💡Alright, so I have this problem here about finding the domain of a function. Let me try to understand what it's asking. The function given is y equals f of (2x plus 1), and its domain is [-1, 0]. I need to find the domain of another function, which is f(x plus 1) divided by (2 to the power of x minus 1). The options are A: [-2, 0], B: [0, 2], C: (0, 2], and D: [-2, 0).Okay, first, I need to recall what the domain of a function means. The domain is all the possible input values (x-values) for which the function is defined. So, for the function y = f(2x + 1), the domain is given as [-1, 0]. That means x can be any value from -1 to 0, inclusive.Now, I need to figure out the domain of f(x). Since f(2x + 1) has a domain of [-1, 0], I can find the corresponding inputs for f by plugging in the endpoints. Let me do that.If x is -1, then 2x + 1 is 2*(-1) + 1 = -2 + 1 = -1.If x is 0, then 2x + 1 is 2*0 + 1 = 0 + 1 = 1.So, the input to f, which is 2x + 1, ranges from -1 to 1. That means the domain of f(x) is [-1, 1]. Because f is defined for all values from -1 to 1.Okay, so f(x) has a domain of [-1, 1]. Now, I need to find the domain of f(x + 1). This is another function where the input to f is (x + 1). So, for f(x + 1) to be defined, the input (x + 1) must lie within the domain of f, which is [-1, 1].So, I can set up the inequality:-1 ≤ x + 1 ≤ 1Now, let's solve for x.Subtracting 1 from all parts:-1 - 1 ≤ x + 1 - 1 ≤ 1 - 1-2 ≤ x ≤ 0So, the domain of f(x + 1) is [-2, 0]. That means x can be any value from -2 to 0, inclusive.But wait, the function we're interested in is f(x + 1) divided by (2^x - 1). So, we also need to consider the denominator here. The denominator is 2^x - 1. For the function to be defined, the denominator cannot be zero because division by zero is undefined.So, let's find when 2^x - 1 equals zero:2^x - 1 = 02^x = 1We know that 2^0 = 1, so x = 0 is the solution.Therefore, x cannot be 0 because that would make the denominator zero, and the function undefined.So, combining this with the domain we found for f(x + 1), which is [-2, 0], we need to exclude x = 0.Therefore, the domain of the function f(x + 1)/(2^x - 1) is all real numbers from -2 to 0, excluding 0. In interval notation, that's [-2, 0).Looking back at the options, option D is [-2, 0), which matches our result.Wait, just to double-check, let me go through the steps again.1. Given y = f(2x + 1) has a domain of [-1, 0]. So, x is between -1 and 0.2. To find the domain of f(x), we plug in the endpoints: - When x = -1, 2x + 1 = -1 - When x = 0, 2x + 1 = 1 So, f(x) is defined for x between -1 and 1.3. Now, for f(x + 1), the input is x + 1. So, x + 1 must be between -1 and 1: -1 ≤ x + 1 ≤ 1 Subtracting 1: -2 ≤ x ≤ 04. Then, considering the denominator 2^x - 1 ≠ 0: 2^x - 1 = 0 => x = 0 So, x cannot be 0.5. Therefore, combining these, the domain is [-2, 0), which is option D.Yes, that seems correct. I think I've covered all the steps and checked for any mistakes. So, I'm confident that the answer is D: [-2, 0).**Final Answer**boxed{D}

💡Okay, so I have this inequality: -1 < ax² + bx + c < 1, and the solution set is given as (-1, 3). I need to find the range of the real number a. Hmm, okay, let me try to figure this out step by step.First, I know that the solution set (-1, 3) means that for all x between -1 and 3, the quadratic expression ax² + bx + c is between -1 and 1. That also implies that outside of this interval, the expression is either less than or equal to -1 or greater than or equal to 1. So, the quadratic must touch or cross the lines y = -1 and y = 1 exactly at x = -1 and x = 3.Wait, actually, since the solution set is (-1, 3), the inequality holds strictly between -1 and 3, but not outside. So, at x = -1 and x = 3, the quadratic must equal either -1 or 1. Hmm, but which one? Let me think.If I plug in x = -1 into the inequality, it should satisfy -1 < a(-1)² + b(-1) + c < 1. Similarly, for x = 3, it should satisfy -1 < a(3)² + b(3) + c < 1. But since the solution set is open, (-1, 3), it doesn't include the endpoints. So, does that mean that at x = -1 and x = 3, the quadratic equals exactly 1 or -1?Wait, actually, if the solution set is (-1, 3), that means the inequality is true for all x in (-1, 3), but not including -1 and 3. So, at x = -1 and x = 3, the quadratic must be equal to 1 or -1. But which one? Let me think about the graph.If the quadratic is above -1 and below 1 between -1 and 3, then at x = -1 and x = 3, the quadratic must touch the lines y = 1 or y = -1. So, depending on whether the parabola opens upwards or downwards, it might touch y = 1 or y = -1 at those points.Wait, maybe I need to consider the direction the parabola opens. If a is positive, the parabola opens upwards, and if a is negative, it opens downwards.Let me consider both cases.Case 1: a > 0. So, the parabola opens upwards. Then, the quadratic will have a minimum point. The solution set (-1, 3) implies that the quadratic is between -1 and 1 in that interval. So, the minimum value of the quadratic must be greater than -1, and it must cross y = 1 at x = -1 and x = 3.Wait, but if it opens upwards, the minimum is in the middle of -1 and 3, which is x = 1. So, at x = 1, the quadratic reaches its minimum. So, the minimum value should be greater than -1, and at x = -1 and x = 3, the quadratic should equal 1.So, let's write down the equations.At x = -1: a(-1)² + b(-1) + c = 1 => a - b + c = 1.At x = 3: a(3)² + b(3) + c = 1 => 9a + 3b + c = 1.Also, the vertex is at x = 1, so the derivative at x = 1 should be zero. The derivative of ax² + bx + c is 2ax + b. So, 2a(1) + b = 0 => 2a + b = 0 => b = -2a.So, now we can substitute b = -2a into the first two equations.First equation: a - (-2a) + c = 1 => a + 2a + c = 1 => 3a + c = 1.Second equation: 9a + 3(-2a) + c = 1 => 9a - 6a + c = 1 => 3a + c = 1.Wait, both equations give 3a + c = 1. So, that means we have only one equation with two variables. Hmm, so perhaps we need another condition.We also know that the minimum value at x = 1 is greater than -1. So, let's compute the value at x = 1.f(1) = a(1)² + b(1) + c = a + b + c.We know that f(1) > -1.But since b = -2a, substitute that in:f(1) = a + (-2a) + c = -a + c > -1.From the earlier equation, 3a + c = 1, so c = 1 - 3a.Substitute c into f(1):-a + (1 - 3a) > -1 => -a + 1 - 3a > -1 => -4a + 1 > -1 => -4a > -2 => 4a < 2 => a < 0.5.So, in this case, a must be less than 0.5. But since we are in the case where a > 0, the range for a is 0 < a < 0.5.Case 2: a < 0. So, the parabola opens downwards. Then, the quadratic will have a maximum point. The solution set (-1, 3) implies that the quadratic is between -1 and 1 in that interval. So, the maximum value of the quadratic must be less than 1, and it must cross y = -1 at x = -1 and x = 3.Wait, similar to before, but now the parabola opens downward, so the maximum is at x = 1, and the quadratic equals -1 at x = -1 and x = 3.So, let's write down the equations.At x = -1: a(-1)² + b(-1) + c = -1 => a - b + c = -1.At x = 3: a(3)² + b(3) + c = -1 => 9a + 3b + c = -1.Also, the vertex is at x = 1, so the derivative at x = 1 is zero. The derivative is 2ax + b, so 2a(1) + b = 0 => 2a + b = 0 => b = -2a.Substitute b = -2a into the first two equations.First equation: a - (-2a) + c = -1 => a + 2a + c = -1 => 3a + c = -1.Second equation: 9a + 3(-2a) + c = -1 => 9a - 6a + c = -1 => 3a + c = -1.Again, both equations give 3a + c = -1. So, same as before, we have c = -1 - 3a.Now, the maximum value at x = 1 is f(1) = a + b + c.Substituting b = -2a and c = -1 - 3a:f(1) = a + (-2a) + (-1 - 3a) = a - 2a -1 -3a = (-4a) -1.Since the parabola opens downward, the maximum value must be less than 1. So:-4a -1 < 1 => -4a < 2 => 4a > -2 => a > -0.5.But since we are in the case where a < 0, the range for a is -0.5 < a < 0.Case 3: a = 0. Then the quadratic becomes linear: bx + c. The inequality becomes -1 < bx + c < 1. The solution set is (-1, 3). So, the linear function must cross y = -1 at x = -1 and y = 1 at x = 3, or vice versa.Wait, but if a = 0, it's a linear function. Let me see.If the solution set is (-1, 3), then for x between -1 and 3, the linear function is between -1 and 1. So, at x = -1, the function is either -1 or 1, and at x = 3, it's the other.But since the solution set is open, it doesn't include the endpoints, so the function must cross y = -1 at x = -1 and y = 1 at x = 3, or vice versa.Let me assume it crosses y = -1 at x = -1 and y = 1 at x = 3.So, f(-1) = -1: b(-1) + c = -1 => -b + c = -1.f(3) = 1: 3b + c = 1.Subtract the first equation from the second:(3b + c) - (-b + c) = 1 - (-1) => 4b = 2 => b = 0.5.Then, from -b + c = -1: -0.5 + c = -1 => c = -0.5.So, the function is 0.5x - 0.5. Let's check if the solution set is indeed (-1, 3).Solve -1 < 0.5x - 0.5 < 1.Add 0.5: -0.5 < 0.5x < 1.5.Multiply by 2: -1 < x < 3.Yes, that works. So, a = 0 is also a valid case.But wait, in the previous cases, when a > 0, we had 0 < a < 0.5, and when a < 0, we had -0.5 < a < 0. So, combining all cases, a can be between -0.5 and 0.5, including 0.Wait, but when a = 0.5, does it still work? Let me check.If a = 0.5, then from case 1, b = -2a = -1, and c = 1 - 3a = 1 - 1.5 = -0.5.So, the quadratic is 0.5x² - x - 0.5.Let me check the value at x = 1: 0.5(1) -1 -0.5 = 0.5 -1 -0.5 = -1. So, f(1) = -1.But the solution set is (-1, 3), which means the inequality is strict. So, at x = 1, f(x) = -1, which is not greater than -1, so it doesn't satisfy the inequality. Therefore, a cannot be equal to 0.5.Similarly, if a = -0.5, let's check.From case 2, b = -2a = 1, and c = -1 -3a = -1 - (-1.5) = 0.5.So, the quadratic is -0.5x² + x + 0.5.At x = 1: -0.5(1) +1 +0.5 = -0.5 +1 +0.5 = 1. So, f(1) = 1.Again, since the solution set is open, the inequality is strict, so f(1) = 1 doesn't satisfy -1 < f(x) < 1. Therefore, a cannot be equal to -0.5.So, the range of a is -0.5 < a < 0.5.Wait, but when a = 0, it's allowed because the function is linear and satisfies the inequality strictly between -1 and 3. So, a can be 0.Therefore, combining all cases, the range of a is -0.5 < a < 0.5.I think that's the answer.

💡First, I need to determine the total amount raised by the school before the 2% administration fees were deducted. This can be calculated by dividing the amount raised after fees by 0.98, as 2% is deducted, leaving 98% of the total amount.Next, I'll find out how much Miss Rollin's class raised. Since Miss Rollin's class raised a third of the total amount, I'll divide the total amount raised before fees by 3.Then, I'll calculate the amount raised by Mrs. Sutton's class. Given that Mrs. Johnson's class raised 2300, which is twice the amount raised by Mrs. Sutton's class, I'll divide 2300 by 2 to find the amount raised by Mrs. Sutton's class.Finally, I'll determine the ratio of the amount raised by Mrs. Sutton's class to the amount raised by Miss Rollin's class by dividing the amount raised by Mrs. Sutton's class by the amount raised by Miss Rollin's class and simplifying the ratio.

💡Okay, so I have this geometry problem about an isosceles trapezoid (ABCD) where (AB = AD = BC), (AB) is parallel to (CD), and (AB) is longer than (CD). I need to prove that the quadrilateral (ANDE) is cyclic, where (E) is the intersection of the diagonals (AC) and (BD), and (N) is the reflection of (B) over (AC).First, let me try to visualize the trapezoid. Since it's isosceles, the non-parallel sides (AD) and (BC) are equal, which they already mentioned as (AB = AD = BC). So, (AB) is the longer base, and (CD) is the shorter base. Diagonals (AC) and (BD) intersect at point (E). Reflecting (B) over (AC) gives point (N). I need to show that (ANDE) is cyclic, meaning all four points lie on a circle.I remember that for a quadrilateral to be cyclic, the sum of its opposite angles must be 180 degrees. So, if I can show that (angle AND + angle AED = 180^circ) or (angle ANE + angle ADE = 180^circ), that would prove it's cyclic.Let me start by drawing the trapezoid. I'll label the vertices (A), (B), (C), (D) such that (AB) is the top base, (CD) is the bottom base, and (AD) and (BC) are the legs. Since it's isosceles, the legs are equal, and the base angles are equal.Now, reflecting (B) over (AC) to get (N). So, (N) is such that (AC) is the perpendicular bisector of (BN), meaning (AN = AB) and (CN = CB). Since (AB = BC), this tells me (AN = BC), so (AN = BC = AB). So, triangle (ABN) is isosceles with (AN = AB).Next, point (E) is the intersection of diagonals (AC) and (BD). In an isosceles trapezoid, the diagonals are equal in length and they intersect each other in the same ratio. So, (AE:EC = BE:ED). Maybe this ratio will come into play later.I think I should assign some variables to the angles to make things clearer. Let me denote (angle CAB = alpha) and (angle DAC = beta). Since (AB parallel CD), the angles at the bases are supplementary. So, (angle ABC + angle BCD = 180^circ), but since it's isosceles, (angle ABC = angle BAD).Wait, actually, in an isosceles trapezoid, the base angles are equal. So, (angle ABC = angle BAD) and (angle BCD = angle ADC). Hmm, maybe I need to consider triangle (ABD) and triangle (BCD). Since (AB = AD = BC), triangles (ABD) and (BCD) are both isosceles.Let me think about triangle (ABD). Since (AB = AD), the base angles are equal, so (angle ABD = angle ADB). Similarly, in triangle (BCD), since (BC = CD), the base angles are equal, so (angle CBD = angle CDB). Wait, but (AB = AD = BC), so maybe these triangles are congruent? Not necessarily, because the sides opposite the equal angles might not be equal.Wait, no, in triangle (ABD), sides (AB = AD), so it's isosceles with base (BD). In triangle (BCD), sides (BC = CD), so it's isosceles with base (BD) as well. So, actually, triangles (ABD) and (BCD) are congruent by SSS, since (AB = BC = AD = CD), and they share the base (BD). Therefore, triangles (ABD) and (BCD) are congruent.This tells me that angles in these triangles are equal. So, (angle ABD = angle CBD) and (angle ADB = angle CDB). Therefore, point (E), being the intersection of diagonals (AC) and (BD), divides (BD) into two equal parts because the triangles are congruent. So, (BE = ED).Wait, but in an isosceles trapezoid, the diagonals are equal and they bisect each other proportionally. So, (AE:EC = BE:ED). Since (AB = AD = BC), maybe (E) is the midpoint of both diagonals? Hmm, not necessarily, because the trapezoid is isosceles but not necessarily a rectangle or square.Wait, let me think again. In an isosceles trapezoid, the diagonals are equal but they don't necessarily bisect each other unless it's a rectangle. So, (AE = EC) only if it's a rectangle, which it's not because (AB > CD). So, (E) divides the diagonals proportionally.Given that (AB = AD = BC), maybe the ratio (AE:EC) can be determined. Let me denote (AE = x) and (EC = y), so (x/y = AB/CD). But since (AB > CD), (x > y). Hmm, but I don't know the exact lengths, so maybe I need another approach.Since (N) is the reflection of (B) over (AC), then (AC) is the perpendicular bisector of (BN). So, (AN = AB) and (CN = CB). Since (AB = BC), this implies (AN = BC = AB). So, triangle (ABN) is isosceles with (AN = AB).Therefore, (angle ABN = angle ANB). Also, since (N) is the reflection, the angles formed by (AC) with (BN) are equal. So, (angle BAC = angle NAC), which is (alpha). Similarly, (angle BCA = angle NCA).Now, looking at quadrilateral (ANDE), I need to show that it's cyclic. So, I need to show that points (A), (N), (D), and (E) lie on a circle. One way to do this is to show that the power of point (E) with respect to the circle through (A), (N), and (D) is zero, meaning (EA cdot EC = ED cdot EB), but I'm not sure.Alternatively, I can show that (angle AND + angle AED = 180^circ). Let me try that.First, let's find (angle AND). Since (N) is the reflection of (B) over (AC), and (AB = AN), triangle (ABN) is isosceles. So, (angle ABN = angle ANB). Let me denote (angle BAN = 2alpha), since (N) is the reflection, so the angle is doubled.Wait, no, actually, reflecting (B) over (AC) would mean that (angle BAC = angle NAC = alpha). So, (angle BAN = 2alpha). Therefore, in triangle (ABN), since (AB = AN), the base angles are equal, so (angle ABN = angle ANB = (180^circ - 2alpha)/2 = 90^circ - alpha).Now, looking at point (D), since (ABCD) is an isosceles trapezoid, (angle ADC = angle BCD). Also, since (AB parallel CD), the consecutive angles are supplementary. So, (angle ADC + angle DAB = 180^circ). Let me denote (angle DAB = theta), so (angle ADC = 180^circ - theta).But I also know that (AB = AD), so triangle (ABD) is isosceles with (AB = AD). Therefore, (angle ABD = angle ADB). Let me denote these angles as (phi). So, in triangle (ABD), we have (angle BAD = theta), and the other two angles are (phi), so (theta + 2phi = 180^circ), which gives (phi = (180^circ - theta)/2).Similarly, in triangle (BCD), since (BC = CD), it's also isosceles with (angle CBD = angle CDB). Let me denote these angles as (psi). Then, (angle BCD = 180^circ - 2psi). But since (AB parallel CD), (angle BCD = angle ADC = 180^circ - theta). Therefore, (180^circ - 2psi = 180^circ - theta), which implies (psi = theta/2).Wait, but earlier I had (phi = (180^circ - theta)/2). Since triangles (ABD) and (BCD) are congruent, their angles should be equal. So, (phi = psi). Therefore, ((180^circ - theta)/2 = theta/2), which implies (180^circ - theta = theta), so (2theta = 180^circ), hence (theta = 90^circ).So, (angle DAB = 90^circ), which means that the trapezoid has a right angle at (A). Therefore, (ABCD) is a right isosceles trapezoid. That simplifies things a bit.So, now, (angle DAB = 90^circ), and since (AB = AD), triangle (ABD) is a right isosceles triangle with legs (AB) and (AD), and hypotenuse (BD). Therefore, (BD = ABsqrt{2}).Similarly, triangle (BCD) is also a right isosceles triangle with legs (BC) and (CD), and hypotenuse (BD). Since (BC = AB), we have (BD = ABsqrt{2}), which is consistent.Now, point (E) is the intersection of diagonals (AC) and (BD). In a right isosceles trapezoid, the diagonals are equal and intersect at (E). Since (AB = AD = BC), and (AB > CD), the ratio (AE:EC) can be determined.Let me denote (AB = AD = BC = s), and (CD = t), where (s > t). Then, the area of the trapezoid can be calculated in two ways: using the height and the bases, or using the diagonals and the angle between them.But maybe it's easier to use coordinate geometry. Let me place the trapezoid on a coordinate system. Let me set point (A) at the origin ((0, 0)). Since (AB) is the top base and (AB = s), let me place point (B) at ((s, 0)). Since (AD = s) and (angle DAB = 90^circ), point (D) will be at ((0, s)). Now, since (AB parallel CD), point (C) must be at ((t, s)), where (t < s).Wait, but (BC = s), so the distance from (B) at ((s, 0)) to (C) at ((t, s)) must be (s). So, the distance formula gives:[sqrt{(t - s)^2 + (s - 0)^2} = s]Simplifying:[(t - s)^2 + s^2 = s^2][(t - s)^2 = 0][t = s]But this contradicts (t < s). Hmm, that can't be right. Maybe my coordinate system is off.Wait, perhaps I should place the trapezoid differently. Let me try again. Let me place (A) at ((0, 0)), (B) at ((b, 0)), (D) at ((0, d)), and (C) at ((c, d)). Since (AB = AD = BC), we have:1. (AB = sqrt{(b - 0)^2 + (0 - 0)^2} = b = s)2. (AD = sqrt{(0 - 0)^2 + (d - 0)^2} = d = s)3. (BC = sqrt{(c - b)^2 + (d - 0)^2} = sqrt{(c - b)^2 + d^2} = s)So, from 1 and 2, (b = d = s). From 3:[sqrt{(c - s)^2 + s^2} = s][(c - s)^2 + s^2 = s^2][(c - s)^2 = 0][c = s]Again, this leads to (c = s), which would make (CD) from ((s, s)) to ((s, 0)), but that's a vertical line, not a base. So, this suggests that my initial assumption of placing (A) at ((0, 0)) and (D) at ((0, s)) might not work because it forces (C) to coincide with (B), which isn't possible.Maybe I need to adjust the coordinate system. Let me instead place (A) at ((0, 0)), (B) at ((b, 0)), (D) at ((d, h)), and (C) at ((c, h)), where (h) is the height of the trapezoid. Since (AB = AD = BC = s), we have:1. (AB = sqrt{(b - 0)^2 + (0 - 0)^2} = b = s)2. (AD = sqrt{(d - 0)^2 + (h - 0)^2} = sqrt{d^2 + h^2} = s)3. (BC = sqrt{(c - b)^2 + (h - 0)^2} = sqrt{(c - b)^2 + h^2} = s)Also, since (AB parallel CD), the slope of (AB) is 0, so the slope of (CD) must also be 0. Therefore, (C) must be at ((c, h)) and (D) at ((d, h)), so (CD) is horizontal.Additionally, since it's an isosceles trapezoid, the legs (AD) and (BC) are equal in length and symmetric with respect to the vertical line through the midpoint of (AB). So, the midpoint of (AB) is at ((b/2, 0)), and the midpoint of (CD) should be at (( (c + d)/2, h)). For the trapezoid to be isosceles, these midpoints should lie on the same vertical line, so:[frac{c + d}{2} = frac{b}{2} implies c + d = b]So, (c = b - d).Now, from equation 2: (d^2 + h^2 = s^2).From equation 3: ((c - b)^2 + h^2 = s^2). Substituting (c = b - d):[(b - d - b)^2 + h^2 = s^2][(-d)^2 + h^2 = s^2][d^2 + h^2 = s^2]Which is the same as equation 2. So, no new information there.Therefore, the coordinates are:- (A(0, 0))- (B(b, 0))- (D(d, h))- (C(b - d, h))With (d^2 + h^2 = s^2), and (b = s).So, (b = s), and (d^2 + h^2 = s^2). Therefore, (d) and (h) are related by (d^2 + h^2 = s^2).Now, let's find the coordinates of point (E), the intersection of diagonals (AC) and (BD).Diagonal (AC) goes from (A(0, 0)) to (C(b - d, h)). Its parametric equations are:[x = t(b - d), quad y = th, quad t in [0, 1]]Diagonal (BD) goes from (B(b, 0)) to (D(d, h)). Its parametric equations are:[x = b + s(d - b), quad y = 0 + s(h - 0) = sh, quad s in [0, 1]]Wait, no, parametric equations for (BD) should be:[x = b + s(d - b), quad y = 0 + s(h - 0) = sh, quad s in [0, 1]]So, to find the intersection (E), we set the coordinates equal:[t(b - d) = b + s(d - b)][th = sh]From the second equation, (th = sh). Assuming (h neq 0), we can divide both sides by (h):[t = s]Substituting into the first equation:[t(b - d) = b + t(d - b)][t(b - d) - t(d - b) = b][t(b - d - d + b) = b][t(2b - 2d) = b][t = frac{b}{2(b - d)}]So, (t = frac{b}{2(b - d)}). Therefore, the coordinates of (E) are:[x = t(b - d) = frac{b}{2(b - d)}(b - d) = frac{b}{2}][y = th = frac{b}{2(b - d)}h]So, (E) is at (left( frac{b}{2}, frac{bh}{2(b - d)} right)).Now, let's find the coordinates of point (N), which is the reflection of (B) over (AC). To find the reflection, I can use the formula for reflection over a line.The line (AC) goes from (A(0, 0)) to (C(b - d, h)). Its slope is (m = frac{h}{b - d}). The equation of line (AC) is (y = frac{h}{b - d}x).To find the reflection of point (B(b, 0)) over line (AC), I can use the formula for reflection over a line (ax + by + c = 0):[x' = x - frac{2a(ax + by + c)}{a^2 + b^2}][y' = y - frac{2b(ax + by + c)}{a^2 + b^2}]First, let's write the equation of (AC) in standard form. From (y = frac{h}{b - d}x), we get:[frac{h}{b - d}x - y = 0][h x - (b - d) y = 0]So, (a = h), (b = -(b - d)), and (c = 0).Now, applying the reflection formula to point (B(b, 0)):[x' = b - frac{2h(h cdot b + (-(b - d)) cdot 0 + 0)}{h^2 + (b - d)^2}][x' = b - frac{2h^2 b}{h^2 + (b - d)^2}][x' = b left( 1 - frac{2h^2}{h^2 + (b - d)^2} right )][x' = b left( frac{h^2 + (b - d)^2 - 2h^2}{h^2 + (b - d)^2} right )][x' = b left( frac{(b - d)^2 - h^2}{h^2 + (b - d)^2} right )]Similarly,[y' = 0 - frac{2(-(b - d))(h cdot b + (-(b - d)) cdot 0 + 0)}{h^2 + (b - d)^2}][y' = frac{2(b - d)h b}{h^2 + (b - d)^2}]So, the coordinates of (N) are:[Nleft( frac{b[(b - d)^2 - h^2]}{h^2 + (b - d)^2}, frac{2b(b - d)h}{h^2 + (b - d)^2} right )]This looks complicated, but maybe we can simplify it using the fact that (d^2 + h^2 = s^2) and (b = s). So, (d^2 + h^2 = b^2).Let me substitute (b^2 = d^2 + h^2) into the expressions for (x') and (y'):First, (x'):[x' = frac{b[(b - d)^2 - h^2]}{h^2 + (b - d)^2}][= frac{b[(b^2 - 2bd + d^2) - h^2]}{h^2 + b^2 - 2bd + d^2}]But (b^2 = d^2 + h^2), so substitute:[= frac{b[(d^2 + h^2 - 2bd + d^2) - h^2]}{h^2 + d^2 + h^2 - 2bd + d^2}]Simplify numerator:[= frac{b[2d^2 - 2bd]}{2h^2 + 2d^2 - 2bd}]Factor numerator and denominator:[= frac{2b(d^2 - bd)}{2(h^2 + d^2 - bd)}]Cancel the 2:[= frac{b(d^2 - bd)}{h^2 + d^2 - bd}]Factor numerator:[= frac{b d(d - b)}{h^2 + d^2 - bd}]Similarly, the denominator:[h^2 + d^2 - bd = (d^2 + h^2) - bd = b^2 - bd = b(b - d)]So, substituting back:[x' = frac{b d(d - b)}{b(b - d)} = frac{b d(d - b)}{b(b - d)} = frac{d(d - b)}{b - d} = frac{-d(b - d)}{b - d} = -d]So, (x' = -d).Now, for (y'):[y' = frac{2b(b - d)h}{h^2 + (b - d)^2}]Again, substitute (b^2 = d^2 + h^2):[= frac{2b(b - d)h}{h^2 + b^2 - 2bd + d^2}][= frac{2b(b - d)h}{(d^2 + h^2) + b^2 - 2bd + d^2}]Wait, that seems messy. Let me try another approach.Wait, denominator is (h^2 + (b - d)^2 = h^2 + b^2 - 2bd + d^2 = (h^2 + d^2) + b^2 - 2bd = b^2 + b^2 - 2bd = 2b^2 - 2bd = 2b(b - d)).So, denominator is (2b(b - d)).Therefore,[y' = frac{2b(b - d)h}{2b(b - d)} = h]So, point (N) has coordinates ((-d, h)).That's a much simpler result! So, (N(-d, h)).Now, let's recap the coordinates:- (A(0, 0))- (B(b, 0))- (C(b - d, h))- (D(d, h))- (Eleft( frac{b}{2}, frac{bh}{2(b - d)} right ))- (N(-d, h))Now, I need to show that quadrilateral (ANDE) is cyclic. So, points (A(0, 0)), (N(-d, h)), (D(d, h)), and (Eleft( frac{b}{2}, frac{bh}{2(b - d)} right )) lie on a circle.One way to show this is to find the equation of the circle passing through three of these points and verify that the fourth lies on it. Let's choose points (A), (N), and (D) to find the circle, then check if (E) lies on it.The general equation of a circle is (x^2 + y^2 + 2gx + 2fy + c = 0). Plugging in point (A(0, 0)):[0 + 0 + 0 + 0 + c = 0 implies c = 0]So, the equation becomes (x^2 + y^2 + 2gx + 2fy = 0).Now, plug in point (N(-d, h)):[(-d)^2 + h^2 + 2g(-d) + 2f(h) = 0][d^2 + h^2 - 2gd + 2fh = 0]Similarly, plug in point (D(d, h)):[d^2 + h^2 + 2g(d) + 2f(h) = 0][d^2 + h^2 + 2gd + 2fh = 0]Now, we have two equations:1. (d^2 + h^2 - 2gd + 2fh = 0)2. (d^2 + h^2 + 2gd + 2fh = 0)Subtracting equation 1 from equation 2:[(d^2 + h^2 + 2gd + 2fh) - (d^2 + h^2 - 2gd + 2fh) = 0 - 0][4gd = 0][gd = 0]Since (d neq 0) (because (CD) is shorter than (AB), so (d) must be positive and less than (b)), we have (g = 0).Substituting (g = 0) back into equation 1:[d^2 + h^2 - 0 + 2fh = 0][d^2 + h^2 + 2fh = 0][2fh = - (d^2 + h^2)][f = -frac{d^2 + h^2}{2h}]But from earlier, (d^2 + h^2 = b^2), so:[f = -frac{b^2}{2h}]Therefore, the equation of the circle passing through (A), (N), and (D) is:[x^2 + y^2 + 0x + 2left(-frac{b^2}{2h}right)y = 0][x^2 + y^2 - frac{b^2}{h} y = 0]Now, let's check if point (Eleft( frac{b}{2}, frac{bh}{2(b - d)} right )) lies on this circle.Substitute (x = frac{b}{2}) and (y = frac{bh}{2(b - d)}) into the equation:[left(frac{b}{2}right)^2 + left(frac{bh}{2(b - d)}right)^2 - frac{b^2}{h} cdot frac{bh}{2(b - d)} = 0]Simplify each term:1. (left(frac{b}{2}right)^2 = frac{b^2}{4})2. (left(frac{bh}{2(b - d)}right)^2 = frac{b^2 h^2}{4(b - d)^2})3. (frac{b^2}{h} cdot frac{bh}{2(b - d)} = frac{b^3 h}{2h(b - d)} = frac{b^3}{2(b - d)})So, the equation becomes:[frac{b^2}{4} + frac{b^2 h^2}{4(b - d)^2} - frac{b^3}{2(b - d)} = 0]Multiply both sides by (4(b - d)^2) to eliminate denominators:[b^2 (b - d)^2 + b^2 h^2 - 2b^3 (b - d) = 0]Let's expand and simplify:First term: (b^2 (b - d)^2 = b^2 (b^2 - 2bd + d^2) = b^4 - 2b^3 d + b^2 d^2)Second term: (b^2 h^2)Third term: (-2b^3 (b - d) = -2b^4 + 2b^3 d)So, combining all terms:[b^4 - 2b^3 d + b^2 d^2 + b^2 h^2 - 2b^4 + 2b^3 d = 0]Simplify:- (b^4 - 2b^4 = -b^4)- (-2b^3 d + 2b^3 d = 0)- (b^2 d^2 + b^2 h^2 = b^2 (d^2 + h^2) = b^2 cdot b^2 = b^4)So, the equation becomes:[-b^4 + b^4 = 0][0 = 0]Which is true. Therefore, point (E) lies on the circle passing through (A), (N), and (D). Hence, quadrilateral (ANDE) is cyclic.I think this proves that (ANDE) is cyclic because all four points lie on the same circle.

💡First, I recognize that Ronald does laundry every 6 days and Tim does laundry every few days. They both did laundry today, and they will both do laundry on the same day again in 18 days.To find out how often Tim does his laundry, I need to determine the number of days between Tim's laundry days, which I'll call ( T ).Since both Ronald and Tim will do laundry together in 18 days, ( T ) must be a factor of 18. The factors of 18 are 1, 2, 3, 6, 9, and 18.I can eliminate 1 and 18 because doing laundry every 1 day or every 18 days doesn't make sense in this context. I can also eliminate 6 because that's Ronald's laundry cycle.This leaves me with the factors 2, 3, and 9. Since 18 days is the first time they do laundry on the same day again, I'm looking for the greatest common factor of 6 and ( T ) that is less than 18.The greatest common factor of 6 and 18 is 6, but since Tim's cycle is different from Ronald's, I can eliminate 6. The next greatest factor of 18 that is less than 6 is 3.If Tim did his laundry every 3 days, then in 18 days, he would have done laundry 6 times (because ( 18 div 3 = 6 )), which aligns with Ronald doing his laundry every 6 days (because ( 18 div 6 = 3 )).Therefore, Tim does his laundry every 3 days.

💡Alright, so I came across this interesting problem today. It goes like this: Ilya invented a number that doesn't contain any zeros. Then he rearranged its digits and added the resulting number to the original one. The question is, can the result be composed entirely of ones?Hmm, okay, let's break this down. First, I need to understand what exactly is being asked. So, we have a number, let's call it N, which doesn't have any zeros in it. Then, Ilya rearranges its digits to form another number, say M. After that, he adds N and M together. The question is whether the sum N + M can be a number that's made up entirely of ones, like 11, 111, 1111, etc.Alright, so my first thought is to consider the properties of such numbers. If N and M are rearrangements of each other, they must have the same digits, just in a different order. That means both N and M have the same number of digits, and the sum of their digits is the same. Let's denote the number of digits as n.Now, if N + M is a number composed entirely of ones, let's call this sum S. So, S is a number like 11...1 with n digits. The key here is to see if such a sum is possible given the constraints.Let me think about the sum of the digits of N and M. Since they are rearrangements, the sum of the digits of N is equal to the sum of the digits of M. Let's denote this sum as D. So, D = sum of digits of N = sum of digits of M.When we add N and M, the sum S will have digits that are the sum of corresponding digits of N and M, possibly with carries. But since S is all ones, each digit in S must be 1. That means, for each digit position, the sum of the digits from N and M, plus any carry from the previous position, must equal 1 or 11 (if there's a carryover to the next position).Wait a minute, that seems tricky. If we have a carryover, the digit sum would be 11, which is two digits. But in S, each digit is just 1, so we can't have a carryover affecting the next digit in a way that would change it from 1 to something else. Hmm, this seems like a potential contradiction.Let me formalize this a bit. Let's denote the digits of N as a1, a2, ..., an, and the digits of M as b1, b2, ..., bn, where each bi is a permutation of the ai's. Then, the sum S would have digits s1, s2, ..., sn, where each si is 1.For each digit position i, we have:ai + bi + carry_in = si + 10 * carry_outBut since si is 1, this becomes:ai + bi + carry_in = 1 + 10 * carry_outNow, carry_in and carry_out can only be 0 or 1 because we're dealing with single-digit additions. So, let's consider the possible cases.Case 1: carry_in = 0Then, ai + bi = 1 + 10 * carry_outBut ai and bi are digits from 1 to 9 (since there are no zeros). The smallest ai + bi can be is 1 + 1 = 2, and the largest is 9 + 9 = 18.So, 2 ≤ ai + bi ≤ 18But 1 + 10 * carry_out must be either 1 or 11, since carry_out can be 0 or 1.If carry_out = 0, then ai + bi = 1, which is impossible because ai and bi are at least 1.If carry_out = 1, then ai + bi = 11.So, in this case, ai + bi must equal 11.Case 2: carry_in = 1Then, ai + bi + 1 = 1 + 10 * carry_outSimplifying, ai + bi = 10 * carry_outAgain, ai and bi are between 1 and 9, so ai + bi is between 2 and 18.If carry_out = 0, then ai + bi = 0, which is impossible.If carry_out = 1, then ai + bi = 10.So, in this case, ai + bi must equal 10.Therefore, for each digit position, either ai + bi = 11 with a carry_out of 1, or ai + bi = 10 with a carry_out of 1, depending on whether there was a carry_in.But wait, this creates a chain reaction because the carry_out from one digit affects the carry_in of the next digit. Let's see how this plays out.Starting from the least significant digit (rightmost), there is no carry_in initially, so we must have ai + bi = 11 with a carry_out of 1.Then, moving to the next digit, there is a carry_in of 1, so ai + bi must equal 10 with a carry_out of 1.This pattern would alternate between requiring ai + bi = 11 and ai + bi = 10 for each subsequent digit.But here's the problem: since N and M are rearrangements of each other, the multiset of digits in N and M are the same. Therefore, the sum of all digits in N is equal to the sum of all digits in M, which is D.When we add N and M, the total sum S is 11...1 with n digits, which is (10^n - 1)/9. The sum of the digits of S is n * 1 = n.But the sum of the digits of N + M is also equal to the sum of the digits of N plus the sum of the digits of M minus 9 times the number of carries (due to carryover). Wait, is that correct?Actually, when adding two numbers, the sum of the digits of the result is equal to the sum of the digits of the addends minus 9 times the number of carries. Because each carryover reduces the digit sum by 9 (since 10 is carried over, but the digit itself is reduced by 10 and increased by 1 in the next higher digit).So, sum of digits of S = sum of digits of N + sum of digits of M - 9 * number of carries.But sum of digits of S is n, and sum of digits of N and M are both D, so:n = 2D - 9 * number of carries.But from our earlier analysis, each digit position either requires a carry or not. Specifically, starting from the right, we have a carry, then no carry, then carry, etc., alternating.But wait, actually, in our earlier analysis, we saw that each digit position must either have ai + bi = 11 or ai + bi = 10, depending on the carry_in.This means that every digit position must produce a carry_out of 1, except possibly the last one if n is even or odd.Wait, no, actually, starting from the right, the first digit must have ai + bi = 11 with a carry_out of 1.Then, the next digit has ai + bi + 1 = 11, so ai + bi = 10 with a carry_out of 1.Then, the next digit has ai + bi + 1 = 11, so ai + bi = 10 with a carry_out of 1.Wait, no, that's not right. Let's correct that.If we have a carry_in of 1, then ai + bi + 1 = 1 + 10 * carry_out.So, if carry_out is 1, then ai + bi + 1 = 11, so ai + bi = 10.If carry_out is 0, then ai + bi + 1 = 1, which is impossible because ai + bi ≥ 2.Therefore, every digit position after the first must have ai + bi = 10 with a carry_out of 1.But wait, starting from the right, the first digit has no carry_in, so ai + bi = 11 with carry_out = 1.Then, the next digit has ai + bi + 1 = 11, so ai + bi = 10 with carry_out = 1.Then, the next digit has ai + bi + 1 = 11, so ai + bi = 10 with carry_out = 1.This pattern continues until the last digit.But wait, if n is the number of digits, then we have n digit positions.Starting from the right, the first digit requires ai + bi = 11.Then, the next n-1 digits require ai + bi = 10.But since N and M are rearrangements, the multiset of digits in N and M are the same. Therefore, the sum of all ai's is D, and the sum of all bi's is also D.Therefore, the total sum of ai + bi over all digits is 2D.But from our earlier analysis, the sum of ai + bi over all digits is 11 + 10*(n-1).So, 2D = 11 + 10*(n-1)Simplifying, 2D = 11 + 10n - 102D = 10n + 1Therefore, D = (10n + 1)/2But D must be an integer because it's the sum of digits.So, (10n + 1)/2 must be an integer.But 10n is even, so 10n + 1 is odd, and dividing by 2 gives a non-integer.This is a contradiction because D must be an integer.Therefore, our assumption that such a number S exists must be false.Hence, it's impossible for the sum N + M to be a number composed entirely of ones.