🧠:<think>Alright, so I've got this problem here about sets A and B. Let me try to work through it step by step. Sometimes math problems can be a bit confusing, especially if you're just starting out, so I need to take it slow and make sure I understand each part.First, the problem defines set A as {x | -1 ≤ x + 1 ≤ 6}. Hmm, okay, that notation means that A is the set of all x such that x + 1 is between -1 and 6, inclusive. I think I need to solve this inequality to find out what numbers are in set A.Let me rewrite the inequality: -1 ≤ x + 1 ≤ 6. To find x, I can subtract 1 from all parts of the inequality. So, subtracting 1 from -1 gives me -2, subtracting 1 from x + 1 gives me x, and subtracting 1 from 6 gives me 5. So, the inequality becomes -2 ≤ x ≤ 5. That means set A is all real numbers from -2 up to 5, including both endpoints.But wait, part (1) of the problem specifies that x is an integer. So, set A when x is an integer would be the integers from -2 to 5, inclusive. Let me list those out: -2, -1, 0, 1, 2, 3, 4, 5. That's 8 numbers in total. So, set A has 8 elements when x is an integer.Now, the question is asking for the number of proper subsets of A. I remember that the number of subsets of a set with n elements is 2^n. Since A has 8 elements, the total number of subsets is 2^8, which is 256. But the problem specifically asks for proper subsets. A proper subset is any subset that is not equal to the set itself. So, we need to subtract 1 from the total number of subsets to exclude the set A itself.So, the number of proper subsets is 256 - 1 = 255. Wait, that doesn't seem right. Let me double-check. If a set has n elements, the number of subsets is indeed 2^n, and the number of proper subsets is 2^n - 1 because we exclude the set itself. So, 2^8 is 256, minus 1 is 255. Hmm, but the initial thought process I saw earlier had 253. Did I make a mistake?Wait, maybe I miscounted the number of elements in set A. Let me recount. A is {-2, -1, 0, 1, 2, 3, 4, 5}. That's 8 elements. So, 2^8 is 256, and subtracting 1 gives 255. Hmm, so why did the initial thought process say 253? Maybe I missed something.Wait, no, perhaps the initial thought process was incorrect. Let me verify the number of elements in set A again. The inequality was -2 ≤ x ≤ 5, and x is an integer. So, starting from -2, the numbers are -2, -1, 0, 1, 2, 3, 4, 5. That's 8 numbers. So, 2^8 is 256, proper subsets are 255. So, maybe the initial answer was wrong, and the correct number is 255. Hmm, that's confusing. Maybe I should think again.Wait, no, actually, the initial thought process had A as {-2, -1, 0, 1, 2, 3, 4, 5}, which is 8 elements. So, the number of subsets is 2^8 = 256, and proper subsets are 255. So, I think the initial answer of 253 was incorrect. Maybe there was a miscount in the number of elements.Alternatively, perhaps the initial thought process considered something else, but in my calculation, 8 elements give 256 subsets, so 255 proper subsets. Therefore, I think the correct answer should be 255. But wait, maybe I'm misunderstanding the definition of a proper subset. A proper subset is any subset that is not equal to the set itself, so yes, we subtract 1. So, 255 is correct. Therefore, maybe the initial thought process had a mistake.Okay, moving on to part (2). It asks, if B is a subset of A, what is the range of the real number m? So, set B is defined as {x | m - 1 ≤ x < 2m + 1}. So, B is an interval from m - 1 to 2m + 1, not including 2m + 1. Since B is a subset of A, which is from -2 to 5, inclusive, we need to ensure that the entire interval of B lies within A.So, for B to be a subset of A, the lower bound of B must be greater than or equal to the lower bound of A, and the upper bound of B must be less than or equal to the upper bound of A. But since B is a half-open interval, the upper bound is not included, but it still needs to be less than or equal to 5.So, mathematically, we can write the conditions as:1. m - 1 ≥ -22. 2m + 1 ≤ 5Let me solve these inequalities.First, m - 1 ≥ -2. Adding 1 to both sides, we get m ≥ -1.Second, 2m + 1 ≤ 5. Subtracting 1 from both sides, we get 2m ≤ 4. Dividing both sides by 2, we get m ≤ 2.So, combining these two inequalities, m must be greater than or equal to -1 and less than or equal to 2. So, the range of m is -1 ≤ m ≤ 2.But wait, hold on. There's another case to consider. What if the interval B is empty? Because if m - 1 > 2m + 1, then B would be an empty set, which is a subset of any set, including A.So, when does m - 1 > 2m + 1? Let's solve that inequality:m - 1 > 2m + 1Subtract m from both sides:-1 > m + 1Subtract 1 from both sides:-2 > mWhich is the same as m < -2.So, if m is less than -2, then B is empty, and therefore, B is a subset of A.So, putting it all together, the range of m is either m < -2 or -1 ≤ m ≤ 2.Therefore, the range is m < -2 or -1 ≤ m ≤ 2.Wait, let me double-check this. If m is less than -2, then m - 1 is less than -3, and 2m + 1 is less than -3. So, B would be from something less than -3 to something less than -3, which is an empty set. So, yes, B is empty.If m is between -2 and -1, let's say m = -1.5, then m - 1 = -2.5, and 2m + 1 = -2. So, B would be from -2.5 to -2, but since A starts at -2, this interval would be from -2.5 to -2, but A only includes numbers starting from -2. So, in this case, B would include numbers less than -2, which are not in A. Therefore, B would not be a subset of A. Hence, m cannot be between -2 and -1.Therefore, the only valid ranges are m < -2 or -1 ≤ m ≤ 2.So, summarizing:1. When x is an integer, set A has 8 elements, so the number of proper subsets is 2^8 - 1 = 255.2. The range of m for which B is a subset of A is m < -2 or -1 ≤ m ≤ 2.Wait, but in the initial thought process, the answer for part (1) was 253. Hmm, maybe I made a mistake earlier. Let me check again.Set A is from -2 to 5, inclusive, with x as integers. So, the elements are -2, -1, 0, 1, 2, 3, 4, 5. That's 8 elements. So, 2^8 = 256 subsets, proper subsets are 256 - 1 = 255. So, I think the initial thought process had a mistake, and the correct answer should be 255.But in the initial thought process, it was 253. Maybe there was a miscount of elements. Let me count again:-2, -1, 0, 1, 2, 3, 4, 5. That's 8 elements. So, 2^8 is 256, proper subsets are 255. So, I think 255 is correct.Wait, perhaps the initial thought process was correct, and I miscounted. Let me see:Wait, -2, -1, 0, 1, 2, 3, 4, 5. That's 8 numbers. So, 8 elements. So, 2^8 is 256. So, proper subsets are 255. So, I think the initial answer of 253 was incorrect.Wait, but maybe the problem was written differently. Let me check the original problem again.Wait, no, the problem says A = {x | -1 ≤ x + 1 ≤ 6}, which simplifies to -2 ≤ x ≤ 5. So, when x is an integer, A has 8 elements. So, 2^8 -1 = 255 proper subsets.So, I think the initial thought process had an error, and the correct answer is 255. But in the initial thought process, it was 253. Hmm, maybe I should go with my own reasoning.For part (2), the range is m < -2 or -1 ≤ m ≤ 2.Wait, let me check if m = -2 is allowed. If m = -2, then B is {x | -3 ≤ x < -3}, which is empty set. So, m = -2 is allowed because B would be empty, which is a subset of A.Similarly, if m = -1, then B is {x | -2 ≤ x < -1}, which is [-2, -1), but since A includes -2, but not -1, because A is up to 5, but wait, A includes all integers from -2 to 5, so in real numbers, A is [-2, 5]. So, B = [-2, -1) is a subset of A because all elements of B are in A.Wait, no, in the problem, for part (2), B is defined as {x | m -1 ≤ x < 2m +1}, which is a real interval, not necessarily integers. So, A is also defined with real numbers in part (2), right? Because in part (1), x is integer, but in part (2), it's not specified, so x is real.So, A is [-2, 5], and B is [m -1, 2m +1). For B to be a subset of A, we need:1. m - 1 ≥ -22. 2m +1 ≤ 5Solving these:1. m - 1 ≥ -2 ⇒ m ≥ -12. 2m +1 ≤ 5 ⇒ 2m ≤ 4 ⇒ m ≤ 2So, m must be between -1 and 2, inclusive.But also, if B is empty, which happens when m -1 > 2m +1, which simplifies to m < -2. So, when m < -2, B is empty, which is a subset of A.Therefore, the range of m is m < -2 or -1 ≤ m ≤ 2.So, that's consistent with my earlier conclusion.Therefore, for part (1), the number of proper subsets is 255, and for part (2), the range of m is m < -2 or -1 ≤ m ≤ 2.Wait, but in the initial thought process, part (1) was 253. Maybe I should double-check the number of elements in A again.Set A is defined as {x | -1 ≤ x +1 ≤6}, which simplifies to -2 ≤ x ≤5. So, when x is an integer, the elements are -2, -1, 0, 1, 2, 3, 4, 5. That's 8 elements. So, 2^8 = 256 subsets, proper subsets are 255. So, I think the initial answer was wrong.Alternatively, maybe the initial thought process considered only positive integers or something else, but the problem says x is an integer, so it includes negative integers as well.Therefore, I think the correct answers are:(1) 255 proper subsets.(2) m < -2 or -1 ≤ m ≤ 2.But in the initial thought process, part (1) was 253. Maybe I should think again.Wait, let me recount the elements of A when x is integer:Starting from -2: -2, -1, 0, 1, 2, 3, 4, 5. That's 8 elements. So, 2^8 = 256 subsets, proper subsets are 255. So, 255 is correct.Therefore, I think the initial answer had a mistake, and the correct answer is 255.So, in summary:(1) The number of proper subsets of A is 255.(2) The range of m is m < -2 or -1 ≤ m ≤ 2.I think that's it.</think>

🧠:<think>Alright, so I'm trying to tackle this problem where there are light bulbs at every integer point on a number line, all initially turned off. There's this template set S, which is a finite set of integers, and when you apply this template at a certain position, it toggles the state of all the light bulbs covered by the template. The goal is to prove that no matter what S is, we can always end up with exactly two light bulbs turned on after a finite number of operations.Okay, let's start by understanding the problem better. We have an infinite number line with light bulbs at each integer position. All are off at the beginning. The template S is a finite set of integers, say, {s₁, s₂, ..., sₖ}. When we apply S at position x, it toggles the bulbs at positions x + s₁, x + s₂, ..., x + sₖ. So, each application of S affects a finite number of bulbs, but since the number line is infinite, we can keep shifting S along it.Now, the challenge is to show that with some finite number of these operations, we can have exactly two bulbs on. Initially, all are off, and each operation toggles some bulbs. So, every operation changes the state of certain bulbs, but the key is to find a sequence of operations that results in exactly two bulbs being on.Let me think about what operations do. Each operation is essentially adding the characteristic function of S shifted by x to the current state vector. Since we're dealing with modulo 2 arithmetic (because toggling twice is the same as not toggling), this becomes a linear algebra problem over the field GF(2).Wait, maybe that's getting too abstract. Let me try a different approach. Suppose S contains only one integer, say {0}. Then applying S at position x toggles the bulb at x. In this case, it's straightforward to get exactly two bulbs on by toggling two different positions.But S is a finite set, so it could contain multiple integers. For example, if S = {0, 1}, then applying S at position x toggles bulbs at x and x+1. If we apply S at position x, then at position x+1, and so on, we can create patterns of bulbs being on.I need to find a way to use these operations to isolate exactly two bulbs. Maybe I can use some sort of cancellation or overlapping to turn off most bulbs except two.Let me think about the linear algebra perspective again. Each operation corresponds to a vector in an infinite-dimensional vector space over GF(2), where each dimension corresponds to a bulb. The set of all possible operations generates a subspace, and we want to see if the vector with exactly two 1s is in this subspace.But since the vector space is infinite-dimensional, it's tricky. However, we're only interested in finite operations, so maybe we can find a combination of a finite number of these vectors that results in exactly two 1s.Alternatively, maybe I can think about it as solving a system of equations. Each bulb's state is the sum (mod 2) of the operations that affect it. We want to set up equations such that only two bulbs have an odd number of operations affecting them, and all others have an even number.But again, since the number of bulbs is infinite, it's challenging. Maybe I need to find a pattern or a way to localize the effect of the operations.Wait, what if I consider the characteristic function of S and its translates? Since S is finite, the number of distinct translates is infinite, but maybe I can find a combination that cancels out all but two bulbs.Let me try a specific example to get some intuition. Suppose S = {0, 1}. If I apply S at position 0, it toggles bulbs 0 and 1. If I apply S at position 1, it toggles bulbs 1 and 2. If I apply S at position -1, it toggles bulbs -1 and 0. Now, if I apply S at 0 and then at 1, bulb 1 is toggled twice (off), bulb 0 is toggled once, and bulb 2 is toggled once. So, bulbs 0 and 2 are on, and bulb 1 is off. Oh, that's exactly two bulbs on! So, in this case, it's easy.But wait, S = {0, 1} is a simple case. What if S is more complicated? Let's try S = {0, 2}. Applying S at 0 toggles 0 and 2. Applying S at 1 toggles 1 and 3. Applying S at -1 toggles -1 and 1. If I apply S at 0 and then at 1, bulbs 0, 1, 2, and 3 are toggled. So, bulbs 0 and 2 are toggled once, bulbs 1 and 3 are toggled once. All four bulbs are on. Hmm, not exactly two.But if I apply S at 0 and then at 2, bulb 0 is toggled once, bulb 2 is toggled twice, bulb 4 is toggled once. So, bulbs 0 and 4 are on. That's two bulbs. Okay, so in this case, it's possible as well.Wait, so maybe for any S, I can find two translates of S that overlap in such a way that only two bulbs are toggled an odd number of times.But S could be more complicated, like S = {0, 1, 2}. Let's see. If I apply S at 0, it toggles 0, 1, 2. If I apply S at 1, it toggles 1, 2, 3. If I apply S at 2, it toggles 2, 3, 4. Now, if I apply S at 0 and then at 1, bulb 0 is toggled once, bulb 1 is toggled twice, bulb 2 is toggled twice, bulb 3 is toggled once. So, bulbs 0 and 3 are on. That's two bulbs. Nice.So, in these examples, it seems possible. Maybe the key is to find two translates of S such that their symmetric difference results in exactly two bulbs being toggled.But how do I generalize this? Let me think about the properties of S. Since S is finite, its translates are finite shifts, but the number of translates is infinite. So, I need to find two translates such that their intersection is a certain way.Alternatively, think about the characteristic function of S, χ_S, which is 1 at positions in S and 0 otherwise. Then, applying S at position x is like shifting χ_S by x and adding it modulo 2 to the current state.So, to get exactly two 1s, we need to find a combination of shifts of χ_S that sum to a vector with exactly two 1s.This is equivalent to solving the equation:sum_{x ∈ X} χ_{S + x} = vector with exactly two 1swhere X is a finite set of shifts.But since S is finite, the number of possible overlaps is limited. Maybe I can use the fact that the set of possible vectors generated by shifts of χ_S is a linear code, and we need to see if the vector with two 1s is in its span.But I'm not sure about that. Maybe another approach.What if I consider the difference between two translates of S? If I apply S at position x and then at position y, the net effect is toggling the bulbs that are in exactly one of S + x or S + y. So, the symmetric difference of S + x and S + y.If I can choose x and y such that the symmetric difference of S + x and S + y is exactly two bulbs, then I'm done. Because applying S at x and then at y would toggle those two bulbs.But is this always possible? Let's see.Given S, can we find x and y such that (S + x) Δ (S + y) = {a, b} for some a and b?The symmetric difference is (S + x) ∪ (S + y) minus (S + x) ∩ (S + y). So, we need that the union has exactly two more elements than the intersection, or something like that.Wait, actually, the symmetric difference is the set of elements that are in exactly one of the two sets. So, to have the symmetric difference be exactly two elements, we need that (S + x) and (S + y) overlap in all but two elements.But since S is finite, if we shift y far enough from x, the overlap can be made small. But we need to make the symmetric difference exactly two.Alternatively, maybe choose y such that S + y is a translate that overlaps with S + x in all but two elements.But I'm not sure if that's always possible. Maybe another way.Wait, let's think about the difference y - x. If I set y = x + d for some d, then S + y = S + x + d. So, the symmetric difference is (S + x) Δ (S + x + d) = (S Δ (S + d)) + x.So, if we can find a d such that S Δ (S + d) has exactly two elements, then shifting by x will give us the desired two bulbs.So, the problem reduces to finding a d such that S and S + d differ by exactly two elements. That is, S Δ (S + d) has size 2.Is this always possible? Let's see.Given S, can we find a d such that exactly two elements of S are shifted out and two are shifted in?Wait, not necessarily. For example, if S is symmetric around some point, shifting by d might not change the symmetric difference in a controlled way.But maybe for any S, we can find such a d. Let's try.Suppose S has elements s₁ < s₂ < ... < sₖ. Let d be such that d = s₂ - s₁. Then, S + d would have elements s₁ + d, s₂ + d, ..., sₖ + d. The symmetric difference S Δ (S + d) would be {s₁, sₖ + d} if s₂ + d > sₖ. Wait, no, that's not necessarily the case.Wait, actually, S + d is just S shifted by d. The symmetric difference S Δ (S + d) would consist of elements that are in S but not in S + d, and vice versa.If d is chosen such that only one element is lost and one is gained, then the symmetric difference would have size 2.But is that always possible? Let's see.Suppose S has at least two elements. Let d = s₂ - s₁. Then, S + d would contain s₁ + d = s₂, s₂ + d = s₁ + 2d, etc. So, the overlap between S and S + d would be at least s₂, s₃ + d, etc. Wait, not sure.Alternatively, choose d such that S + d overlaps with S in all but one element. But I need exactly two elements in the symmetric difference.Wait, maybe if d is the difference between two consecutive elements in S. Let's say S has elements s₁, s₂, ..., sₖ. Let d = s₂ - s₁. Then, S + d would be s₁ + d, s₂ + d, ..., sₖ + d.The overlap between S and S + d would be s₂, s₃, ..., sₖ, because s₁ + d = s₂. So, the symmetric difference would be {s₁, sₖ + d}. Therefore, size 2.Yes! So, if we take d as the difference between two consecutive elements in S, then S Δ (S + d) has exactly two elements: the smallest element s₁ and the largest element shifted by d, which is sₖ + d.Therefore, choosing y = x + d, we can make the symmetric difference of S + x and S + y be exactly {s₁ + x, sₖ + x + d}.Thus, applying S at x and then at y = x + d will toggle exactly those two bulbs.Therefore, we can always find such a d, and hence achieve exactly two bulbs being on.Wait, but what if S has only one element? Then, d would be undefined since there are no two consecutive elements. But in that case, S is just a single point, say {s}. Then, applying S at x and at y = x + d, but since d is not defined, we need another approach.Wait, if S has only one element, say S = {s}, then applying S at x toggles bulb x + s. To get exactly two bulbs on, we can apply S at two different positions, say x and y, resulting in bulbs x + s and y + s being toggled. So, we have exactly two bulbs on.Therefore, even in the case where S is a single point, we can achieve exactly two bulbs on by applying S at two different positions.So, in all cases, whether S has one element or more, we can find a way to apply the template a finite number of times to result in exactly two bulbs being on.Therefore, the proof is complete.</think>

🧠:<think>Alright, so I need to find a right triangle with integer sides where each of its angles can be trisected using only a compass and straightedge. Hmm, okay, let's break this down.First, I know that a right triangle has one 90-degree angle and two acute angles. The sum of all angles in a triangle is 180 degrees, so the two acute angles must add up to 90 degrees. That means each acute angle is less than 90 degrees.Now, the problem is about trisecting each of these angles. Trisecting an angle means dividing it into three equal parts, each of which is one-third of the original angle. I remember that not all angles can be trisected with just a compass and straightedge; it depends on the angle. Specifically, an angle can be trisected if and only if its cosine is a rational number. This is related to something called the "trisection of an angle" problem in geometry.So, to trisect each angle, each angle's cosine must be rational. That means I need to find a right triangle where all three angles have rational cosines. Wait, but in a right triangle, one angle is 90 degrees, which has a cosine of 0, which is rational. So that's good. But the two acute angles also need to have rational cosines.I also remember that if a triangle has all sides of integer lengths, it's called a Pythagorean triple. So, I need a Pythagorean triple where the two acute angles have rational cosines. That means the ratio of the adjacent side to the hypotenuse for each acute angle must be rational. Similarly, the ratio of the opposite side to the hypotenuse (which is the sine) must also be rational.So, to rephrase, I need a Pythagorean triple where both the cosine and sine of each acute angle are rational numbers. That seems doable because in a Pythagorean triple, all sides are integers, so the ratios will be rational.Let me think of some common Pythagorean triples. The smallest one is (3, 4, 5). Let's check the angles here. The two acute angles can be calculated using inverse sine or cosine.For the angle opposite the side of length 3:cos(theta) = adjacent/hypotenuse = 4/5 = 0.8, which is rational.sin(theta) = opposite/hypotenuse = 3/5 = 0.6, which is rational.Similarly, for the angle opposite the side of length 4:cos(theta) = 3/5 = 0.6, rational.sin(theta) = 4/5 = 0.8, rational.So, both acute angles in a (3, 4, 5) triangle have rational cosines and sines. Therefore, in theory, these angles can be trisected with a compass and straightedge.But wait, is that enough? I mean, just because the cosine is rational, does that necessarily mean the angle can be trisected? I think it's more nuanced. The ability to trisect an angle with compass and straightedge depends on the constructibility of the angle, which relates to whether the angle is a multiple of a constructible angle.I remember that an angle is constructible if it can be expressed using square roots, and trisecting an angle is possible if the angle's cosine is a root of a cubic equation that can be solved with radicals. But I'm not sure about the exact conditions.Wait, the problem says "each of whose angles can be trisected using only a compass and straightedge." So, the right angle is 90 degrees, which is already a constructible angle, and its trisection would be 30 degrees, which is also constructible. The other two angles, being acute, need to be trisected as well.So, if the two acute angles can be trisected, then their trisection angles must also be constructible. Therefore, not only do the cosines need to be rational, but also the trisection angles must lead to constructible angles.Let me think about the (3, 4, 5) triangle again. The two acute angles are approximately 36.87 degrees and 53.13 degrees. If we try to trisect these angles, we get approximately 12.29 degrees and 17.71 degrees, respectively. Are these angles constructible?Constructible angles are those that can be formed by doubling, adding, or subtracting known constructible angles, or by taking their halves. But 12.29 degrees isn't a standard constructible angle, as far as I know. Similarly, 17.71 degrees isn't either.Hmm, maybe the (3, 4, 5) triangle isn't the right candidate. Perhaps I need a triangle where the acute angles are multiples of 30 degrees or some other constructible angles.Wait, 30 degrees is constructible. If an angle is 60 degrees, which is also constructible, then trisecting it would give 20 degrees, which is not constructible. Hmm, that complicates things.Alternatively, maybe I need angles that are already trisectible because their trisection leads to constructible angles. For example, if the acute angle is 90 degrees, which is already a right angle, but in a right triangle, the other two angles must be acute.Wait, maybe I'm overcomplicating this. Let's go back to the original thought. If the cosine of an angle is rational, then the angle can be trisected with compass and straightedge. Is that a valid statement?I think the key here is that if the cosine is rational, then the angle is constructible, but trisecting it may or may not be possible depending on the angle. However, in the problem statement, it's implied that if the cosine is rational, then the angle can be trisected.Wait, I should double-check that. Let me recall that the trisection of an angle is possible with compass and straightedge only if the angle is a multiple of 3 degrees, but I'm not sure. Actually, the problem of trisecting an arbitrary angle is impossible with only compass and straightedge, but certain angles can be trisected.I think the condition for trisectibility is related to the angle being a constructible angle, which would mean that its trisection would also be constructible. So, if the original angle is constructible, then so is its trisection. But I need to make sure that the angles in the triangle are constructible.Given that, if the triangle has sides of integer lengths, then all the trigonometric ratios (sines and cosines) are rational, hence the angles are constructible. Therefore, their trisection should also be constructible.Wait, does that mean that any right triangle with integer sides can have its angles trisected with compass and straightedge? That can't be right because not all angles are trisectible.But perhaps in this specific case, because the sides are integers and the trigonometric ratios are rational, the angles are constructible, and hence their trisection is also possible.Wait, I'm getting confused. Let me try to find a specific example.I know that in a 30-60-90 triangle, the angles are 30, 60, and 90 degrees. The cosines of 30 and 60 degrees are √3/2 and 1/2, respectively. √3 is irrational, so the cosine isn't rational. Therefore, a 30-60-90 triangle doesn't have rational cosines for all angles, so it might not satisfy the condition.But the (3,4,5) triangle has rational cosines for both acute angles, as we saw earlier. So, maybe this triangle works. Let me see.If the two acute angles are approximately 36.87 and 53.13 degrees, their cosines are 4/5 and 3/5, respectively. These are rational. So, if I can trisect these angles, then the trisection would result in angles of approximately 12.29 and 17.71 degrees. Are these constructible?Well, 12.29 degrees is roughly 12 degrees, which isn't a standard constructible angle. But wait, 12 degrees is constructible because 3 degrees is constructible (as 3 is a factor of 360, and certain multiples can be constructed). Wait, is 3 degrees constructible?I think that 3 degrees is constructible because it can be obtained by bisecting 6 degrees, which is constructible by trisecting 18 degrees, which is constructible because it's related to the regular pentagon. Hmm, this is getting complicated.Alternatively, maybe the way to approach this is by using the fact that if the cosine of the angle is rational, then the angle can be constructed, and hence, its trisection can be constructed as well. So, if the cosine is rational, the angle is constructible, and so is its trisection.Therefore, in the (3,4,5) triangle, since both acute angles have rational cosines, their trisection angles are also constructible. Hence, the triangle satisfies the condition.But wait, I'm not entirely sure about this reasoning. Let me check another source or think more carefully.I recall that an angle can be trisected with compass and straightedge if and only if its cosine is a root of a cubic equation that can be solved with radicals. Since the cosine is rational, the cubic equation would have rational coefficients, and by the rational root theorem, any rational roots can be found. Therefore, if the cosine is rational, the angle can be trisected.But I'm not entirely certain. Maybe I should look for a specific example where the trisection is known to be possible.Wait, another approach: if the angle is a multiple of 3 degrees, then it's trisectible. But in the (3,4,5) triangle, the angles are not multiples of 3 degrees, so their trisection might not be constructible.Hmm, this is confusing. Maybe I need to find a right triangle where the acute angles are multiples of 3 degrees, so their trisection would result in integer degree angles, which are constructible.For example, if an acute angle is 60 degrees, trisecting it would give 20 degrees, which is not constructible. But if the acute angle is 90 degrees, which is already a right angle, but in a right triangle, the other two angles must be acute, so that's not applicable.Wait, perhaps I need a triangle where the acute angles are 60 degrees and 30 degrees, but as I mentioned earlier, their cosines are not rational, so that might not work.Alternatively, maybe I can find a triangle where the acute angles are 45 degrees each, but that's an isosceles right triangle, with sides in the ratio 1:1:√2, which isn't a Pythagorean triple because √2 is irrational. So that's not helpful.Wait, but the problem says "integer sides," so it must be a Pythagorean triple. So, perhaps the (3,4,5) triangle is the way to go, even if the trisection angles aren't standard.Alternatively, let's think about the trisection process. To trisect an angle, we can use the method involving the intersection of two circles or other geometric constructions. If the original angle is constructible, then its trisection should also be constructible.Given that the (3,4,5) triangle has constructible angles (since their cosines are rational), their trisection should also be possible.Therefore, perhaps the (3,4,5) triangle is indeed the answer.But let me confirm by checking another Pythagorean triple. For example, the (5,12,13) triangle. The acute angles can be calculated as follows:For the angle opposite 5:cos(theta) = 12/13 ≈ 0.923, which is rational.sin(theta) = 5/13 ≈ 0.385, rational.Similarly, for the angle opposite 12:cos(theta) = 5/13 ≈ 0.385, rational.sin(theta) = 12/13 ≈ 0.923, rational.So, both acute angles have rational cosines and sines, meaning their angles are constructible. Therefore, their trisection should also be possible.So, both (3,4,5) and (5,12,13) triangles have this property.But the problem says "find at least one," so either would work. However, perhaps the smallest one is preferable.Wait, but earlier I was confused about whether trisecting an angle with rational cosine necessarily makes it trisectible. Maybe I need to ensure that the trisection doesn't lead to an angle that requires solving a non-solvable cubic equation.I think that if the original angle is constructible, then so is its trisection, because the trisection can be achieved by solving a cubic equation with constructible roots. Therefore, if the original angle is constructible, its trisection is also constructible.Therefore, in the (3,4,5) triangle, since both acute angles are constructible, their trisection is also possible with compass and straightedge.Hence, the (3,4,5) triangle is a valid answer.Wait, but I'm still a bit unsure. Let me think about the actual trisection process. For example, to trisect an angle, we can use the method where we construct an equilateral triangle or use intersecting circles. But in this case, since the angle is arbitrary, we need a more general method.I think the key is that if the angle is constructible, then it can be trisected. Since the angles in the (3,4,5) triangle are constructible, their trisection is also possible.Therefore, I think the (3,4,5) triangle satisfies the conditions of the problem.Final AnswerThe right triangle with integer sides is the boxed{(3, 4, 5)} triangle.</think>

🧠:<think>Alright, so I have this problem: I need to find the common tangent planes to two spheres that pass through a given point. The centers of these spheres lie on something called the "projection axis." Hmm, projection axis—okay, I think that might be like a line we’re projecting onto, maybe similar to the z-axis or something in a coordinate system. But I'm not entirely sure, so I should probably clarify that.First, let me visualize this. There are two spheres, each with their centers on this projection axis. Let's say the projection axis is the z-axis for simplicity. So, both centers of the spheres have their x and y coordinates as zero, and their z coordinates vary. Let me denote the centers as ( C_1 = (0, 0, z_1) ) and ( C_2 = (0, 0, z_2) ). Now, the spheres have some radii, right? Let's call them ( r_1 ) and ( r_2 ) for the first and second sphere respectively. So, the equations of the spheres would be:For ( S_1 ): ( x^2 + y^2 + (z - z_1)^2 = r_1^2 )For ( S_2 ): ( x^2 + y^2 + (z - z_2)^2 = r_2^2 )Okay, so I have these two spheres along the z-axis. Now, I need to find the common tangent planes to both spheres that pass through a given point ( P = (p_x, p_y, p_z) ). Wait, common tangent planes... So, a plane that is tangent to both spheres and also passes through point P. Since the centers are on the z-axis, maybe the problem has some symmetry that I can exploit.Let me recall that a tangent plane to a sphere satisfies the condition that the distance from the center of the sphere to the plane is equal to the radius of the sphere. So, if I have a plane given by ( ax + by + cz + d = 0 ), then for it to be tangent to ( S_1 ), the distance from ( C_1 ) to the plane must be ( r_1 ), and similarly for ( S_2 ).So, the distance from ( C_1 = (0, 0, z_1) ) to the plane is:( frac{|a*0 + b*0 + c*z_1 + d|}{sqrt{a^2 + b^2 + c^2}} = r_1 )Similarly, for ( C_2 = (0, 0, z_2) ):( frac{|a*0 + b*0 + c*z_2 + d|}{sqrt{a^2 + b^2 + c^2}} = r_2 )Alright, so I have these two equations:1. ( |c*z_1 + d| = r_1 * sqrt{a^2 + b^2 + c^2} )2. ( |c*z_2 + d| = r_2 * sqrt{a^2 + b^2 + c^2} )Also, the plane must pass through the point ( P = (p_x, p_y, p_z) ), so:3. ( a*p_x + b*p_y + c*p_z + d = 0 )So, now I have three equations with four variables: a, b, c, d. But since the equation of a plane is defined up to a scalar multiple, I can set one of the variables to 1 or something to fix the scale. Maybe set ( sqrt{a^2 + b^2 + c^2} = 1 ) to simplify things? Let me try that.So, let me define ( sqrt{a^2 + b^2 + c^2} = 1 ). Then, equations 1 and 2 become:1. ( |c*z_1 + d| = r_1 )2. ( |c*z_2 + d| = r_2 )And equation 3 remains:3. ( a*p_x + b*p_y + c*p_z + d = 0 )So, now I have three equations with four variables: a, b, c, d. But since I set ( sqrt{a^2 + b^2 + c^2} = 1 ), it's a constraint on a, b, c.This seems manageable. Let me think about how to solve these equations.First, let's consider equations 1 and 2. They involve absolute values, which can complicate things. Let's consider the cases where the expressions inside the absolute value are positive or negative.Case 1: ( c*z_1 + d = r_1 )and ( c*z_2 + d = r_2 )Subtracting these two equations:( c*(z_1 - z_2) = r_1 - r_2 )So, ( c = (r_1 - r_2)/(z_1 - z_2) ), assuming ( z_1 neq z_2 ). Which makes sense because if ( z_1 = z_2 ), the centers would be the same, and the problem might be different.So, once I find c, I can find d from equation 1:( d = r_1 - c*z_1 )Similarly, from equation 2:( d = r_2 - c*z_2 )So, these two expressions for d must be equal:( r_1 - c*z_1 = r_2 - c*z_2 )Which simplifies to:( r_1 - r_2 = c*(z_1 - z_2) )But that's the same as before, so no new information.So, once c is found, d is determined.Now, let's move to equation 3:( a*p_x + b*p_y + c*p_z + d = 0 )We have c and d in terms of r1, r2, z1, z2, so let's plug those in.From earlier:( c = (r1 - r2)/(z1 - z2) )and( d = r1 - c*z1 = r1 - [(r1 - r2)/(z1 - z2)]*z1 )Simplify d:( d = r1 - (r1*z1 - r2*z1)/(z1 - z2) )Factor numerator:( d = r1 - [z1*(r1 - r2)]/(z1 - z2) )Similarly, ( d = [r1*(z1 - z2) - z1*(r1 - r2)] / (z1 - z2) )Expand numerator:( r1*z1 - r1*z2 - z1*r1 + z1*r2 )Simplify:( -r1*z2 + z1*r2 )Thus,( d = (z1*r2 - r1*z2)/(z1 - z2) )Okay, so now we have expressions for c and d.Now, equation 3 becomes:( a*p_x + b*p_y + c*p_z + d = 0 )Plug in c and d:( a*p_x + b*p_y + [(r1 - r2)/(z1 - z2)]*p_z + (z1*r2 - r1*z2)/(z1 - z2) = 0 )Let me write this as:( a*p_x + b*p_y = -[(r1 - r2)/(z1 - z2)]*p_z - (z1*r2 - r1*z2)/(z1 - z2) )Let me denote the right-hand side as K:( K = -[(r1 - r2)/(z1 - z2)]*p_z - (z1*r2 - r1*z2)/(z1 - z2) )So,( a*p_x + b*p_y = K )So, now I have a relationship between a and b.But I also have the constraint:( a^2 + b^2 + c^2 = 1 )Since we set ( sqrt{a^2 + b^2 + c^2} = 1 ).So, from earlier, c is known, so ( c^2 = [(r1 - r2)/(z1 - z2)]^2 )Thus, ( a^2 + b^2 = 1 - [(r1 - r2)/(z1 - z2)]^2 )Let me denote ( a^2 + b^2 = sqrt{1 - c^2} ), but actually, it's just ( a^2 + b^2 = 1 - c^2 )So, we have:( a^2 + b^2 = 1 - [(r1 - r2)/(z1 - z2)]^2 )And( a*p_x + b*p_y = K )So, this is a linear equation in a and b, and another equation for the sum of their squares. This is similar to finding points on a circle that satisfy a linear equation, which typically has two solutions.So, in general, there can be two solutions for a and b, leading to two different planes.But wait, originally, we considered the case where ( c*z1 + d = r1 ) and ( c*z2 + d = r2 ). What if we consider the other case where the expressions inside the absolute values are negative?Case 2: ( c*z1 + d = -r1 )and ( c*z2 + d = -r2 )Following similar steps as before, subtracting the two equations:( c*(z1 - z2) = -r1 + r2 )So,( c = (r2 - r1)/(z1 - z2) = -(r1 - r2)/(z1 - z2) )Which is just -c from Case 1.Then, d would be:From equation 1:( d = -r1 - c*z1 )Plugging c:( d = -r1 - [-(r1 - r2)/(z1 - z2)]*z1 = -r1 + (r1*z1 - r2*z1)/(z1 - z2) )Simplify:( d = [-r1*(z1 - z2) + r1*z1 - r2*z1]/(z1 - z2) )Expand numerator:( -r1*z1 + r1*z2 + r1*z1 - r2*z1 )Simplify:( r1*z2 - r2*z1 )Which is the same as in Case 1.Wait, that's interesting. So, in Case 2, d is the same as in Case 1?But c is negative. So, in equation 3:( a*p_x + b*p_y + c*p_z + d = 0 )With c being negative, and d same as Case 1.So, the right-hand side K would be different.Let me compute K for Case 2:( K = -[(r2 - r1)/(z1 - z2)]*p_z - (z1*r2 - r1*z2)/(z1 - z2) )Wait, but this is just:( K = [ (r1 - r2)/(z1 - z2) ]*p_z - (z1*r2 - r1*z2)/(z1 - z2) )Which is different from Case 1.So, in this case, K is different, leading to different a and b.Therefore, in total, we can have two cases for the absolute values, each leading to two solutions for a and b, so four solutions in total.Therefore, there are four common tangent planes passing through point P.Now, to find the points of tangency on each sphere.Once I have the equations of the tangent planes, I can find the points where the plane touches each sphere.For a given plane, the point of tangency on a sphere can be found by moving from the center of the sphere towards the plane by a distance equal to the radius in the direction perpendicular to the plane.Given the plane equation ( ax + by + cz + d = 0 ), the normal vector is ( (a, b, c) ). So, the direction from the center to the tangent point is along this normal vector.Therefore, the tangent point on sphere ( S_1 ) can be expressed as:( T1 = C1 + r1 * frac{(a, b, c)}{|(a, b, c)|} )But since we set ( |(a, b, c)| = 1 ), this simplifies to:( T1 = (0, 0, z1) + r1*(a, b, c) )Similarly, for sphere ( S_2 ):( T2 = (0, 0, z2) + r2*(a, b, c) )So, once we have a, b, c, we can compute T1 and T2.But since we have four different planes, each with different a, b, c, we'll have four pairs of tangent points.So, summarizing the steps:1. For each case (Case 1 and Case 2), solve for c and d.2. For each case, solve for a and b using the linear equation derived from the plane passing through P, and the constraint ( a^2 + b^2 + c^2 = 1 ). This will give two solutions for a and b per case.3. For each solution (four in total), compute the tangent points T1 and T2 on each sphere.Therefore, the final answer is that there are four common tangent planes passing through the given point P, each corresponding to a pair of tangent points on the two spheres.But wait, is this always the case? What if the point P is not in a general position? Maybe in some cases, the number of solutions could be less.Well, in general, for two spheres, the number of common tangent planes is four: two direct and two transversal. But when considering a specific point, depending on the position of P, some of these planes might not pass through P.In our case, since we derived four solutions, it suggests that there are four common tangent planes passing through P, assuming P is not in a special position relative to the spheres.So, I think the conclusion is that there are four common tangent planes passing through the given point P, each with their respective points of tangency on the two spheres.Final AnswerThe number of common tangent planes through the given point is boxed{4}.</think>