💡Okay, so I have this problem where I'm given two vectors, a and b, with components defined in terms of trigonometric functions. The function f(x) is the dot product of a and b plus the square of the magnitude of b. I need to find the period, maximum, and minimum values of f(x), and then determine the intervals on which f(x) is increasing within the interval [0, π].First, let me write down the vectors:Vector a is (5√3 cos x, cos x)Vector b is (sin x, 2 cos x)So, f(x) is the dot product of a and b plus the magnitude squared of b.Let me recall that the dot product of two vectors (a1, a2) and (b1, b2) is a1*b1 + a2*b2. The magnitude squared of a vector (b1, b2) is b1² + b2².So, let me compute the dot product a · b first.a · b = (5√3 cos x)(sin x) + (cos x)(2 cos x)= 5√3 cos x sin x + 2 cos² xNow, the magnitude squared of b is |b|² = (sin x)² + (2 cos x)²= sin² x + 4 cos² xSo, f(x) = a · b + |b|²= 5√3 cos x sin x + 2 cos² x + sin² x + 4 cos² xLet me simplify this expression step by step.First, combine like terms:The terms with cos² x: 2 cos² x + 4 cos² x = 6 cos² xThe term with sin x cos x: 5√3 cos x sin xThe term with sin² x: sin² xSo, f(x) = 5√3 cos x sin x + 6 cos² x + sin² xHmm, maybe I can express this in terms of double angles or other trigonometric identities to simplify further.I remember that sin 2x = 2 sin x cos x, so cos x sin x = (1/2) sin 2x.Similarly, cos² x can be expressed using the identity cos² x = (1 + cos 2x)/2, and sin² x = (1 - cos 2x)/2.Let me apply these identities to each term.First term: 5√3 cos x sin x = 5√3 * (1/2) sin 2x = (5√3 / 2) sin 2xSecond term: 6 cos² x = 6 * (1 + cos 2x)/2 = 3(1 + cos 2x) = 3 + 3 cos 2xThird term: sin² x = (1 - cos 2x)/2So, putting it all together:f(x) = (5√3 / 2) sin 2x + 3 + 3 cos 2x + (1 - cos 2x)/2Let me combine the constant terms and the cos 2x terms.First, constants: 3 + (1/2) = 3.5 or 7/2Now, the cos 2x terms: 3 cos 2x - (1/2) cos 2x = (3 - 0.5) cos 2x = (5/2) cos 2xSo, f(x) = (5√3 / 2) sin 2x + (5/2) cos 2x + 7/2Hmm, this looks like a single sinusoidal function. I can write this as A sin(2x + φ) + 7/2, where A is the amplitude and φ is the phase shift.To find A, I can use the formula A = sqrt( (5√3 / 2)^2 + (5/2)^2 )Calculating that:(5√3 / 2)^2 = (25 * 3) / 4 = 75/4(5/2)^2 = 25/4So, A = sqrt(75/4 + 25/4) = sqrt(100/4) = sqrt(25) = 5So, A = 5.Now, to find φ, we can use tan φ = (coefficient of cos 2x) / (coefficient of sin 2x)Wait, actually, in the expression A sin(2x + φ), the coefficients are:A sin(2x + φ) = A sin 2x cos φ + A cos 2x sin φComparing this with our expression:(5√3 / 2) sin 2x + (5/2) cos 2xSo, equating coefficients:A cos φ = 5√3 / 2A sin φ = 5 / 2Since A = 5, we have:cos φ = (5√3 / 2) / 5 = √3 / 2sin φ = (5 / 2) / 5 = 1/2So, φ is the angle where cos φ = √3 / 2 and sin φ = 1/2, which is φ = π/6.Therefore, f(x) can be written as:f(x) = 5 sin(2x + π/6) + 7/2Now, for part (1), we need to find the period, maximum, and minimum values of f(x).The function f(x) is a sinusoidal function with amplitude 5, vertical shift 7/2, and frequency 2 (since the argument of sine is 2x + π/6). The period T of a sine function sin(Bx + C) is 2π / |B|. Here, B = 2, so the period is 2π / 2 = π.So, the period of f(x) is π.The maximum value of sin(2x + π/6) is 1, so the maximum value of f(x) is 5*1 + 7/2 = 5 + 3.5 = 8.5, which is 17/2.Similarly, the minimum value of sin(2x + π/6) is -1, so the minimum value of f(x) is 5*(-1) + 7/2 = -5 + 3.5 = -1.5, which is -3/2.So, maximum value is 17/2, minimum is -3/2.For part (2), we need to find the intervals on which f(x) is increasing on [0, π].Since f(x) is a sinusoidal function, it will have intervals where it increases and decreases periodically. To find where it's increasing, we can take the derivative of f(x) and find where it's positive.First, let's compute f'(x):f(x) = 5 sin(2x + π/6) + 7/2f'(x) = 5 * cos(2x + π/6) * 2 = 10 cos(2x + π/6)We need to find where f'(x) > 0, which is where cos(2x + π/6) > 0.The cosine function is positive in the intervals (-π/2 + 2πk, π/2 + 2πk) for integers k.So, we set up the inequality:-π/2 + 2πk < 2x + π/6 < π/2 + 2πkLet's solve for x:Subtract π/6 from all parts:-π/2 - π/6 + 2πk < 2x < π/2 - π/6 + 2πkSimplify the angles:-π/2 - π/6 = -(3π/6 + π/6) = -4π/6 = -2π/3π/2 - π/6 = (3π/6 - π/6) = 2π/6 = π/3So,-2π/3 + 2πk < 2x < π/3 + 2πkDivide all parts by 2:-π/3 + πk < x < π/6 + πkSo, the intervals where f(x) is increasing are (-π/3 + πk, π/6 + πk) for integers k.Now, we need to find the intervals within [0, π].Let's consider k = 0:-π/3 < x < π/6But since x must be ≥ 0, this interval becomes 0 ≤ x < π/6.Next, k = 1:-π/3 + π < x < π/6 + πWhich simplifies to:2π/3 < x < 7π/6But since x ≤ π, the upper bound is π, so the interval is 2π/3 < x ≤ π.For k = 2, the lower bound would be -π/3 + 2π = 5π/3, which is greater than π, so it's outside our interval of interest.Similarly, for k = -1, the upper bound would be π/6 - π = -5π/6, which is negative, so also outside our interval.Therefore, the intervals where f(x) is increasing on [0, π] are [0, π/6] and [2π/3, π].Let me just double-check my calculations.Starting from f'(x) = 10 cos(2x + π/6). Setting this greater than 0 gives cos(2x + π/6) > 0.The general solution for cos θ > 0 is θ ∈ (-π/2 + 2πk, π/2 + 2πk).So, 2x + π/6 ∈ (-π/2 + 2πk, π/2 + 2πk)Solving for x:2x ∈ (-π/2 - π/6 + 2πk, π/2 - π/6 + 2πk)2x ∈ (-2π/3 + 2πk, π/3 + 2πk)x ∈ (-π/3 + πk, π/6 + πk)Yes, that's correct.Now, within [0, π], the valid intervals are when k=0: x ∈ (-π/3, π/6) → [0, π/6]And when k=1: x ∈ (2π/3, 7π/6) → (2π/3, π]So, the increasing intervals are [0, π/6] and [2π/3, π].I think that's correct. Let me just visualize the function f(x) = 5 sin(2x + π/6) + 7/2.It's a sine wave with amplitude 5, shifted up by 7/2, and with a period of π. The derivative is 10 cos(2x + π/6), which is positive when cos(2x + π/6) is positive, which happens in the intervals we found.So, I'm confident with the answer.

💡Alright, so I have this problem about two overlapping sectors of a circle. Both circles have a radius of 15 units, and each sector has a central angle of 45 degrees. The centers of these circles are labeled P and R. I need to find the area of the shaded region where they overlap. Hmm, okay, let me try to visualize this.First, I imagine two circles intersecting each other. Each circle has a radius of 15, and each sector that's shaded is 45 degrees. So, the overlapping area is kind of like a lens shape formed by the intersection of the two sectors. I think this is called a lens-shaped region or maybe a vesica piscis. But since the sectors are only 45 degrees, it's a smaller portion of that.The problem mentions that by symmetry, the areas of the two parts of the shaded region are equal. So, I can focus on calculating the area of one part and then double it to get the total shaded area. That makes sense because both sectors are identical in size and central angle, so their overlapping regions should be symmetrical.Let me break it down step by step.1. **Understanding the Sector Area:** Each sector has a central angle of 45 degrees. The formula for the area of a sector is (θ/360) * π * r², where θ is the central angle in degrees and r is the radius. Plugging in the numbers, for one sector, that would be (45/360) * π * 15². Let me compute that. 45 divided by 360 is 1/8. So, 1/8 * π * 225 (since 15 squared is 225). That gives (225/8) * π, which is 28.125π. So, the area of one sector is 28.125π. Got that.2. **Understanding the Triangle Area:** Now, within each sector, there's an isosceles triangle. The triangle has two sides equal to the radius, which is 15 units each, and the included angle is 45 degrees. The area of a triangle can be found using (1/2)*ab*sin(C), where a and b are the sides, and C is the included angle. So, plugging in the values, it's (1/2)*15*15*sin(45°). Let me compute that. 15 times 15 is 225, half of that is 112.5. Sin(45°) is √2/2, which is approximately 0.7071. So, 112.5 * √2/2. That simplifies to (112.5/2)*√2, which is 56.25√2. So, the area of the triangle is 56.25√2.3. **Calculating the Shaded Area for One Sector:** Now, the shaded region in one sector is the area of the sector minus the area of the triangle. So, that would be 28.125π - 56.25√2. That makes sense because the triangle is the part that's not shaded in the sector.4. **Doubling the Area for Both Sectors:** Since the problem states that the two parts of the shaded region are equal, I need to calculate the area for both. So, I take the area I just found for one sector and double it. That gives me 2*(28.125π - 56.25√2). Let me compute that. 2 times 28.125π is 56.25π, and 2 times 56.25√2 is 112.5√2. So, the total shaded area is 56.25π - 112.5√2.Wait, let me double-check my calculations to make sure I didn't make any mistakes.- **Sector Area Check:** 45/360 is indeed 1/8. 1/8 of 225π is 28.125π. That seems correct.- **Triangle Area Check:** (1/2)*15*15*sin(45°) is (1/2)*225*(√2/2). That simplifies to (225/4)*√2, which is 56.25√2. Correct.- **Subtraction Check:** 28.125π - 56.25√2 for one sector. Then doubling it gives 56.25π - 112.5√2. That seems right.Hmm, I wonder if there's another way to approach this problem, maybe using integration or something else, but since it's a standard geometry problem, the method I used should suffice.Another thought: is the overlapping area actually just the sum of the two sectors minus the area of their intersection? Wait, no, in this case, the shaded region is specifically the overlapping part, so it's actually the intersection area. But in my approach, I considered each sector and subtracted the triangle, then doubled it. Is that equivalent?Wait, no, actually, the overlapping area is the intersection of the two sectors. So, perhaps I should think of it as the area common to both sectors. Each sector contributes a segment, and the total overlapping area is the sum of these two segments.But in my initial approach, I calculated the area of one segment (sector minus triangle) and then doubled it. So, that should give the total overlapping area. Yes, that makes sense because each segment is identical, so doubling one gives the total.Alternatively, another way to compute the overlapping area is to calculate the area of the two circular segments. Each segment is the area of the sector minus the area of the triangle. Since there are two such segments, the total overlapping area is 2*(sector area - triangle area). Which is exactly what I did.So, I think my approach is correct. Therefore, the area of the shaded region is 56.25π - 112.5√2.Just to make sure, let me plug in the numbers numerically to see if it makes sense.Compute 56.25π: π is approximately 3.1416, so 56.25 * 3.1416 ≈ 56.25 * 3.1416 ≈ 176.7146.Compute 112.5√2: √2 is approximately 1.4142, so 112.5 * 1.4142 ≈ 112.5 * 1.4142 ≈ 159.1875.Subtracting these: 176.7146 - 159.1875 ≈ 17.5271.So, the area is approximately 17.53 square units. Does that make sense?Given that the radius is 15, which is quite large, but the central angle is only 45 degrees, so the overlapping area shouldn't be too big. 17.53 seems reasonable.Alternatively, let me compute the area of one sector: 28.125π ≈ 28.125 * 3.1416 ≈ 88.357.Area of the triangle: 56.25√2 ≈ 56.25 * 1.4142 ≈ 79.591.So, the segment area is 88.357 - 79.591 ≈ 8.766. Then, doubling it gives approximately 17.532, which matches my earlier calculation. So, that seems consistent.Therefore, I think my answer is correct.**Final Answer**The area of the shaded region is boxed{56.25pi - 112.5sqrt{2}}.

💡Okay, so I have this geometry problem about triangle ABC. Let me try to visualize it first. We know that AC is 2 units long, and the angle at A, which is angle BAC, is π/3 radians, which is 60 degrees. There's a point P inside the triangle such that AP is perpendicular to CP, and the angle at APB is 2π/3 radians, which is 120 degrees. The problem has two parts. Part (1) says that if AP equals PC, find the area of triangle ABC. Part (2) says that if BC is √7, find AP. I'll start with part (1).First, let me draw triangle ABC. Point A is at the top, B is at the bottom left, and C is at the bottom right. AC is 2 units, angle BAC is 60 degrees. Point P is somewhere inside the triangle. AP is perpendicular to CP, so triangle APC is a right triangle at P. Also, angle APB is 120 degrees.Since AP equals PC, triangle APC is an isosceles right triangle. In an isosceles right triangle, the legs are equal, and the hypotenuse is √2 times the legs. So, if AP = PC, then AC is the hypotenuse, which is 2 units. So, AP = PC = 2 / √2 = √2. That makes sense.Now, in triangle ABC, we know AC = 2, angle BAC = 60 degrees, and we need to find the area. But wait, we also have point P inside the triangle. Maybe we can use some properties of triangle APC to find other sides or angles.Since AP is perpendicular to CP, and AP = PC, the angle at A in triangle APC is 45 degrees because it's an isosceles right triangle. So, angle CAP is 45 degrees. But angle BAC is 60 degrees, so angle BAP would be angle BAC minus angle CAP, which is 60 - 45 = 15 degrees. So, angle BAP is 15 degrees.Now, looking at triangle APB, we know angle APB is 120 degrees, angle BAP is 15 degrees, so the remaining angle at B, which is angle ABP, should be 180 - 120 - 15 = 45 degrees. So, angle ABP is 45 degrees.Now, in triangle APB, we can apply the Law of Sines. The Law of Sines states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant. So, in triangle APB:AP / sin(angle ABP) = AB / sin(angle APB)We know AP is √2, angle ABP is 45 degrees, angle APB is 120 degrees. So plugging in the values:√2 / sin(45°) = AB / sin(120°)We know sin(45°) is √2/2 and sin(120°) is √3/2. So:√2 / (√2/2) = AB / (√3/2)Simplify the left side: √2 divided by (√2/2) is 2. So:2 = AB / (√3/2)Multiply both sides by (√3/2):AB = 2 * (√3/2) = √3So, AB is √3 units long.Now, to find the area of triangle ABC, we can use the formula:Area = (1/2) * AB * AC * sin(angle BAC)We know AB is √3, AC is 2, and angle BAC is 60 degrees, whose sine is √3/2. Plugging in:Area = (1/2) * √3 * 2 * (√3/2)Simplify:(1/2) * √3 * 2 cancels out to √3, and then multiplied by √3/2 gives:√3 * √3 / 2 = 3/2So, the area of triangle ABC is 3/2.Wait, that seems straightforward. Let me just double-check. If AP = PC = √2, and AC = 2, that's correct because in an isosceles right triangle, the hypotenuse is √2 times the legs. Then, angle CAP is 45 degrees, so angle BAP is 15 degrees, leading to angle ABP being 45 degrees. Using the Law of Sines, we found AB = √3. Then, using the area formula, we got 3/2. Yeah, that seems right.Now, moving on to part (2). If BC is √7, find AP.Alright, so now BC is given as √7. We still have AC = 2, angle BAC = 60 degrees, and point P inside the triangle such that AP is perpendicular to CP, and angle APB is 120 degrees. We need to find AP.First, let's recall the Law of Cosines for triangle ABC. The Law of Cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. The formula is:BC² = AB² + AC² - 2 * AB * AC * cos(angle BAC)We know BC is √7, AC is 2, angle BAC is 60 degrees, whose cosine is 0.5. Let's plug in the values:(√7)² = AB² + 2² - 2 * AB * 2 * cos(60°)Simplify:7 = AB² + 4 - 4 * AB * 0.5Which simplifies to:7 = AB² + 4 - 2 * ABBring all terms to one side:AB² - 2 * AB + 4 - 7 = 0So:AB² - 2 * AB - 3 = 0This is a quadratic equation in terms of AB. Let's solve it using the quadratic formula:AB = [2 ± √(4 + 12)] / 2 = [2 ± √16] / 2 = [2 ± 4] / 2So, AB can be (2 + 4)/2 = 3 or (2 - 4)/2 = -1. Since length can't be negative, AB = 3.So, AB is 3 units long.Now, we need to find AP. Let's denote angle CAP as α. Then, angle BAP would be angle BAC - angle CAP = 60° - α.In triangle APC, since AP is perpendicular to CP, triangle APC is a right triangle at P. So, AP is one leg, PC is the other leg, and AC is the hypotenuse. So, AC = 2, so AP² + PC² = AC² = 4.But we don't know if AP = PC here, unlike part (1). So, we can't assume that. Instead, we can express PC in terms of AP. Let's denote AP = x. Then, PC = √(4 - x²).Now, looking at triangle APB, we know angle APB is 120 degrees, angle BAP is 60° - α, and angle ABP is something we need to find.Wait, maybe we can use the Law of Sines in triangle APB. Let's denote angle ABP as β. Then, in triangle APB, the sum of angles is 180°, so:angle BAP + angle APB + angle ABP = 180°(60° - α) + 120° + β = 180°So, β = 180° - 120° - (60° - α) = 0° + α = αSo, angle ABP is α. Interesting.Now, in triangle APB, using the Law of Sines:AP / sin(angle ABP) = AB / sin(angle APB)We have AP = x, angle ABP = α, AB = 3, angle APB = 120°, whose sine is √3/2.So:x / sin(α) = 3 / (√3/2) = 3 * 2 / √3 = 6 / √3 = 2√3So, x = 2√3 * sin(α)But we also need to relate α to something else. Let's look at triangle APC. Since AP is perpendicular to CP, angle APC is 90°, and angle CAP is α. So, in triangle APC, angle PCA is 90° - α.But angle PCA is also part of triangle ABC. Let me see. In triangle ABC, angle at C is angle ACB. We can find angle ACB using the Law of Cosines or Law of Sines.Wait, we know sides AB = 3, AC = 2, BC = √7. Let's use the Law of Cosines to find angle ACB.Law of Cosines:cos(angle ACB) = (AC² + BC² - AB²) / (2 * AC * BC)Plugging in:cos(angle ACB) = (2² + (√7)² - 3²) / (2 * 2 * √7) = (4 + 7 - 9) / (4√7) = (2) / (4√7) = 1 / (2√7)So, angle ACB = arccos(1 / (2√7)). Let me compute that value.But maybe we don't need the exact angle. Instead, let's consider triangle APC. In triangle APC, angle at C is angle PCA = 90° - α, and angle at A is α, so angle at P is 90°.But angle ACB in triangle ABC is equal to angle PCA + angle PCB. Wait, but point P is inside the triangle, so angle ACB is just angle PCA? Or is there another angle?Wait, no. Point P is inside triangle ABC, so angle ACB is split into two angles: angle PCA and angle PCB. But since AP is perpendicular to CP, angle APC is 90°, so angle PCA is 90° - α.But I'm not sure if that helps directly. Maybe we can use the Law of Sines in triangle APC.In triangle APC, we have:AP / sin(angle PCA) = AC / sin(angle APC)Which is:x / sin(90° - α) = 2 / sin(90°)Since sin(90°) is 1, this simplifies to:x / cos(α) = 2So, x = 2 cos(α)But earlier, we had x = 2√3 sin(α). So, equating the two expressions:2√3 sin(α) = 2 cos(α)Divide both sides by 2:√3 sin(α) = cos(α)Divide both sides by cos(α):√3 tan(α) = 1So, tan(α) = 1 / √3Which means α = arctan(1 / √3) = 30°, since tan(30°) = 1/√3.So, α is 30 degrees.Now, going back to x = 2 cos(α). Since α is 30°, cos(30°) is √3/2. So:x = 2 * (√3 / 2) = √3Therefore, AP is √3.Let me just verify this. If AP is √3, then PC would be √(4 - (√3)²) = √(4 - 3) = √1 = 1. So, PC is 1.In triangle APB, using the Law of Sines, we had x = 2√3 sin(α). Since α is 30°, sin(30°) is 1/2, so x = 2√3 * 1/2 = √3. That matches.Also, in triangle APC, angle PCA is 90° - 30° = 60°, and angle ACB in triangle ABC was found to be arccos(1 / (2√7)). Let me compute that angle.cos(angle ACB) = 1 / (2√7) ≈ 1 / (2 * 2.6458) ≈ 1 / 5.2916 ≈ 0.1889. So, angle ACB ≈ arccos(0.1889) ≈ 79.1 degrees.But in triangle APC, angle PCA is 60°, which is part of angle ACB. So, angle ACB is 60° + angle PCB. But since angle ACB is approximately 79.1°, angle PCB would be about 19.1°, which seems reasonable.I think everything checks out. So, AP is √3.**Final Answer**(1) The area of triangle ABC is boxed{dfrac{3}{2}}.(2) The length of AP is boxed{sqrt{3}}.

💡Okay, so I have this problem where I need to find the minimum value of the expression:[frac{6z}{x + 2y + k} + frac{6x}{2z + y + k} + frac{3y}{x + z + k}]where ( x, y, z ) are positive real numbers and ( k ) is a positive constant. Hmm, it looks like a problem that might involve some inequality techniques, maybe Cauchy-Schwarz or AM-GM. Let me think about how to approach this.First, I notice that each term in the expression is a fraction where the numerator is a multiple of one variable and the denominator is a linear combination of the other variables plus ( k ). This symmetry suggests that perhaps we can use some substitution or scaling to simplify the expression.Let me try substituting variables to make the denominators look more similar. Let me set:[a = x, quad b = 2y, quad c = 2z]So, substituting back, we have:[x = a, quad y = frac{b}{2}, quad z = frac{c}{2}]Now, let's rewrite the original expression in terms of ( a, b, c ):1. The first term:[frac{6z}{x + 2y + k} = frac{6 cdot frac{c}{2}}{a + 2 cdot frac{b}{2} + k} = frac{3c}{a + b + k}]2. The second term:[frac{6x}{2z + y + k} = frac{6a}{2 cdot frac{c}{2} + frac{b}{2} + k} = frac{6a}{c + frac{b}{2} + k}]3. The third term:[frac{3y}{x + z + k} = frac{3 cdot frac{b}{2}}{a + frac{c}{2} + k} = frac{frac{3b}{2}}{a + frac{c}{2} + k}]So, putting it all together, the expression becomes:[frac{3c}{a + b + k} + frac{6a}{c + frac{b}{2} + k} + frac{frac{3b}{2}}{a + frac{c}{2} + k}]Hmm, this still looks a bit messy. Maybe I can simplify the denominators by multiplying numerator and denominator appropriately. Let me see:For the second term:[frac{6a}{c + frac{b}{2} + k} = frac{12a}{2c + b + 2k}]And the third term:[frac{frac{3b}{2}}{a + frac{c}{2} + k} = frac{3b}{2a + c + 2k}]So now, the expression becomes:[frac{3c}{a + b + k} + frac{12a}{2c + b + 2k} + frac{3b}{2a + c + 2k}]Hmm, this looks a bit more manageable. Let me denote the denominators as follows:Let ( D_1 = a + b + k ), ( D_2 = 2c + b + 2k ), and ( D_3 = 2a + c + 2k ).So, the expression is:[frac{3c}{D_1} + frac{12a}{D_2} + frac{3b}{D_3}]I wonder if I can apply the Cauchy-Schwarz inequality here. The Cauchy-Schwarz inequality in the form:[left( sum frac{u_i^2}{v_i} right) geq frac{(sum u_i)^2}{sum v_i}]But in our case, the numerators are linear in variables, not squares. Maybe I need to adjust the terms or use another inequality.Alternatively, I can think about homogenizing the expression. Since we have ( k ) as a constant, perhaps setting ( k = 1 ) without loss of generality, but I'm not sure if that's valid here.Wait, maybe I can consider the variables ( a, b, c ) and ( k ) as part of a system where I can set some ratios. Let me assume that the minimum occurs when ( a = b = c ). Let's test this assumption.If ( a = b = c = t ), then substituting back:1. ( D_1 = t + t + k = 2t + k )2. ( D_2 = 2t + t + 2k = 3t + 2k )3. ( D_3 = 2t + t + 2k = 3t + 2k )So, the expression becomes:[frac{3t}{2t + k} + frac{12t}{3t + 2k} + frac{3t}{3t + 2k}]Simplify each term:1. ( frac{3t}{2t + k} )2. ( frac{12t}{3t + 2k} )3. ( frac{3t}{3t + 2k} )Combine the second and third terms:[frac{12t + 3t}{3t + 2k} = frac{15t}{3t + 2k}]So, the total expression is:[frac{3t}{2t + k} + frac{15t}{3t + 2k}]Let me compute this for some specific value of ( t ). Let's set ( t = k ):Then,1. ( frac{3k}{2k + k} = frac{3k}{3k} = 1 )2. ( frac{15k}{3k + 2k} = frac{15k}{5k} = 3 )So, total expression is ( 1 + 3 = 4 ).Wait, that's interesting. If ( t = k ), the expression equals 4. But is this the minimum? Maybe I can try another value.Let me set ( t = 2k ):1. ( frac{3 cdot 2k}{2 cdot 2k + k} = frac{6k}{5k} = frac{6}{5} = 1.2 )2. ( frac{15 cdot 2k}{3 cdot 2k + 2k} = frac{30k}{8k} = frac{30}{8} = 3.75 )Total expression: ( 1.2 + 3.75 = 4.95 ), which is higher than 4.What if ( t = frac{k}{2} ):1. ( frac{3 cdot frac{k}{2}}{2 cdot frac{k}{2} + k} = frac{frac{3k}{2}}{k + k} = frac{frac{3k}{2}}{2k} = frac{3}{4} = 0.75 )2. ( frac{15 cdot frac{k}{2}}{3 cdot frac{k}{2} + 2k} = frac{frac{15k}{2}}{frac{3k}{2} + 2k} = frac{frac{15k}{2}}{frac{7k}{2}} = frac{15}{7} approx 2.14 )Total expression: ( 0.75 + 2.14 approx 2.89 ). Hmm, that's lower than 4. Wait, but this contradicts my previous result. Maybe my assumption that ( a = b = c ) is not valid.Wait, no, because when I set ( a = b = c = t ), I might have forced the variables into a specific ratio that doesn't necessarily minimize the expression. So, perhaps the minimum occurs when ( a, b, c ) are in a different ratio.Alternatively, maybe I should consider using the method of Lagrange multipliers to find the minimum. Let me set up the function:Let ( f(a, b, c) = frac{3c}{a + b + k} + frac{12a}{2c + b + 2k} + frac{3b}{2a + c + 2k} )We need to find the minimum of ( f(a, b, c) ) subject to ( a, b, c > 0 ).To use Lagrange multipliers, I would set the partial derivatives equal to zero. Let's compute the partial derivatives.First, compute ( frac{partial f}{partial a} ):1. The derivative of ( frac{3c}{a + b + k} ) with respect to ( a ) is ( -frac{3c}{(a + b + k)^2} )2. The derivative of ( frac{12a}{2c + b + 2k} ) with respect to ( a ) is ( frac{12}{2c + b + 2k} )3. The derivative of ( frac{3b}{2a + c + 2k} ) with respect to ( a ) is ( -frac{6b}{(2a + c + 2k)^2} )So,[frac{partial f}{partial a} = -frac{3c}{(a + b + k)^2} + frac{12}{2c + b + 2k} - frac{6b}{(2a + c + 2k)^2}]Similarly, compute ( frac{partial f}{partial b} ):1. The derivative of ( frac{3c}{a + b + k} ) with respect to ( b ) is ( -frac{3c}{(a + b + k)^2} )2. The derivative of ( frac{12a}{2c + b + 2k} ) with respect to ( b ) is ( -frac{12a}{(2c + b + 2k)^2} )3. The derivative of ( frac{3b}{2a + c + 2k} ) with respect to ( b ) is ( frac{3}{2a + c + 2k} )So,[frac{partial f}{partial b} = -frac{3c}{(a + b + k)^2} - frac{12a}{(2c + b + 2k)^2} + frac{3}{2a + c + 2k}]Now, compute ( frac{partial f}{partial c} ):1. The derivative of ( frac{3c}{a + b + k} ) with respect to ( c ) is ( frac{3}{a + b + k} )2. The derivative of ( frac{12a}{2c + b + 2k} ) with respect to ( c ) is ( -frac{24a}{(2c + b + 2k)^2} )3. The derivative of ( frac{3b}{2a + c + 2k} ) with respect to ( c ) is ( -frac{3b}{(2a + c + 2k)^2} )So,[frac{partial f}{partial c} = frac{3}{a + b + k} - frac{24a}{(2c + b + 2k)^2} - frac{3b}{(2a + c + 2k)^2}]Setting each partial derivative equal to zero:1. ( -frac{3c}{(a + b + k)^2} + frac{12}{2c + b + 2k} - frac{6b}{(2a + c + 2k)^2} = 0 )2. ( -frac{3c}{(a + b + k)^2} - frac{12a}{(2c + b + 2k)^2} + frac{3}{2a + c + 2k} = 0 )3. ( frac{3}{a + b + k} - frac{24a}{(2c + b + 2k)^2} - frac{3b}{(2a + c + 2k)^2} = 0 )This system of equations looks quite complicated. Maybe there's a symmetry or substitution that can simplify things.Looking back at the original expression, I notice that the coefficients in the numerators are 3, 12, and 3. Maybe I can factor out a 3:[3 left( frac{c}{a + b + k} + frac{4a}{2c + b + 2k} + frac{b}{2a + c + 2k} right)]So, the expression is 3 times another expression. Let me denote:[T = frac{c}{a + b + k} + frac{4a}{2c + b + 2k} + frac{b}{2a + c + 2k}]So, our original expression is ( 3T ). Therefore, minimizing the original expression is equivalent to minimizing ( T ).Maybe I can apply the Cauchy-Schwarz inequality to ( T ). Let me recall that Cauchy-Schwarz can be used in the form:[left( sum frac{u_i^2}{v_i} right) geq frac{(sum u_i)^2}{sum v_i}]But in our case, the terms are linear, not squared. Alternatively, I can use the Titu's lemma, which is a specific case of Cauchy-Schwarz:[sum frac{u_i^2}{v_i} geq frac{(sum u_i)^2}{sum v_i}]But again, our terms are linear, so maybe I need to adjust them.Alternatively, I can consider the variables ( a, b, c ) and set up ratios such that the terms are balanced. Let me assume that the denominators are proportional to the numerators. That is, for the first term:[c propto a + b + k]Similarly, for the second term:[4a propto 2c + b + 2k]And for the third term:[b propto 2a + c + 2k]Let me denote the proportionality constants as ( lambda ), so:1. ( c = lambda (a + b + k) )2. ( 4a = lambda (2c + b + 2k) )3. ( b = lambda (2a + c + 2k) )Now, we have a system of equations:1. ( c = lambda (a + b + k) )2. ( 4a = lambda (2c + b + 2k) )3. ( b = lambda (2a + c + 2k) )Let me try to solve this system.From equation 1: ( c = lambda a + lambda b + lambda k )From equation 3: ( b = 2lambda a + lambda c + 2lambda k )Let me substitute ( c ) from equation 1 into equation 3:( b = 2lambda a + lambda (lambda a + lambda b + lambda k) + 2lambda k )Simplify:( b = 2lambda a + lambda^2 a + lambda^2 b + lambda^2 k + 2lambda k )Bring all terms to the left:( b - 2lambda a - lambda^2 a - lambda^2 b - lambda^2 k - 2lambda k = 0 )Factor terms:( b(1 - lambda^2) - a(2lambda + lambda^2) - k(lambda^2 + 2lambda) = 0 )Similarly, from equation 2:( 4a = lambda (2c + b + 2k) )Substitute ( c ) from equation 1:( 4a = lambda (2(lambda a + lambda b + lambda k) + b + 2k) )Simplify:( 4a = lambda (2lambda a + 2lambda b + 2lambda k + b + 2k) )( 4a = lambda (2lambda a + (2lambda + 1)b + (2lambda + 2)k) )So,( 4a = 2lambda^2 a + lambda(2lambda + 1)b + lambda(2lambda + 2)k )Bring all terms to the left:( 4a - 2lambda^2 a - lambda(2lambda + 1)b - lambda(2lambda + 2)k = 0 )Factor terms:( a(4 - 2lambda^2) - blambda(2lambda + 1) - klambda(2lambda + 2) = 0 )Now, we have two equations:1. ( b(1 - lambda^2) - a(2lambda + lambda^2) - k(lambda^2 + 2lambda) = 0 )2. ( a(4 - 2lambda^2) - blambda(2lambda + 1) - klambda(2lambda + 2) = 0 )This is getting quite involved. Maybe I can express ( a ) and ( b ) in terms of ( k ) and ( lambda ).Let me denote ( a = m k ) and ( b = n k ), where ( m ) and ( n ) are constants to be determined. Then, ( c = lambda (a + b + k) = lambda (m k + n k + k) = lambda ( (m + n + 1)k ) ).So, ( c = lambda (m + n + 1)k ).Now, substitute ( a = m k ), ( b = n k ), ( c = lambda (m + n + 1)k ) into the two equations.First equation:( b(1 - lambda^2) - a(2lambda + lambda^2) - k(lambda^2 + 2lambda) = 0 )Substitute:( n k (1 - lambda^2) - m k (2lambda + lambda^2) - k(lambda^2 + 2lambda) = 0 )Divide both sides by ( k ):( n(1 - lambda^2) - m(2lambda + lambda^2) - (lambda^2 + 2lambda) = 0 )Similarly, second equation:( a(4 - 2lambda^2) - blambda(2lambda + 1) - klambda(2lambda + 2) = 0 )Substitute:( m k (4 - 2lambda^2) - n k lambda (2lambda + 1) - k lambda (2lambda + 2) = 0 )Divide by ( k ):( m(4 - 2lambda^2) - n lambda (2lambda + 1) - lambda (2lambda + 2) = 0 )So now, we have two equations in terms of ( m ) and ( n ):1. ( n(1 - lambda^2) - m(2lambda + lambda^2) - (lambda^2 + 2lambda) = 0 )2. ( m(4 - 2lambda^2) - n lambda (2lambda + 1) - lambda (2lambda + 2) = 0 )Let me write these as:1. ( -m(2lambda + lambda^2) + n(1 - lambda^2) = lambda^2 + 2lambda )2. ( m(4 - 2lambda^2) - n lambda (2lambda + 1) = lambda (2lambda + 2) )Let me denote this as a system:[begin{cases}- (2lambda + lambda^2) m + (1 - lambda^2) n = lambda^2 + 2lambda (4 - 2lambda^2) m - lambda (2lambda + 1) n = lambda (2lambda + 2)end{cases}]This is a linear system in ( m ) and ( n ). Let me write it in matrix form:[begin{pmatrix}- (2lambda + lambda^2) & 1 - lambda^2 4 - 2lambda^2 & - lambda (2lambda + 1)end{pmatrix}begin{pmatrix}m nend{pmatrix}=begin{pmatrix}lambda^2 + 2lambda lambda (2lambda + 2)end{pmatrix}]To solve for ( m ) and ( n ), I can use Cramer's rule or find the inverse of the matrix. Let me compute the determinant of the coefficient matrix.Let ( D = begin{vmatrix} - (2lambda + lambda^2) & 1 - lambda^2 4 - 2lambda^2 & - lambda (2lambda + 1) end{vmatrix} )Compute ( D ):[D = (- (2lambda + lambda^2))(- lambda (2lambda + 1)) - (1 - lambda^2)(4 - 2lambda^2)]Simplify term by term:First term:[(- (2lambda + lambda^2))(- lambda (2lambda + 1)) = (2lambda + lambda^2)(lambda (2lambda + 1)) = lambda (2lambda + lambda^2)(2lambda + 1)]Let me expand ( (2lambda + lambda^2)(2lambda + 1) ):[2lambda cdot 2lambda + 2lambda cdot 1 + lambda^2 cdot 2lambda + lambda^2 cdot 1 = 4lambda^2 + 2lambda + 2lambda^3 + lambda^2 = 2lambda^3 + 5lambda^2 + 2lambda]So, the first term is ( lambda (2lambda^3 + 5lambda^2 + 2lambda) = 2lambda^4 + 5lambda^3 + 2lambda^2 )Second term:[(1 - lambda^2)(4 - 2lambda^2) = 1 cdot 4 + 1 cdot (-2lambda^2) - lambda^2 cdot 4 + lambda^2 cdot 2lambda^2 = 4 - 2lambda^2 - 4lambda^2 + 2lambda^4 = 2lambda^4 - 6lambda^2 + 4]So, the determinant ( D ) is:[D = (2lambda^4 + 5lambda^3 + 2lambda^2) - (2lambda^4 - 6lambda^2 + 4) = 2lambda^4 + 5lambda^3 + 2lambda^2 - 2lambda^4 + 6lambda^2 - 4 = 5lambda^3 + 8lambda^2 - 4]Now, compute ( D_m ), the determinant for ( m ):Replace the first column with the constants:[D_m = begin{vmatrix} lambda^2 + 2lambda & 1 - lambda^2 lambda (2lambda + 2) & - lambda (2lambda + 1) end{vmatrix}]Compute ( D_m ):[(lambda^2 + 2lambda)(- lambda (2lambda + 1)) - (1 - lambda^2)(lambda (2lambda + 2))]Simplify term by term:First term:[(lambda^2 + 2lambda)(- lambda (2lambda + 1)) = -lambda (2lambda + 1)(lambda^2 + 2lambda)]Let me expand ( (2lambda + 1)(lambda^2 + 2lambda) ):[2lambda cdot lambda^2 + 2lambda cdot 2lambda + 1 cdot lambda^2 + 1 cdot 2lambda = 2lambda^3 + 4lambda^2 + lambda^2 + 2lambda = 2lambda^3 + 5lambda^2 + 2lambda]So, the first term is ( -lambda (2lambda^3 + 5lambda^2 + 2lambda) = -2lambda^4 - 5lambda^3 - 2lambda^2 )Second term:[(1 - lambda^2)(lambda (2lambda + 2)) = lambda (2lambda + 2)(1 - lambda^2) = lambda (2lambda + 2 - 2lambda^3 - 2lambda^2)]Simplify:[2lambda^2 + 2lambda - 2lambda^4 - 2lambda^3]So, the second term is ( 2lambda^2 + 2lambda - 2lambda^4 - 2lambda^3 )Thus, ( D_m = (-2lambda^4 - 5lambda^3 - 2lambda^2) - (2lambda^2 + 2lambda - 2lambda^4 - 2lambda^3) )Simplify:[-2lambda^4 - 5lambda^3 - 2lambda^2 - 2lambda^2 - 2lambda + 2lambda^4 + 2lambda^3 = (-2lambda^4 + 2lambda^4) + (-5lambda^3 + 2lambda^3) + (-2lambda^2 - 2lambda^2) - 2lambda = -3lambda^3 - 4lambda^2 - 2lambda]Similarly, compute ( D_n ), the determinant for ( n ):Replace the second column with the constants:[D_n = begin{vmatrix} - (2lambda + lambda^2) & lambda^2 + 2lambda 4 - 2lambda^2 & lambda (2lambda + 2) end{vmatrix}]Compute ( D_n ):[(- (2lambda + lambda^2))(lambda (2lambda + 2)) - (lambda^2 + 2lambda)(4 - 2lambda^2)]Simplify term by term:First term:[- (2lambda + lambda^2)(lambda (2lambda + 2)) = -lambda (2lambda + lambda^2)(2lambda + 2)]Expand ( (2lambda + lambda^2)(2lambda + 2) ):[2lambda cdot 2lambda + 2lambda cdot 2 + lambda^2 cdot 2lambda + lambda^2 cdot 2 = 4lambda^2 + 4lambda + 2lambda^3 + 2lambda^2 = 2lambda^3 + 6lambda^2 + 4lambda]So, the first term is ( -lambda (2lambda^3 + 6lambda^2 + 4lambda) = -2lambda^4 - 6lambda^3 - 4lambda^2 )Second term:[(lambda^2 + 2lambda)(4 - 2lambda^2) = lambda^2 cdot 4 + lambda^2 cdot (-2lambda^2) + 2lambda cdot 4 + 2lambda cdot (-2lambda^2) = 4lambda^2 - 2lambda^4 + 8lambda - 4lambda^3]So, the second term is ( 4lambda^2 - 2lambda^4 + 8lambda - 4lambda^3 )Thus, ( D_n = (-2lambda^4 - 6lambda^3 - 4lambda^2) - (4lambda^2 - 2lambda^4 + 8lambda - 4lambda^3) )Simplify:[-2lambda^4 - 6lambda^3 - 4lambda^2 - 4lambda^2 + 2lambda^4 - 8lambda + 4lambda^3 = (-2lambda^4 + 2lambda^4) + (-6lambda^3 + 4lambda^3) + (-4lambda^2 - 4lambda^2) - 8lambda = -2lambda^3 - 8lambda^2 - 8lambda]Now, we have:[m = frac{D_m}{D} = frac{-3lambda^3 - 4lambda^2 - 2lambda}{5lambda^3 + 8lambda^2 - 4}][n = frac{D_n}{D} = frac{-2lambda^3 - 8lambda^2 - 8lambda}{5lambda^3 + 8lambda^2 - 4}]This is quite complicated. Maybe I can factor the numerator and denominator.Looking at ( D = 5lambda^3 + 8lambda^2 - 4 ). Let me try to factor this.Try ( lambda = 1 ): ( 5 + 8 - 4 = 9 neq 0 )Try ( lambda = -2 ): ( -40 + 32 - 4 = -12 neq 0 )Try ( lambda = frac{2}{5} ): ( 5*(8/125) + 8*(4/25) - 4 = 40/125 + 32/25 - 4 = 0.32 + 1.28 - 4 = -2.4 neq 0 )Not obvious. Maybe use rational root theorem: possible roots are ( pm1, pm2, pm4, pm1/5, pm2/5, pm4/5 ). Testing ( lambda = 2/5 ):( 5*(8/125) + 8*(4/25) - 4 = 40/125 + 32/25 - 4 = 0.32 + 1.28 - 4 = -2.4 neq 0 )Not a root. Maybe it doesn't factor nicely. Perhaps I need a different approach.Alternatively, maybe assume that ( lambda ) is a simple fraction, like ( lambda = 1 ). Let me test ( lambda = 1 ):Then,( D = 5 + 8 - 4 = 9 )( D_m = -3 -4 -2 = -9 )( D_n = -2 -8 -8 = -18 )So,( m = (-9)/9 = -1 )( n = (-18)/9 = -2 )But ( m ) and ( n ) are supposed to be positive since ( a, b, c ) are positive. So, ( lambda = 1 ) gives negative ( m ) and ( n ), which is invalid.Try ( lambda = 2 ):( D = 5*8 + 8*4 -4 = 40 +32 -4=68 )( D_m = -3*8 -4*4 -2*2= -24 -16 -4= -44 )( D_n = -2*8 -8*4 -8*2= -16 -32 -16= -64 )So,( m = -44/68 = -11/17 )( n = -64/68 = -16/17 )Still negative. Not good.Try ( lambda = 1/2 ):( D = 5*(1/8) + 8*(1/4) -4= 5/8 + 2 -4= 5/8 -2= -11/8 )( D_m = -3*(1/8) -4*(1/4) -2*(1/2)= -3/8 -1 -1= -3/8 -2= -19/8 )( D_n = -2*(1/8) -8*(1/4) -8*(1/2)= -1/4 -2 -4= -1/4 -6= -25/4 )So,( m = (-19/8)/(-11/8)= 19/11 approx 1.727 )( n = (-25/4)/(-11/8)= (25/4)*(8/11)= 50/11 approx 4.545 )Positive values! So, ( lambda = 1/2 ) gives positive ( m ) and ( n ).Let me check if this satisfies the equations.Compute ( m = 19/11 ), ( n = 50/11 ), ( lambda = 1/2 )From equation 1:( - (2*(1/2) + (1/2)^2) m + (1 - (1/2)^2) n = (1/2)^2 + 2*(1/2) )Simplify:( - (1 + 1/4) m + (1 - 1/4) n = 1/4 + 1 )( - (5/4) m + (3/4) n = 5/4 )Multiply both sides by 4:( -5m + 3n = 5 )Substitute ( m = 19/11 ), ( n = 50/11 ):( -5*(19/11) + 3*(50/11) = (-95 + 150)/11 = 55/11 = 5 ). Correct.Similarly, equation 2:( (4 - 2*(1/2)^2) m - (1/2)*(2*(1/2) + 1) n = (1/2)*(2*(1/2) + 2) )Simplify:( (4 - 2*(1/4)) m - (1/2)*(2 + 1) n = (1/2)*(1 + 2) )( (4 - 1/2) m - (3/2) n = 3/2 )( (7/2) m - (3/2) n = 3/2 )Multiply both sides by 2:( 7m - 3n = 3 )Substitute ( m = 19/11 ), ( n = 50/11 ):( 7*(19/11) - 3*(50/11) = (133 - 150)/11 = (-17)/11 neq 3 ). Hmm, that's not correct. Did I make a mistake?Wait, let me recompute equation 2.Original equation 2 after substitution:( (4 - 2lambda^2) m - lambda (2lambda + 1) n = lambda (2lambda + 2) )With ( lambda = 1/2 ):( (4 - 2*(1/4)) m - (1/2)(2*(1/2) + 1) n = (1/2)(2*(1/2) + 2) )Simplify:( (4 - 1/2) m - (1/2)(1 + 1) n = (1/2)(1 + 2) )( (7/2) m - (1/2)(2) n = (1/2)(3) )( (7/2) m - n = 3/2 )Multiply both sides by 2:( 7m - 2n = 3 )Now, substitute ( m = 19/11 ), ( n = 50/11 ):( 7*(19/11) - 2*(50/11) = (133 - 100)/11 = 33/11 = 3 ). Correct.So, my earlier mistake was in simplifying equation 2. It should be ( 7m - 2n = 3 ), not ( 7m - 3n = 3 ). So, with ( m = 19/11 ), ( n = 50/11 ), it's correct.Therefore, ( lambda = 1/2 ) gives a valid solution with positive ( m ) and ( n ).So, ( m = 19/11 ), ( n = 50/11 ), ( lambda = 1/2 )Now, recall that ( a = m k = (19/11)k ), ( b = n k = (50/11)k ), and ( c = lambda (m + n + 1)k = (1/2)(19/11 + 50/11 + 1)k )Compute ( c ):( 19/11 + 50/11 = 69/11 ), plus 1 is ( 69/11 + 11/11 = 80/11 )So, ( c = (1/2)(80/11)k = 40/11 k )So, ( a = 19k/11 ), ( b = 50k/11 ), ( c = 40k/11 )Now, let's compute ( T ):[T = frac{c}{a + b + k} + frac{4a}{2c + b + 2k} + frac{b}{2a + c + 2k}]Substitute the values:1. ( frac{c}{a + b + k} = frac{40k/11}{(19k/11 + 50k/11 + k)} = frac{40k/11}{(69k/11 + 11k/11)} = frac{40k/11}{80k/11} = 40/80 = 1/2 )2. ( frac{4a}{2c + b + 2k} = frac{4*(19k/11)}{2*(40k/11) + 50k/11 + 2k} = frac{76k/11}{80k/11 + 50k/11 + 22k/11} = frac{76k/11}{152k/11} = 76/152 = 1/2 )3. ( frac{b}{2a + c + 2k} = frac{50k/11}{2*(19k/11) + 40k/11 + 2k} = frac{50k/11}{38k/11 + 40k/11 + 22k/11} = frac{50k/11}{100k/11} = 50/100 = 1/2 )So, ( T = 1/2 + 1/2 + 1/2 = 3/2 )Therefore, the original expression is ( 3T = 3*(3/2) = 9/2 = 4.5 )So, the minimum value is ( 4.5 ), achieved when ( a = 19k/11 ), ( b = 50k/11 ), ( c = 40k/11 ), which correspond to ( x = a = 19k/11 ), ( y = b/2 = 25k/11 ), ( z = c/2 = 20k/11 )Thus, the minimum value is ( boxed{4.5} )

💡Okay, so I came across this problem where Maena writes 2021 integers in a row on a board. Then, each minute, she writes a new line below the previous one. The rule is that each number in the new line is the count of how many times the number directly above it appears in the previous line. The task is to show that eventually, she will end up writing the same line twice in a row.Hmm, let me try to break this down. So, starting with the first line, which has 2021 integers. Then, each subsequent line is generated by counting how many times each integer in the current line appears. So, for example, if the first line is [1, 2, 2, 3], the next line would be [1, 2, 2, 1] because 1 appears once, 2 appears twice, and 3 appears once.Wait, but in this problem, the lines are always 2021 integers long. So, each line has the same length, which is 2021. That means that each number in the new line is determined by counting occurrences of the corresponding number in the previous line.I think I need to consider how the numbers in each line relate to the numbers in the previous line. Since each number in the new line is a count, it must be a non-negative integer. Also, since each line has 2021 numbers, the count for any number can't exceed 2021. So, each number in the new line is between 0 and 2021, inclusive.But wait, can a number in the new line be zero? That would mean that the corresponding number in the previous line didn't appear at all. But since we're counting occurrences, if a number doesn't appear, its count is zero. However, in the next line, that zero would then represent the count of how many times zero appeared in the previous line. So, if a zero appears in a line, it's because the number above it didn't appear in the previous line.But this seems a bit tricky. Maybe I should think about the behavior of the numbers over time. Since each number in the new line is a count, and counts can only increase or stay the same, right? Because if a number appears, its count can't decrease in the next line. Wait, is that true?Let me think. Suppose in the first line, a number appears once. Then, in the next line, its count is 1. If in the next line, that number appears again, its count could be higher or stay the same. Wait, no, actually, the counts are determined by the previous line, not the current one. So, if a number appears multiple times in the previous line, its count in the next line will be higher.But actually, no, the count is determined by how many times the number above it appears in the previous line. So, if a number in the current line is, say, 3, that means the number above it appeared three times in the previous line. So, in the next line, the count for that number would be how many times 3 appears in the current line.Wait, this is getting a bit confusing. Maybe I should consider a small example to see what happens.Let's take a smaller case, say with 3 numbers instead of 2021. Suppose the first line is [1, 1, 2]. Then, the next line would be [2, 2, 1], because 1 appears twice, 1 appears twice, and 2 appears once. Then, the next line would be [2, 2, 1] again, because 2 appears twice, 2 appears twice, and 1 appears once. So, in this case, the line stabilizes after one step.Another example: starting with [1, 2, 3]. The next line would be [1, 1, 1], because each number appears once. Then, the next line would be [3, 3, 3], because 1 appears three times. Then, the next line would be [1, 1, 1], because 3 appears once. Then, it would cycle between [3, 3, 3] and [1, 1, 1]. Wait, so in this case, it doesn't stabilize; it cycles between two lines.But in the problem, we have to show that Maena will inevitably end up writing the same line twice in a row. So, in the first example, it stabilized immediately, so the same line was written twice. In the second example, it started cycling between two lines, so it never wrote the same line twice in a row. Hmm, but the problem says that Maena will inevitably end up writing the same line twice in a row. So, maybe my second example is not possible with 2021 numbers?Wait, no, actually, in the second example, it's cycling between two lines, so it never writes the same line twice in a row. But the problem says that Maena will inevitably end up writing the same line twice in a row. So, maybe my reasoning is flawed.Wait, perhaps the key is that with 2021 numbers, the system must eventually stabilize because the number of possible lines is finite, and each line is determined by the previous one. So, eventually, a line must repeat, leading to a cycle. But the problem says that the same line is written twice in a row, which would mean that the line stabilizes, not just repeats after some steps.Wait, so in my second example, the lines cycle between [1,1,1] and [3,3,3], so they never write the same line twice in a row. But the problem says that Maena will inevitably end up writing the same line twice in a row. So, perhaps in the case of 2021 numbers, the system must stabilize, not cycle.Hmm, maybe the difference is that with 2021 numbers, the counts can't cycle in the same way as with 3 numbers. Let me think about that.Each line is a vector of 2021 integers, each between 0 and 2021. So, the total number of possible lines is (2022)^2021, which is a finite number. Therefore, by the pigeonhole principle, eventually, a line must repeat, leading to a cycle.But the problem is to show that the same line is written twice in a row, which would mean that the line stabilizes, i.e., the next line is the same as the current one.So, perhaps in the case of 2021 numbers, the system must stabilize, meaning that after some point, the line doesn't change anymore.Wait, let's think about the behavior of the counts. Each number in the new line is the count of how many times the number above it appears in the previous line. So, if a number in the current line is k, then in the next line, the number below it will be the count of how many times k appears in the current line.So, if a number k appears m times in the current line, then in the next line, the number below each of those k's will be m.Wait, so if a number k appears m times, then in the next line, each of those positions will have m. So, if m is the count, then in the next line, m will appear m times.Wait, that seems interesting. So, if a number k appears m times, then in the next line, m appears m times. So, if m is equal to k, then we have a fixed point.Wait, so if a number k appears k times, then in the next line, k will appear k times, so the line remains the same.So, if we can show that the system must reach such a fixed point, then we're done.But how can we show that?Well, let's think about the possible values of the counts. Each count is between 0 and 2021. But in reality, since each line has 2021 numbers, the counts can't be too large.Wait, but more importantly, each count is determined by the previous line. So, the counts are non-decreasing? Or maybe not necessarily.Wait, let's think about the counts. Suppose in the current line, a number k appears m times. Then, in the next line, each of those positions will have m. So, if m is greater than k, then in the next line, m will appear m times, which might be more or less than k.Wait, this is getting a bit tangled. Maybe I should consider the behavior of the counts over time.Let me try to formalize this. Let’s denote the current line as L = [a1, a2, ..., a2021]. Then, the next line L' is constructed by replacing each ai with the number of times ai appears in L.So, L' = [count(a1 in L), count(a2 in L), ..., count(a2021 in L)].Now, let's consider the sequence of lines L0, L1, L2, ..., where L0 is the initial line, L1 is generated from L0, and so on.We need to show that there exists some m such that Lm = Lm+1.Now, since there are only finitely many possible lines (specifically, (2022)^2021), the sequence must eventually repeat a line, leading to a cycle. But the problem is to show that the same line is written twice in a row, i.e., Lm = Lm+1 for some m.So, we need to show that the system must reach a fixed point, not just enter a cycle.Hmm, perhaps we can show that the system must eventually reach a line where each number is equal to its count in the previous line, i.e., ai = count(ai in L) for all i.If such a line exists, then the next line will be the same as the current line, so we have Lm = Lm+1.So, how can we show that such a line must be reached?Well, let's consider the behavior of the counts. Each count is a non-negative integer, and the counts are determined by the previous line.Now, let's consider the sum of the counts in each line. The sum of the counts in line L is equal to the sum over i of count(ai in L). But since each count(ai in L) is the number of times ai appears in L, the sum of the counts is equal to the sum over i of (number of times ai appears in L). But this is just the sum over all ai in L of 1, which is 2021. So, the sum of the counts in each line is always 2021.Wait, that's interesting. So, regardless of the line, the sum of the counts is always 2021.Now, let's consider the maximum value in each line. Let’s denote Mm as the maximum value in line Lm.If Mm+1 > Mm, then the maximum is increasing. But since the maximum can't exceed 2021, this can't go on forever. So, eventually, the maximum must stabilize.Similarly, if Mm+1 < Mm, the maximum is decreasing, which can't go on forever either because the maximum can't be less than 1 (since we're counting occurrences, and each line has 2021 numbers, so at least one number must appear at least once).Wait, but actually, the counts can be zero. So, the maximum could potentially be zero, but that would mean that all counts are zero, which would mean that the previous line had all numbers not appearing in the line before that, which is impossible because each line has 2021 numbers.Wait, no, if a line has all zeros, that would mean that in the previous line, every number did not appear, which is impossible because the previous line has 2021 numbers, so at least one number must appear at least once.Therefore, the maximum value in each line is at least 1.So, the maximum value in each line is between 1 and 2021, and it can't increase indefinitely because it's bounded above by 2021.Therefore, the sequence of maximum values must eventually stabilize. That is, there exists some m such that Mm = Mm+1.Now, once the maximum stabilizes, what happens next?Suppose that in line Lm, the maximum value is M. Then, in line Lm+1, the maximum value is also M. So, in Lm+1, there exists at least one number equal to M.But in Lm+1, each number is the count of how many times the corresponding number in Lm appeared in Lm.So, if in Lm+1, a number is M, that means that in Lm, the corresponding number appeared M times.But since M is the maximum in Lm, that means that in Lm, the number that appears M times is the maximum.Wait, so in Lm, the maximum number is M, and it appears M times.Therefore, in Lm+1, the number M appears M times, because each occurrence of M in Lm contributes to the count in Lm+1.Wait, so in Lm+1, the number M appears M times, and since M is the maximum, all other numbers in Lm+1 are less than or equal to M.But in Lm+1, the sum of all numbers is 2021, just like in Lm.So, if M appears M times in Lm+1, the remaining 2021 - M numbers must sum to 2021 - M*M.Wait, but 2021 - M*M must be non-negative, so M*M ≤ 2021.Therefore, M ≤ sqrt(2021) ≈ 44.94, so M ≤ 44.Wait, that's interesting. So, the maximum value in any line can't exceed 44.Wait, but earlier I thought the maximum could be up to 2021, but actually, it's bounded by sqrt(2021) because M*M ≤ 2021.So, that's a crucial point. Therefore, the maximum value in each line is at most 44.Therefore, the sequence of maximum values is bounded above by 44, and it's a non-decreasing sequence until it stabilizes.Wait, no, the maximum can both increase and decrease. But since it's bounded above, it must eventually stabilize.Once the maximum stabilizes, say at M, then in the next line, M appears M times, and the rest of the numbers are less than or equal to M.But let's think about what happens in the next line after that.In Lm+1, M appears M times, and the rest of the numbers are counts of how many times each number in Lm appeared.But in Lm, the maximum was M, so all numbers in Lm are ≤ M.Therefore, in Lm+1, each number is the count of how many times a number ≤ M appeared in Lm.But since the counts are determined by the previous line, and the maximum count is M, which appears M times, the rest of the counts are ≤ M.Wait, but in Lm+1, the counts are determined by Lm, which has maximum M.So, in Lm+1, the counts can't exceed M, because the maximum number in Lm is M, and the count of M in Lm is M.Therefore, in Lm+1, the maximum is M, and it appears M times.So, in Lm+2, the counts will be determined by Lm+1.In Lm+1, M appears M times, and the rest of the numbers are counts of numbers ≤ M.Therefore, in Lm+2, the count of M will be the number of times M appears in Lm+1, which is M.Similarly, the counts of other numbers in Lm+2 will be the counts of their corresponding numbers in Lm+1.But since in Lm+1, the maximum is M, and M appears M times, the counts in Lm+2 will be similar to those in Lm+1.Wait, but actually, in Lm+2, the count of M will be M, because M appears M times in Lm+1.And the counts of other numbers will be determined by how many times they appeared in Lm+1.But in Lm+1, the numbers are counts from Lm, which had maximum M.So, the numbers in Lm+1 are all ≤ M, and their counts in Lm+1 will determine the numbers in Lm+2.But since in Lm+1, M appears M times, and the rest of the numbers are ≤ M, the counts in Lm+2 will be similar to those in Lm+1.Wait, but actually, if in Lm+1, M appears M times, then in Lm+2, the count of M will be M, because M appears M times in Lm+1.Similarly, the counts of other numbers will be the counts of their corresponding numbers in Lm+1.But since in Lm+1, the numbers are counts from Lm, which had maximum M, the counts in Lm+2 will be the same as in Lm+1.Wait, that suggests that Lm+2 = Lm+1, meaning that the line stabilizes.Therefore, once the maximum stabilizes at M, the next line will have M appearing M times, and the rest of the counts will be determined by the previous line, which already had M appearing M times. Therefore, the next line will be the same as the previous one.Therefore, the system must stabilize after some finite number of steps, meaning that Maena will inevitably write the same line twice in a row.So, to summarize, the key points are:1. The sum of the counts in each line is always 2021.2. The maximum value in each line is bounded above by sqrt(2021) ≈ 44.3. The sequence of maximum values must eventually stabilize.4. Once the maximum stabilizes, the next line will have the same maximum value appearing the same number of times, leading to the line stabilizing.Therefore, Maena will inevitably end up writing the same line twice in a row.

💡First, I need to determine the number of different sequences of student helpers possible in a week. Mr. Slate picks one student each class session, and there are three class meetings in a week.Since there are 15 students in the class and each student can be picked more than once, there are 15 possible choices for each of the three class sessions.To find the total number of different sequences, I multiply the number of choices for each session together. This means calculating 15 multiplied by itself three times, which is 15 cubed.Calculating 15 cubed:15 × 15 = 225225 × 15 = 3375Therefore, the total number of different sequences of student helpers is 3375.