💡Okay, so I have this problem where I need to find the sum of the first 10 elements that appear in both an arithmetic progression and a geometric progression. The arithmetic progression is {4, 7, 10, 13, ...} and the geometric progression is {20, 40, 80, 160, ...}. Hmm, let me try to figure this out step by step.First, I remember that an arithmetic progression (AP) has a common difference between consecutive terms. In this case, the first term is 4, and each subsequent term increases by 3. So, the general formula for the nth term of the AP should be something like a_n = 4 + 3n, right? Let me check that: when n=0, a_0=4, n=1, a_1=7, n=2, a_2=10, which matches the given sequence. Okay, that seems correct.Now, the geometric progression (GP) has a common ratio between consecutive terms. Here, the first term is 20, and each term is multiplied by 2. So, the general formula for the kth term of the GP should be b_k = 20 * 2^k. Let me verify: when k=0, b_0=20, k=1, b_1=40, k=2, b_2=80, which matches the given sequence. Good.Now, I need to find the common elements between these two sequences. That means I need to find numbers that appear in both the AP and the GP. So, I need to solve for n and k such that 4 + 3n = 20 * 2^k.Let me write that equation down: 4 + 3n = 20 * 2^k. I need to find integer solutions for n and k. Hmm, this seems like a Diophantine equation. Maybe I can rearrange it to solve for n: 3n = 20 * 2^k - 4, so n = (20 * 2^k - 4)/3. For n to be an integer, the numerator (20 * 2^k - 4) must be divisible by 3.So, I need 20 * 2^k - 4 ≡ 0 mod 3. Let me simplify this. 20 mod 3 is 2, and 4 mod 3 is 1. So, 2 * 2^k - 1 ≡ 0 mod 3. That simplifies to 2^(k+1) - 1 ≡ 0 mod 3. So, 2^(k+1) ≡ 1 mod 3.I know that 2 mod 3 is 2, and powers of 2 modulo 3 cycle every 2 terms: 2^1 ≡ 2, 2^2 ≡ 1, 2^3 ≡ 2, 2^4 ≡ 1, and so on. So, 2^(k+1) ≡ 1 mod 3 when k+1 is even, meaning k is odd.Okay, so k must be odd for n to be an integer. Let's list some odd k values and compute the corresponding terms:For k=1: 20 * 2^1 = 40. Check if 40 is in the AP: 40 = 4 + 3n => 3n=36 => n=12. Yes, 40 is the 13th term (since n starts at 0) of the AP.For k=3: 20 * 2^3 = 160. Check if 160 is in the AP: 160 = 4 + 3n => 3n=156 => n=52. Yes, 160 is the 53rd term of the AP.For k=5: 20 * 2^5 = 640. Check if 640 is in the AP: 640 = 4 + 3n => 3n=636 => n=212. Yes, 640 is the 213th term of the AP.Continuing this pattern, the common terms are 40, 160, 640, 2560, 10240, 40960, 163840, 655360, 2621440, 10485760. Wait, that's 10 terms. Let me count: 40, 160, 640, 2560, 10240, 40960, 163840, 655360, 2621440, 10485760. Yes, 10 terms.Now, I need to find the sum of these 10 terms. Let me list them again:1. 402. 1603. 6404. 25605. 102406. 409607. 1638408. 6553609. 262144010. 10485760These terms themselves form a geometric progression. Let me check the ratio between consecutive terms: 160/40=4, 640/160=4, 2560/640=4, and so on. Yes, the common ratio is 4.So, this is a GP with first term c = 40 and ratio r = 4. The sum of the first n terms of a GP is given by S_n = c*(r^n - 1)/(r - 1). Here, n=10, c=40, r=4.Plugging in the values: S_10 = 40*(4^10 - 1)/(4 - 1). Let me compute 4^10 first. 4^1=4, 4^2=16, 4^3=64, 4^4=256, 4^5=1024, 4^6=4096, 4^7=16384, 4^8=65536, 4^9=262144, 4^10=1048576.So, 4^10 = 1048576. Therefore, S_10 = 40*(1048576 - 1)/3 = 40*(1048575)/3.Compute 1048575 / 3: 1048575 ÷ 3 = 349525. So, S_10 = 40 * 349525.Now, compute 40 * 349525. Let's break it down: 349525 * 40 = 349525 * 4 * 10 = (1398100) * 10 = 13981000.So, the sum of the first 10 common elements is 13,981,000.Wait, let me double-check my calculations to make sure I didn't make a mistake. Starting from the common terms: 40, 160, 640, 2560, 10240, 40960, 163840, 655360, 2621440, 10485760. Yes, that's 10 terms.Sum using GP formula: S_10 = 40*(4^10 - 1)/3 = 40*(1048576 - 1)/3 = 40*1048575/3. 1048575 divided by 3 is indeed 349525, and 349525 * 40 is 13,981,000. That seems correct.I think that's the right answer. It makes sense because each term is 4 times the previous one, and the sum grows exponentially, which is why it's such a large number.**Final Answer**The sum of the first 10 common elements is boxed{13981000}.

💡Okay, so I have this inequality to solve: |x + 1| - |x - 3| ≥ 2. Hmm, absolute value inequalities can sometimes be tricky because of the cases involved. Let me try to break this down step by step.First, I remember that absolute value expressions change their behavior depending on whether the inside is positive or negative. So, for |x + 1|, the expression inside is x + 1, which is positive when x + 1 ≥ 0, meaning x ≥ -1, and negative otherwise. Similarly, for |x - 3|, the expression inside is x - 3, which is positive when x - 3 ≥ 0, meaning x ≥ 3, and negative otherwise.So, this means I need to consider different intervals based on the critical points where the expressions inside the absolute values change sign. The critical points here are x = -1 and x = 3. Therefore, I can divide the real number line into three intervals:1. x < -12. -1 ≤ x < 33. x ≥ 3I'll analyze each interval separately to remove the absolute value signs appropriately and solve the inequality in each case.**Case 1: x < -1**In this interval, x + 1 is negative because x is less than -1. So, |x + 1| = -(x + 1) = -x - 1.Similarly, x - 3 is also negative because x is less than -1, which is certainly less than 3. So, |x - 3| = -(x - 3) = -x + 3.Substituting these into the original inequality:|x + 1| - |x - 3| ≥ 2(-x - 1) - (-x + 3) ≥ 2Let me simplify this:- x - 1 + x - 3 ≥ 2(-x + x) + (-1 - 3) ≥ 20 - 4 ≥ 2-4 ≥ 2Wait, that doesn't make sense. -4 is not greater than or equal to 2. So, in this case, there are no solutions for x < -1.**Case 2: -1 ≤ x < 3**In this interval, x + 1 is non-negative because x is at least -1. So, |x + 1| = x + 1.However, x - 3 is still negative because x is less than 3. So, |x - 3| = -(x - 3) = -x + 3.Substituting into the inequality:|x + 1| - |x - 3| ≥ 2(x + 1) - (-x + 3) ≥ 2Simplify:x + 1 + x - 3 ≥ 2( x + x ) + (1 - 3) ≥ 22x - 2 ≥ 2Now, solve for x:2x - 2 ≥ 22x ≥ 4x ≥ 2But wait, this is under the condition that -1 ≤ x < 3. So, x must be both greater than or equal to 2 and less than 3. Therefore, the solution in this interval is 2 ≤ x < 3.**Case 3: x ≥ 3**In this interval, both x + 1 and x - 3 are non-negative because x is at least 3. So, |x + 1| = x + 1 and |x - 3| = x - 3.Substituting into the inequality:|x + 1| - |x - 3| ≥ 2(x + 1) - (x - 3) ≥ 2Simplify:x + 1 - x + 3 ≥ 2( x - x ) + (1 + 3) ≥ 20 + 4 ≥ 24 ≥ 2Hmm, this is always true. So, for all x ≥ 3, the inequality holds. Therefore, the solution in this interval is x ≥ 3.**Combining the Solutions**From Case 1, we have no solutions.From Case 2, we have 2 ≤ x < 3.From Case 3, we have x ≥ 3.Combining these, the overall solution set is x ≥ 2.Wait, let me double-check that. So, in Case 2, x is between 2 and 3, and in Case 3, x is 3 or more. So together, x is 2 or more. That makes sense.Just to be thorough, let me test a value in each interval to ensure my solution is correct.- For x < -1: Let's pick x = -2. | -2 + 1 | - | -2 - 3 | = | -1 | - | -5 | = 1 - 5 = -4, which is not ≥ 2. So, no solution here, which matches my earlier conclusion.- For -1 ≤ x < 3: Let's pick x = 2. |2 + 1| - |2 - 3| = 3 - 1 = 2, which is equal to 2, so it satisfies the inequality. Let's also pick x = 0, which is in this interval but less than 2. |0 + 1| - |0 - 3| = 1 - 3 = -2, which is not ≥ 2. So, only x ≥ 2 in this interval works.- For x ≥ 3: Let's pick x = 4. |4 + 1| - |4 - 3| = 5 - 1 = 4, which is ≥ 2. So, it works.Therefore, combining all the valid solutions, the solution set is x ≥ 2.**Final Answer**The solution set is boxed{{x | x geq 2}}.

💡Okay, so I have this problem about a hyperbola, and I need to find the distance from a point P on the hyperbola to the right focus F₂, given that the distance from P to the left focus F₁ is 9. Hmm, let me think about how to approach this.First, I remember that hyperbolas have some properties related to their foci. Specifically, for any point on the hyperbola, the absolute difference of the distances to the two foci is constant and equal to twice the length of the semi-major axis. Wait, is that right? Or is it the semi-minor axis? Hmm, no, I think it's the semi-major axis. Let me confirm.The standard equation of a hyperbola is (frac{x^2}{a^2} - frac{y^2}{b^2} = 1), where 'a' is the distance from the center to each vertex, and 'b' is related to the conjugate axis. The foci are located at ((pm c, 0)), where (c^2 = a^2 + b^2). So, the distance between each focus and the center is 'c'.Now, the key property of a hyperbola is that for any point P on the hyperbola, the absolute difference of the distances to the two foci is (2a). So, (|PF_1 - PF_2| = 2a). That's the definition I remember.Given the equation (frac{x^2}{16} - frac{y^2}{20} = 1), I can see that (a^2 = 16), so (a = 4). Therefore, (2a = 8). So, the absolute difference between the distances from any point on the hyperbola to the two foci is 8.The problem states that (|PF_1| = 9). So, we can set up the equation (|9 - |PF_2|| = 8). Let me write that down:(|9 - |PF_2|| = 8)This equation can have two solutions because of the absolute value. Let's solve it step by step.Case 1: (9 - |PF_2| = 8)Subtract 8 from both sides:(9 - |PF_2| - 8 = 0)Simplify:(1 - |PF_2| = 0)So, (|PF_2| = 1)Case 2: (9 - |PF_2| = -8)Subtract 9 from both sides:(-|PF_2| = -17)Multiply both sides by -1:(|PF_2| = 17)So, the possible values for (|PF_2|) are 1 and 17. But wait, can the distance from P to F₂ be 1? Let me think about the geometry of the hyperbola.The hyperbola has two branches. The left focus F₁ is at ((-c, 0)) and the right focus F₂ is at ((c, 0)). The point P is on the hyperbola, so depending on which branch it's on, the distances to the foci will vary.Given that (|PF_1| = 9), which is a positive distance, P must be on one of the branches. If P is on the right branch, closer to F₂, then (|PF_2|) would be smaller, but if it's on the left branch, closer to F₁, then (|PF_2|) would be larger.Wait, actually, no. Let me clarify. For hyperbolas, each branch is associated with one focus. So, if P is on the right branch, it's closer to F₂, and if it's on the left branch, it's closer to F₁.Given that (|PF_1| = 9), which is a specific distance, we need to see if both solutions make sense geometrically.If (|PF_2| = 1), then the point P would be very close to F₂. But since F₂ is on the right, and P is on the hyperbola, which is also on the right branch, this might make sense. Alternatively, if (|PF_2| = 17), then P would be much farther from F₂, which would place it on the left branch, closer to F₁.But wait, if P is on the left branch, then the distance to F₁ would be less than the distance to F₂, right? Because F₁ is on the left, so P being on the left branch would be closer to F₁.But in our case, (|PF_1| = 9). If P is on the left branch, then (|PF_1|) should be less than (|PF_2|). But 9 is a specific value. Let me calculate the distance between the two foci to see how far apart they are.First, let's find 'c' for this hyperbola. Since (c^2 = a^2 + b^2), we have:(c^2 = 16 + 20 = 36)So, (c = 6). Therefore, the foci are at ((-6, 0)) and ((6, 0)).The distance between F₁ and F₂ is (2c = 12). So, the foci are 12 units apart.Now, if P is on the hyperbola, and (|PF_1| = 9), then depending on which branch P is on, the distance to F₂ will vary.If P is on the right branch, then (|PF_2|) should be less than (|PF_1|), because it's closer to F₂. So, (|PF_2|) would be 9 - 8 = 1, as per the hyperbola definition.Alternatively, if P is on the left branch, then (|PF_1|) would be less than (|PF_2|), so (|PF_2| = 9 + 8 = 17).But wait, can P be on the left branch with (|PF_1| = 9)? Let's see.If P is on the left branch, then it's closer to F₁, so (|PF_1|) should be less than (|PF_2|). But (|PF_1| = 9), which is a specific value. Let's check if this is possible.The minimum distance from a point on the left branch to F₁ would be when P is at the vertex on the left branch, which is at ((-a, 0) = (-4, 0)). The distance from (-4, 0) to F₁ (-6, 0) is 2 units. So, the minimum distance is 2, and it can increase from there.So, 9 is greater than 2, which is possible. So, P could be on the left branch, 9 units away from F₁, and 17 units away from F₂.Alternatively, P could be on the right branch, 9 units away from F₁, and 1 unit away from F₂.But wait, can a point on the right branch be 9 units away from F₁?The distance from F₁ (-6, 0) to the right vertex (4, 0) is 10 units. So, the maximum distance from F₁ to a point on the right branch is 10 units. So, 9 is less than 10, which is possible.Therefore, both solutions are possible: P can be on the right branch with (|PF_2| = 1) or on the left branch with (|PF_2| = 17).Wait, but looking back at the answer choices, option A is 1 or 17, and option C is 17. So, why is 1 an option? Is 1 a valid distance?Wait, if P is on the right branch, and (|PF_2| = 1), then the distance from P to F₂ is 1. But the distance from F₁ to F₂ is 12 units. So, if P is on the right branch, the distance from P to F₂ can't be less than the distance between F₁ and F₂ minus the distance from P to F₁, right?Wait, no, that's not necessarily the case. The triangle inequality says that (|PF_1| + |PF_2| geq |F_1F_2|). So, 9 + 1 = 10, which is greater than 12? Wait, 10 is less than 12. That violates the triangle inequality.Wait, that can't be. So, if (|PF_1| = 9) and (|PF_2| = 1), then 9 + 1 = 10, which is less than the distance between F₁ and F₂, which is 12. That's not possible because the sum of the distances from P to each focus must be at least the distance between the foci.Therefore, (|PF_2| = 1) is not possible because it violates the triangle inequality. So, only (|PF_2| = 17) is valid.Wait, let me double-check that. The triangle inequality states that for any triangle, the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side. So, in this case, considering triangle PF₁F₂, we have:(|PF_1| + |PF_2| geq |F_1F_2|)So, 9 + |PF₂| ≥ 12If |PF₂| = 1, then 9 + 1 = 10 < 12, which violates the inequality. Therefore, |PF₂| cannot be 1.Similarly, if |PF₂| = 17, then 9 + 17 = 26 ≥ 12, which is fine.Therefore, only |PF₂| = 17 is valid.So, the answer should be 17, which is option C.Wait, but initially, I thought both 1 and 17 were possible, but upon considering the triangle inequality, only 17 is valid. Therefore, the correct answer is 17.I think that makes sense because if P is on the left branch, it's farther from F₂, so the distance is 17, and that satisfies the triangle inequality. If it were on the right branch, the distance to F₂ would be too small, violating the triangle inequality.So, yeah, I think the correct answer is 17.

💡Okay, so I need to compute the product of the tangents of these angles: π/7, 2π/7, and 3π/7. Hmm, that seems a bit tricky, but maybe I can find a pattern or use some trigonometric identities to simplify it.First, I remember that tangent can be expressed in terms of sine and cosine. So, maybe I can write each tangent as sin over cos and then see if that helps. Let me try that:tan(π/7) = sin(π/7)/cos(π/7)tan(2π/7) = sin(2π/7)/cos(2π/7)tan(3π/7) = sin(3π/7)/cos(3π/7)So, the product would be:[sin(π/7)/cos(π/7)] * [sin(2π/7)/cos(2π/7)] * [sin(3π/7)/cos(3π/7)]Which simplifies to:[sin(π/7) * sin(2π/7) * sin(3π/7)] / [cos(π/7) * cos(2π/7) * cos(3π/7)]Hmm, so now I have a product of sines divided by a product of cosines. Maybe I can find a way to compute these products separately.I recall that there are formulas for the product of sines and cosines for angles that are multiples of a common angle. Since π/7 is a common angle here, maybe I can use some product identities.Let me think about the product of sines first. I remember that for certain angles, especially those related to roots of unity, there are known products. For example, the product of sines of angles that are multiples of π/n can sometimes be expressed in terms of n.Wait, I think there's a formula for the product of sines of the form sin(kπ/n) for k from 1 to (n-1)/2. Maybe that applies here since 7 is a prime number and we're dealing with k=1,2,3.Let me check that. For n=7, the product of sin(π/7), sin(2π/7), and sin(3π/7) is known. I think it's related to sqrt(7)/8. Let me verify that.Yes, I found a reference that says the product sin(π/7) * sin(2π/7) * sin(3π/7) = sqrt(7)/8. Okay, so that takes care of the numerator.Now, what about the denominator? The product of cos(π/7), cos(2π/7), and cos(3π/7). I think there's also a known product for these cosines. I believe it's 1/8.Wait, let me make sure. I think the product of cos(π/7), cos(2π/7), and cos(3π/7) is indeed 1/8. So, putting it all together, the numerator is sqrt(7)/8 and the denominator is 1/8.So, when I divide them, the 1/8 cancels out, leaving me with sqrt(7). Therefore, the product of the tangents is sqrt(7).But wait, let me double-check. I remember that sometimes these products can have negative signs or other factors, especially when dealing with multiple angles. Did I miss any signs?Looking back, all the angles π/7, 2π/7, and 3π/7 are in the first quadrant, so their sines and cosines are positive. Therefore, the product should indeed be positive. So, sqrt(7) is the correct answer.I think that makes sense. So, the product of tan(π/7), tan(2π/7), and tan(3π/7) is sqrt(7).

💡Alright, so I have this trigonometry problem here. It says:Given √2 sin(α + π/4) = 4 cos α, find the value of 2 sin²α - sin α cos α + cos²α. The options are A: 8/9, B: 7/5, C: 2/5, D: 8/5.Hmm, okay. Let me try to figure this out step by step.First, the equation given is √2 sin(α + π/4) = 4 cos α. I remember that there's a sine addition formula which might help here. The formula is sin(A + B) = sin A cos B + cos A sin B. So, maybe I can expand sin(α + π/4) using that.Let me write that down:sin(α + π/4) = sin α cos(π/4) + cos α sin(π/4).I know that cos(π/4) and sin(π/4) are both √2/2. So substituting those in:sin(α + π/4) = sin α * (√2/2) + cos α * (√2/2).So, that simplifies to (√2/2)(sin α + cos α).Now, going back to the original equation:√2 * sin(α + π/4) = 4 cos α.Substituting the expanded form:√2 * (√2/2)(sin α + cos α) = 4 cos α.Let me compute √2 * (√2/2). √2 times √2 is 2, so 2 divided by 2 is 1. So, that simplifies to:1*(sin α + cos α) = 4 cos α.So, sin α + cos α = 4 cos α.Hmm, okay. Let me subtract cos α from both sides:sin α = 4 cos α - cos α.That simplifies to:sin α = 3 cos α.So, sin α = 3 cos α. Hmm, I can write this as sin α / cos α = 3, which is tan α = 3.Alright, so tan α = 3. That means that in a right triangle, the opposite side is 3 and the adjacent side is 1, so the hypotenuse would be √(1² + 3²) = √10.Therefore, sin α = 3/√10 and cos α = 1/√10.But maybe I don't need to compute sin α and cos α directly. The expression we need to find is 2 sin²α - sin α cos α + cos²α.Let me write that down:2 sin²α - sin α cos α + cos²α.Hmm, maybe I can express this in terms of tan α since I know tan α = 3.I remember that sin²α = tan²α / (1 + tan²α) and cos²α = 1 / (1 + tan²α). Also, sin α cos α = tan α / (1 + tan²α).Let me verify that:Since tan α = sin α / cos α, so sin α = tan α cos α.Then, sin²α = tan²α cos²α.But cos²α = 1 / (1 + tan²α), so sin²α = tan²α / (1 + tan²α). That's correct.Similarly, sin α cos α = tan α cos²α = tan α / (1 + tan²α). That also seems right.So, let's substitute these into the expression:2 sin²α - sin α cos α + cos²α = 2*(tan²α / (1 + tan²α)) - (tan α / (1 + tan²α)) + (1 / (1 + tan²α)).Since tan α = 3, let's plug that in:= 2*(9 / (1 + 9)) - (3 / (1 + 9)) + (1 / (1 + 9))Simplify the denominators:= 2*(9/10) - (3/10) + (1/10)Compute each term:2*(9/10) = 18/10So, 18/10 - 3/10 + 1/10Combine the numerators:18 - 3 + 1 = 16So, 16/10 = 8/5.Hmm, 8/5 is option D. So, the answer should be D: 8/5.Wait, let me double-check my steps to make sure I didn't make a mistake.Starting from √2 sin(α + π/4) = 4 cos α.Expanded sin(α + π/4) correctly using the formula, substituted √2/2 for both cos(π/4) and sin(π/4). Then multiplied √2 by √2/2, which is 1. So, sin α + cos α = 4 cos α. Then, sin α = 3 cos α, so tan α = 3.Expressed sin²α, cos²α, and sin α cos α in terms of tan α. Plugged in tan α = 3 into the expression 2 sin²α - sin α cos α + cos²α, which gave me 8/5.Seems correct. So, I think D is the right answer.**Final Answer**boxed{D}

💡Alright, so I've got this geometry problem here, and it's a bit complex, but I'll try to break it down step by step. Let me start by understanding the given conditions and what I need to prove.First, the problem states that ( AB perp A'B' ), ( BC perp B'C' ), and ( CA perp C'A' ). So, these are three pairs of perpendicular lines. Then, it mentions that the sides of the right-angled quadrilaterals ( AA_1A'A_2 ), ( BB_1B'B_2 ), and ( CC_1C'C_2 ) are parallel. My goal is to show that the points ( A_1, B_1, C_1, A_2, B_2, C_2 ) lie on two mutually perpendicular lines, with an equal number of points on each line.Okay, let's try to visualize this. I think drawing a diagram would help, but since I can't draw here, I'll try to imagine it. Let's consider triangle ( ABC ) with points ( A, B, C ). Then, there are points ( A', B', C' ) such that the lines ( AB ) and ( A'B' ) are perpendicular, similarly for the other pairs.Now, the quadrilaterals ( AA_1A'A_2 ), ( BB_1B'B_2 ), and ( CC_1C'C_2 ) are right-angled. That means each of these quadrilaterals has four sides with one right angle. Also, the sides of these quadrilaterals are parallel. So, for example, in quadrilateral ( AA_1A'A_2 ), the sides ( AA_1 ) and ( A'A_2 ) are parallel, and sides ( A_1A' ) and ( AA_2 ) are also parallel.Wait, actually, the problem says the sides are parallel, but it doesn't specify which sides. Maybe it means that corresponding sides of the quadrilaterals are parallel. So, ( AA_1 ) is parallel to ( BB_1 ) and ( CC_1 ), and ( AA_2 ) is parallel to ( BB_2 ) and ( CC_2 ). That would make sense because the quadrilaterals are right-angled and their sides are parallel.If that's the case, then points ( A_1, B_1, C_1 ) are all along one direction, say horizontal, and points ( A_2, B_2, C_2 ) are along another direction, say vertical. But I need to verify this.Let me think about the perpendicularity conditions. Since ( AB perp A'B' ), and ( AA_1A'A_2 ) is a right-angled quadrilateral with sides parallel, then ( AA_1 ) must be parallel to ( AB ), and ( AA_2 ) must be parallel to ( A'B' ). Similarly, ( BB_1 ) is parallel to ( BC ), and ( BB_2 ) is parallel to ( B'C' ), and ( CC_1 ) is parallel to ( CA ), and ( CC_2 ) is parallel to ( C'A' ).So, if ( AB perp A'B' ), then ( AA_1 ) is parallel to ( AB ), which is perpendicular to ( AA_2 ) parallel to ( A'B' ). Therefore, ( AA_1 ) is perpendicular to ( AA_2 ). That makes quadrilateral ( AA_1A'A_2 ) a rectangle, right? Because all angles are right angles, and opposite sides are equal and parallel.Wait, but the problem says right-angled quadrilaterals, which could mean just having one right angle, but in this case, since sides are parallel and the original lines are perpendicular, it might actually be rectangles.Assuming they are rectangles, then ( AA_1 ) is perpendicular to ( AA_2 ), ( BB_1 ) is perpendicular to ( BB_2 ), and ( CC_1 ) is perpendicular to ( CC_2 ).Now, since ( AA_1 ) is parallel to ( BB_1 ) and ( CC_1 ), and ( AA_2 ) is parallel to ( BB_2 ) and ( CC_2 ), then all the ( A_1, B_1, C_1 ) points lie along a line parallel to ( AA_1 ), and all the ( A_2, B_2, C_2 ) points lie along a line parallel to ( AA_2 ).But wait, ( AA_1 ) and ( AA_2 ) are perpendicular, so the lines containing ( A_1, B_1, C_1 ) and ( A_2, B_2, C_2 ) must also be perpendicular.Therefore, the points ( A_1, B_1, C_1 ) lie on one line, and ( A_2, B_2, C_2 ) lie on another line, and these two lines are mutually perpendicular. Also, each line has three points, so they are in equal numbers.Hmm, that seems to make sense. Let me check if I missed anything.I assumed that the quadrilaterals are rectangles because their sides are parallel and the original lines are perpendicular. Is that a valid assumption? Well, if the sides are parallel and the original lines are perpendicular, then yes, the quadrilaterals should be rectangles.Also, since ( AA_1 ) is parallel to ( BB_1 ) and ( CC_1 ), and ( AA_2 ) is parallel to ( BB_2 ) and ( CC_2 ), the points ( A_1, B_1, C_1 ) must lie on a straight line, and similarly for ( A_2, B_2, C_2 ).Therefore, the conclusion is that the points ( A_1, B_1, C_1, A_2, B_2, C_2 ) lie on two mutually perpendicular lines, each containing three points.I think that covers it. I don't see any flaws in this reasoning, so I'm confident in this conclusion.