t x1, we can substitute:y - t x1 = (-x1 / (t x1 - 1)) (x - x1)Simplify denominator:t x1 - 1So,y - t x1 = (-x1 / (t x1 - 1)) (x - x1)Multiply both sides by (t x1 - 1):(y - t x1)(t x1 - 1) = -x1 (x - x1)Let me expand the left side:y (t x1 - 1) - t x1 (t x1 - 1) = -x1 x + x1²So,y t x1 - y - t² x1² + t x1 = -x1 x + x1²Bring all terms to one side:y t x1 - y - t² x1² + t x1 + x1 x - x1² = 0But this seems complicated. Maybe there's a better way.Alternatively, since we have the point of tangency (x1, y1), we can use the equation of the tangent to the circle at that point.For a circle x² + (y - 1)² = 1, the tangent at (x1, y1) is:x x1 + (y - 1)(y1 - 1) = 1Yes, that's the standard equation for the tangent to a circle at a given point.So, the tangent line at (x1, y1) is:x x1 + (y - 1)(y1 - 1) = 1So, plugging in x1 = 2 t / (1 + t²) and y1 = 2 t² / (1 + t²):x * (2 t / (1 + t²)) + (y - 1)( (2 t² / (1 + t²)) - 1 ) = 1Simplify the second term:(2 t² / (1 + t²) - 1) = (2 t² - (1 + t²)) / (1 + t²) = (t² - 1) / (1 + t²)So, the equation becomes:(2 t / (1 + t²)) x + (y - 1)( (t² - 1) / (1 + t²) ) = 1Multiply both sides by (1 + t²) to eliminate denominators:2 t x + (y - 1)(t² - 1) = 1 + t²Expand the second term:2 t x + (y - 1)(t² - 1) = 1 + t²So,2 t x + y (t² - 1) - (t² - 1) = 1 + t²Bring all terms to one side:2 t x + y (t² - 1) - (t² - 1) - 1 - t² = 0Simplify:2 t x + y (t² - 1) - t² + 1 - 1 - t² = 0Combine like terms:2 t x + y (t² - 1) - 2 t² = 0So,2 t x + (t² - 1) y - 2 t² = 0This is the equation of the tangent from A(t, 0) to the circle.Similarly, we can find the equation of the tangent from B(t + 5, 0) to the circle.Let me denote point B as (s, 0) where s = t + 5. Then, following similar steps as above, the equation of the tangent from B(s, 0) to the circle will be:2 s x + (s² - 1) y - 2 s² = 0So, substituting s = t + 5:2 (t + 5) x + ((t + 5)² - 1) y - 2 (t + 5)² = 0Simplify:2 (t + 5) x + (t² + 10 t + 25 - 1) y - 2 (t² + 10 t + 25) = 0Which simplifies to:2 (t + 5) x + (t² + 10 t + 24) y - 2 t² - 20 t - 50 = 0Now, we have two equations:1. 2 t x + (t² - 1) y - 2 t² = 02. 2 (t + 5) x + (t² + 10 t + 24) y - 2 t² - 20 t - 50 = 0We need to solve this system of equations to find the coordinates (x, y) of point C.Let me write them again:Equation (1): 2 t x + (t² - 1) y = 2 t²Equation (2): 2 (t + 5) x + (t² + 10 t + 24) y = 2 t² + 20 t + 50Let me denote Equation (1) as:A1 x + B1 y = C1Where A1 = 2 t, B1 = t² - 1, C1 = 2 t²Similarly, Equation (2):A2 x + B2 y = C2Where A2 = 2 (t + 5), B2 = t² + 10 t + 24, C2 = 2 t² + 20 t + 50We can solve this system using substitution or elimination. Let's use elimination.First, let's make the coefficients of x the same in both equations.Multiply Equation (1) by A2 / A1:(A2 / A1) * Equation (1):(2 (t + 5) / 2 t) * (2 t x + (t² - 1) y) = (2 (t + 5) / 2 t) * 2 t²Simplify:(t + 5)/t * (2 t x + (t² - 1) y) = (t + 5)/t * 2 t²Which simplifies to:(t + 5) (2 t x + (t² - 1) y) / t = 2 t (t + 5)But this seems messy. Maybe instead, let's solve for x from Equation (1) and substitute into Equation (2).From Equation (1):2 t x = 2 t² - (t² - 1) ySo,x = [2 t² - (t² - 1) y] / (2 t)Now, substitute this into Equation (2):2 (t + 5) * [ (2 t² - (t² - 1) y) / (2 t) ] + (t² + 10 t + 24) y = 2 t² + 20 t + 50Simplify term by term:First term:2 (t + 5) * [ (2 t² - (t² - 1) y) / (2 t) ] = (t + 5) * [ (2 t² - (t² - 1) y) / t ]= (t + 5) [ 2 t² / t - (t² - 1) y / t ]= (t + 5) [ 2 t - (t² - 1) y / t ]= (t + 5) * 2 t - (t + 5) * (t² - 1) y / t= 2 t (t + 5) - [ (t + 5)(t² - 1) / t ] ySecond term:(t² + 10 t + 24) ySo, putting it all together:2 t (t + 5) - [ (t + 5)(t² - 1) / t ] y + (t² + 10 t + 24) y = 2 t² + 20 t + 50Simplify the left side:2 t (t + 5) + [ - (t + 5)(t² - 1) / t + (t² + 10 t + 24) ] y = 2 t² + 20 t + 50Let me compute the coefficient of y:- (t + 5)(t² - 1) / t + (t² + 10 t + 24)Let me write both terms with denominator t:= [ - (t + 5)(t² - 1) + t (t² + 10 t + 24) ] / tExpand numerator:- (t + 5)(t² - 1) + t (t² + 10 t + 24)First term:- (t * t² + t * (-1) + 5 * t² + 5 * (-1)) = - (t³ - t + 5 t² - 5) = -t³ + t - 5 t² + 5Second term:t (t² + 10 t + 24) = t³ + 10 t² + 24 tCombine both terms:(-t³ + t - 5 t² + 5) + (t³ + 10 t² + 24 t) =(-t³ + t³) + (-5 t² + 10 t²) + (t + 24 t) + 5 =0 + 5 t² + 25 t + 5So, the coefficient of y is (5 t² + 25 t + 5) / tTherefore, the equation becomes:2 t (t + 5) + [ (5 t² + 25 t + 5) / t ] y = 2 t² + 20 t + 50Compute 2 t (t + 5):2 t² + 10 tSo, the equation is:2 t² + 10 t + (5 t² + 25 t + 5)/t * y = 2 t² + 20 t + 50Subtract 2 t² + 10 t from both sides:(5 t² + 25 t + 5)/t * y = 10 t + 50Factor numerator:5 t² + 25 t + 5 = 5(t² + 5 t + 1)So,[5(t² + 5 t + 1)/t] y = 10 t + 50Divide both sides by 5:(t² + 5 t + 1)/t * y = 2 t + 10Multiply both sides by t:(t² + 5 t + 1) y = 2 t² + 10 tTherefore,y = (2 t² + 10 t) / (t² + 5 t + 1)Simplify numerator:2 t (t + 5)Denominator:t² + 5 t + 1So,y = [2 t (t + 5)] / (t² + 5 t + 1)Now, let's find x from Equation (1):x = [2 t² - (t² - 1) y] / (2 t)Substitute y:x = [2 t² - (t² - 1) * (2 t (t + 5))/(t² + 5 t + 1)] / (2 t)Let me compute the numerator:2 t² - (t² - 1)(2 t (t + 5))/(t² + 5 t + 1)Factor out 2 t:= 2 t [ t - (t² - 1)(t + 5)/(t² + 5 t + 1) ]Let me compute the term inside the brackets:t - (t² - 1)(t + 5)/(t² + 5 t + 1)Let me write t as t (t² + 5 t + 1)/(t² + 5 t + 1) to have a common denominator:= [ t (t² + 5 t + 1) - (t² - 1)(t + 5) ] / (t² + 5 t + 1)Expand numerator:t (t² + 5 t + 1) = t³ + 5 t² + t(t² - 1)(t + 5) = t³ + 5 t² - t - 5Subtract:(t³ + 5 t² + t) - (t³ + 5 t² - t - 5) = t³ + 5 t² + t - t³ - 5 t² + t + 5 = 2 t + 5So, the term inside the brackets is (2 t + 5)/(t² + 5 t + 1)Therefore, numerator of x is 2 t * (2 t + 5)/(t² + 5 t + 1)So,x = [2 t (2 t + 5) / (t² + 5 t + 1)] / (2 t) = (2 t + 5)/(t² + 5 t + 1)Therefore, the coordinates of point C are:x = (2 t + 5)/(t² + 5 t + 1)y = [2 t (t + 5)] / (t² + 5 t + 1)So, point C is ((2 t + 5)/(t² + 5 t + 1), [2 t (t + 5)] / (t² + 5 t + 1))Now, as I thought earlier, the area of triangle ABC is (1/2)*base*height, where base AB is 5, and height is the y-coordinate of point C.Therefore, area = (1/2)*5*y = (5/2)*ySo, area = (5/2)*[2 t (t + 5)] / (t² + 5 t + 1) = (5/2)*(2 t (t + 5))/(t² + 5 t + 1) = 5 t (t + 5)/(t² + 5 t + 1)Simplify:Area = 5 t (t + 5)/(t² + 5 t + 1)We need to find the minimum value of this area for t ∈ [-4, -1]So, let me denote:A(t) = 5 t (t + 5)/(t² + 5 t + 1)We need to find the minimum of A(t) over t ∈ [-4, -1]First, let's analyze the function A(t).Note that t is between -4 and -1, so t is negative, t + 5 is between 1 and 4, so positive.Denominator: t² + 5 t + 1Let me compute the denominator for t ∈ [-4, -1]:At t = -4: (-4)^2 + 5*(-4) + 1 = 16 - 20 + 1 = -3At t = -1: (-1)^2 + 5*(-1) + 1 = 1 - 5 + 1 = -3Wait, so denominator is negative throughout the interval?Wait, let me compute the discriminant of the denominator:t² + 5 t + 1Discriminant D = 25 - 4 = 21 > 0, so it has two real roots.Roots are at t = [-5 ± √21]/2Approximately, √21 ≈ 4.5837So, roots are at (-5 + 4.5837)/2 ≈ (-0.4163)/2 ≈ -0.208And (-5 - 4.5837)/2 ≈ (-9.5837)/2 ≈ -4.7918So, the denominator t² + 5 t + 1 is positive outside the interval (-4.7918, -0.208) and negative inside.Since our t is in [-4, -1], which is inside (-4.7918, -0.208), so denominator is negative.Therefore, denominator is negative, numerator:5 t (t + 5)t is negative, t + 5 is positive, so numerator is negative.So, A(t) = negative / negative = positive.So, A(t) is positive in this interval.Now, to find the minimum of A(t), we can take derivative and find critical points.Let me compute A(t) = 5 t (t + 5)/(t² + 5 t + 1)Let me denote f(t) = t (t + 5) = t² + 5 tg(t) = t² + 5 t + 1So, A(t) = 5 f(t)/g(t)Compute derivative A’(t):A’(t) = 5 [f’(t) g(t) - f(t) g’(t)] / [g(t)]²Compute f’(t) = 2 t + 5g’(t) = 2 t + 5So,A’(t) = 5 [ (2 t + 5)(t² + 5 t + 1) - (t² + 5 t)(2 t + 5) ] / (t² + 5 t + 1)^2Simplify numerator:(2 t + 5)(t² + 5 t + 1) - (t² + 5 t)(2 t + 5)Factor out (2 t + 5):= (2 t + 5)[(t² + 5 t + 1) - (t² + 5 t)]= (2 t + 5)(1)= 2 t + 5Therefore,A’(t) = 5 (2 t + 5) / (t² + 5 t + 1)^2Set derivative equal to zero:5 (2 t + 5) / (t² + 5 t + 1)^2 = 0The denominator is always positive (since squared), so numerator must be zero:2 t + 5 = 0 => t = -5/2 = -2.5So, critical point at t = -2.5, which is within our interval [-4, -1].Therefore, we need to evaluate A(t) at t = -4, t = -2.5, and t = -1.Compute A(-4):A(-4) = 5*(-4)*(-4 + 5)/[(-4)^2 + 5*(-4) + 1] = 5*(-4)*(1)/(16 - 20 + 1) = 5*(-4)/(-3) = 20/3 ≈ 6.6667Compute A(-2.5):First, t = -2.5Compute numerator: 5*(-2.5)*(-2.5 + 5) = 5*(-2.5)*(2.5) = 5*(-6.25) = -31.25Denominator: (-2.5)^2 + 5*(-2.5) + 1 = 6.25 - 12.5 + 1 = -5.25So, A(-2.5) = (-31.25)/(-5.25) = 31.25 / 5.25 ≈ 5.9524But let me compute it exactly:Numerator: 5*(-5/2)*( -5/2 + 5 ) = 5*(-5/2)*(5/2) = 5*(-25/4) = -125/4Denominator: (25/4) + 5*(-5/2) + 1 = 25/4 - 25/2 + 1 = 25/4 - 50/4 + 4/4 = (-21/4)So,A(-2.5) = (-125/4) / (-21/4) = (125/4) / (21/4) = 125/21 ≈ 5.9524Compute A(-1):A(-1) = 5*(-1)*(-1 + 5)/[(-1)^2 + 5*(-1) + 1] = 5*(-1)*(4)/(1 - 5 + 1) = 5*(-4)/(-3) = 20/3 ≈ 6.6667So, A(-4) = A(-1) = 20/3 ≈ 6.6667A(-2.5) = 125/21 ≈ 5.9524Therefore, the minimum value of A(t) is 125/21, achieved at t = -2.5.Therefore, the minimum area is 125/21.But let me verify if this is indeed the minimum.Since A’(t) = 5(2 t + 5)/(denominator)^2For t < -2.5, say t = -3:A’(-3) = 5(2*(-3) + 5)/(denominator)^2 = 5*(-6 + 5)/(denominator)^2 = 5*(-1)/(positive) = negativeSo, function is decreasing before t = -2.5For t > -2.5, say t = -2:A’(-2) = 5(2*(-2) + 5)/(denominator)^2 = 5*(-4 + 5)/(denominator)^2 = 5*(1)/(positive) = positiveSo, function is increasing after t = -2.5Therefore, t = -2.5 is indeed a local minimum, and since it's the only critical point in the interval, it's the global minimum.Therefore, the minimum area is 125/21.**Final Answer**(1) The equation of circle ( M ) is (boxed{x^2 + (y - 1)^2 = 1}).(2) The minimum value of the area of (triangle ABC) is (boxed{dfrac{125}{21}}).

💡First, I need to determine how many green balloons Fred gave to Sandy. Fred initially had 709 green balloons and now has 488 green balloons.To find the number of balloons given to Sandy, I can subtract the number of balloons Fred has now from the original number of balloons he had.So, I subtract 488 from 709.709 minus 488 equals 221.Therefore, Fred gave 221 green balloons to Sandy.

💡Okay, so I have this problem where ABCD is a rectangle, and points E and F are on sides BC and CD respectively. Triangle AEF is equilateral, and I need to show that the area of triangle CEF is equal to the sum of the areas of triangles ABE and AFD. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, let me visualize the rectangle ABCD. Let me label the rectangle with A at the bottom-left corner, B at the bottom-right, C at the top-right, and D at the top-left. So, AB is the bottom side, BC is the right side, CD is the top side, and DA is the left side. Now, E is somewhere on BC and F is somewhere on CD. Triangle AEF is equilateral, meaning all its sides are equal and all its angles are 60 degrees.I think it might help to assign some coordinates to the rectangle to make things more concrete. Let me place point A at the origin (0,0). Then, if the rectangle has length 'a' and width 'b', point B would be at (a,0), point C at (a,b), and point D at (0,b). Now, point E is on BC, so its coordinates would be (a, y) where y is between 0 and b. Similarly, point F is on CD, so its coordinates would be (x, b) where x is between 0 and a.Since triangle AEF is equilateral, the distances AE, EF, and FA must all be equal. Let me write down the coordinates:- A is (0,0)- E is (a, y)- F is (x, b)So, the distance AE is the distance from A to E, which is sqrt[(a - 0)^2 + (y - 0)^2] = sqrt(a^2 + y^2).Similarly, the distance AF is the distance from A to F, which is sqrt[(x - 0)^2 + (b - 0)^2] = sqrt(x^2 + b^2).The distance EF is the distance from E to F, which is sqrt[(x - a)^2 + (b - y)^2].Since triangle AEF is equilateral, all these distances must be equal:sqrt(a^2 + y^2) = sqrt(x^2 + b^2) = sqrt[(x - a)^2 + (b - y)^2]Hmm, that gives me two equations:1. sqrt(a^2 + y^2) = sqrt(x^2 + b^2)2. sqrt(a^2 + y^2) = sqrt[(x - a)^2 + (b - y)^2]Let me square both sides of the first equation to eliminate the square roots:a^2 + y^2 = x^2 + b^2So, rearranged:x^2 - y^2 = a^2 - b^2That's one equation.Now, let's square the second equation:a^2 + y^2 = (x - a)^2 + (b - y)^2Expanding the right-hand side:(x - a)^2 = x^2 - 2ax + a^2(b - y)^2 = b^2 - 2by + y^2So, adding them together:x^2 - 2ax + a^2 + b^2 - 2by + y^2So, the equation becomes:a^2 + y^2 = x^2 - 2ax + a^2 + b^2 - 2by + y^2Simplify both sides:Left side: a^2 + y^2Right side: x^2 - 2ax + a^2 + b^2 - 2by + y^2Subtract left side from both sides:0 = x^2 - 2ax + b^2 - 2bySo,x^2 - 2ax + b^2 - 2by = 0But from the first equation, we have x^2 - y^2 = a^2 - b^2, so x^2 = y^2 + a^2 - b^2Let me substitute x^2 from the first equation into the second equation:(y^2 + a^2 - b^2) - 2ax + b^2 - 2by = 0Simplify:y^2 + a^2 - b^2 - 2ax + b^2 - 2by = 0The -b^2 and +b^2 cancel out:y^2 + a^2 - 2ax - 2by = 0Hmm, so now I have:y^2 + a^2 - 2ax - 2by = 0I wonder if I can express x in terms of y or vice versa.From the first equation, x^2 = y^2 + a^2 - b^2So, x = sqrt(y^2 + a^2 - b^2)But that might complicate things. Alternatively, maybe I can express a in terms of x and y, but that might not be helpful.Alternatively, perhaps I can use some trigonometric relationships since triangle AEF is equilateral, which has angles of 60 degrees.Let me think about the angles. Since AEF is equilateral, angle at A is 60 degrees. So, the angle between vectors AE and AF is 60 degrees.Wait, vectors. Maybe I can use vectors to find some relationships.Vector AE is from A(0,0) to E(a,y), so vector AE is (a, y).Vector AF is from A(0,0) to F(x,b), so vector AF is (x, b).The angle between vectors AE and AF is 60 degrees, so the dot product formula can be used:AE · AF = |AE| |AF| cos(theta)Where theta is 60 degrees.So,(a)(x) + (y)(b) = |AE| |AF| cos(60°)But since triangle AEF is equilateral, |AE| = |AF|, so let's denote |AE| = |AF| = d.So,ax + by = d * d * (1/2) = (d^2)/2But from earlier, d^2 = a^2 + y^2 = x^2 + b^2So,ax + by = (a^2 + y^2)/2So,2ax + 2by = a^2 + y^2Which is the same as the equation I had earlier: y^2 + a^2 - 2ax - 2by = 0So, that's consistent.Hmm, maybe another approach. Since ABCD is a rectangle, opposite sides are equal and all angles are 90 degrees. Maybe I can use coordinate geometry to find the areas.The area of triangle ABE: points A(0,0), B(a,0), E(a,y). So, it's a right triangle with base AB = a and height BE = y. So, area is (1/2)*a*y.Similarly, the area of triangle AFD: points A(0,0), F(x,b), D(0,b). So, it's a right triangle with base AD = b and height AF_x = x. Wait, actually, AF is from A(0,0) to F(x,b), so the triangle AFD is formed by points A(0,0), F(x,b), D(0,b). Hmm, to find its area, maybe using coordinates.Using the formula for the area of a triangle given coordinates:Area = (1/2)| (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) |For triangle AFD: A(0,0), F(x,b), D(0,b)Plugging in:Area = (1/2)| 0*(b - b) + x*(b - 0) + 0*(0 - b) | = (1/2)|0 + xb + 0| = (1/2)|xb| = (1/2)xbSo, area of AFD is (1/2)xb.Similarly, area of ABE is (1/2)ay.Now, area of triangle CEF: points C(a,b), E(a,y), F(x,b). Let me compute this area.Using the same area formula:Area = (1/2)| x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) |Plugging in C(a,b), E(a,y), F(x,b):Area = (1/2)| a*(y - b) + a*(b - b) + x*(b - y) |Simplify:= (1/2)| a(y - b) + a(0) + x(b - y) |= (1/2)| a(y - b) + x(b - y) |Factor out (y - b):= (1/2)| (y - b)(a - x) |Since area is positive, we can drop the absolute value:= (1/2)| (b - y)(a - x) | because (y - b) = -(b - y)So, Area of CEF = (1/2)(b - y)(a - x)Now, according to the problem, we need to show that Area of CEF = Area of ABE + Area of AFD.So,(1/2)(b - y)(a - x) = (1/2)ay + (1/2)xbMultiply both sides by 2:(b - y)(a - x) = ay + xbExpand the left side:ab - bx - ay + xy = ay + xbBring all terms to one side:ab - bx - ay + xy - ay - xb = 0Combine like terms:ab - 2bx - 2ay + xy = 0Hmm, so ab - 2bx - 2ay + xy = 0Is this equation true? Let's see if we can derive it from previous relationships.Earlier, we had:From the first equation: x^2 - y^2 = a^2 - b^2From the second equation: y^2 + a^2 - 2ax - 2by = 0Wait, let me write down the second equation again:y^2 + a^2 - 2ax - 2by = 0From here, maybe I can express something in terms of other variables.Let me rearrange the second equation:y^2 = 2ax + 2by - a^2Now, plug this into the first equation:x^2 - (2ax + 2by - a^2) = a^2 - b^2Simplify:x^2 - 2ax - 2by + a^2 = a^2 - b^2Subtract a^2 from both sides:x^2 - 2ax - 2by = -b^2So,x^2 - 2ax - 2by + b^2 = 0Hmm, this is similar to the equation we had earlier when trying to equate the areas.Wait, from the area equation, we had:ab - 2bx - 2ay + xy = 0So, let me write both equations:1. x^2 - 2ax - 2by + b^2 = 02. ab - 2bx - 2ay + xy = 0Hmm, perhaps I can solve these two equations together.Let me denote equation 1 as:x^2 - 2ax - 2by + b^2 = 0 --> equation (1)And equation 2 as:ab - 2bx - 2ay + xy = 0 --> equation (2)Let me try to express y from equation (2) in terms of x and substitute into equation (1).From equation (2):ab - 2bx - 2ay + xy = 0Let me collect terms with y:(-2a + x)y + ab - 2bx = 0So,y(-2a + x) = 2bx - abThus,y = (2bx - ab)/(-2a + x) = (ab - 2bx)/(2a - x)So, y = (ab - 2bx)/(2a - x)Now, plug this into equation (1):x^2 - 2ax - 2b*( (ab - 2bx)/(2a - x) ) + b^2 = 0This looks complicated, but let me try to simplify step by step.First, compute the term with y:-2b*( (ab - 2bx)/(2a - x) ) = (-2b)(ab - 2bx)/(2a - x) = (-2ab + 4b^2x)/(2a - x)So, equation (1) becomes:x^2 - 2ax + (-2ab + 4b^2x)/(2a - x) + b^2 = 0Let me write all terms over the denominator (2a - x):[ x^2(2a - x) - 2ax(2a - x) - 2ab + 4b^2x + b^2(2a - x) ] / (2a - x) = 0So, numerator must be zero:x^2(2a - x) - 2ax(2a - x) - 2ab + 4b^2x + b^2(2a - x) = 0Let me expand each term:1. x^2(2a - x) = 2a x^2 - x^32. -2ax(2a - x) = -4a^2x + 2a x^23. -2ab4. 4b^2x5. b^2(2a - x) = 2a b^2 - b^2xNow, combine all these:2a x^2 - x^3 - 4a^2x + 2a x^2 - 2ab + 4b^2x + 2a b^2 - b^2x = 0Combine like terms:- x^3+ (2a x^2 + 2a x^2) = 4a x^2- 4a^2x+ (4b^2x - b^2x) = 3b^2x- 2ab + 2a b^2So, the equation becomes:- x^3 + 4a x^2 - 4a^2x + 3b^2x - 2ab + 2a b^2 = 0Hmm, this is a cubic equation in x. It might be difficult to solve directly, but perhaps we can factor it.Let me rearrange the terms:- x^3 + 4a x^2 - 4a^2x + 3b^2x - 2ab + 2a b^2 = 0Let me group terms:(-x^3 + 4a x^2) + (-4a^2x + 3b^2x) + (-2ab + 2a b^2) = 0Factor out common terms:- x^2(x - 4a) - x(4a^2 - 3b^2) - 2ab(1 - b) = 0Hmm, not sure if that helps. Maybe another approach.Alternatively, perhaps I can assume specific values for a and b to simplify the problem. Since the problem is general, but maybe choosing a = b would make it a square, which might simplify things.Let me try setting a = b. So, the rectangle becomes a square with side length a.So, a = b.Then, from the first equation:x^2 - y^2 = a^2 - a^2 = 0 => x^2 = y^2 => x = y or x = -yBut since x and y are coordinates on the sides, they must be positive, so x = y.From the second equation:y^2 + a^2 - 2ax - 2by = 0But since a = b and x = y,y^2 + a^2 - 2a y - 2a y = 0Simplify:y^2 + a^2 - 4a y = 0Which is:y^2 - 4a y + a^2 = 0This is a quadratic in y:y = [4a ± sqrt(16a^2 - 4a^2)] / 2 = [4a ± sqrt(12a^2)] / 2 = [4a ± 2a sqrt(3)] / 2 = 2a ± a sqrt(3)Since y must be less than a (because E is on BC which has length a), we take the smaller root:y = 2a - a sqrt(3) ≈ 2a - 1.732a = 0.268aSo, y = a(2 - sqrt(3)) ≈ 0.268aSimilarly, x = y = a(2 - sqrt(3))Now, let's compute the areas.Area of ABE: (1/2) a y = (1/2) a * a(2 - sqrt(3)) = (1/2) a^2 (2 - sqrt(3)) = a^2 (1 - (sqrt(3)/2))Area of AFD: (1/2) x b = (1/2) x a = (1/2) a(2 - sqrt(3)) * a = (1/2) a^2 (2 - sqrt(3)) = same as ABE, so also a^2 (1 - (sqrt(3)/2))So, total area of ABE + AFD = 2 * a^2 (1 - (sqrt(3)/2)) = 2a^2 - a^2 sqrt(3)Now, area of CEF: (1/2)(b - y)(a - x) = (1/2)(a - y)(a - x)But since x = y,= (1/2)(a - y)^2= (1/2)(a - a(2 - sqrt(3)))^2= (1/2)(a(1 - (2 - sqrt(3))))^2= (1/2)(a(-1 + sqrt(3)))^2= (1/2)(a^2 (1 - 2 sqrt(3) + 3))= (1/2)(a^2 (4 - 2 sqrt(3)))= (1/2)(4a^2 - 2a^2 sqrt(3))= 2a^2 - a^2 sqrt(3)Which is equal to the sum of areas of ABE and AFD.So, in the case where a = b, the statement holds true.But the problem is for a general rectangle, not necessarily a square. So, does this relationship hold for any rectangle?Alternatively, maybe I can use similarity of triangles or some other geometric properties.Let me think about the angles. Since triangle AEF is equilateral, angle at A is 60 degrees. So, angle EAF is 60 degrees.In rectangle ABCD, angle at A is 90 degrees. So, the remaining angle between AF and AD is 30 degrees, because 90 - 60 = 30.Similarly, the angle between AE and AB would also be related.Wait, maybe using trigonometry.Let me denote angle BAE as theta, so angle between BA and AE is theta. Then, since angle EAF is 60 degrees, angle FAD would be 90 - theta - 60 = 30 - theta.Wait, maybe not. Let me think again.In rectangle ABCD, angle at A is 90 degrees. Triangle AEF is equilateral, so angle EAF is 60 degrees. Therefore, the remaining angle between AF and AD is 90 - 60 = 30 degrees. So, angle FAD is 30 degrees.Similarly, angle between AE and AB is also 30 degrees because the total angle at A is 90 degrees, and 90 - 60 = 30.Wait, is that correct? Let me see.If angle EAF is 60 degrees, then the angle between AE and AF is 60 degrees. But in the rectangle, the angle between AB and AD is 90 degrees. So, if I consider the angles from AB to AE and from AF to AD, their sum should be 90 - 60 = 30 degrees.Wait, maybe it's better to use coordinates and vectors again.Let me denote vector AE as (a, y) and vector AF as (x, b). The angle between them is 60 degrees, so the dot product is |AE||AF|cos(60°).We already used this earlier, but perhaps we can find a relationship between x and y.From earlier, we have:ax + by = (1/2)(a^2 + y^2)And also:x^2 - y^2 = a^2 - b^2Hmm, maybe I can solve for one variable in terms of the other.From x^2 - y^2 = a^2 - b^2, we can write x^2 = y^2 + a^2 - b^2Let me plug this into the equation ax + by = (1/2)(a^2 + y^2)So,ax + by = (1/2)(a^2 + y^2)But x = sqrt(y^2 + a^2 - b^2). Hmm, but that might complicate things.Alternatively, let me express a in terms of x and y.From x^2 - y^2 = a^2 - b^2, we have a^2 = x^2 - y^2 + b^2Plug this into ax + by = (1/2)(a^2 + y^2):ax + by = (1/2)(x^2 - y^2 + b^2 + y^2) = (1/2)(x^2 + b^2)So,ax + by = (1/2)(x^2 + b^2)Multiply both sides by 2:2ax + 2by = x^2 + b^2Rearrange:x^2 - 2ax + b^2 - 2by = 0Wait, this is the same equation we had earlier. So, we're back to where we started.Hmm, maybe I need to find another relationship.Alternatively, perhaps using the areas.We have Area of CEF = (1/2)(b - y)(a - x)And Area of ABE + AFD = (1/2)ay + (1/2)xbSo, we need to show:(1/2)(b - y)(a - x) = (1/2)ay + (1/2)xbMultiply both sides by 2:(b - y)(a - x) = ay + xbExpand left side:ab - bx - ay + xy = ay + xbBring all terms to left:ab - bx - ay + xy - ay - xb = 0Simplify:ab - 2bx - 2ay + xy = 0So, same equation as before.So, we need to show that ab - 2bx - 2ay + xy = 0But from earlier, we have:From the first equation: x^2 - y^2 = a^2 - b^2From the second equation: y^2 + a^2 - 2ax - 2by = 0So, let me try to manipulate these two equations to get ab - 2bx - 2ay + xy = 0From equation (1): x^2 - y^2 = a^2 - b^2From equation (2): y^2 + a^2 - 2ax - 2by = 0Let me add equations (1) and (2):x^2 - y^2 + y^2 + a^2 - 2ax - 2by = a^2 - b^2 + 0Simplify:x^2 + a^2 - 2ax - 2by = a^2 - b^2Subtract a^2 from both sides:x^2 - 2ax - 2by = -b^2So,x^2 - 2ax - 2by + b^2 = 0Hmm, that's similar to the area equation, but not exactly the same.Wait, the area equation is ab - 2bx - 2ay + xy = 0Let me see if I can relate x^2 - 2ax - 2by + b^2 = 0 to ab - 2bx - 2ay + xy = 0Let me denote equation (3): x^2 - 2ax - 2by + b^2 = 0And equation (4): ab - 2bx - 2ay + xy = 0Let me try to express equation (4) in terms of equation (3).From equation (3):x^2 = 2ax + 2by - b^2Plug this into equation (4):ab - 2bx - 2ay + (2ax + 2by - b^2)y = 0Wait, no, equation (4) is ab - 2bx - 2ay + xy = 0So, if I can express x in terms of y or vice versa, maybe I can substitute.Alternatively, perhaps I can solve for one variable.From equation (3):x^2 - 2ax - 2by + b^2 = 0Let me write this as:x^2 - 2ax = 2by - b^2From equation (4):ab - 2bx - 2ay + xy = 0Let me solve for ab:ab = 2bx + 2ay - xyNow, from equation (3):x^2 - 2ax = 2by - b^2Let me express b from equation (4):From ab = 2bx + 2ay - xy, solve for b:b(a - 2x) = 2ay - xySo,b = (2ay - xy)/(a - 2x)Now, plug this into equation (3):x^2 - 2ax = 2*( (2ay - xy)/(a - 2x) )*y - ( (2ay - xy)/(a - 2x) )^2This is getting very complicated. Maybe there's a better approach.Alternatively, perhaps using complex numbers or rotation.Since triangle AEF is equilateral, point F can be obtained by rotating point E around point A by 60 degrees.In complex plane terms, if we represent points as complex numbers, then F = A + (E - A) * e^{iπ/3}But since A is at (0,0), this simplifies to F = E * e^{iπ/3}Let me denote E as (a, y), so in complex number, E = a + yiThen, F = (a + yi) * (cos60° + i sin60°) = (a + yi)*(1/2 + i√3/2)Multiply this out:= a*(1/2) + a*(i√3/2) + yi*(1/2) + yi*(i√3/2)= (a/2) + (a√3/2)i + (y/2)i + (y√3/2)i^2Since i^2 = -1,= (a/2) + (a√3/2 + y/2)i - y√3/2So, F = (a/2 - y√3/2) + (a√3/2 + y/2)iWhich gives coordinates:x = a/2 - y√3/2b = a√3/2 + y/2Wait, but in our rectangle, b is the height, which is fixed. So, from this, we can solve for y in terms of a and b.From the imaginary part:b = (a√3)/2 + y/2So,y/2 = b - (a√3)/2Thus,y = 2b - a√3Similarly, from the real part:x = a/2 - (y√3)/2Plug y = 2b - a√3 into this:x = a/2 - ( (2b - a√3)√3 ) / 2Simplify:= a/2 - (2b√3 - a*3)/2= a/2 - (2b√3)/2 + (3a)/2= (a/2 + 3a/2) - b√3= 2a - b√3So, x = 2a - b√3Now, let's check if these satisfy the earlier equations.From equation (1): x^2 - y^2 = a^2 - b^2Compute x^2 - y^2:x = 2a - b√3, so x^2 = (2a)^2 - 2*(2a)*(b√3) + (b√3)^2 = 4a^2 - 4ab√3 + 3b^2y = 2b - a√3, so y^2 = (2b)^2 - 2*(2b)*(a√3) + (a√3)^2 = 4b^2 - 4ab√3 + 3a^2Thus, x^2 - y^2 = (4a^2 - 4ab√3 + 3b^2) - (4b^2 - 4ab√3 + 3a^2) = 4a^2 - 4ab√3 + 3b^2 - 4b^2 + 4ab√3 - 3a^2 = (4a^2 - 3a^2) + (-4ab√3 + 4ab√3) + (3b^2 - 4b^2) = a^2 - b^2Which matches equation (1). Good.Now, let's compute the areas.Area of ABE: (1/2) a y = (1/2) a (2b - a√3) = ab - (a^2√3)/2Area of AFD: (1/2) x b = (1/2) (2a - b√3) b = ab - (b^2√3)/2Sum of areas: (ab - (a^2√3)/2) + (ab - (b^2√3)/2) = 2ab - (a^2√3 + b^2√3)/2 = 2ab - (√3/2)(a^2 + b^2)Now, area of CEF: (1/2)(b - y)(a - x)Compute b - y = b - (2b - a√3) = b - 2b + a√3 = -b + a√3Compute a - x = a - (2a - b√3) = a - 2a + b√3 = -a + b√3Thus, area of CEF = (1/2)(-b + a√3)(-a + b√3)Multiply the terms:(-b + a√3)(-a + b√3) = (-b)(-a) + (-b)(b√3) + (a√3)(-a) + (a√3)(b√3)= ab - b^2√3 - a^2√3 + ab*3= ab - b^2√3 - a^2√3 + 3ab= 4ab - √3(a^2 + b^2)Thus, area of CEF = (1/2)(4ab - √3(a^2 + b^2)) = 2ab - (√3/2)(a^2 + b^2)Which is equal to the sum of areas of ABE and AFD.Therefore, the area of triangle CEF is equal to the sum of the areas of triangles ABE and AFD.**Final Answer**boxed{text{The area of triangle CEF equals the sum of the areas of triangles ABE and AFD.}}