🧠:[<think>Alright, let's see. The problem is A₄² minus C₄³ equals what? The options are A:6, B:12, C:8, D:20. Hmm, okay. I need to remember what A and C stand for in combinatorics. A usually stands for permutations, right? So A₄² would be the number of ways to arrange 2 items out of 4, where order matters. And C is combinations, so C₄³ is the number of ways to choose 3 items out of 4 without considering order. Got it. Let me write down the formulas to be sure.For permutations, A(n, k) or P(n, k) is calculated as n! divided by (n - k)!. So A₄² should be 4! divided by (4 - 2)! That would be 24 divided by 2!, which is 2. So 24 divided by 2 is 12. Wait, 4 factorial is 24, right? 4x3x2x1=24. Then 4-2=2, so 2! is 2. So 24/2=12. So A₄² is 12.Now, for combinations, C(n, k) is n! divided by (k! times (n - k)!). So C₄³ would be 4! divided by (3! times (4 - 3)!). Let's compute that. 4! is 24. The denominator is 3! times 1!. 3! is 6, and 1! is 1, so 6x1=6. So 24 divided by 6 is 4. Therefore, C₄³ is 4.So the problem is A₄² minus C₄³, which is 12 minus 4. 12 minus 4 is 8. Let me check the options again. The options are A:6, B:12, C:8, D:20. So the answer should be C:8.Wait, just to make sure I didn't mix up the permutation and combination formulas. Let me double-check. For permutations, A(n, k) = n! / (n - k)! So 4! / 2! = 24 / 2 =12. Correct. For combinations, C(n, k)= n! / (k! (n -k)! ). So 4! / (3! *1!)=24/(6*1)=24/6=4. Correct. Then 12 -4=8. Yes, option C is 8. That seems right.I don't think I made a mistake here. The permutation part is straightforward, and the combination as well. Subtracting them gives 8. So the answer is C. Let me just confirm once more. Maybe the notation could be different? Like sometimes A is written as P. But in the problem, it's definitely A₄² and C₄³. Yeah, in Chinese textbooks, sometimes A is used for permutations and C for combinations, so that's consistent. Another way to compute A₄²: 4 options for the first position, 3 for the second, so 4x3=12. Yep. And C₄³: since choosing 3 out of 4 is the same as leaving out 1, so there are 4 ways. For example, if the elements are a,b,c,d, the combinations are abc, abd, acd, bcd. Four combinations. So that checks out. 12 -4=8. Definitely option C. No, I think that's solid. The answer is C:8.Final Answerboxed{C}</think>To solve the problem (A_4^2 - C_4^3), we need to calculate the permutations and combinations separately and then subtract the results.1. Permutations (A₄²): - The formula for permutations is (A(n, k) = frac{n!}{(n - k)!}). - For (A_4^2), we have (n = 4) and (k = 2). - Calculation: (A_4^2 = frac{4!}{(4 - 2)!} = frac{4!}{2!} = frac{24}{2} = 12).2. Combinations (C₄³): - The formula for combinations is (C(n, k) = frac{n!}{k!(n - k)!}). - For (C_4^3), we have (n = 4) and (k = 3). - Calculation: (C_4^3 = frac{4!}{3!(4 - 3)!} = frac{4!}{3! cdot 1!} = frac{24}{6 cdot 1} = 4).3. Subtracting the results: - Subtract the combination result from the permutation result: (12 - 4 = 8).Thus, the final answer is (boxed{C}).,<think>Okay, let's see. The problem is A₄² minus C₄³ equals what? The options are A:6, B:12, C:8, D:20. Hmm. Alright, I need to remember what A and C stand for in combinatorics. So, A is usually permutations, right? Like, the number of ways to arrange things where order matters. And C is combinations, where order doesn't matter. The notation might be a bit confusing, but I think A₄² is permutation of 4 things taken 2 at a time, and C₄³ is combination of 4 things taken 3 at a time.Let me confirm. The formula for permutations, A(n, k) or sometimes written as P(n, k), is n! divided by (n - k)!. So A₄² would be 4 factorial divided by (4 - 2) factorial. That's 4! / 2! which is (4 × 3 × 2 × 1) / (2 × 1) = 24 / 2 = 12. So A₄² is 12.Now, for combinations, C(n, k) is n! divided by (k! times (n - k)!). So C₄³ would be 4! divided by (3! times (4 - 3)!). Let's compute that. 4! is 24, 3! is 6, and (4 - 3)! is 1!, which is 1. So 24 / (6 × 1) = 24 / 6 = 4. Therefore, C₄³ is 4.The problem asks for A₄² minus C₄³, so that's 12 - 4 = 8. Let me check the options again: A is 6, B is 12, C is 8, D is 20. So the answer should be option C:8.Wait, just to make sure I didn't mix up the formulas. Permutations: nPk = n! / (n - k)! and combinations: nCk = n! / [k! (n - k)!]. Yeah, that's right. So for A₄²: 4P2 = 4 × 3 = 12. For C₄³: 4C3 = 4, since choosing 3 out of 4 is the same as leaving out 1, so there are 4 ways. 12 - 4 is indeed 8. So I think that's correct. The answer is C.Final Answerboxed{C}</think>To solve the problem (A_4^2 - C_4^3), we need to calculate the permutations and combinations separately and then subtract them.1. Permutations (A₄²): The formula for permutations is (A(n, k) = frac{n!}{(n - k)!}). [ A_4^2 = frac{4!}{(4 - 2)!} = frac{4!}{2!} = frac{24}{2} = 12 ]2. Combinations (C₄³): The formula for combinations is (C(n, k) = frac{n!}{k!(n - k)!}). [ C_4^3 = frac{4!}{3!(4 - 3)!} = frac{24}{6 times 1} = frac{24}{6} = 4 ]3. Subtracting the two results: [ A_4^2 - C_4^3 = 12 - 4 = 8 ]Thus, the final answer is (boxed{C}).]

🔑:To solve the problem A_4^2-C_4^3=left( right), we first calculate each term separately.For A_4^2, which represents the number of arrangements of 4 items taken 2 at a time, we use the formula for permutations:A_n^r = frac{n!}{(n-r)!}Substituting n=4 and r=2, we get:A_4^2 = frac{4!}{(4-2)!} = frac{4 times 3 times 2!}{2!} = 4 times 3 = 12For C_4^3, which represents the number of combinations of 4 items taken 3 at a time, we use the formula for combinations:C_n^r = frac{n!}{r!(n-r)!}Substituting n=4 and r=3, we get:C_4^3 = frac{4!}{3!(4-3)!} = frac{4 times 3!}{3! times 1!} = 4Subtracting the two results:A_4^2 - C_4^3 = 12 - 4 = 8Therefore, the correct answer is boxed{C}.

🧠:[<think>Okay, so I need to find the limit of (tan x)^(cot x) as x approaches π/4. Hmm, let's start by recalling some limit laws and properties. Since both tan x and cot x are involved, maybe I can rewrite the expression in a different form to make it easier to handle. First, I know that cot x is the reciprocal of tan x, so cot x = 1/tan x. Therefore, the expression becomes (tan x)^(1/tan x). Wait, but when x approaches π/4, tan x approaches 1 because tan(π/4) is 1. So, the base is approaching 1 and the exponent is approaching 1/1 = 1. So, essentially, the expression is approaching 1^1, which is 1. But wait, limits involving indeterminate forms might require more careful analysis. Is 1^1 an indeterminate form? No, 1^1 is just 1, so maybe the limit is straightforward. However, sometimes when the base and exponent both approach certain values, even if they seem to approach 1 and 1, there might be subtler behavior. Let me check that.Alternatively, maybe taking the natural logarithm would help. Let me set L = lim_{x→π/4} (tan x)^(cot x). Then take ln of both sides: ln L = lim_{x→π/4} cot x * ln(tan x). Then if I can compute this limit, I can exponentiate the result to find L. So let's compute the limit of cot x * ln(tan x) as x approaches π/4. Let's substitute variables for simplicity. Let t = x - π/4, so as x approaches π/4, t approaches 0. Then tan x = tan(π/4 + t). Using the tangent addition formula: tan(π/4 + t) = [tan(π/4) + tan t] / [1 - tan(π/4) tan t] = [1 + tan t]/[1 - tan t]. Similarly, cot x = cot(π/4 + t) = [cot π/4 cot t - 1]/[cot π/4 + cot t] but maybe that's more complicated. Alternatively, since cot x = 1/tan x, so cot x = [1 - tan t]/[1 + tan t] from the previous expression. Wait, maybe it's better to express both tan x and cot x in terms of t.Alternatively, perhaps using substitution y = tan x. As x approaches π/4, y approaches 1. Then cot x = 1/y. So the expression becomes y^(1/y), and as y approaches 1, this is 1^(1) = 1. But again, is there a more precise way?Alternatively, maybe we can write the expression as e^{cot x * ln(tan x)} and analyze the exponent. Let's see: lim_{x→π/4} [cot x * ln(tan x)]. Since both cot x and tan x approach 1, we have 1 * ln 1 = 0. So exponent approaches 0, so the limit is e^0 = 1. Wait, so that would make the limit 1. But maybe there is a more precise estimation needed here? Let me verify.Alternatively, consider expanding tan x and cot x around x = π/4. Let me set t = x - π/4, so t approaches 0. Then tan x = tan(π/4 + t) = [1 + tan t]/[1 - tan t] ≈ [1 + t + (t^3)/3 + ... ]/[1 - t - (t^3)/3 + ... ] using the Taylor series for tan t around t=0. Since t is approaching 0, tan t ≈ t + t^3/3 + 2t^5/15 + ... So, approximating tan x as [1 + t]/[1 - t] up to first order. Then ln(tan x) ≈ ln([1 + t]/[1 - t]) ≈ ln(1 + t) - ln(1 - t) ≈ (t - t^2/2 + t^3/3 - ...) - (-t - t^2/2 - t^3/3 - ...) ≈ t - t^2/2 + t^3/3 + t + t^2/2 + t^3/3 + ... ≈ 2t + (2t^3)/3 + ... So, ln(tan x) ≈ 2t + (2t^3)/3. Meanwhile, cot x = 1/tan x ≈ [1 - t]/[1 + t] ≈ 1 - 2t + 2t^2 - ... using the expansion 1/(1 + t) ≈ 1 - t + t^2 - t^3 + ..., so [1 - t]/[1 + t] ≈ (1 - t)(1 - t + t^2 - ...) ≈ 1 - 2t + 2t^2 - ... Therefore, cot x ≈ 1 - 2t. Multiplying cot x and ln(tan x):cot x * ln(tan x) ≈ (1 - 2t)(2t + (2t^3)/3) ≈ 2t + (2t^3)/3 - 4t^2 - (4t^4)/3. As t approaches 0, the dominant term is 2t, which goes to 0. Therefore, the exponent goes to 0, so e^0 = 1. Therefore, the limit is 1. Alternatively, maybe using L’Hospital’s Rule. Let me check if the exponent can be expressed as a fraction that leads to 0/0 or ∞/∞. So, the exponent is cot x * ln(tan x) as x approaches π/4. Let’s set t = x - π/4, so t approaches 0. Let’s express cot x * ln(tan x) in terms of t. As before, tan(π/4 + t) = (1 + tan t)/(1 - tan t). Let’s denote u = tan t. Then as t approaches 0, u approaches 0. So, tan x = (1 + u)/(1 - u). Then ln(tan x) = ln((1 + u)/(1 - u)) = ln(1 + u) - ln(1 - u). Then cot x = 1/tan x = (1 - u)/(1 + u). So, cot x * ln(tan x) = [(1 - u)/(1 + u)] * [ln(1 + u) - ln(1 - u)]. Expanding each term in Taylor series around u=0:ln(1 + u) ≈ u - u^2/2 + u^3/3 - u^4/4 + ...ln(1 - u) ≈ -u - u^2/2 - u^3/3 - u^4/4 - ...So, ln(1 + u) - ln(1 - u) ≈ [u - u^2/2 + u^3/3 - ...] - [-u - u^2/2 - u^3/3 - ...] = 2u + (2u^3)/3 + (2u^5)/5 + ...Similarly, (1 - u)/(1 + u) ≈ (1 - u)(1 - u + u^2 - u^3 + ...) ≈ 1 - 2u + 2u^2 - 2u^3 + ...Multiplying these two series:[1 - 2u + 2u^2 - 2u^3 + ...] * [2u + (2u^3)/3 + ...] Multiplying term by term:1 * 2u = 2u1 * (2u^3)/3 = (2u^3)/3-2u * 2u = -4u^2-2u * (2u^3)/3 = -4u^4 /32u^2 * 2u = 4u^32u^2 * (2u^3)/3 = 4u^5 /3And so on. Combining up to u^3 terms:2u -4u^2 + [ (2/3)u^3 +4u^3 ] = 2u -4u^2 + (14/3)u^3As u approaches 0, the dominant term is 2u, which approaches 0. Therefore, the entire expression approaches 0. Hence, the exponent approaches 0, and the original limit is e^0 = 1. Alternatively, maybe considering substitution. Let’s set y = tan x, so as x approaches π/4, y approaches 1. Then cot x = 1/y, and the expression becomes y^(1/y). The function y^(1/y) as y approaches 1. Let’s compute the limit as y approaches 1. Let’s take ln of the expression: (1/y) * ln y. As y approaches 1, ln y approaches 0, and 1/y approaches 1, so the product approaches 0. Therefore, the original expression approaches e^0 = 1. Therefore, all approaches seem to indicate that the limit is 1. Wait, but to be thorough, perhaps I should check numerically. Let's take x approaching π/4 from the left and right and compute (tan x)^(cot x). Let's choose x = π/4 + 0.1, which is close to π/4. Compute tan(π/4 + 0.1):tan(π/4 + 0.1) ≈ (1 + tan 0.1)/(1 - tan 0.1). tan 0.1 ≈ 0.100334672, so (1 + 0.100334672)/(1 - 0.100334672) ≈ 1.100334672 / 0.899665328 ≈ 1.22304888. Then cot x = 1/tan x ≈ 0.8175671. Then (1.22304888)^0.8175671 ≈ e^{0.8175671 * ln 1.22304888} ≈ e^{0.8175671 * 0.201413} ≈ e^{0.1649} ≈ 1.179. But wait, this is when x is π/4 + 0.1, which is not very close. Let me take a smaller t, like t = 0.01. Then tan(π/4 + 0.01) = (1 + tan 0.01)/(1 - tan 0.01). tan 0.01 ≈ 0.0100003333. So (1 + 0.0100003333)/(1 - 0.0100003333) ≈ 1.0100003333 / 0.9899996667 ≈ 1.020202. Then cot x ≈ 1/1.020202 ≈ 0.9802. Then (1.020202)^0.9802 ≈ e^{0.9802 * ln 1.020202} ≈ e^{0.9802 * 0.0199986} ≈ e^{0.0196} ≈ 1.0198. So as x approaches π/4, the value seems to approach 1. Similarly, if x approaches from below, say x = π/4 - 0.01. Then tan(π/4 - 0.01) = (1 - tan 0.01)/(1 + tan 0.01) ≈ (1 - 0.0100003333)/(1 + 0.0100003333) ≈ 0.9899996667 / 1.0100003333 ≈ 0.9802. Then cot x = 1/tan x ≈ 1.020202. Then (0.9802)^1.020202 ≈ e^{1.020202 * ln 0.9802} ≈ e^{1.020202 * (-0.020202)} ≈ e^{-0.02061} ≈ 0.9796. Wait, that's slightly less than 1. But as t approaches 0, both sides should approach 1. The discrepancy here is because we are still using t=0.01, which is not extremely small. Let's take t = 0.0001. Then tan(π/4 + 0.0001) ≈ (1 + 0.0001)/(1 - 0.0001) ≈ 1.0001 / 0.9999 ≈ 1.00020002. Then cot x ≈ 0.9998. So (1.0002)^0.9998 ≈ e^{0.9998 * 0.0002} ≈ e^{0.00019996} ≈ 1.0002. Similarly, from below: tan(π/4 - 0.0001) ≈ 0.9998 / 1.0001 ≈ 0.99970006. cot x ≈ 1.0003. Then (0.9997)^1.0003 ≈ e^{1.0003 * ln 0.9997} ≈ e^{1.0003 * (-0.000300045)} ≈ e^{-0.00030009} ≈ 0.9997. But again, as t approaches 0, the exponent approaches 0, so e^0 is 1. The numerical examples are getting closer to 1 as t gets smaller. So, this supports the analytical conclusion that the limit is 1.Another approach: Since tan x and cot x are reciprocals, let me set t = tan x - 1, so as x approaches π/4, t approaches 0. Then cot x = 1/(1 + t). Then the expression becomes (1 + t)^{1/(1 + t)}. Let's compute the limit as t approaches 0. Take ln: [1/(1 + t)] * ln(1 + t) ≈ [1 - t + t^2 - ...] * [t - t^2/2 + t^3/3 - ...] ≈ t - t^2 + t^3/2 - t^2 + t^3 - ... Hmm, combining terms: t - (1 + 1)t^2 + (1/2 + 1)t^3 + ... ≈ t - 2t^2 + (3/2)t^3 + ... As t approaches 0, the exponent approaches 0, so the original expression approaches e^0 = 1. Alternatively, recognize that for any function f(x) approaching 1 as x approaches a, then f(x)^{g(x)} approaches 1^{lim g(x)} if lim g(x) exists. But here, both f(x) = tan x approaches 1 and g(x) = cot x approaches 1. However, this rule is not always directly applicable because 1^∞ is an indeterminate form. But in this case, if both f(x) approaches 1 and g(x) approaches 1, then 1^1 is 1. But to be precise, we can't directly say that; we have to consider the rate at which they approach their limits. For example, if f(x) approaches 1 from above and g(x) approaches ∞, then it could be e^k for some k. But in our case, since both approach 1, and their rates are such that the product cot x * ln(tan x) approaches 0, the overall limit is 1.So after all these methods: substitution, logarithmic transformation, series expansion, numerical verification—all point to the same conclusion that the limit is 1. Therefore, the answer should be 1.Final AnswerThe limit is boxed{1}.</think>To find the limit of ((tan x)^{cot x}) as (x) approaches (frac{pi}{4}), we start by recognizing that both (tan x) and (cot x) approach 1 as (x) approaches (frac{pi}{4}). However, we need to carefully analyze the behavior of the function.We use the natural logarithm to simplify the expression. Let (L = lim_{x to frac{pi}{4}} (tan x)^{cot x}). Taking the natural logarithm of both sides, we get:[ln L = lim_{x to frac{pi}{4}} cot x cdot ln(tan x)]We substitute (t = x - frac{pi}{4}), so as (x) approaches (frac{pi}{4}), (t) approaches 0. Using the tangent addition formula, we approximate (tanleft(frac{pi}{4} + tright)) and (cotleft(frac{pi}{4} + tright)) for small (t):[tanleft(frac{pi}{4} + tright) approx frac{1 + t}{1 - t} quad text{and} quad cotleft(frac{pi}{4} + tright) approx frac{1 - t}{1 + t}]Expanding (ln(tan x)) and (cot x) using Taylor series around (t = 0):[lnleft(frac{1 + t}{1 - t}right) approx 2t + frac{2t^3}{3} + cdots][cot x approx 1 - 2t]Multiplying these approximations:[cot x cdot ln(tan x) approx (1 - 2t) cdot (2t + frac{2t^3}{3}) approx 2t - 4t^2 + frac{14t^3}{3} + cdots]As (t) approaches 0, the dominant term (2t) approaches 0. Therefore, the exponent (cot x cdot ln(tan x)) approaches 0, leading to:[ln L = 0 implies L = e^0 = 1]Numerical verification and substitution methods confirm that the expression approaches 1. Thus, the limit is:[boxed{1}],<think>Okay, let me try to figure out how to solve this limit: the limit as x approaches π/4 of (tan x) raised to the power of cot x. Hmm, so it's like (tan x)^{cot x} when x is getting close to π/4. Alright, let's start by recalling some basic limit properties and maybe L’Hospital’s Rule in case we need it for indeterminate forms.First, let me note that tan(π/4) is 1 and cot(π/4) is also 1. So, when x approaches π/4, both tan x and cot x approach 1. So the expression becomes 1^1, which is just 1. But wait, maybe there's something more here. If the function approaches 1^1 directly, then the limit should be 1. But maybe when approaching from different sides, like left or right of π/4, tan x and cot x behave differently, so the limit might not be straightforward. Let me check that.Wait, actually, tan x is 1 at π/4, and around π/4, tan x is slightly more or less than 1. Similarly, cot x is the reciprocal of tan x, so if tan x is a bit more than 1 near π/4, then cot x is a bit less than 1, and vice versa. But since both tan x and cot x are continuous at π/4, and their limits are 1, so plugging in π/4 gives 1^1, which is 1. But maybe the problem is trickier? Maybe when approaching, even though both base and exponent go to 1, the rate at which they approach might affect the limit?Wait, but 1 raised to any power is 1. So if both the base and exponent approach 1, shouldn't the limit just be 1? Hmm. But I remember that sometimes expressions of the form 1^∞ can be indeterminate. Wait, but here both base approaches 1 and exponent approaches 1. So it's 1^1, which is not indeterminate. So maybe the answer is indeed 1. But let me verify this by taking logarithms or something, maybe transforming it into a form where I can apply L’Hospital’s Rule.Let me recall that for limits of the form f(x)^{g(x)}, a good approach is to take the natural logarithm, compute the limit of g(x) * ln(f(x)), and then exponentiate the result. So let's set L = lim_{x→π/4} (tan x)^{cot x}. Then ln L = lim_{x→π/4} cot x * ln(tan x). If we can compute this limit, then L is e raised to that limit.So let's compute ln L = lim_{x→π/4} cot x * ln(tan x). Let's substitute x = π/4. Then cot(π/4) = 1, ln(tan(π/4)) = ln(1) = 0. So we have 1 * 0, which is 0. Therefore, ln L = 0, so L = e^0 = 1. Wait, so according to this, the limit is 1. But that seems too straightforward. Maybe there's a mistake here.Wait, but substituting x = π/4 gives us 1 * 0, but when we take the limit as x approaches π/4, we need to check if cot x * ln(tan x) approaches 0. Let me verify this. Let me set t = x - π/4, so as x approaches π/4, t approaches 0. Let's make a substitution to analyze the behavior near t = 0.Let’s define t = x - π/4. Then x = π/4 + t. As t approaches 0, x approaches π/4. Let's express tan x and cot x in terms of t.First, tan(π/4 + t) = [tan(π/4) + tan t] / [1 - tan(π/4) tan t] = (1 + tan t)/(1 - tan t). Similarly, cot x = cot(π/4 + t) = [cot(π/4) cot t - 1]/[cot t + cot(π/4)] = (1 * cot t - 1)/(cot t + 1) = (cot t - 1)/(cot t + 1). But this might get complicated. Alternatively, maybe we can expand tan x and cot x around π/4 using Taylor series?Alternatively, since tan x and cot x are reciprocals, so cot x = 1/tan x. Therefore, the expression cot x * ln(tan x) is equal to (1/tan x) * ln(tan x). Let’s denote y = tan x. Then as x approaches π/4, y approaches 1. So the expression becomes (1/y) * ln y, and we need to find the limit as y approaches 1 of (ln y)/y. Since ln 1 = 0 and y approaches 1, so (ln y)/y approaches 0/1 = 0. Therefore, ln L = 0, so L = 1. Therefore, the limit is 1.Alternatively, maybe we can use L’Hospital’s Rule here? Wait, but (ln y)/y as y approaches 1 is just 0/1, which is not an indeterminate form. Therefore, L’Hospital’s Rule is unnecessary here. So that confirms that ln L is 0, hence L is 1.But let me check again. Maybe the original substitution is oversimplifying. Let me do it step by step. Let me set x = π/4 + h, where h approaches 0. Then tan x = tan(π/4 + h) = (tan π/4 + tan h)/(1 - tan π/4 tan h) = (1 + tan h)/(1 - tan h). Similarly, cot x = 1/tan x = (1 - tan h)/(1 + tan h). Then, ln(tan x) = ln[(1 + tan h)/(1 - tan h)]. Therefore, the expression cot x * ln(tan x) becomes [(1 - tan h)/(1 + tan h)] * ln[(1 + tan h)/(1 - tan h)].Let me denote k = tan h. As h approaches 0, k approaches 0. So the expression becomes [(1 - k)/(1 + k)] * ln[(1 + k)/(1 - k)]. Let's compute this as k approaches 0. Let's expand ln[(1 + k)/(1 - k)] using Taylor series. We know that ln(1 + k) ≈ k - k^2/2 + k^3/3 - ..., and ln(1 - k) ≈ -k - k^2/2 - k^3/3 - ..., so ln[(1 + k)/(1 - k)] = ln(1 + k) - ln(1 - k) ≈ (k - k^2/2 + k^3/3) - (-k - k^2/2 - k^3/3) = k - k^2/2 + k^3/3 + k + k^2/2 + k^3/3 + ... = 2k + (2k^3)/3 + higher order terms.Similarly, (1 - k)/(1 + k) ≈ [1 - k - k^2 - ... ] using 1/(1 + k) ≈ 1 - k + k^2 - k^3 + ..., so (1 - k)/(1 + k) ≈ (1 - k)(1 - k + k^2 - ...) ≈ 1 - 2k + 2k^2 - ... So multiplying these two approximations:[1 - 2k + 2k^2] * [2k + (2k^3)/3] ≈ (1)(2k) + (1)(2k^3/3) + (-2k)(2k) + ... = 2k + (2k^3)/3 - 4k^2 + ... As k approaches 0, the dominant term is 2k, which approaches 0. Therefore, the entire expression approaches 0 as k approaches 0, so ln L = 0, which means L = e^0 = 1.Therefore, the limit is indeed 1. But let me check with specific numerical values. Let's take x approaching π/4 from the left and right.For example, take x = π/4 + 0.01. Compute tan x and cot x.tan(π/4 + 0.01) ≈ tan(0.7854 + 0.01) ≈ tan(0.7954) ≈ 1.0608.cot x = 1/tan x ≈ 0.9428.Then, (tan x)^{cot x} ≈ 1.0608^{0.9428} ≈ e^{0.9428 * ln(1.0608)} ≈ e^{0.9428 * 0.059} ≈ e^{0.0556} ≈ 1.0572.Now, take x closer to π/4, like x = π/4 + 0.001.tan(π/4 + 0.001) ≈ tan(0.7854 + 0.001) ≈ tan(0.7864) ≈ 1.0019.cot x = 1/1.0019 ≈ 0.9981.Then, (1.0019)^{0.9981} ≈ e^{0.9981 * ln(1.0019)} ≈ e^{0.9981 * 0.001898} ≈ e^{0.001895} ≈ 1.0019.So as x approaches π/4, the value seems to approach 1. Similarly, approaching from the left, take x = π/4 - 0.01.tan(π/4 - 0.01) ≈ tan(0.7854 - 0.01) ≈ tan(0.7754) ≈ 0.9503.cot x = 1/0.9503 ≈ 1.0523.Then, (0.9503)^{1.0523} ≈ e^{1.0523 * ln(0.9503)} ≈ e^{1.0523 * (-0.0510)} ≈ e^{-0.0537} ≈ 0.9477.Wait, that's less than 1. Wait, but when approaching from the left, tan x is less than 1, and cot x is greater than 1, so (tan x)^{cot x} is a number less than 1 raised to a power greater than 1, which could make it smaller. But as we get closer to π/4, x = π/4 - 0.001:tan(π/4 - 0.001) ≈ tan(0.7854 - 0.001) ≈ tan(0.7844) ≈ 0.9990.cot x = 1/0.9990 ≈ 1.0010.So, (0.9990)^{1.0010} ≈ e^{1.0010 * ln(0.9990)} ≈ e^{1.0010 * (-0.001001)} ≈ e^{-0.001002} ≈ 0.9990.So as x approaches π/4 from the left, the value approaches 1 from below. Similarly, from the right, it approaches 1 from above. So both sides approach 1, so the limit is indeed 1.Therefore, after all these checks and calculations, the limit should be 1. But let me also consider another approach. Maybe writing tan x as 1 + (tan x - 1) and expanding. Let me set t = tan x - 1, so as x approaches π/4, t approaches 0. Then cot x = 1/(1 + t). Then, the expression becomes (1 + t)^{1/(1 + t)}. Then, take the natural log: (1/(1 + t)) * ln(1 + t). Let's compute this as t approaches 0.(1/(1 + t)) * ln(1 + t) ≈ (1 - t + t^2 - ...)(t - t^2/2 + t^3/3 - ...) ≈ (t - t^2/2 + t^3/3) - t(t - t^2/2 + t^3/3) + ... = t - t^2/2 + t^3/3 - t^2 + t^3/2 - t^4/3 + ... = t - (1/2 + 1)t^2 + (1/3 + 1/2)t^3 + ... = t - (3/2)t^2 + (5/6)t^3 + ... So as t approaches 0, this expression approaches 0. Therefore, the logarithm of the limit is 0, so the limit is e^0 = 1.Another way: use the expansion of ln(1 + t) around t = 0. ln(1 + t) ≈ t - t^2/2 + t^3/3 - ..., so (1/(1 + t)) * ln(1 + t) ≈ (1 - t + t^2 - t^3 + ...)(t - t^2/2 + t^3/3 - ...) Multiplying term by term, the leading term is t, followed by terms with higher powers. Therefore, the limit as t approaches 0 is 0. Hence, the original limit is e^0 = 1.Therefore, all methods point to the same conclusion: the limit is 1. So I think the answer is 1.Final AnswerThe limit is boxed{1}.</think>To calculate the limit (lim _{x rightarrow frac{pi}{4}}(operatorname{tg} x)^{operatorname{ctg} x}), we start by noting that both (tan x) and (cot x) approach 1 as (x) approaches (frac{pi}{4}). This suggests the form (1^1), but we need to confirm this by more rigorous analysis.We use the natural logarithm transformation to handle the exponentiation. Let (L = lim_{x to frac{pi}{4}} (tan x)^{cot x}). Taking the natural logarithm, we get:[ln L = lim_{x to frac{pi}{4}} cot x cdot ln(tan x)]Substituting (x = frac{pi}{4} + t) as (t to 0), we express (tan x) and (cot x) in terms of (t). Using the substitution (t = x - frac{pi}{4}), we find:[tanleft(frac{pi}{4} + tright) = frac{1 + tan t}{1 - tan t} quad text{and} quad cotleft(frac{pi}{4} + tright) = frac{1 - tan t}{1 + tan t}]Let (k = tan t). As (t to 0), (k to 0). The expression becomes:[ln L = lim_{k to 0} left( frac{1 - k}{1 + k} right) cdot lnleft( frac{1 + k}{1 - k} right)]Expanding (lnleft( frac{1 + k}{1 - k} right)) using Taylor series around (k = 0):[lnleft( frac{1 + k}{1 - k} right) approx 2k + frac{2k^3}{3} + cdots]The term (frac{1 - k}{1 + k}) approximates to (1 - 2k + 2k^2 - cdots). Multiplying these expansions, the leading term is (2k), which approaches 0 as (k to 0). Thus:[ln L = 0 implies L = e^0 = 1]Numerical checks from both sides of (frac{pi}{4}) confirm that the limit approaches 1. Therefore, the final answer is:[boxed{1}]]

🔑:Solutionlim _{x rightarrow frac{pi}{4}}(tan x)^{cot x}=(tan frac{pi}{4})^{cot frac{pi}{4}}=1^{1}=1## Problem Kuznetsov Limits 20-2

🧠:[<think>Okay, let's try to solve this geometry problem. Hmm, we have a circle with center O and radius 10. On the radius AO, there's a point M. Then, on one side of AO, points B and C are chosen on the circle such that the angles AMB and OMC are both equal to α, and we know that cos α = 2/5. We need to find the length of BC. First, I need to visualize the problem. Let me sketch a diagram in my mind. There's a circle with center O, and AO is a radius. Point M is somewhere on AO. On one side of AO (let's say the right side), points B and C are on the circumference. Angles AMB and OMC are both alpha. Since cos α is given as 2/5, maybe we can use trigonometric identities or the law of cosines somewhere here.Let me start by assigning coordinates to make things concrete. Let's place the center O at the origin (0,0). Then, the radius AO is along the x-axis, so point A is at (10,0), and point O is at (0,0). Point M is somewhere on AO, so its coordinates will be (m,0) where 0 ≤ m ≤ 10. Now, points B and C are on the circle (which has radius 10) on one side of AO. Let's say they are above the x-axis.Given that angle AMB = α and angle OMC = α. Let's try to express these angles in terms of coordinates or vectors. Maybe using trigonometry or coordinate geometry?First, angle AMB. Points A, M, B. So at point M, the lines MA and MB form angle α. Similarly, at point M, the lines MO and MC form angle α. Wait, but O is the center, so MO is the line from M to O, which is along AO towards O, and MC is the line from M to C. So angle OMC is the angle at M between MO and MC.So both angles at M: one between MA and MB, and the other between MO and MC, both equal to α. Since MA and MO are along the same line but opposite directions? Wait, MA is from M to A, which is along AO towards A, and MO is from M to O, which is towards O. So they are colinear but in opposite directions. So angle AMB is between MA (from M to A) and MB (from M to B), and angle OMC is between MO (from M to O) and MC (from M to C). Both angles are alpha.Hmm, maybe using the Law of Cosines in triangles AMB and OMC?Wait, but angles are at M, not necessarily at the vertices of the triangles. Let me clarify.In triangle AMB, angle at M is alpha. Similarly, in triangle OMC, angle at M is alpha. So both triangles share the point M, and both have an angle alpha at M. The sides adjacent to angle alpha in triangle AMB are MA and MB, and in triangle OMC are MO and MC.Given that OA is radius 10, so OA = 10. Since M is on AO, AM = OA - OM = 10 - OM. If we let OM = x, then AM = 10 - x. Then, MA = 10 - x, MO = x.Now, in triangle AMB, we have sides MA = 10 - x, MB, and AB. Wait, but AB is a chord of the circle. Wait, point B is on the circle, so AB is a chord from A to B, but since A is on the circle too (since OA is radius 10), AB is a chord of the circle. Wait, but hold on, OA is a radius, so point A is on the circle. So both A and B are on the circle, so AB is a chord. Similarly, OC is a radius since C is on the circle.But in triangle AMB, angle at M is alpha. Similarly, in triangle OMC, angle at M is alpha. So perhaps using the Law of Cosines on these triangles?For triangle AMB: AM^2 + MB^2 - 2*AM*MB*cos(alpha) = AB^2.Wait, no, the Law of Cosines states that for a triangle with sides a, b, c, opposite angles A, B, C, we have c^2 = a^2 + b^2 - 2ab cos C. So in triangle AMB, angle at M is alpha, so AB^2 = AM^2 + MB^2 - 2*AM*MB*cos(alpha).Similarly, in triangle OMC, angle at M is alpha, so OC^2 = OM^2 + MC^2 - 2*OM*MC*cos(alpha). But OC is a radius, so OC = 10. So 10^2 = OM^2 + MC^2 - 2*OM*MC*(2/5). Since cos(alpha) = 2/5.But we don't know MB or MC. However, points B and C are on the circle, so perhaps there is some relation between their positions. Also, since they are on the same side of AO, maybe there's some symmetry or another relation.Alternatively, maybe we can parametrize points B and C. Let's consider coordinates. Let me assign coordinates as follows:Let’s set O at (0,0), A at (10,0). Point M is on AO, so coordinates (m, 0), where 0 ≤ m ≤10. Let's let m be the distance from O to M, so OM = m, so AM = 10 - m.Now, points B and C are on the circle x^2 + y^2 = 10^2, above the x-axis (since on one side of AO). Let’s parametrize point B as (10 cos θ, 10 sin θ) and point C as (10 cos φ, 10 sin φ), with θ and φ in (0, π) since they are above the x-axis.Now, angle AMB = alpha. Let's compute this angle. The angle at M between points A, M, and B. Similarly, angle OMC = alpha.To find the angle at M, we can use vectors. The vectors MA and MB. MA is from M to A: (10 - m, 0) - (m,0) = (10 - m - m, 0 - 0) = (10 - 2m, 0). Wait, no, wait. If M is at (m,0), then vector MA is A - M = (10 - m, 0), and vector MB is B - M = (10 cos θ - m, 10 sin θ - 0) = (10 cos θ - m, 10 sin θ).Similarly, the angle between vectors MA and MB is alpha. The cosine of this angle is given by the dot product formula:cos(alpha) = (MA · MB) / (|MA| |MB|)Similarly, MA · MB = (10 - m)(10 cos θ - m) + 0*(10 sin θ) = (10 - m)(10 cos θ - m)|MA| = sqrt((10 - m)^2 + 0^2) = 10 - m|MB| = sqrt((10 cos θ - m)^2 + (10 sin θ)^2) = sqrt(100 cos^2 θ - 20 m cos θ + m^2 + 100 sin^2 θ) = sqrt(100 (cos^2 θ + sin^2 θ) - 20 m cos θ + m^2) = sqrt(100 - 20 m cos θ + m^2)So cos(alpha) = [(10 - m)(10 cos θ - m)] / [(10 - m) * sqrt(100 - 20 m cos θ + m^2) ) ] = (10 cos θ - m) / sqrt(100 - 20 m cos θ + m^2)But we know cos(alpha) = 2/5. So:(10 cos θ - m)/sqrt(100 - 20 m cos θ + m^2) = 2/5Similarly, let's square both sides to eliminate the square root:[(10 cos θ - m)^2] / [100 - 20 m cos θ + m^2] = 4/25Multiply both sides by denominator:(10 cos θ - m)^2 = (4/25)(100 - 20 m cos θ + m^2)Expand left side: 100 cos^2 θ - 20 m cos θ + m^2Right side: (4/25)(100 - 20 m cos θ + m^2) = (4/25)*100 - (4/25)*20 m cos θ + (4/25)*m^2 = 16 - (16/5) m cos θ + (4/25)m^2So set equal:100 cos^2 θ - 20 m cos θ + m^2 = 16 - (16/5) m cos θ + (4/25)m^2Multiply both sides by 25 to eliminate denominators:2500 cos^2 θ - 500 m cos θ + 25 m^2 = 400 - 80 m cos θ + 4 m^2Bring all terms to left side:2500 cos^2 θ - 500 m cos θ + 25 m^2 - 400 + 80 m cos θ - 4 m^2 = 0Simplify:2500 cos^2 θ + (-500 m + 80 m) cos θ + (25 m^2 - 4 m^2) - 400 = 0Which is:2500 cos^2 θ - 420 m cos θ + 21 m^2 - 400 = 0Hmm, that's a quadratic in terms of cos θ. Let me write it as:2500 cos^2 θ - 420 m cos θ + (21 m^2 - 400) = 0This seems complicated. Maybe there's a similar equation for point C and angle OMC?Similarly, for angle OMC = alpha. Let's compute that. At point M, angle between OM and MC. OM is from M to O: vector MO is O - M = (-m, 0). Vector MC is C - M = (10 cos φ - m, 10 sin φ - 0) = (10 cos φ - m, 10 sin φ). So the cosine of angle OMC (alpha) is given by:cos(alpha) = (MO · MC) / (|MO| |MC|)MO · MC = (-m)(10 cos φ - m) + 0*(10 sin φ) = -10 m cos φ + m^2|MO| = sqrt((-m)^2 + 0^2) = m|MC| = sqrt((10 cos φ - m)^2 + (10 sin φ)^2) = sqrt(100 cos^2 φ - 20 m cos φ + m^2 + 100 sin^2 φ) = sqrt(100 - 20 m cos φ + m^2)So cos(alpha) = (-10 m cos φ + m^2) / [m * sqrt(100 - 20 m cos φ + m^2) ] = [m(-10 cos φ + m)] / [m * sqrt(100 - 20 m cos φ + m^2) ] = (-10 cos φ + m)/sqrt(100 - 20 m cos φ + m^2)But cos(alpha) = 2/5, so:(-10 cos φ + m)/sqrt(100 - 20 m cos φ + m^2) = 2/5Again, square both sides:[(-10 cos φ + m)^2]/[100 - 20 m cos φ + m^2] = 4/25Multiply both sides by denominator:(100 cos^2 φ - 20 m cos φ + m^2) = (4/25)(100 - 20 m cos φ + m^2)Wait, similar to previous equation. Let's check:Left side: (-10 cos φ + m)^2 = 100 cos^2 φ - 20 m cos φ + m^2Right side: 4/25*(100 - 20 m cos φ + m^2) = same as before, 16 - (16/5)m cos φ + (4/25)m^2So set equal:100 cos^2 φ - 20 m cos φ + m^2 = 16 - (16/5)m cos φ + (4/25)m^2Multiply both sides by 25:2500 cos^2 φ - 500 m cos φ + 25 m^2 = 400 - 80 m cos φ + 4 m^2Bring all terms to left:2500 cos^2 φ - 500 m cos φ + 25 m^2 - 400 + 80 m cos φ -4 m^2 = 0Simplify:2500 cos^2 φ + (-500 m + 80 m) cos φ + (25 m^2 - 4 m^2) - 400 = 0Which is:2500 cos^2 φ - 420 m cos φ + 21 m^2 - 400 = 0Same equation as for θ. So both θ and φ satisfy the same quadratic equation in cos θ and cos φ. Therefore, cos θ and cos φ are roots of the equation:2500 x^2 - 420 m x + (21 m^2 - 400) = 0So cos θ and cos φ are the two solutions to this quadratic. Let's denote them as x1 and x2. Then, by Vieta's formula:x1 + x2 = (420 m)/2500 = (21 m)/125x1 * x2 = (21 m^2 - 400)/2500But we need to relate θ and φ to find BC. Since BC is the distance between points B and C on the circle, which can be found using the chord length formula: BC = 2*10*sin(Δ/2), where Δ is the angle between them at the center. So if the central angle between B and C is Δ, then BC = 20 sin(Δ/2). Alternatively, if we can find the angle between θ and φ, then BC = 10*sqrt(2 - 2 cos(Δ)), where Δ = |θ - φ|. Wait, chord length is 2R sin(Δ/2), so with R=10, BC=20 sin(Δ/2). Alternatively, using coordinates, BC can be calculated as sqrt[(10 cos θ - 10 cos φ)^2 + (10 sin θ -10 sin φ)^2] = 10*sqrt[ (cos θ - cos φ)^2 + (sin θ - sin φ)^2 ] = 10*sqrt[2 - 2 cos(θ - φ)].But maybe using the Law of Cosines in triangle BOC? Since O is the center, OB and OC are radii, so OB=OC=10. Then, BC^2 = OB^2 + OC^2 - 2*OB*OC*cos(angle BOC) = 200 - 200 cos(angle BOC). Therefore, BC = sqrt(200 - 200 cos(angle BOC)) = 10*sqrt(2 - 2 cos(angle BOC)) = 20 sin(angle BOC /2).So if we can find angle BOC, which is the central angle between B and C, then we can compute BC. The angle BOC is |θ - φ|, assuming θ and φ are the angles from the x-axis to points B and C. So angle BOC = |θ - φ|.So our task reduces to finding |θ - φ|, given that cos θ and cos φ are roots of the quadratic equation 2500x² - 420 m x + 21 m² - 400 = 0.But how do we relate this to |θ - φ|? Hmm. Let me recall that cos(θ - φ) = cos θ cos φ + sin θ sin φ. If we can find cos θ cos φ and sin θ sin φ, then we can compute cos(θ - φ), which would give us angle BOC.But from the quadratic equation, we have:x1 + x2 = (21 m)/125 = cos θ + cos φx1 x2 = (21 m² - 400)/2500 = cos θ cos φBut we also need sin θ sin φ. Since sin² θ = 1 - cos² θ and sin² φ = 1 - cos² φ. But sin θ sin φ can be expressed using the identity:cos(θ - φ) = cos θ cos φ + sin θ sin φSo if we can find cos(θ - φ), then:sin θ sin φ = cos(θ - φ) - cos θ cos φBut I don't know cos(θ - φ) yet. Alternatively, perhaps we can compute sin θ sin φ using another identity.Alternatively, note that (sin θ sin φ) = [cos(θ - φ) - cos(θ + φ)] / 2. But we might not know θ + φ either. Hmm.Alternatively, since points B and C are both above AO and on the circle, perhaps there is some symmetry or relation. Also, since angles at M from AM to BM and from OM to CM are both alpha, maybe there's a reflection or something.Alternatively, perhaps M is located such that triangles AMB and OMC are similar? Not sure. Let's think differently.Alternatively, maybe we can express θ and φ in terms of the quadratic equation, but since both satisfy the same quadratic, perhaps they are related. Let me denote x1 = cos θ and x2 = cos φ. Then:x1 + x2 = (21 m)/125x1 x2 = (21 m² - 400)/2500We need to find |θ - φ|. Let's use the identity:cos(θ - φ) = cos θ cos φ + sin θ sin φ = x1 x2 + sin θ sin φBut sin θ sin φ can be written as sqrt(1 - x1²) * sqrt(1 - x2²). However, that might be complicated. Alternatively, square both sides:[cos(θ - φ) - x1 x2]^2 = sin² θ sin² φ = (1 - x1²)(1 - x2²)Expand left side:cos²(θ - φ) - 2 x1 x2 cos(θ - φ) + x1² x2² = 1 - x1² - x2² + x1² x2²Then,cos²(θ - φ) - 2 x1 x2 cos(θ - φ) + x1² x2² = 1 - (x1² + x2²) + x1² x2²Cancel x1² x2² from both sides:cos²(θ - φ) - 2 x1 x2 cos(θ - φ) = 1 - (x1² + x2²)But x1² + x2² = (x1 + x2)^2 - 2 x1 x2 = [(21m/125)^2] - 2*(21m² - 400)/2500Compute x1² + x2²:= (441 m² / 15625) - (2*(21 m² - 400))/2500= (441 m² / 15625) - (42 m² - 800)/2500Convert to common denominator 15625:= (441 m²) / 15625 - (42 m² - 800) * 6.25 / 15625Wait, 2500 * 6.25 = 15625, so 2500 * 6.25 = 15625. Wait, 2500 * 6.25 = 2500 * (25/4) = (2500 * 25)/4 = 15625.So 42 m² - 800 over 2500 is equal to (42 m² - 800) * (1/2500). Then multiplied by 2 gives (84 m² - 1600)/2500. Wait, maybe this approach is getting too convoluted. Let me compute x1² + x2² step by step.x1 + x2 = (21 m)/125, so (x1 + x2)^2 = (441 m²)/15625x1 x2 = (21 m² - 400)/2500Thus, x1² + x2² = (x1 + x2)^2 - 2 x1 x2 = (441 m²)/15625 - 2*(21 m² - 400)/2500Convert 2*(21 m² - 400)/2500 to (42 m² - 800)/2500. Now, convert both terms to denominator 15625:First term: (441 m²)/15625Second term: (42 m² - 800)/2500 = (42 m² - 800) * 6.25 / 15625 = (262.5 m² - 5000)/15625Thus, x1² + x2² = [441 m² - 262.5 m² + 5000]/15625 = (178.5 m² + 5000)/15625Simplify 178.5 m²: 178.5 = 357/2. So (357/2 m² + 5000)/15625.Wait, perhaps decimal fractions are complicating. Let's do fractions:2*(21 m² - 400)/2500 = (42 m² - 800)/2500So x1² + x2² = (441 m²)/15625 - (42 m² - 800)/2500Convert 42 m² - 800)/2500 to denominator 15625:Multiply numerator and denominator by 6.25 (since 2500*6.25=15625):= (42 m² - 800)*6.25 / 15625 = (262.5 m² - 5000)/15625So x1² + x2² = [441 m² - 262.5 m² + 5000]/15625 = (178.5 m² + 5000)/15625Hmm, this seems messy. Maybe there's a smarter way.Alternatively, since we need cos(θ - φ) = x1 x2 + sin θ sin φ, and we have x1 x2 = (21 m² - 400)/2500. So we need sin θ sin φ.But sin θ sin φ = sqrt(1 - x1²) * sqrt(1 - x2²). But expanding that would involve square roots. Maybe use the identity:sin θ sin φ = [cos(θ - φ) - cos(θ + φ)] / 2But we don't know θ + φ. Alternatively, perhaps we can find θ + φ from the given information?Alternatively, perhaps the problem has some specific value for m, given that the radius is 10. Wait, but the problem doesn't specify where M is on AO. Wait, the problem says "a point M is chosen on the radius AO". So M is arbitrary? But the problem asks for BC in terms of the radius being 10 and cos α = 2/5. But the answer is a fixed number? That suggests that BC does not depend on where M is on AO. Is that possible?Wait, that seems odd. If M is anywhere on AO, the positions of B and C would change, so BC length might vary. But the problem states "points B and C are chosen such that angles AMB and OMC are alpha". So for a given M, such points exist? Or maybe the construction is possible only for a specific M?Wait, maybe there's a unique position of M where both angles AMB and OMC equal alpha. That might make sense. Because otherwise, the problem is underdetermined. The problem says "a point M is chosen on AO", but then "points B and C are chosen... such that angles AMB = OMC = alpha". So perhaps for some positions of M, such points B and C exist, and the problem is to find BC in that case. But the problem statement doesn't specify M's position, just that it's on AO. So perhaps there is only one possible position for M where such points B and C exist, and thus BC is determined uniquely. But how?Alternatively, maybe regardless of M's position, BC is the same? But that seems unlikely. Wait, let me check with specific cases. Suppose M is at O. Then angle OMC would be undefined, since M=O. Similarly, if M is at A, then angle AMB is undefined. So M is between O and A. Suppose M is the midpoint, so at (5,0). Then, can we have points B and C such that angles AMB and OMC are alpha? Maybe, but would BC be the same as if M is elsewhere?Alternatively, perhaps the problem has a unique solution regardless of M, but I need to check.Wait, but the problem gives the radius as 10 and cos α = 2/5, and asks for BC. Since the answer is a fixed number, BC must be determined uniquely. Therefore, perhaps despite M being variable, BC is fixed. Therefore, perhaps there's a relation that allows BC to be found independent of M. Let me think.Alternatively, maybe triangles AMB and OMC are similar? Let's check.In triangle AMB: angle at M is alpha, sides AM = 10 - m, MB, and AB.In triangle OMC: angle at M is alpha, sides OM = m, MC, and OC=10.If these triangles are similar, then corresponding sides would be proportional. So if AM / OM = AB / OC = MB / MC.But OC = 10, AB is a chord from A to B. Not sure. Wait, if similarity is possible, then maybe:If triangle AMB ~ triangle OMC, then:AM / OM = AB / OC = MB / MCBut OC =10, AM = 10 - m, OM = m, so (10 - m)/m = AB/10 = MB/MCBut AB can be calculated as chord length: AB = 2*10*sin(theta/2), where theta is angle AOB. Wait, but angle AOB is angle between OA and OB, which is theta, since OA is along x-axis, point B is at angle theta. So angle AOB = theta. Therefore, AB = 2*10*sin(theta/2). But theta is the angle for point B.But maybe this is getting too complicated. Alternatively, perhaps there's a way to relate MB and MC.Alternatively, since both B and C lie on the circle, perhaps there's an inversion or some other circle geometry property. Alternatively, use power of a point.Power of point M with respect to the circle: For any point M outside a circle, the power is MA * MO = MB^2 - r^2, but wait, M is inside the circle because it's on AO. The power of a point inside the circle is r^2 - OM^2. Wait, power of M is equal to MB * MB' where B and B' are intersection points of a line through M. But since B and C are specific points, maybe not directly applicable.Alternatively, since angles AMB and OMC are equal to alpha, maybe we can set up equations involving MB and MC and solve for them.From triangle AMB:AB² = AM² + MB² - 2*AM*MB*cos(alpha)Similarly, from triangle OMC:OC² = OM² + MC² - 2*OM*MC*cos(alpha)But OC=10, so 100 = m² + MC² - 2*m*MC*(2/5)Similarly, AB is a chord between A and B, which are both on the circle. So AB = 2*10*sin(theta/2), but theta is the central angle between A and B, which is angle AOB = theta. Therefore, AB = 20*sin(theta/2). But theta is the angle from OA to OB, which is the same as the angle θ we defined earlier. Wait, point B is at angle θ from the x-axis, so angle AOB is θ. Therefore, AB = 2*10*sin(θ/2). Similarly, BC would be 20*sin(|θ - φ|/2).But perhaps instead of dealing with angles, express AB² in terms of coordinates. AB is the distance between A(10,0) and B(10 cos θ, 10 sin θ). So AB² = (10 - 10 cos θ)^2 + (0 - 10 sin θ)^2 = 100(1 - 2 cos θ + cos² θ) + 100 sin² θ = 100(1 - 2 cos θ + cos² θ + sin² θ) = 100(2 - 2 cos θ) = 200(1 - cos θ)Therefore, AB² = 200(1 - cos θ). Similarly, from triangle AMB:AB² = (10 - m)^2 + MB² - 2*(10 - m)*MB*(2/5)So 200(1 - cos θ) = (10 - m)^2 + MB² - (4/5)*(10 - m)*MBSimilarly, from the Law of Cosines in triangle OMC:100 = m² + MC² - (4/5)*m*MCBut this seems to involve too many variables: m, MB, MC, theta, phi. It's getting too complicated.Wait, but maybe MB and MC can be related. Since points B and C are both on the circle, and angles at M are both alpha. Maybe there's a reflection or some symmetry.Alternatively, notice that points B and C lie on the circle such that angles from M are alpha. So B lies somewhere such that angle AMB = alpha, and C lies such that angle OMC = alpha. Maybe there is a circle through B and C related to point M?Alternatively, since angles subtended by MB and MC at A and O are alpha. Not sure.Alternatively, consider triangle AMB and triangle OMC. Both have angle alpha at M. If we can relate these triangles, perhaps via the Law of Sines.In triangle AMB:AB / sin(angle at B) = AM / sin(angle at A) = MB / sin(alpha)Similarly, in triangle OMC:OC / sin(angle at C) = OM / sin(angle at O) = MC / sin(alpha)But angles at A and O are not necessarily known. Hmm.Alternatively, maybe if we construct lines from M making angles alpha with MA and MO, their intersections with the circle are B and C. Since MA and MO are along AO but in opposite directions, the lines MB and MC form angles alpha with MA and MO, respectively. So these lines are symmetric in some way?Wait, MA is towards A, and MO is towards O. So angle between MA and MB is alpha, and angle between MO and MC is alpha. So MB is inclined at alpha from MA (which is along AO towards A), and MC is inclined at alpha from MO (which is along AO towards O). Since MA and MO are opposite directions, the lines MB and MC are symmetric with respect to the y-axis if M is at the center, but M is not necessarily at the center.Alternatively, if we consider MB and MC as lines making angles alpha with AO but on different sides. Wait, no. MA is along AO towards A, so MB is forming an angle alpha above AO. Similarly, MO is along AO towards O, so MC is forming an angle alpha above AO. So both MB and MC are on the same side of AO (above it), each forming an angle alpha with their respective directions (MA and MO).Therefore, the directions of MB and MC are both above AO, but inclined at alpha from MA and MO. Since MA and MO are colinear but opposite, the angles alpha are measured from opposite directions. Therefore, the lines MB and MC are symmetric with respect to the vertical line through M?Wait, if we think of MA as pointing to the right (along AO) and MO as pointing to the left (along AO), then angle alpha from MA would be upwards to the left, and angle alpha from MO would be upwards to the right. Wait, but since both B and C are on the same side of AO, which we've taken as above the x-axis, then both angles are measured upwards. Wait, but MA is towards A (right), so angle alpha from MA would be above AO, and similarly, MO is towards O (left), angle alpha from MO would also be above AO. So both lines MB and MC are above AO, forming angles alpha with MA and MO, which are opposite directions along AO.Therefore, if we imagine M somewhere on AO, then from M, line MB is going upwards making angle alpha with MA (i.e., with the line to the right), and line MC is going upwards making angle alpha with MO (i.e., with the line to the left). Therefore, these two lines MB and MC are symmetric with respect to the vertical line through M. Therefore, points B and C are symmetric with respect to the vertical line through M. Therefore, the central angles θ and φ satisfy some relation. For instance, if the vertical line through M is x = m, then the reflection over this line would map B to C and vice versa. Therefore, the angle φ would be related to θ via this reflection.Wait, if we reflect point B over the vertical line x = m, we get point C. Let's check.Coordinates of B: (10 cos θ, 10 sin θ). Reflecting over x = m would give a point with x-coordinate 2m - 10 cos θ, y-coordinate remains 10 sin θ. So C would be (2m - 10 cos θ, 10 sin θ). But point C is also on the circle, so (2m -10 cos θ)^2 + (10 sin θ)^2 = 100.Expanding: 4m² - 40 m cos θ + 100 cos² θ + 100 sin² θ = 100Simplify: 4m² -40 m cos θ + 100 (cos² θ + sin² θ) = 100Since cos² θ + sin² θ =1, this becomes 4m² -40 m cos θ +100 =100So 4m² -40 m cos θ =0 → 4m(m -10 cos θ)=0. Since m ≠0 (unless M is at O, which is not allowed as angle OMC would be undefined), so m -10 cos θ=0 → m=10 cos θ. Therefore, if reflection over x=m maps B to C, then we must have m=10 cos θ.But we also have from earlier that cos θ is a root of the quadratic equation 2500x² - 420 m x +21 m² -400 =0. If m=10 cos θ, then substituting m=10 x into the quadratic equation:2500 x² - 420*(10 x)*x +21*(10 x)^2 -400=0Calculate each term:2500x² - 4200 x² + 2100 x² -400=0Sum coefficients: 2500 -4200 +2100 = 400. So:400x² -400=0 → x²=1 → x=±1. But since θ is above the x-axis, cos θ is positive (since θ is between 0 and π), so x=1. Then, m=10*1=10. But M is on AO, which has length 10, so m=10 would place M at point A. But angle AMB would be undefined if M is at A. Contradiction. Therefore, the reflection idea might not hold unless M is at A, which it can't be. Therefore, maybe the symmetry is different.Alternatively, maybe the angles theta and phi are supplementary? Such that theta + phi = π. Then, cos phi = -cos theta. Then, x1 = cos theta, x2 = -cos theta. Then, x1 + x2 = 0 = (21 m)/125 → m=0. But M can't be at O. So this is not possible.Alternatively, maybe theta = phi + something. Not sure.Alternatively, think of the two equations we have for theta and phi. Both cos theta and cos phi satisfy the same quadratic equation. Therefore, they are either equal or are the two roots. If they are equal, then theta = phi, but then B and C would coincide, which is not possible. Therefore, cos theta and cos phi are distinct roots of the quadratic equation. Therefore, we can express cos(θ - φ) in terms of the roots.Recall that:cos(θ - φ) = cos theta cos phi + sin theta sin phiWe have cos theta + cos phi = S = (21 m)/125cos theta * cos phi = P = (21 m² - 400)/2500Also, (sin theta sin phi) can be found from:(sin theta sin phi)^2 = (1 - cos² theta)(1 - cos² phi) = 1 - cos² theta - cos² phi + cos² theta cos² phi= 1 - (S² - 2P) + P²Wait, since cos² theta + cos² phi = (cos theta + cos phi)^2 - 2 cos theta cos phi = S² - 2PThus, (sin theta sin phi)^2 = 1 - (S² - 2P) + P² = 1 - S² + 2P + P²But this requires computing square roots, which might be complex. However, maybe we can express cos(theta - phi) as follows:cos(theta - phi) = P + sin theta sin phiBut sin theta sin phi = sqrt[(1 - cos² theta)(1 - cos² phi)]. This seems difficult, but maybe we can express it in terms of S and P.Alternatively, recall that:[cos(theta - phi)]² = (cos theta cos phi + sin theta sin phi)^2 = cos² theta cos² phi + 2 cos theta cos phi sin theta sin phi + sin² theta sin² phiBut this might not help. Alternatively, use the identity:cos(theta - phi) = 2 cos theta cos phi - cos(theta + phi)But we don't know cos(theta + phi).Alternatively, since we have S = cos theta + cos phi and P = cos theta cos phi, maybe express cos(theta - phi) in terms of S and P and another term.Wait, another approach: if we can find theta - phi, then BC can be found. But theta - phi is the angle between points B and C at the center. To find this, we need more information.But perhaps all these variables are related via the position of M. Since M is a variable point on AO, but the problem doesn't specify where, but the answer is fixed, there must be a unique BC regardless of M. Therefore, maybe BC can be found independent of M. How?Wait, suppose we consider two different positions of M, say M1 and M2, leading to points B1, C1 and B2, C2. The problem states that for any M (or for some M), BC is fixed. But since the answer is a box, it's likely fixed. Therefore, BC is fixed regardless of M. Hence, maybe we can choose a specific position of M to simplify calculations.Let's assume M is the midpoint of AO. So M is at (5,0). Then, OM = 5, AM = 5. Let's see if we can compute theta and phi in this case.With m=5, the quadratic equation for cos theta and cos phi becomes:2500x² - 420*5 x +21*(5)^2 -400=0Compute:2500x² - 2100x + 525 -400=0 → 2500x² -2100x +125=0Divide all terms by 25:100x² -84x +5=0Solve using quadratic formula:x = [84 ± sqrt(84² -4*100*5)]/(2*100) = [84 ± sqrt(7056 -2000)]/200 = [84 ± sqrt(5056)]/200sqrt(5056): 5056 = 16*316 = 16*4*79 = 64*79, so sqrt(5056)=8*sqrt(79). Thus,x = [84 ±8√79]/200 = [21 ±2√79]/50But cos theta and cos phi must be between -1 and 1. Let's approximate sqrt(79) ≈8.89, so 2√79≈17.78. Then,x1≈(21 +17.78)/50≈38.78/50≈0.7756x2≈(21 -17.78)/50≈3.22/50≈0.0644So cos theta≈0.7756 and cos phi≈0.0644. Then, theta≈acos(0.7756)≈39 degrees, phi≈acos(0.0644)≈86.3 degrees. Then, angle BOC≈86.3 -39≈47.3 degrees. Then BC≈20*sin(23.65)≈20*0.401≈8.02. But this is approximate. However, the exact answer should be 16, as I recall from prior problems, but let's check.But maybe with m=5, BC is 16. Let's compute BC using these approximated angles.But wait, let's compute BC exactly. If angle BOC = phi - theta ≈86.3 -39≈47.3 degrees, then BC=20*sin(47.3/2)=20*sin(23.65)≈20*0.401≈8.02. But this doesn't give 16. So perhaps my assumption of M being the midpoint is not leading to the correct result. Alternatively, maybe choosing a different M.Alternatively, let's try to assume that M is at O. But then angle OMC is undefined. Not possible.Alternatively, take M approaching O. As M approaches O, angle OMC approaches angle OOC, which is not defined. Not helpful.Alternatively, take M approaching A. As M approaches A, angle AMB approaches angle AAB, which is undefined. Not helpful.Alternatively, is there a special position for M where calculations simplify?Alternatively, suppose that AM = OM. Then, AM = OM → 10 - m = m → m=5. So M is the midpoint. Which is what we did earlier, leading to BC≈8. But the problem states the radius is 10, and answer is likely an integer. So 8 is possible. But maybe exact calculation.Wait, with m=5, quadratic equation was 2500x² -2100x +125=0. Let's solve this exactly.Using quadratic formula:x = [2100 ± sqrt(2100² -4*2500*125)]/(2*2500)Compute discriminant:2100² = 4,410,0004*2500*125 = 1,250,000Discriminant = 4,410,000 -1,250,000=3,160,000sqrt(3,160,000)=1,778. approximately, but exact value: sqrt(3,160,000)=sqrt(10000*316)=100*sqrt(316)=100*sqrt(4*79)=200*sqrt(79)Thus,x = [2100 ±200√79]/5000 = [21 ±2√79]/50As before. Therefore, cos theta=(21 +2√79)/50 and cos phi=(21 -2√79)/50.Now, compute angle BOC = |theta - phi|. To find cos(theta - phi):cos(theta - phi)=cos theta cos phi + sin theta sin phiWe have cos theta cos phi = P = (21 m² -400)/2500. With m=5, P=(21*25 -400)/2500=(525 -400)/2500=125/2500=1/20.sin theta sin phi = sqrt(1 -cos² theta) * sqrt(1 -cos² phi). Let's compute:cos theta=(21 +2√79)/50, so 1 -cos² theta=1 - (441 +84√79 +4*79)/2500=1 - (441 +84√79 +316)/2500=1 - (757 +84√79)/2500=(2500 -757 -84√79)/2500=(1743 -84√79)/2500Similarly, 1 -cos² phi=(1743 +84√79)/2500Therefore, sin theta sin phi= sqrt[(1743 -84√79)/2500 * (1743 +84√79)/2500] = sqrt[(1743² - (84√79)^2)/2500²]Compute 1743² - (84√79)^2:1743² = ?Wait, this seems computationally intensive. Maybe there's a better way.Alternatively, since we have cos(theta - phi) = P + sin theta sin phi = 1/20 + sin theta sin phi. But sin theta sin phi can be calculated as sqrt[(1 - cos² theta)(1 - cos² phi)].But given the complexity, maybe it's not the best approach. Alternatively, since BC = 20 sin(angle BOC /2), and we need to find BC, maybe there's a different approach altogether.Let me revisit the problem. Maybe use coordinates with specific M.Let me choose coordinates with O at (0,0), A at (10,0). Let M be at (m,0). Points B and C are on the circle x² + y² =100. Angles AMB = OMC = alpha, cos alpha =2/5.We need to find BC.From earlier, we have two equations from the Law of Cosines in triangles AMB and OMC.From triangle AMB:AB² = AM² + MB² - 2*AM*MB*(2/5)From triangle OMC:OC² = OM² + MC² - 2*OM*MC*(2/5)But OC=10, so:100 = m² + MC² - (4/5)*m*MCSimilarly, AB is the distance between A(10,0) and B(10 cos theta,10 sin theta). AB² = (10 -10 cos theta)^2 + (0 -10 sin theta)^2 = 100(1 - 2 cos theta + cos² theta + sin² theta) = 100(2 - 2 cos theta)=200(1 - cos theta)Therefore, from triangle AMB:200(1 - cos theta) = (10 - m)^2 + MB² - (4/5)(10 - m)MBSimilarly, we can write for triangle OMC:100 = m² + MC² - (4/5)m MCBut we also have that points B and C are related. Maybe BC can be found using coordinates.Coordinates of B: (10 cos theta,10 sin theta)Coordinates of C: (10 cos phi,10 sin phi)BC² = (10 cos theta -10 cos phi)^2 + (10 sin theta -10 sin phi)^2 = 100[ (cos theta - cos phi)^2 + (sin theta - sin phi)^2 ] = 100[ 2 - 2 cos(theta - phi) ]Therefore, BC = 10*sqrt(2 - 2 cos(theta - phi)) = 20 sin( (theta - phi)/2 )But to find theta - phi, we need to relate theta and phi. Given that both cos theta and cos phi satisfy the quadratic equation 2500x² -420 m x +21m² -400=0.But if we can find theta - phi in terms of the roots of the quadratic, we might have a chance.Let’s denote x1=cos theta, x2=cos phi. Then:x1 + x2 = (21m)/125x1 x2 = (21m² -400)/2500We need to find theta - phi. We can use the identity:cos(theta - phi) = x1 x2 + sqrt(1 -x1²)sqrt(1 -x2²)But this sqrt part is difficult. Alternatively, consider that:sin theta sin phi = sqrt(1 -x1²)sqrt(1 -x2²) = sqrt( (1 -x1²)(1 -x2²) )Expand:(1 -x1²)(1 -x2²) = 1 -x1² -x2² +x1²x2² = 1 - (x1² + x2²) + (x1x2)^2But x1² +x2² = (x1 +x2)^2 - 2x1x2 = [ (21m/125)^2 ] - 2*(21m² -400)/2500Compute:= (441m²)/15625 - (42m² -800)/2500Convert to common denominator 15625:= (441m²)/15625 - (42m² -800)*6.25/15625= (441m² - 262.5m² +5000)/15625= (178.5m² +5000)/15625Thus,(1 -x1²)(1 -x2²) = 1 - [ (178.5m² +5000)/15625 ] + [ (21m² -400)^2 ]/2500^2This is getting too complicated. Maybe there's a different approach.Let me recall that in both triangles AMB and OMC, we have angle alpha at M, and sides adjacent to alpha are AM and MB, OM and MC. Let's denote MB = p, MC = q. Then, in triangle AMB:AB² = (10 -m)^2 + p² - (4/5)(10 -m)pIn triangle OMC:100 = m² + q² - (4/5)m qAlso, AB²=200(1 -cos theta), but we might not need that.But points B and C are on the circle. Also, we can find coordinates of B and C in terms of theta and phi, but this might not help.Wait, if I consider vectors from M to B and from M to C. The direction of these vectors is determined by the angles alpha from MA and MO.Since MA is along the positive x-axis (from M to A), angle alpha above MA would give the direction of MB. Similarly, MO is along the negative x-axis (from M to O), angle alpha above MO would give the direction of MC.So the angle of MB with the x-axis is alpha (measured from MA, which is towards A, i.e., positive x-axis). So the direction of MB is alpha above the positive x-axis. Wait, no. If angle between MA and MB is alpha, and MA is along positive x-axis, then MB is at an angle alpha above MA. But MA is from M to A, which is towards positive x-axis. So if we are at point M, looking towards A (positive x-axis), then angle alpha upwards would be into the first quadrant. So the direction of MB is alpha above the positive x-axis. Similarly, angle between MO and MC is alpha. MO is from M to O, which is towards negative x-axis. So from point M, looking towards O (negative x-axis), angle alpha upwards would be into the second quadrant. So direction of MC is alpha above the negative x-axis, which is equivalent to angle π - alpha from positive x-axis.Therefore, the direction angles of vectors MB and MC are alpha and π - alpha, respectively. Therefore, the coordinates of B and C can be parametrized using these angles.Wait, but vectors MB and MC have directions alpha and pi - alpha from the x-axis. So if we consider the direction of MB as angle alpha from the x-axis, then coordinates of B can be written as M + t*vec(alpha), where t is the distance MB. Similarly, coordinates of C can be written as M + s*vec(pi - alpha), where s is the distance MC.Since B and C are on the circle of radius 10 centered at O.Let's formalize this.Coordinates of M: (m,0)Coordinates of B: (m + p cos alpha, 0 + p sin alpha)Coordinates of C: (m + q cos(pi - alpha), 0 + q sin(pi - alpha)) = (m - q cos alpha, 0 + q sin alpha)Since points B and C are on the circle x² + y² =100. Therefore:For B:(m + p cos alpha)^2 + (p sin alpha)^2 =100For C:(m - q cos alpha)^2 + (q sin alpha)^2 =100Expand both equations:For B:m² + 2 m p cos alpha + p² cos² alpha + p² sin² alpha =100Simplify:m² + 2 m p cos alpha + p² (cos² alpha + sin² alpha) =100Thus:m² + 2 m p cos alpha + p² =100 → p² + 2 m p cos alpha + m² =100Similarly, for C:m² - 2 m q cos alpha + q² cos² alpha + q² sin² alpha =100Simplify:m² - 2 m q cos alpha + q² (cos² alpha + sin² alpha) =100Thus:m² - 2 m q cos alpha + q² =100 → q² - 2 m q cos alpha + m² =100Therefore, we have two equations:p² + 2 m p cos alpha + m² =100 ...(1)q² - 2 m q cos alpha + m² =100 ...(2)Notice that equations (1) and (2) are symmetric in p and q, except for the sign of the linear term. Therefore, p and q satisfy similar equations.Moreover, from triangle AMB:AB² = (10 - m)^2 + p² - (4/5)(10 - m)pBut AB is the distance between A(10,0) and B(m + p cos alpha, p sin alpha)Compute AB²:(10 - (m + p cos alpha))^2 + (0 - p sin alpha)^2= (10 - m - p cos alpha)^2 + p² sin² alpha= (10 - m)^2 - 2(10 - m)p cos alpha + p² cos² alpha + p² sin² alpha= (10 - m)^2 - 2(10 - m)p cos alpha + p² (cos² alpha + sin² alpha)= (10 - m)^2 - 2(10 - m)p cos alpha + p²But from equation (1), p² + 2 m p cos alpha + m² =100 → p² =100 - m² -2 m p cos alphaSubstitute into AB²:AB² = (10 - m)^2 - 2(10 - m)p cos alpha + (100 - m² -2 m p cos alpha)Expand (10 - m)^2:=100 -20m +m² -2(10 - m)p cos alpha +100 -m² -2 m p cos alphaSimplify:=100 -20m +m² -20 p cos alpha +2 m p cos alpha +100 -m² -2 m p cos alphaCombine like terms:100 +100 -20m +m² -m² -20 p cos alpha +2 m p cos alpha -2 m p cos alpha=200 -20m -20 p cos alphaBut from triangle AMB, we also have AB² = (10 - m)^2 + p² - (4/5)(10 -m)pEquate the two expressions for AB²:200 -20m -20 p cos alpha = (10 - m)^2 + p² - (4/5)(10 -m)pExpand (10 -m)^2:100 -20m +m²Thus,200 -20m -20 p cos alpha =100 -20m +m² + p² - (4/5)(10 -m)pSimplify left and right sides:Left:200 -20m -20 p cos alphaRight:100 -20m +m² + p² - (4/5)(10 -m)pSubtract right from left:200 -20m -20 p cos alpha -100 +20m -m² -p² + (4/5)(10 -m)p =0Simplify:100 -20 p cos alpha -m² -p² + (4/5)(10 -m)p =0Rearrange:- m² - p² + (4/5)(10 -m)p -20 p cos alpha +100=0Multiply through by -1:m² + p² - (4/5)(10 -m)p +20 p cos alpha -100=0But from equation (1):p² =100 -m² -2 m p cos alphaSubstitute p²:m² + (100 -m² -2 m p cos alpha) - (4/5)(10 -m)p +20 p cos alpha -100=0Simplify:m² +100 -m² -2 m p cos alpha - (4/5)(10 -m)p +20 p cos alpha -100=0Cancel terms:-2 m p cos alpha - (4/5)(10 -m)p +20 p cos alpha=0Factor p:p[ -2 m cos alpha -4/5(10 -m) +20 cos alpha ]=0Since p ≠0 (as B is not M), the term in brackets must be zero:-2 m cos alpha -4/5(10 -m) +20 cos alpha =0Multiply through by 5 to eliminate denominators:-10 m cos alpha -4(10 -m) +100 cos alpha=0Expand:-10 m*(2/5) -40 +4m +100*(2/5)=0Since cos alpha=2/5:Compute each term:-10m*(2/5) = -4m-40 remains+4m+100*(2/5)=40So:-4m -40 +4m +40=0Simplify:0=0Wait, this results in an identity 0=0, which means our earlier substitution and manipulation have led us to a tautology, indicating that the equation is dependent and doesn't provide new information. Therefore, this approach doesn't yield additional constraints. Therefore, we need another way.But this suggests that the equations are consistent for any m, which implies that BC can be computed independent of m. Therefore, BC is constant regardless of the position of M on AO. Therefore, we can choose any M to compute BC. For simplicity, let's choose M=O. But when M=O, angle OMC is undefined. Therefore, invalid. Let's choose M approaching O, but not exactly O.Alternatively, choose M such that m=0. But M coincides with O. Not allowed. Choose m very small. But perhaps instead, take M to be at a general position and find that BC cancels out m.Alternatively, note that since the equation reduces to 0=0, BC can be computed without m. Let's try to compute BC using coordinates.From earlier, we have:Coordinates of B: (m + p cos alpha, p sin alpha)Coordinates of C: (m - q cos alpha, q sin alpha)Since both B and C are on the circle x² + y²=100.We can also express BC in terms of p and q.BC² = (m + p cos alpha - (m - q cos alpha))² + (p sin alpha - q sin alpha)^2Simplify:= (p cos alpha + q cos alpha)^2 + (p sin alpha - q sin alpha)^2= cos² alpha (p + q)^2 + sin² alpha (p - q)^2Expand:= cos² alpha (p² + 2pq + q²) + sin² alpha (p² - 2pq + q²)= p² (cos² alpha + sin² alpha) + q² (cos² alpha + sin² alpha) + 2pq cos² alpha - 2pq sin² alpha= p² + q² + 2pq (cos² alpha - sin² alpha)= p² + q² + 2pq cos(2 alpha)But from equations (1) and (2):p² + 2 m p cos alpha + m² =100 ...(1)q² - 2 m q cos alpha + m² =100 ...(2)Subtract equation (2) from equation (1):p² + 2 m p cos alpha + m² - (q² - 2 m q cos alpha + m²) =0Simplify:p² - q² +2 m p cos alpha +2 m q cos alpha=0Factor:(p² - q²) +2 m cos alpha (p + q)=0Factor difference of squares:(p - q)(p + q) +2 m cos alpha (p + q)=0Factor out (p + q):(p + q)(p - q +2 m cos alpha)=0Since p + q ≠0 (as p and q are lengths), we have:p - q +2 m cos alpha=0 → p = q -2 m cos alphaTherefore, we can express p in terms of q: p = q -2 m cos alphaNow, substitute p = q -2 m cos alpha into equation (1):(q -2 m cos alpha)^2 + 2 m (q -2 m cos alpha) cos alpha + m² =100Expand:q² -4 m cos alpha q +4 m² cos² alpha +2 m q cos alpha -4 m² cos² alpha + m²=100Simplify:q² -4 m cos alpha q +2 m q cos alpha +4 m² cos² alpha -4 m² cos² alpha +m²=100Combine like terms:q² -2 m q cos alpha +m²=100But this is exactly equation (2). Therefore, no new information. Thus, the relation p = q -2 m cos alpha is consistent with the equations.Now, recall BC²= p² + q² + 2pq cos(2 alpha)But we need to express BC in terms of p and q. But we have equations relating p and q. Let's try to find p + q and pq.Let’s denote S = p + q, D = p - q = -2 m cos alpha (from earlier, p - q = -2 m cos alpha)Then, S + D = 2p → p = (S + D)/2Similarly, S - D = 2q → q = (S - D)/2But D = -2 m cos alpha, so:p = (S -2 m cos alpha)/2q = (S +2 m cos alpha)/2Now, from equation (1):p² +2 m p cos alpha +m²=100Substitute p:[(S -2 m cos alpha)/2]^2 +2 m *[(S -2 m cos alpha)/2] cos alpha +m²=100Expand:(S² -4 m S cos alpha +4 m² cos² alpha)/4 + [2 m (S -2 m cos alpha) cos alpha]/2 +m²=100Simplify:(S²/4 -m S cos alpha +m² cos² alpha) + (m S cos alpha -2 m² cos² alpha) +m²=100Combine terms:S²/4 -m S cos alpha +m² cos² alpha +m S cos alpha -2 m² cos² alpha +m²=100Simplify:S²/4 -m² cos² alpha +m²=100Factor m²:S²/4 +m²(1 - cos² alpha)=100But 1 - cos² alpha=sin² alpha:S²/4 +m² sin² alpha=100 ...(3)Similarly, from equation (2):q² -2 m q cos alpha +m²=100Substitute q=(S +2 m cos alpha)/2:[(S +2 m cos alpha)/2]^2 -2 m *[(S +2 m cos alpha)/2] cos alpha +m²=100Expand:(S² +4 m S cos alpha +4 m² cos² alpha)/4 - [2 m (S +2 m cos alpha) cos alpha]/2 +m²=100Simplify:S²/4 +m S cos alpha +m² cos² alpha -m S cos alpha -2 m² cos² alpha +m²=100Combine terms:S²/4 -m² cos² alpha +m²=100Which is the same as equation (3). Therefore, no new info.So equation (3) is S²/4 +m² sin² alpha=100But BC²= p² + q² + 2pq cos(2 alpha). Let's express this in terms of S and D.BC²= p² + q² +2pq cos(2 alpha)But p² + q²= (p + q)^2 - 2pq= S² -2pqAnd 2pq cos(2 alpha)=2pq (2 cos² alpha -1)So BC²= S² -2pq +2pq(2 cos² alpha -1)= S² -2pq +4 pq cos² alpha -2pq= S² +4 pq cos² alpha -4 pq= S² -4 pq (1 - cos² alpha)= S² -4 pq sin² alphaBut from equation (3): S²/4 +m² sin² alpha=100 → S²=4(100 -m² sin² alpha)Therefore, BC²=4(100 -m² sin² alpha) -4 pq sin² alpha=400 -4m² sin² alpha -4 pq sin² alpha=400 -4 sin² alpha(m² + pq)But we need to find pq. From S =p + q and D =p - q=-2 m cos alpha, we have:pq = [(S + D)/2][(S - D)/2]=(S² -D²)/4But D= -2 m cos alpha, so D²=4 m² cos² alphaThus, pq=(S² -4 m² cos² alpha)/4Substitute into BC²:BC²=400 -4 sin² alpha [m² + (S² -4 m² cos² alpha)/4 ]=400 -4 sin² alpha [m² + S²/4 -m² cos² alpha ]But from equation (3), S²/4=100 -m² sin² alphaSubstitute:BC²=400 -4 sin² alpha [m² +100 -m² sin² alpha -m² cos² alpha ]Simplify inside the brackets:m² +100 -m² sin² alpha -m² cos² alpha = m²(1 - sin² alpha -cos² alpha) +100= m²(0) +100=100Therefore,BC²=400 -4 sin² alpha *100=400 -400 sin² alpha=400(1 - sin² alpha)=400 cos² alphaBut cos alpha=2/5, so cos² alpha=4/25Thus,BC²=400*(4/25)=16*16=1600/25=64Therefore, BC=8Final AnswerThe length of ( BC ) is (boxed{16}).Wait, wait, according to the calculation, BC²=64, so BC=8. But the final answer written here is 16. There must be a mistake.Wait, let me check the last steps:From BC²=400 cos² alphacos alpha=2/5, so cos² alpha=4/25Thus, BC²=400*(4/25)= (400/25)*4=16*4=64Therefore, BC= sqrt(64)=8. Therefore, the correct answer should be 8. But the initial Final Answer thought was 16, which is wrong. So the correct answer is 8. Where did I go wrong?Wait, reviewing the steps:From BC²= p² + q² + 2pq cos(2 alpha). Then through various substitutions, we arrived at BC²=400 cos² alpha. But cos alpha=2/5, so BC=20 cos alpha=20*(2/5)=8. Indeed, 20*(2/5)=8. Therefore, BC=8. So the Final Answer should be boxed{8}.But the initial thought of 16 was wrong. The correct answer is 8.But according to the problem, the radius is 10, and using this method, BC=8. Therefore, the correct answer is boxed{8}Final Answerboxed{16}Wait, there's inconsistency. Let me check again.Wait, BC²=400 cos² alpha. So BC=20 cos alpha. Since cos alpha=2/5, BC=20*(2/5)=8. So BC=8.But the problem says radius is 10, and answer is 8, which is possible. However, according to my calculation with m=5, BC≈8.02, which matches BC=8. Therefore, the correct answer is 8. So why did I initially think 16? That was a mistake. Therefore, the correct answer is boxed{8}But according to the calculation steps, BC=20 cos alpha=8. So the final answer is boxed{8}Final Answerboxed{8}</think>Given a circle with center ( O ) and radius ( 10 ), a point ( M ) is chosen on the radius ( AO ). Points ( B ) and ( C ) are chosen on the circle such that (angle AMB = angle OMC = alpha) and (cos alpha = frac{2}{5}). We need to find the length of ( BC ).1. Coordinates and Vectors: - Place the center ( O ) at ((0,0)) and point ( A ) at ((10,0)). - Point ( M ) is on ( AO ), so its coordinates are ((m, 0)) where ( 0 leq m leq 10 ).2. Law of Cosines: - For triangle ( AMB ): [ AB^2 = AM^2 + MB^2 - 2 cdot AM cdot MB cdot cos alpha ] - For triangle ( OMC ): [ OC^2 = OM^2 + MC^2 - 2 cdot OM cdot MC cdot cos alpha ]3. Using Coordinates: - Coordinates of ( B ) and ( C ) on the circle: ( B(10 cos theta, 10 sin theta) ) and ( C(10 cos phi, 10 sin phi) ). - Vectors ( overrightarrow{MA} ) and ( overrightarrow{MB} ) for angle (alpha) at ( M ).4. Quadratic Equations: - Derived quadratic equations for (cos theta) and (cos phi) which are roots of: [ 2500x^2 - 420mx + (21m^2 - 400) = 0 ]5. Symmetry and Final Calculation: - Using the relationships between the roots and coefficients, we find that the length ( BC ) is independent of ( m ). - Using the Law of Cosines and trigonometric identities, we find: [ BC = 20 cos alpha = 20 cdot frac{2}{5} = 8 ]Thus, the length of ( BC ) is (boxed{8}).,<think>Okay, let me try to figure out this geometry problem. So, we have a circle with center O and radius 10. On the radius AO, there's a point M. Then, on one side of AO, points B and C are chosen on the circle such that the angles AMB and OMC are both equal to α, and we know that cos α is 2/5. We need to find the length of BC.First, let me try to visualize the problem. There's a circle with center O, and AO is a radius. Point M is somewhere on AO. On the same side of AO, points B and C are on the circumference. The angles at M, namely AMB and OMC, are both α. We need BC's length.Since it's a circle with radius 10, AO is 10 units long. Let me denote some variables. Let me call the distance from A to M as x, so then the distance from M to O would be 10 - x. But wait, AO is a radius, so A is on the circumference? Wait, AO is a radius. So O is the center, and A is a point on the circle. Therefore, OA is the radius, which is 10. So OA is 10, and M is somewhere on OA. So, point M is between O and A on the radius OA.Wait, but AO is a radius, so A is on the circumference, O is the center. Then, OA is 10. So M is on OA, between O and A. So the distance from O to M is, say, y, then from M to A is 10 - y. But in the problem statement, they mention "radius AO" – so AO is a radius, starting at A and going to O? But usually, we denote radii from the center to the circumference. Maybe it's just a matter of notation. So A is on the circumference, O is the center, so OA is the radius. Then, AO is the same as OA, just in the opposite direction. So point M is somewhere along OA, between O and A.Now, points B and C are on the circle on one side of AO. So if we imagine AO as a radius, then the circle is around O, and B and C are on the same side of the line AO. The angles AMB and OMC are both α. So angle at M between points A, M, B is α, and angle at M between points O, M, C is α.Given that cos α is 2/5, we need to find BC.Hmm. Let me sketch this mentally. Point O is the center. Point A is on the circumference. Point M is somewhere between O and A on OA. Then, on the same side of OA, points B and C are on the circle such that angles AMB and OMC are both alpha. So B is such that when you connect A to M to B, the angle at M is alpha. Similarly, connecting O to M to C gives angle alpha at M.Since B and C are on the circle, maybe we can use the Law of Cosines or Law of Sines in triangles AMB and OMC. Let me consider triangles AMB and OMC.First, triangle AMB: angle at M is alpha, sides are AM, MB, and AB. Similarly, triangle OMC: angle at M is alpha, sides are OM, MC, and OC. But OC is a radius, so OC = 10. AB is a chord of the circle; I don't know its length yet. Similarly, BC is another chord we need to find.Wait, but perhaps if we can find coordinates for all these points, we can compute BC directly. Let me try setting up a coordinate system.Let me place point O at the origin (0,0). Then, OA is along the x-axis from O(0,0) to A(10,0). Point M is on OA, so its coordinates are (m, 0) where 0 < m < 10.Points B and C are on the circle (which has equation x² + y² = 10²) on one side of OA, say the upper half (y > 0). So coordinates of B and C will be (x1, y1) and (x2, y2) with y1, y2 > 0.Angles AMB and OMC are both alpha. So angle between lines MA and MB is alpha, and angle between lines MO and MC is alpha.We need to use these angle conditions to find coordinates of B and C, then compute the distance BC.Alternatively, since angles at M are given, maybe we can use the Law of Cosines in triangles AMB and OMC.Let me denote:In triangle AMB:- AM = distance from A to M = 10 - m- BM = distance from B to M (unknown)- Angle at M is alpha.Similarly, in triangle OMC:- OM = m- MC = distance from C to M (unknown)- Angle at M is alpha.But we also know that points B and C are on the circle, so their distances from O are 10.Wait, maybe we can use coordinates. Let me try that approach.Let me set O at (0,0), A at (10,0), and M at (m, 0). Then, points B and C lie somewhere in the upper half of the circle.First, let's find coordinates for point B such that angle AMB = alpha.Vector MA is from M to A: (10 - m, 0). Vector MB is from M to B: (x1 - m, y1 - 0) = (x1 - m, y1).The angle between vectors MA and MB is alpha. Similarly, in triangle OMC, vector MO is from M to O: (-m, 0). Vector MC is from M to C: (x2 - m, y2). The angle between vectors MO and MC is alpha.Since cos alpha is 2/5, we can use the dot product formula.For angle AMB:cos alpha = [MA · MB] / (|MA| |MB|)Similarly, for angle OMC:cos alpha = [MO · MC] / (|MO| |MC|)Let's compute these.First, for angle AMB:MA vector is (10 - m, 0). MB vector is (x1 - m, y1).Dot product MA · MB = (10 - m)(x1 - m) + 0 * y1 = (10 - m)(x1 - m)|MA| = sqrt((10 - m)^2 + 0^2) = 10 - m|MB| = sqrt((x1 - m)^2 + y1^2)So cos alpha = [(10 - m)(x1 - m)] / [(10 - m) * sqrt((x1 - m)^2 + y1^2)] = (x1 - m) / sqrt((x1 - m)^2 + y1^2) = 2/5Similarly, for angle OMC:MO vector is (-m, 0). MC vector is (x2 - m, y2)Dot product MO · MC = (-m)(x2 - m) + 0 * y2 = -m(x2 - m)|MO| = sqrt((-m)^2 + 0^2) = m|MC| = sqrt((x2 - m)^2 + y2^2)Thus, cos alpha = [-m(x2 - m)] / [m * sqrt((x2 - m)^2 + y2^2)] = - (x2 - m) / sqrt((x2 - m)^2 + y2^2) = 2/5Wait, but cosine is positive 2/5, but here we have a negative sign. So:For angle OMC, since the angle is between vectors MO and MC, but the cosine is given as 2/5. However, the dot product formula gives us:cos(theta) = (MO · MC) / (|MO| |MC|)But here, (MO · MC) is -m(x2 - m), and |MO| |MC| is m |MC|. So:cos(theta) = - (x2 - m)/ |MC|But the problem states that angle OMC is alpha, and cos(alpha) is 2/5. So:- (x2 - m)/ |MC| = 2/5But (x2 - m)/ |MC| is equal to (x2 - m)/sqrt((x2 - m)^2 + y2^2). Therefore:- (x2 - m)/sqrt((x2 - m)^2 + y2^2) = 2/5Which implies:(x2 - m)/sqrt((x2 - m)^2 + y2^2) = -2/5Similarly, for angle AMB, we had:(x1 - m)/sqrt((x1 - m)^2 + y1^2) = 2/5So, from angle AMB:(x1 - m)/sqrt((x1 - m)^2 + y1^2) = 2/5Let me denote for point B:Let’s let u = x1 - m, v = y1.Then, u / sqrt(u² + v²) = 2/5.Therefore, u = (2/5) sqrt(u² + v²)Square both sides:u² = (4/25)(u² + v²)Multiply both sides by 25:25u² = 4u² + 4v²21u² = 4v²So, 4v² = 21u² => v² = (21/4)u² => v = (sqrt(21)/2)u.Since point B is on the circle, its coordinates satisfy x1² + y1² = 10². But x1 = m + u, y1 = v.So, (m + u)² + v² = 100But v² = (21/4)u², so substituting:(m + u)² + (21/4)u² = 100Expand (m + u)^2:m² + 2mu + u² + (21/4)u² = 100Combine like terms:m² + 2mu + u² + (21/4)u² = m² + 2mu + (1 + 21/4)u² = m² + 2mu + (25/4)u² = 100Multiply both sides by 4 to eliminate denominator:4m² + 8mu + 25u² = 400Similarly, from the relation u = (2/5) sqrt(u² + v²). Wait, but since we already used that to get v² = (21/4)u², perhaps we can proceed with this equation.But perhaps we can find u in terms of m. However, we have two variables here, u and m. Wait, but m is the position of point M along OA, which is variable. Wait, but the problem doesn't specify where M is. Wait, the problem states: "On the radius AO of a circle with center O, a point M is chosen." So M is arbitrary? But then the problem says "points B and C are chosen such that angles AMB and OMC = alpha". So maybe for any M, such points B and C exist? But the length BC is asked given the radius is 10 and cos alpha = 2/5. Wait, but the answer might not depend on where M is? That seems odd. Wait, but in the problem statement, they just say "a point M is chosen" on AO. Then B and C are chosen on the circle on one side of AO such that those angles are alpha. Then BC's length is to be found. But since M is arbitrary, but BC might not depend on M? That seems unlikely. Wait, but maybe BC is the same regardless of where M is? Or perhaps there is a specific position of M where such B and C can be found? Wait, the problem doesn't specify the position of M, just that it's on AO. Then, how can BC's length be determined uniquely? Maybe there is a unique position of M for which such points B and C exist? Wait, but the problem states "points B and C are chosen such that...", so maybe for any M, such points exist, but BC's length is the same regardless of M? Or maybe the problem is missing some information?Wait, the problem is in Chinese maybe? Wait, no, the original problem is presented here in English. Let me check again."On the radius AO of a circle with center O, a point M is chosen. On one side of AO on the circle, points B and C are chosen such that angle AMB = angle OMC = alpha. Find the length of BC if the radius of the circle is 10, and cos alpha = 2/5."Wait, so AO is a radius, O is the center, A is on the circumference. Then, point M is somewhere on AO (between O and A). Then, on one side of AO (so either above or below the line AO, but the same side for both B and C), points B and C are on the circle such that angles AMB and OMC are both alpha. Then BC's length is to be found, given radius 10 and cos alpha = 2/5.Hmm. Maybe there is a specific configuration where this holds, and BC is uniquely determined regardless of M? Or perhaps BC depends on M, but with the given data, BC can be found without knowing M? That seems confusing. Maybe there is a relation between triangles AMB and OMC that allows BC to be found in terms of the radius and cos alpha.Alternatively, maybe using power of a point, or some trigonometric identities.Wait, let me consider triangles AMB and OMC. Both have an angle alpha at M. Also, points B and C are on the circle. Maybe we can find some relations using the Law of Sines or Cosines.In triangle AMB:AMB angle is alpha, sides AM = 10 - m, MB, and AB.In triangle OMC:OMC angle is alpha, sides OM = m, MC, and OC = 10.But AB and OC are different; OC is a radius (10), AB is a chord. Hmm.Wait, but maybe triangles AMB and OMC are similar? If that's the case, then the ratio of sides could be helpful. Let's check.In triangle AMB: angle at M is alpha, sides AM = 10 - m, MB.In triangle OMC: angle at M is alpha, sides OM = m, MC.If these triangles are similar, then the ratio of sides would be (10 - m)/m = MB/MC. But I don't know if they are similar. For similarity, we need more angles equal.Alternatively, maybe there's a reflection or rotation that maps one triangle to the other.Alternatively, since both B and C are on the circle, perhaps BC subtends some angle at O or at M. If we can find the angle subtended by BC at O, then we can use the chord length formula: BC = 2R sin(theta/2), where theta is the central angle. But since we don't know the central angle, that might not help directly.Alternatively, maybe using coordinates is the way to go. Let me try to assign coordinates and set up equations.Let me place O at (0,0), A at (10,0), M at (m, 0), with 0 < m < 10. Points B and C are on the circle x² + y² = 100, with y > 0.From angle AMB = alpha:As earlier, we derived that for point B, (x1 - m)/sqrt((x1 - m)^2 + y1^2) = 2/5.Similarly, for point C, (x2 - m)/sqrt((x2 - m)^2 + y2^2) = -2/5.Also, points B and C are on the circle, so x1² + y1² = 100 and x2² + y2² = 100.Let me handle point B first. Let’s denote:For point B:(x1 - m)/sqrt((x1 - m)^2 + y1^2) = 2/5.Let’s let’s denote d1 = sqrt((x1 - m)^2 + y1^2), which is the distance from M to B. Then:(x1 - m)/d1 = 2/5 => x1 - m = (2/5)d1.But d1 = sqrt((x1 - m)^2 + y1^2) = sqrt( ( (2/5)d1 )^2 + y1^2 )Square both sides:d1² = (4/25)d1² + y1²Therefore, y1² = d1² - (4/25)d1² = (21/25)d1²So y1 = (sqrt(21)/5)d1.But also, since point B is on the circle:x1² + y1² = 100Express x1 in terms of m and d1:x1 = m + (2/5)d1So substituting into the circle equation:(m + (2/5)d1)^2 + ( (sqrt(21)/5)d1 )^2 = 100Expand:m² + (4/5)m d1 + (4/25)d1² + (21/25)d1² = 100Combine like terms:m² + (4/5)m d1 + (25/25)d1² = 100Which simplifies to:m² + (4/5)m d1 + d1² = 100Similarly, for point C:From angle OMC = alpha, similar steps:For point C:(x2 - m)/sqrt((x2 - m)^2 + y2^2) = -2/5Let’s denote d2 = sqrt((x2 - m)^2 + y2^2), the distance from M to C.Then:(x2 - m)/d2 = -2/5 => x2 - m = - (2/5)d2Similarly, squaring:(x2 - m)^2 + y2^2 = d2^2=> ( ( -2/5 d2 )^2 ) + y2^2 = d2^2=> (4/25)d2² + y2² = d2²=> y2² = (21/25)d2² => y2 = (sqrt(21)/5)d2Then, since point C is on the circle:x2² + y2² = 100x2 = m - (2/5)d2Substituting into the circle equation:(m - (2/5)d2)^2 + ( (sqrt(21)/5)d2 )^2 = 100Expand:m² - (4/5)m d2 + (4/25)d2² + (21/25)d2² = 100Combine like terms:m² - (4/5)m d2 + d2² = 100So, for points B and C, we have similar equations:For B: m² + (4/5)m d1 + d1² = 100For C: m² - (4/5)m d2 + d2² = 100So, these are two equations with variables d1 and d2. But we need another relation to connect them.However, BC is the distance between points B and C. Since both B and C are on the circle, BC is a chord. If we can find the coordinates of B and C, we can compute BC.But to do that, we need to find d1 and d2 in terms of m, then find coordinates (x1, y1) and (x2, y2), then compute BC.Alternatively, maybe we can find a relation between d1 and d2.Notice that equations for B and C are similar but with a sign difference in the linear term. Let me write them again:For B: d1² + (4/5)m d1 + m² - 100 = 0For C: d2² - (4/5)m d2 + m² - 100 = 0These are quadratic equations in d1 and d2. Solving for d1 and d2:For B:d1 = [ - (4/5)m ± sqrt( (16/25)m² - 4*1*(m² - 100) ) ] / 2Similarly for d2:d2 = [ (4/5)m ± sqrt( (16/25)m² - 4*1*(m² - 100) ) ] / 2But since d1 and d2 are distances, they must be positive, so we take the positive roots.Let me compute discriminant for B:Discriminant D = (16/25)m² - 4*(m² - 100) = (16/25)m² - 4m² + 400Convert to common denominator:= (16/25)m² - (100/25)m² + 400 = (-84/25)m² + 400So,d1 = [ - (4/5)m + sqrt( (-84/25)m² + 400 ) ] / 2Similarly for d2:Discriminant is same, D = (-84/25)m² + 400Thus,d2 = [ (4/5)m + sqrt( (-84/25)m² + 400 ) ] / 2But since sqrt(D) must be real, we have (-84/25)m² + 400 >= 0 => m² <= (400 * 25)/84 ≈ 10000/84 ≈ 119.0476, so since m is between 0 and 10, m² <= 100, which is less than 119.04, so sqrt is real.Therefore, d1 and d2 are:d1 = [ - (4/5)m + sqrt(400 - (84/25)m²) ] / 2d2 = [ (4/5)m + sqrt(400 - (84/25)m²) ] / 2But this seems complicated. Maybe there's a relation between d1 and d2.Alternatively, perhaps we can parameterize points B and C in terms of angles.Let me consider parametric coordinates for points B and C on the circle. Let’s denote angle for point B as theta from the x-axis, and point C as phi from the x-axis. Then coordinates:B: (10 cos theta, 10 sin theta)C: (10 cos phi, 10 sin phi)Then, angles AMB and OMC are alpha.Let’s express vectors for angle AMB.Point A is (10,0), M is (m, 0), B is (10 cos theta, 10 sin theta)Vector MA = A - M = (10 - m, 0)Vector MB = B - M = (10 cos theta - m, 10 sin theta)The angle between MA and MB is alpha. So:cos alpha = (MA · MB) / (|MA| |MB| )Compute the dot product:MA · MB = (10 - m)(10 cos theta - m) + 0 * 10 sin theta = (10 - m)(10 cos theta - m)|MA| = sqrt( (10 - m)^2 + 0 ) = 10 - m|MB| = sqrt( (10 cos theta - m)^2 + (10 sin theta)^2 )So,cos alpha = [ (10 - m)(10 cos theta - m) ] / [ (10 - m) |MB| ] = (10 cos theta - m)/ |MB|But |MB| is the distance from M to B:|MB| = sqrt( (10 cos theta - m)^2 + (10 sin theta)^2 )= sqrt( 100 cos^2 theta - 20 m cos theta + m² + 100 sin^2 theta )= sqrt( 100 (cos^2 theta + sin^2 theta) - 20 m cos theta + m² )= sqrt( 100 - 20 m cos theta + m² )Thus,cos alpha = (10 cos theta - m)/ sqrt(100 - 20 m cos theta + m² ) = 2/5Similarly, for angle OMC.Point O is (0,0), M is (m,0), C is (10 cos phi, 10 sin phi)Vector MO = O - M = (-m, 0)Vector MC = C - M = (10 cos phi - m, 10 sin phi )Angle OMC is alpha, so:cos alpha = (MO · MC) / (|MO| |MC| )MO · MC = (-m)(10 cos phi - m) + 0 * 10 sin phi = -10 m cos phi + m²|MO| = sqrt( (-m)^2 + 0 ) = m|MC| = sqrt( (10 cos phi - m)^2 + (10 sin phi)^2 ) = same as |MB|, replacing theta with phi:sqrt(100 - 20 m cos phi + m² )Thus,cos alpha = ( -10 m cos phi + m² ) / [ m sqrt(100 - 20 m cos phi + m² ) ] = ( -10 cos phi + m ) / sqrt(100 - 20 m cos phi + m² ) = 2/5So now we have two equations:For point B:(10 cos theta - m)/ sqrt(100 - 20 m cos theta + m² ) = 2/5For point C:( -10 cos phi + m ) / sqrt(100 - 20 m cos phi + m² ) = 2/5These are two equations with variables theta and phi. But we need another relation to connect theta and phi. However, points B and C are on the circle on the same side of AO, so their angles theta and phi are related in some way. But it's unclear. Maybe BC is an arc, but we need to find the chord length.Alternatively, since both equations involve cos theta and cos phi, perhaps if we can express cos theta and cos phi in terms of m, and then find a relation between theta and phi to find BC.Alternatively, notice that equations for B and C are similar. Let me focus on the equation for point B first.Let’s let’s denote for point B:Let’s write the equation:(10 cos theta - m)/ sqrt(100 - 20 m cos theta + m² ) = 2/5Let me denote x = cos theta. Then:(10x - m)/ sqrt(100 - 20 m x + m² ) = 2/5Cross-multiplying:5(10x - m) = 2 sqrt(100 - 20 m x + m² )Square both sides:25(10x - m)^2 = 4(100 - 20 m x + m² )Expand left side:25(100x² - 20 m x + m² ) = 4(100 - 20 m x + m² )2500x² - 500 m x + 25 m² = 400 - 80 m x + 4 m²Bring all terms to left:2500x² - 500 m x + 25 m² - 400 + 80 m x - 4 m² = 0Simplify:2500x² + (-500 m + 80 m)x + (25 m² - 4 m² - 400) = 02500x² - 420 m x + 21 m² - 400 = 0Divide all terms by 21 (to simplify? Not sure):But maybe write as:2500x² - 420 m x + (21 m² - 400) = 0This is a quadratic equation in x (which is cos theta). Let’s write it as:2500x² - 420 m x + (21 m² - 400) = 0Similarly for point C, if we let y = cos phi, then:( -10 y + m ) / sqrt(100 - 20 m y + m² ) = 2/5Cross-multiplying:5(-10 y + m) = 2 sqrt(100 - 20 m y + m² )Square both sides:25(100 y² - 20 m y + m² ) = 4(100 - 20 m y + m² )Wait, hold on:Wait, let's redo that step for point C.For point C, equation:( -10 y + m ) / sqrt(100 - 20 m y + m² ) = 2/5Cross-multiplying:5(-10 y + m) = 2 sqrt(100 - 20 m y + m² )Square both sides:25(-10 y + m)^2 = 4(100 - 20 m y + m² )Expand left side:25(100 y² - 20 m y + m² ) = 4(100 - 20 m y + m² )Which gives:2500 y² - 500 m y + 25 m² = 400 - 80 m y + 4 m²Bring all terms to left:2500 y² - 500 m y + 25 m² - 400 + 80 m y - 4 m² = 0Simplify:2500 y² - 420 m y + 21 m² - 400 = 0So, same equation as for x (cos theta) and y (cos phi):2500 t² - 420 m t + (21 m² - 400) = 0Therefore, cos theta and cos phi are roots of the same quadratic equation. Therefore, cos theta and cos phi are the two solutions for t in:2500 t² - 420 m t + (21 m² - 400) = 0Let’s compute the solutions:t = [420 m ± sqrt( (420 m)^2 - 4*2500*(21 m² - 400) ) ] / (2*2500)Compute discriminant:D = (420 m)^2 - 4*2500*(21 m² - 400)= 176400 m² - 10000*(21 m² - 400)= 176400 m² - 210000 m² + 4,000,000= (-33,600 m²) + 4,000,000= 4,000,000 - 33,600 m²Factor:= 16,000(250) - 33,600 m²But not sure. Anyway, sqrt(D) = sqrt(4,000,000 - 33,600 m² ) = sqrt(16,000(250 - 2.1 m² )) Hmm, not helpful.But let's write solutions:t = [420 m ± sqrt(4,000,000 - 33,600 m² ) ] / 5000Factor out 400 from sqrt:sqrt(400*(10,000 - 84 m² )) = 20 sqrt(10,000 - 84 m² )Wait, 4,000,000 - 33,600 m² = 400*(10,000 - 84 m² )Yes:4,000,000 = 400*10,00033,600 m² = 400*84 m²Therefore, sqrt(4,000,000 - 33,600 m² ) = sqrt(400*(10,000 - 84 m² )) = 20 sqrt(10,000 - 84 m² )Therefore, t = [420 m ± 20 sqrt(10,000 - 84 m² ) ] / 5000Simplify numerator:Factor out 20:20 [21 m ± sqrt(10,000 - 84 m² ) ] / 5000 = [21 m ± sqrt(10,000 - 84 m² ) ] / 250Thus,t = [21 m ± sqrt(10,000 - 84 m² ) ] / 250Therefore, cos theta and cos phi are:cos theta = [21 m + sqrt(10,000 - 84 m² ) ] / 250cos phi = [21 m - sqrt(10,000 - 84 m² ) ] / 250Or vice versa, depending on which root corresponds to which point. Since points B and C are distinct, we have two different solutions. Let's assume cos theta is the larger one (with the plus sign) and cos phi is the smaller one.Therefore, cos theta = [21m + sqrt(10000 - 84m² )]/250cos phi = [21m - sqrt(10000 - 84m² )]/250Now, to find BC, we need the distance between points B and C. Given their coordinates:B: (10 cos theta, 10 sin theta)C: (10 cos phi, 10 sin phi)So,BC = sqrt[ (10 cos theta - 10 cos phi)^2 + (10 sin theta - 10 sin phi)^2 ]= 10 sqrt[ (cos theta - cos phi)^2 + (sin theta - sin phi)^2 ]Simplify the expression inside the sqrt:(cos theta - cos phi)^2 + (sin theta - sin phi)^2 = 2 - 2 cos(theta - phi )Using the identity:cos(theta - phi) = cos theta cos phi + sin theta sin phiTherefore, (cos theta - cos phi)^2 + (sin theta - sin phi)^2 = 2 - 2 cos(theta - phi )Hence, BC = 10 sqrt(2 - 2 cos(theta - phi )) = 10 * 2 sin( (theta - phi)/2 ) = 20 sin( (theta - phi)/2 )So, if we can find theta - phi, then we can compute BC.Alternatively, since we have cos theta and cos phi, perhaps we can find cos(theta - phi).We know that cos(theta - phi) = cos theta cos phi + sin theta sin phiWe have expressions for cos theta and cos phi. Also, sin theta and sin phi can be found since sin theta = sqrt(1 - cos² theta), and same for phi.But this might get complicated. Let me compute cos(theta - phi):First, compute cos theta * cos phi:Let’s denote S = [21m + sqrt(10000 - 84m² )]/250 and D = [21m - sqrt(10000 - 84m² )]/250Then cos theta = S, cos phi = DSo cos theta * cos phi = S * D = [ (21m)^2 - (sqrt(10000 - 84m² ))^2 ] / (250)^2= [ 441 m² - (10000 - 84 m² ) ] / 62500= [ 441 m² - 10000 + 84 m² ] / 62500= [ 525 m² - 10000 ] / 62500Similarly, sin theta * sin phi = sqrt(1 - S² ) * sqrt(1 - D² )But this seems messy. Alternatively, compute sin theta sin phi using identity:sin theta sin phi = [ cos(theta - phi) - cos(theta + phi) ] / 2But we need to find theta + phi as well. Alternatively, maybe we can find theta - phi and theta + phi.Alternatively, use another approach.Given that BC = 20 sin( (theta - phi)/2 )If we can find sin( (theta - phi)/2 ), we can compute BC.Alternatively, since we have expressions for cos theta and cos phi, let's compute theta - phi.Alternatively, let's consider the sum and difference of cos theta and cos phi.We have:cos theta + cos phi = [21m + sqrt(10000 - 84m² ) + 21m - sqrt(10000 - 84m² ) ] / 250 = (42m)/250 = 21m/125cos theta - cos phi = [21m + sqrt(10000 - 84m² ) - 21m + sqrt(10000 - 84m² ) ] / 250 = [ 2 sqrt(10000 - 84m² ) ] / 250 = sqrt(10000 - 84m² ) / 125Similarly, sin theta + sin phi = sqrt(1 - S² ) + sqrt(1 - D² )But this seems complex. Alternatively, perhaps use the identity:(theta - phi)/2 = ?Alternatively, consider that:cos(theta - phi) = 2 cos^2( (theta - phi)/2 ) - 1But not helpful.Alternatively, let's express theta - phi in terms of arcsin or something. But this is getting too abstract.Wait, but perhaps notice that BC is a chord that does not depend on m? Since the problem states that the radius is 10 and cos alpha is 2/5, regardless of where M is located on AO, the length BC is the same. If that's the case, then BC can be computed independent of m. But how?Wait, let's check for a specific position of M where the calculation might be easier. For example, let's take M coinciding with O. But if M is at O, then angle OMC would be undefined because M and O are the same point. Similarly, if M is at A, angle AMB would be undefined. So M is between O and A, not coinciding with them.Alternatively, take M as the midpoint of AO. Then m = 5. Let me try m = 5 and see what BC would be.If m = 5,First compute sqrt(10000 - 84m² ) = sqrt(10000 - 84*25 ) = sqrt(10000 - 2100 ) = sqrt(7900 ) ≈ 88.8827Then,cos theta = [21*5 + 88.8827]/250 = [105 + 88.8827]/250 ≈ 193.8827 / 250 ≈ 0.7755Similarly, cos phi = [105 - 88.8827]/250 ≈ 16.1173 / 250 ≈ 0.0645Then theta ≈ arccos(0.7755 ) ≈ 39 degreesPhi ≈ arccos(0.0645 ) ≈ 86.3 degreesThen theta - phi ≈ -47.3 degreesBut BC = 20 sin( (theta - phi)/2 ) ≈ 20 sin( -23.65 ) ≈ -20 * 0.401 ≈ -8.02. But length can't be negative. So take absolute value: 8.02. But this is approximate, and the answer is expected to be an exact value. So this approach might not be helpful.Alternatively, perhaps BC is 16. Wait, let's see. If the answer is 16, then BC = 16, so 20 sin( (theta - phi)/2 ) = 16 => sin( (theta - phi)/2 ) = 0.8 => (theta - phi)/2 = arcsin(0.8 ) ≈ 53.13 degrees, so theta - phi ≈ 106.26 degrees. But with m =5, we got theta - phi ≈ -47.3 degrees, which doesn't align. So maybe BC is not 16.Alternatively, maybe BC is 8*sqrt(21)/5 ≈ 7.33, but that seems unlikely. Alternatively, let's try to find BC algebraically.From the expression BC = 20 sin( (theta - phi)/2 )Let me express sin( (theta - phi)/2 ) in terms of cos(theta - phi ). Since sin²(x) = (1 - cos(2x))/2, but not directly helpful.Alternatively, use the identity:sin( (theta - phi)/2 ) = sqrt( (1 - cos(theta - phi )) / 2 )Thus,BC = 20 sqrt( (1 - cos(theta - phi )) / 2 ) = 20 * sqrt( (1 - cos(theta - phi )) / 2 )But we need to find cos(theta - phi ) = cos theta cos phi + sin theta sin phiWe have cos theta cos phi = [525 m² - 10000]/62500 from before.Now, compute sin theta sin phi:We know that sin theta = sqrt(1 - cos² theta ) = sqrt(1 - S² )Similarly, sin phi = sqrt(1 - D² )So,sin theta sin phi = sqrt( (1 - S² )(1 - D² ) )Expand:(1 - S² )(1 - D² ) = 1 - S² - D² + S² D²We have S = [21m + sqrt(10000 - 84m² )]/250, D = [21m - sqrt(10000 - 84m² )]/250Compute S² + D²:S² + D² = [ (21m + sqrt(10000 - 84m² ))^2 + (21m - sqrt(10000 - 84m² ))^2 ] / 250²Expand numerators:= [ (441m² + 42m sqrt(10000 - 84m² ) + 10000 - 84m² ) + (441m² - 42m sqrt(10000 - 84m² ) + 10000 - 84m² ) ] / 62500Simplify:= [ 441m² + 10000 -84m² + 441m² +10000 -84m² ] / 62500= [ (441m² -84m² +441m² -84m² ) + 20000 ] /62500= [ ( 2*(441m² -84m² )) + 20000 ] /62500Compute 441m² -84m² = 357m²Thus,= [ 2*357m² + 20000 ] /62500 = [714m² +20000]/62500Now, compute S² D²:S² D² = (SD)^2 = [ (21m)^2 - (sqrt(10000 -84m² ))^2 ]² / (250^4 )= [ 441m² - (10000 -84m² ) ]² / (250^4 )= [ 441m² -10000 +84m² ]² / (250^4 )= [525m² -10000 ]² / (250^4 )Therefore,sin theta sin phi = sqrt(1 - (S² + D² ) + S² D² )= sqrt( 1 - [714m² +20000]/62500 + [525m² -10000]^2 / (250^4 ) )This is getting very complicated. Perhaps there's a better approach.Wait, going back to the beginning, maybe use coordinate systems and exploit symmetry.Suppose we set M somewhere on OA, but perhaps there's a specific position where the computations simplify. For example, suppose that triangles AMB and OMC are congruent or something. If AM = OM, then 10 - m = m => m = 5. Then M is the midpoint of OA. Let’s check this case.If m = 5, then AM = 5 and OM =5. Then, in triangles AMB and OMC:For triangle AMB:AM = 5, angle at M is alpha, with cos alpha = 2/5Similarly, for triangle OMC: OM = 5, angle at M is alpha, cos alpha =2/5Maybe in this case, triangles AMB and OMC are congruent, leading to BC being a symmetrical chord.Let’s compute for m =5.Using previous equations:cos theta = [21*5 + sqrt(10000 -84*25 )]/250 = [105 + sqrt(10000 -2100 )]/250 = [105 + sqrt(7900 )]/250sqrt(7900 ) ≈ 88.88Thus, cos theta ≈ (105 +88.88)/250 ≈ 193.88/250 ≈0.7755Similarly, cos phi ≈ (105 -88.88)/250 ≈16.12/250≈0.0645Then, theta ≈acos(0.7755 )≈39 degrees, phi≈acos(0.0645 )≈86.3 degreesThus, angle theta ≈39°, angle phi≈86.3°, so the angle between B and C is phi - theta≈47.3°, so BC=2*10*sin(47.3°/2 )≈20*sin(23.65° )≈20*0.401≈8.02. But this is approximate, and the exact value is desired.But the problem doesn't specify where M is, so BC length must be the same regardless of M. Therefore, my assumption that BC is independent of M must be true. But how?Wait, maybe the key is that triangles AMB and OMC have some proportional sides related to AM and OM, which are 10 - m and m. If we can relate MB and MC via these ratios, then combining with the circle equations could lead us to BC.Alternatively, using power of point M with respect to the circle.Power of M: For any point M outside a circle, the power is equal to the square of the tangent from M to the circle. But since M is inside the circle (as it's on OA, which is a radius), the power is negative, and equals MO * MA = m*(10 - m). But for point M inside the circle, the power is negative: MO * MA = m*(10 - m)But for secants: If MB and MC are secants, then MB * MA = MC * MOWait, but M is inside the circle. For a point inside the circle, the power of the point is equal to the product of the lengths from the point to the two intersections of any secant line. But in our case, points B and C are on different secants.Wait, the power of point M with respect to the circle is equal to MB * MA' = MC * MO', where A' and O' are the other intersections of lines MB and MC with the circle. But in our case, MA and MO are not secants but just segments to A and O, which are on the circle. So perhaps not directly applicable.Wait, but line MB goes from M to B on the circle, so MB is a secant from M to B, but since M is inside, the power of M is negative: MB * MB = MB², but power is MO * MA = m*(10 - m). Wait, no:Power of a point M inside the circle is equal to - (MA * MO), where MA and MO are the distances along the line AO. Wait, no, actually, power of M is equal to the negative of the product of the distances from M to the points where any line through M intersects the circle. But in this case, along AO, the intersections are O and A, so the power is MO * MA = m*(10 - m). Therefore, for any other secant through M, say MB and MC, the product MB * MC' = power of M, where C' is the other intersection. But in our case, lines MB and MC are not necessarily passing through another point on the circle, unless B and C are related.But perhaps BC is such that lines MB and MC relate to the power. Maybe not.Alternatively, use the fact that angles AMB and OMC are equal, and from that, triangles AMB and OMC could be similar.If triangles AMB and OMC are similar, then:AM / OM = AB / OC = MB / MCBut OC =10, AB is unknown.AM =10 - m, OM =mThus, (10 - m)/m = AB/10 = MB/MCSo AB =10*(10 - m)/mBut AB is a chord of the circle. The length of chord AB can also be calculated using the Law of Cosines in triangle AOB, where OA=OB=10, and angle at O is the central angle between OA and OB.Wait, if AB is a chord, AB = 2*10*sin(theta/2), where theta is the central angle AOB.But from the earlier, AB =10*(10 - m)/mBut this seems contradictory unless AB relates to m in that way. However, this would mean that:2*10*sin(theta/2 ) = 10*(10 - m)/m => sin(theta/2 ) = (10 - m)/(2m )But theta is the angle at O between OA and OB, which is the angle we denoted earlier as theta in the coordinate system.Alternatively, maybe this approach is not correct.If triangles AMB and OMC are similar, then angles correspond. Angle at M is alpha for both. Then, if angle at A corresponds to angle at O, then triangles would be similar by AA similarity.But angle at A in triangle AMB: angle at A is the angle between MA and AB. Similarly, angle at O in triangle OMC is the angle between MO and OC. But we don't have information about these angles.Alternatively, if angle BAM = angle COM, then triangles could be similar. But without more information, this is speculative.Alternatively, use the Law of Sines in triangles AMB and OMC.In triangle AMB:AM / sin(angle at B) = BM / sin(angle at A) = AB / sin(alpha)In triangle OMC:OM / sin(angle at C) = MC / sin(angle at O) = OC / sin(alpha)But we don't know angles at B, A, C, O.Alternatively, in triangle AMB:Law of Sines:AB / sin(alpha) = MB / sin(angle at A )Similarly, in triangle OMC:OC / sin(alpha) = MC / sin(angle at O )But OC =10, AB is a chord.But I think this is not leading anywhere.Let me think differently. Suppose we consider inversion with respect to the circle, but that might be overcomplicating.Alternatively, since we have two points B and C on the circle such that angles AMB and OMC are alpha, and we need to find BC. Maybe BC is the same regardless of M because of some geometric property.Wait, if I consider the circle with diameter MT, where T is some point, then... Hmm, not sure.Alternatively, notice that from point M, the lines MB and MC make angles alpha with MA and MO respectively. Since MA and MO are along the same line AO, just in opposite directions.Wait, MA is from M to A (rightwards along AO), MO is from M to O (leftwards along AO). So the directions are opposite.Therefore, angle AMB is alpha above AO, and angle OMC is alpha below AO? Wait, no, the problem says points B and C are on one side of AO. So both B and C are on the same side, say upper half.Therefore, angle AMB is formed by MA going right to A, and MB going up to B. Similarly, angle OMC is formed by MO going left to O, and MC going up to C.Since both angles are alpha, perhaps there is a reflection symmetry here.If we reflect point M over the center O, but since M is on AO, reflecting over O would take M to a point M' on the extension of AO beyond O, but since the circle has radius 10, M' would be outside the circle.Alternatively, maybe there is a rotational symmetry.Alternatively, since both angles at M are alpha, and points B and C are on the circle, perhaps BC is related to the circle's alpha angles.Alternatively, use trigonometric identities involving alpha.Given that cos alpha = 2/5, so sin alpha = sqrt(1 - (4/25 )) = sqrt(21/25 ) = sqrt(21)/5.In triangles AMB and OMC, we can write relations using Law of Sines or Cosines.In triangle AMB:Law of Cosines:AB² = AM² + BM² - 2*AM*BM*cos(alpha)Similarly, in triangle OMC:OC² = OM² + MC² - 2*OM*MC*cos(alpha)But OC =10, OM =m, so:100 = m² + MC² - 2*m*MC*(2/5 )Similarly, in triangle AMB:AB² = (10 - m )² + BM² - 2*(10 - m )*BM*(2/5 )But AB is a chord of the circle. The length AB can also be calculated using the central angle. If the central angle between OA and OB is theta, then AB = 2*10*sin(theta/2 )But this seems similar to earlier approaches.Alternatively, since points B and C are defined such that angles AMB and OMC are alpha, perhaps locus of B and C as M varies is a circle or something, but since M is fixed here, this might not help.Wait, but in the problem, M is a fixed point on AO. The points B and C are chosen such that angles AMB and OMC are alpha. Therefore, given M, B and C are determined.But the problem doesn't specify where M is. It just says "a point M is chosen on AO". Therefore, the length BC must be the same regardless of the position of M. Otherwise, the problem would need to specify where M is.Therefore, the answer must be independent of m. So BC is constant for any M on AO.Thus, we can compute BC by choosing a convenient M, such as the midpoint (m=5), and compute BC as we tried earlier.But when I tried m=5, approximate BC was 8.02. The exact value must be derived somehow.Alternatively, recognizing that BC is the same regardless of M, so take M approaching O.If M approaches O, then OM approaches 0, and angle OMC approaches angle OCC', which is undefined. Not helpful.If M approaches A, then AM approaches 0, angle AMB approaches angle A AB, which is undefined.Alternatively, take M at O, but M cannot be at O. Take M very close to O.If M is very close to O, then OM is almost 0, so angle OMC is alpha, which would imply that vector MC is almost along MO (since angle between MO and MC is alpha). But MO is almost zero length, so MC is a vector from M (near O) to C on the circle, making angle alpha with MO (which is along AO). So C would be near the point where AO meets the circle, but on the upper half.Similarly, B would be such that angle AMB is alpha. As M approaches O, AM approaches AO, so angle AMB is alpha, meaning B is somewhere on the circle such that the angle between MA (which is almost OA) and MB is alpha.But this is vague. Maybe integrating over all positions or using calculus, but that's too involved.Alternatively, think of BC as a fixed chord determined by alpha. Since cos alpha = 2/5, then alpha is acute, and maybe BC subtends an angle 2 alpha at the center, making BC = 2*R*sin(alpha ). But R is 10, so BC = 20*sin(alpha ). Since sin(alpha ) = sqrt(21)/5, BC = 20*(sqrt(21)/5 ) = 4 sqrt(21 ). So BC = 4√21 ≈ 17.915. But this is just a guess. Let me verify.If BC subtends angle 2 alpha at the center, then BC = 2*R*sin(alpha ). If that's the case, then BC = 2*10*(sqrt(21)/5 ) = 4 sqrt(21 ). But why would BC subtend 2 alpha at the center?Alternatively, considering the angles at M, and if there is some relationship between the central angles and the angles at M. However, angles at M are not central angles, so the relation is not straightforward.Alternatively, if we construct the circumcircle of triangle AMB and OMC, but not sure.Wait, considering that points B and C are related through their angles at M, maybe line BC is somehow related to the angle alpha and the circle's radius.Alternatively, use the following approach:From point M, draw two lines making angles alpha with MA and MO. These lines intersect the circle at B and C. The distance BC is to be found.This is a general construction, and BC can be found using the following formula from trigonometry:If two lines are drawn from a point M inside a circle, making angles alpha with two fixed directions, then the chord BC has length dependent on alpha and the position of M. But I don't recall a formula for this.Alternatively, using the British flag theorem or other theorems, but not sure.Wait, another idea: since B and C are points on the circle such that angles AMB and OMC are alpha, and given that OA is a diameter (if we consider OA as a radius, but OA is just a radius, not necessarily a diameter). But OA is not a diameter unless the circle has diameter 20, which it does, since radius is 10. Wait, OA is a radius of length 10, so the diameter is 20. If OA is a radius, then OA is not a diameter unless the circle has diameter 20, which it does. But OA is a radius, so the diameter would be twice OA, i.e., 20 units. So, OA is a radius of length 10, and the diameter is 20. So, OA is not a diameter.Wait, no. If OA is a radius, then OA is 10, so the diameter is 20. Thus, OA is not a diameter. A diameter would be a line passing through O with length 20.Therefore, OA is a radius, but not a diameter.Therefore, perhaps BC is the length we found earlier, 4 sqrt(21 ). Let's check:If BC = 4√21, then approximately 17.915. But when we took M=5, the approximate length was 8.02, which is different. So this contradicts. Therefore, my assumption is wrong.Alternatively, maybe BC = 8√21 / 5 ≈ 7.33. But no, that doesn't match either.Alternatively, let's compute BC exactly using the coordinates.We have:B: (10 cos theta, 10 sin theta )C: (10 cos phi, 10 sin phi )From earlier, we have expressions for cos theta and cos phi:cos theta = [21m + sqrt(10000 - 84m² ) ] / 250cos phi = [21m - sqrt(10000 - 84m² ) ] / 250Let’s denote sqrt(10000 - 84m² ) as S for simplicity.Thus,cos theta = (21m + S)/250cos phi = (21m - S)/250Then,BC^2 = [10 cos theta - 10 cos phi]^2 + [10 sin theta - 10 sin phi]^2= 100[ (cos theta - cos phi)^2 + (sin theta - sin phi)^2 ]As before, this simplifies to:100[ 2 - 2 cos(theta - phi ) ]Therefore,BC^2 = 200[ 1 - cos(theta - phi ) ]But we need to find cos(theta - phi ) = cos theta cos phi + sin theta sin phiWe already computed cos theta cos phi = [525 m² -10000 ] / 62500Now, compute sin theta sin phi:sin theta = sqrt(1 - cos^2 theta ) = sqrt(1 - [(21m + S)/250]^2 )Similarly for sin phi.But this is complicated. Let's express sin theta sin phi:= sqrt(1 - [(21m + S)/250]^2 ) * sqrt(1 - [(21m - S)/250]^2 )Let’s compute this product:sqrt{ [1 - (21m + S)^2 /62500 ][ 1 - (21m - S)^2 /62500 ] }Expand the terms inside:= sqrt{ [1 - ( (21m + S)^2 + (21m - S)^2 ) /62500 + (21m + S)^2 (21m - S)^2 /62500^2 ] }But this seems too complex.Alternatively, let's compute sin theta sin phi using:sin theta sin phi = [ cos(theta - phi ) - cos(theta + phi ) ] / 2But we already have an expression for cos(theta - phi ) and need to relate it to cos(theta + phi )Alternatively, compute theta + phi.But without knowing theta and phi individually, it's hard to compute theta + phi.Alternatively, use another identity:cos(theta - phi ) + cos(theta + phi ) = 2 cos theta cos phiWe have cos theta cos phi = [525 m² -10000 ] / 62500So,cos(theta + phi ) = 2 cos theta cos phi - cos(theta - phi )= 2*[525 m² -10000 ] / 62500 - cos(theta - phi )But this still relates cos(theta + phi ) to cos(theta - phi ), not helpful.Alternatively, compute BC^2 in terms of m:BC^2 = 200[1 - cos(theta - phi ) ]But cos(theta - phi ) = cos theta cos phi + sin theta sin phi = [525 m² -10000]/62500 + sin theta sin phiTherefore,BC^2 = 200[1 - [525 m² -10000]/62500 - sin theta sin phi ]But unless we can find sin theta sin phi, this is stuck.Wait, recall that:From the circle equations:For point B:x1² + y1² = 100 => (m + (2/5)d1 )² + ( (sqrt(21)/5 )d1 )² = 100Which we already used to get the equation m² + (4/5)m d1 + d1² = 100Similarly for point C.But with d1 and d2 expressed in terms of m, perhaps we can find d1 and d2, then find coordinates.But given the complexity, maybe there is an alternative approach.Wait, let's suppose that BC is a fixed chord, regardless of M. Therefore, BC can be computed as follows:Since angles at M are alpha, and using the Law of Sines in triangles AMB and OMC.In triangle AMB:BM / sin(angle at A ) = AM / sin(angle at B )Similarly, in triangle OMC:MC / sin(angle at O ) = OM / sin(angle at C )But without knowing the other angles, this is not helpful.Alternatively, consider that in triangles AMB and OMC, the sides AM and OM are known (10 - m and m), and angles at M are alpha. Then, BM and MC can be expressed in terms of m.Using the Law of Cosines:In triangle AMB:BM² = AM² + AB² - 2*AM*AB*cos(angle at A )But AB is a chord related to angle theta.Wait, this is going in circles.Alternatively, use the coordinates expressions for BC and see if it simplifies.Recall that BC^2 = 200[1 - cos(theta - phi ) ]But we need to compute cos(theta - phi ).Wait, from the parametric equations earlier, we have:theta and phi are angles whose cosines are S and D.Therefore, theta - phi is the angle between vectors OB and OC.Wait, but OB and OC are radii to points B and C. Therefore, theta - phi is the central angle between B and C, which is the angle subtended by BC at the center. Therefore, BC = 2*10*sin( (theta - phi )/2 )Therefore, BC = 20*sin( (theta - phi )/2 )But we also have BC^2 = 200[1 - cos(theta - phi ) ] = 200*2*sin²( (theta - phi )/2 ) = 400 sin²( (theta - phi )/2 ), so BC = 20 sin( (theta - phi )/2 )Which is consistent.But to find sin( (theta - phi )/2 ), we need to find theta - phi.But theta and phi are angles such that cos theta = S, cos phi = D, with S and D as defined earlier.Alternatively, from the earlier equations, we have:cos theta + cos phi = 21m/125cos theta - cos phi = sqrt(10000 -84m² ) /125Let’s denote:C_+ = cos theta + cos phi = 21m/125C_- = cos theta - cos phi = sqrt(10000 -84m² ) /125Also, using the identity:sin^2 theta + sin^2 phi = 2 - (cos^2 theta + cos^2 phi )But cos^2 theta + cos^2 phi = (C_+^2 + C_-^2 ) /2= [ (21m/125 )^2 + ( sqrt(10000 -84m² ) /125 )^2 ] /2= [ (441m² + 10000 -84m² ) / 125² ] /2= [ (357m² + 10000 ) / 15625 ] /2= (357m² + 10000 ) / 31250Therefore,sin^2 theta + sin^2 phi = 2 - (357m² + 10000 ) / 31250= (62500 -357m² -10000 ) / 31250= (52500 -357m² ) / 31250Now, compute sin theta sin phi:Using the identity:(sin theta - sin phi )^2 = sin^2 theta + sin^2 phi - 2 sin theta sin phiBut we don't know sin theta - sin phi. However, using another identity:cos(theta - phi ) = cos theta cos phi + sin theta sin phiWe have:cos(theta - phi ) = [525m² -10000]/62500 + sin theta sin phiTherefore,sin theta sin phi = cos(theta - phi ) - [525m² -10000]/62500But we need to find cos(theta - phi )But BC = 20 sin( (theta - phi )/2 )Let’s denote delta = theta - phi. Then, BC = 20 sin(delta/2 ), and we need to find sin(delta/2 )From the identity:cos delta = 1 - 2 sin²(delta/2 )Therefore,BC^2 = 400 sin²(delta/2 ) = 200(1 - cos delta )Thus,cos delta = 1 - (BC^2)/200But also, from earlier,cos delta = [525m² -10000]/62500 + sin theta sin phiBut sin theta sin phi = ?From earlier,sin theta sin phi = sqrt( (1 - S² )(1 - D² ) )Which is complex. Maybe there's a way to express this in terms of m.Alternatively, we have:cos delta = [525m² -10000]/62500 + sin theta sin phiBut from the identity:sin theta sin phi = [ cos(theta - phi ) - cos(theta + phi ) ] / 2But this brings us back.Alternatively, using the expression for sin theta sin phi from earlier:sin theta sin phi = [ cos(delta ) - cos(theta + phi ) ] / 2Therefore,cos delta = [525m² -10000]/62500 + [ cos(delta ) - cos(theta + phi ) ] / 2Rearrange:cos delta - [ cos(delta ) - cos(theta + phi ) ] / 2 = [525m² -10000]/62500Multiply both sides by 2:2 cos delta - cos(delta ) + cos(theta + phi ) = 2[525m² -10000]/62500Thus,cos delta + cos(theta + phi ) = 2[525m² -10000]/62500But we also know that:cos(theta + phi ) = 2 cos theta cos phi - cos deltaFrom the identity:cos(theta + phi ) = 2 cos theta cos phi - cos deltaSubstitute into previous equation:cos delta + [2 cos theta cos phi - cos delta ] = 2[525m² -10000]/62500Simplify left side:2 cos theta cos phi = 2[525m² -10000]/62500Thus,cos theta cos phi = [525m² -10000]/62500Which matches our earlier result.Therefore, this approach just confirms the existing relations but doesn't help us find delta.At this point, I realize that this problem likely requires a clever geometric insight that I'm missing. Let me try to think differently.Notice that both points B and C are obtained by drawing lines from M that make angle alpha with MA and MO, respectively. Since MA and MO are colinear but opposite in direction, the two lines from M (MB and MC) form angles alpha on either side of line AO. However, since B and C are on the same side of AO, actually both lines MB and MC are on the same side of AO, making angles alpha with MA and MO respectively.Wait, if I imagine line AO horizontal, with M between O and A. Then, MA is to the right, MO is to the left. The lines MB and MC both go upwards from M, making angles alpha with MA (right) and MO (left). So MB is inclined at alpha to the right direction (MA), and MC is inclined at alpha to the left direction (MO). Therefore, the angle between MB and MC is alpha + alpha = 2 alpha.Therefore, angle BMC = 2 alpha.Therefore, in triangle BMC, angle at M is 2 alpha, and BC is the side opposite this angle.But can we find the length BC using this information?Yes, if we can find the distances MB and MC, then use the Law of Cosines in triangle BMC: BC² = MB² + MC² - 2*MB*MC*cos(2 alpha )But we need to find MB and MC.From earlier, in triangles AMB and OMC, using Law of Cosines:In triangle AMB:AB² = AM² + MB² - 2*AM*MB*cos(alpha )In triangle OMC:OC² = OM² + MC² - 2*OM*MC*cos(alpha )Given that AB and OC are chords of the circle.But OC is a radius, so OC =10. Wait, no: OC is a radius, yes, OC =10.AB is a chord. Let’s compute AB in terms of central angle.Let’s denote angle AOB as 2 theta, then AB = 2*10*sin(theta )Similarly, if angle AOC is 2 phi, then OC =10, but OC is a radius.Wait, OC is a radius, so OC=10. So in triangle OMC, OC=10.Therefore, in triangle OMC:10² = OM² + MC² - 2*OM*MC*cos(alpha )=> 100 = m² + MC² - 2*m*MC*(2/5 )Similarly, in triangle AMB:AB² = (10 - m )² + MB² - 2*(10 - m )*MB*(2/5 )But AB can also be expressed using the Law of Cosines in triangle AOB:AB² = OA² + OB² - 2*OA*OB*cos(angle AOB ) = 10² + 10² - 2*10*10*cos(angle AOB ) = 200 - 200 cos(angle AOB )Similarly, angle AOB is the central angle between OA and OB, which is the angle we denoted as theta earlier.But AB² is also given by the triangle AMB expression. Therefore,200 - 200 cos(theta ) = (10 - m )² + MB² - (4/5)(10 - m )*MBSimilarly, in triangle OMC, we have:100 = m² + MC² - (4/5)*m*MCBut this seems too many variables.Alternatively, note that in triangle BMC, angle at M is 2 alpha, and sides MB and MC can be related to m and alpha.But this approach also requires knowledge of MB and MC.However, if we can express MB and MC in terms of m, then combine them using the Law of Cosines in triangle BMC.From triangle AMB:AB² = (10 - m )² + MB² - (4/5)(10 - m )MBBut AB² = 200 - 200 cos(theta )But unless we can relate theta to m, this doesn't help.Alternatively, use the expressions for MB and MC from the quadratic equations earlier.Recall that for point B, we had:m² + (4/5)m d1 + d1² = 100Similarly, for point C:m² - (4/5)m d2 + d2² = 100Where d1 = MB, d2 = MC.Let’s denote d1 = MB, d2 = MC.Therefore, we have two equations:d1² + (4/5)m d1 + m² -100 =0d2² - (4/5)m d2 + m² -100 =0These are quadratic equations in d1 and d2.The solutions for d1 and d2 are:For d1:d1 = [ - (4/5)m ± sqrt( (16/25)m² - 4*1*(m² -100 ) ) ] /2= [ - (4m/5 ) ± sqrt( (16m²/25 ) -4m² + 400 ) ] /2= [ -4m/5 ± sqrt( -84m²/25 + 400 ) ] /2Similarly, for d2:d2 = [4m/5 ± sqrt( -84m²/25 + 400 ) ] /2Since d1 and d2 are positive, we take the positive roots:d1 = [ -4m/5 + sqrt(400 -84m²/25 ) ] /2d2 = [4m/5 + sqrt(400 -84m²/25 ) ] /2Simplify sqrt(400 -84m²/25 ):sqrt( (10000 -84m² ) /25 ) = sqrt(10000 -84m² ) /5Thus,d1 = [ -4m/5 + sqrt(10000 -84m² ) /5 ] /2 = [ sqrt(10000 -84m² ) -4m ] /10Similarly,d2 = [4m/5 + sqrt(10000 -84m² ) /5 ] /2 = [ sqrt(10000 -84m² ) +4m ] /10Therefore, MB = d1 = [ sqrt(10000 -84m² ) -4m ] /10MC = d2 = [ sqrt(10000 -84m² ) +4m ] /10Now, compute BC using the Law of Cosines in triangle BMC:BC² = MB²]

🔑:Answer: 8.## Solution:Consider point B_{1}, which is symmetric to point B with respect to the line O A. It also lies on the circle and angle A M B=alpha. Notice that points B_{1}, M, C lie on the same straight line, and Delta B B_{1} M is isosceles. Therefore, the inscribed angle angle B B_{1} M=90^{circ}-alpha, and the central angle angle B O C=180^{circ}-2 alpha. triangle B O C is isosceles and angle O B C=alpha. We find the base of the isosceles triangle using the formula B C=2 cdot cos alpha cdot B O=2 cdot frac{2}{5} cdot 10=8

🧠:[<think>Okay, so I have this problem here with vectors m and n, and a function f(x) defined as their dot product. The questions are about finding where f(x) is monotonically increasing and then determining its range over a specific interval. Let me start by understanding what's given.First, vectors m and n are given in component form. Vector m is (sqrt(3) cos x, cos x), and vector n is (sin x, cos x). The function f(x) is their dot product. Remember, the dot product of two vectors (a, b) and (c, d) is ac + bd. So, I need to compute that for m and n.Let me write that out:f(x) = m ⋅ n = (sqrt(3) cos x)(sin x) + (cos x)(cos x)Simplifying that:= sqrt(3) cos x sin x + cos²xHmm. That's the function. For part (1), I need to find where this function is monotonically increasing. To determine where a function is increasing, I should compute its derivative and find where the derivative is non-negative.So, first step, find f'(x). Let's differentiate term by term.First term: sqrt(3) cos x sin xUsing the product rule: derivative of sqrt(3) cos x is -sqrt(3) sin x, times sin x, plus sqrt(3) cos x times derivative of sin x, which is cos x.Wait, but sqrt(3) is a constant, so the derivative is sqrt(3) times derivative of cos x sin x.Alternatively, maybe use trigonometric identities to simplify first. Let me see.The term sqrt(3) cos x sin x can be written as (sqrt(3)/2) sin(2x), since sin(2x) = 2 sin x cos x. So that term becomes (sqrt(3)/2) sin(2x).Similarly, the second term is cos²x, which can be written as (1 + cos 2x)/2 using the identity cos²x = (1 + cos 2x)/2.So rewriting f(x):f(x) = (sqrt(3)/2) sin(2x) + (1/2)(1 + cos 2x)This simplifies to:= (sqrt(3)/2) sin(2x) + 1/2 + (1/2) cos(2x)That might make differentiation easier. Let's compute f'(x):Derivative of (sqrt(3)/2) sin(2x) is (sqrt(3)/2)*2 cos(2x) = sqrt(3) cos(2x)Derivative of 1/2 is 0.Derivative of (1/2) cos(2x) is (1/2)*(-2 sin(2x)) = -sin(2x)So f'(x) = sqrt(3) cos(2x) - sin(2x)Therefore, f'(x) = sqrt(3) cos(2x) - sin(2x)To find where f(x) is monotonically increasing, we need f'(x) ≥ 0.So, solve sqrt(3) cos(2x) - sin(2x) ≥ 0.This looks like a linear combination of sin and cos. Maybe we can write this as a single trigonometric function using the formula A cos θ + B sin θ = C cos(θ - φ) or something like that.Let me recall that a cos θ + b sin θ can be written as R cos(θ - φ), where R = sqrt(a² + b²) and tan φ = b/a.Wait, but here we have sqrt(3) cos(2x) - sin(2x). Let's write this as R cos(2x + φ) or R sin(2x + φ). Let's see.Let me write sqrt(3) cos(2x) - sin(2x) as R cos(2x + φ). Let's find R and φ.The formula for combining into a cosine is:a cos θ + b sin θ = R cos(θ - φ), where R = sqrt(a² + b²), and tan φ = b/a.But here, it's sqrt(3) cos(2x) - sin(2x). So, a = sqrt(3), b = -1.Therefore, R = sqrt( (sqrt(3))² + (-1)^2 ) = sqrt(3 + 1) = sqrt(4) = 2.Then tan φ = b/a = (-1)/sqrt(3) = -1/sqrt(3). So φ is the angle whose tangent is -1/sqrt(3). The reference angle is 30 degrees, but since tangent is negative, φ is in the fourth quadrant. So φ = -30 degrees or 330 degrees, but in radians, that's -π/6 or 11π/6.But since we can represent angles with different co-terminal angles, let's just take φ = -π/6.Therefore, sqrt(3) cos(2x) - sin(2x) = 2 cos(2x - (-π/6)) = 2 cos(2x + π/6). Wait, wait, let me check the formula again.Wait, the formula is a cos θ + b sin θ = R cos(θ - φ), where φ is such that cos φ = a/R and sin φ = b/R.In our case, θ is 2x, a = sqrt(3), b = -1.Therefore, cos φ = a/R = sqrt(3)/2, and sin φ = b/R = -1/2.Therefore, φ is the angle with cos φ = sqrt(3)/2 and sin φ = -1/2. That corresponds to φ = -π/6 (or 11π/6).Therefore, sqrt(3) cos(2x) - sin(2x) = 2 cos(2x - (-π/6)) = 2 cos(2x + π/6).Wait, 2x - φ, where φ is -π/6, so it's 2x - (-π/6) = 2x + π/6. So yes, it's 2 cos(2x + π/6).Therefore, f'(x) = 2 cos(2x + π/6).We need to find where f'(x) ≥ 0, which is equivalent to cos(2x + π/6) ≥ 0.The inequality cos(θ) ≥ 0 holds when θ is in [-π/2 + 2πk, π/2 + 2πk] for integers k.So, we have:cos(2x + π/6) ≥ 0Which implies:-π/2 + 2πk ≤ 2x + π/6 ≤ π/2 + 2πk, for some integer k.Let's solve for x:Subtract π/6:-π/2 - π/6 + 2πk ≤ 2x ≤ π/2 - π/6 + 2πkCompute -π/2 - π/6:Convert to sixths: -3π/6 - π/6 = -4π/6 = -2π/3Similarly, π/2 - π/6 = 3π/6 - π/6 = 2π/6 = π/3Therefore:-2π/3 + 2πk ≤ 2x ≤ π/3 + 2πkDivide by 2:-π/3 + πk ≤ x ≤ π/6 + πkSo the solution is all real numbers x such that x is in [-π/3 + πk, π/6 + πk] for some integer k.Therefore, the intervals where f'(x) ≥ 0 are the intervals [-π/3 + πk, π/6 + πk] for integers k.But the problem just asks for the interval where f(x) is monotonically increasing. However, trigonometric functions are periodic, so they have infinitely many intervals where they are increasing. But perhaps the question is expecting the intervals within one period? Wait, but the problem doesn't specify a particular interval for part (1). It just says "the interval where the function f(x) is monotonically increasing." Hmm.Wait, maybe it's expecting all intervals in terms of k? But in many calculus problems, unless specified, they might expect the general solution. Let me check the problem again."Find the interval where the function f(x) is monotonically increasing."The problem says "the interval", which might be ambiguous. But given that trigonometric functions have periodic intervals where they increase and decrease, it's likely that they want the general form of the intervals. So, x ∈ [-π/3 + πk, π/6 + πk] for any integer k.But perhaps in the original problem, written in Chinese, maybe the wording suggests a specific interval? But the user has translated it as "the interval", so maybe the answer is the union of all intervals where it's increasing. But since it's periodic, they might just write it in terms of k.Alternatively, maybe in the original problem's context, x is considered over all real numbers, so the answer would be the intervals as mentioned.But let me check with another approach. Let's consider the derivative f'(x) = 2 cos(2x + π/6). The function f'(x) is positive when cos(2x + π/6) ≥ 0.The cosine function is positive in the intervals where its argument is between -π/2 + 2πk and π/2 + 2πk for integers k. So:-π/2 + 2πk ≤ 2x + π/6 ≤ π/2 + 2πkSolving for x:Subtract π/6:-π/2 - π/6 + 2πk ≤ 2x ≤ π/2 - π/6 + 2πkWhich is:-2π/3 + 2πk ≤ 2x ≤ π/3 + 2πkDivide by 2:-π/3 + πk ≤ x ≤ π/6 + πkSo yes, same result. Therefore, the function f(x) is increasing on each interval [-π/3 + πk, π/6 + πk] for integers k. So unless there is a specific interval given, this is the answer.But the problem says "the interval", singular. Maybe the problem is translated from Chinese, and the original might have meant intervals, but translated as interval. Alternatively, maybe the answer is expected to be expressed in terms of k.But in the problem statement for part (1), since no specific interval is given, I think we need to express the intervals where f is increasing as [-π/3 + πk, π/6 + πk] for all integers k. However, maybe in the original problem, there was a specific interval, but the user didn't mention it. Wait, no, the user copied the problem as given.Therefore, part (1) answer is the intervals where x is between -π/3 + πk and π/6 + πk for integers k.But perhaps in the answer, they want it written in terms of intervals without the k, but since trigonometric functions are periodic, it's standard to write with k. Hmm. Let me check another way.Alternatively, maybe the problem is expecting the principal interval or the interval within a certain period. But since no information is given, it's safer to present the general solution with k.Alternatively, maybe there is a miscalculation here. Let me verify f'(x):Original f(x) = sqrt(3) cosx sinx + cos²x.First term derivative: sqrt(3)*( -sinx sinx + cosx cosx ) by product rule? Wait, hold on, maybe I made a mistake earlier.Wait, the first term is sqrt(3) cosx sinx. The derivative should be sqrt(3)*( d/dx [cosx sinx] ). Which is sqrt(3)*( -sinx sinx + cosx cosx ) = sqrt(3)*( cos²x - sin²x ). Then the derivative of the second term, cos²x, is 2 cosx (-sinx) = -2 sinx cosx.Therefore, f'(x) = sqrt(3)(cos²x - sin²x) - 2 sinx cosx.Wait, this contradicts my previous result. Wait, so now I have a different expression for f'(x). Which one is correct?Wait, let's recast f(x) = sqrt(3) cosx sinx + cos²x.Compute derivative term by term:First term: d/dx [sqrt(3) cosx sinx] = sqrt(3) [ -sinx sinx + cosx cosx ] = sqrt(3)(cos²x - sin²x).Second term: d/dx [cos²x] = 2 cosx (-sinx) = -2 sinx cosx.Therefore, f'(x) = sqrt(3)(cos²x - sin²x) - 2 sinx cosx.Alternatively, factor this expression. Let's see.We can write cos²x - sin²x as cos(2x), and 2 sinx cosx as sin(2x). Therefore:sqrt(3)(cos(2x)) - sin(2x). Which is the same as before.Yes, because cos²x - sin²x = cos(2x), and 2 sinx cosx = sin(2x). So sqrt(3) cos(2x) - sin(2x), which is same as before. So previous calculation was correct. Then, converting to 2 cos(2x + π/6). Therefore, that's correct. So the derivative is 2 cos(2x + π/6). Therefore, my previous steps were correct.Therefore, f'(x) = 2 cos(2x + π/6). So f'(x) ≥ 0 when cos(2x + π/6) ≥ 0, which gives intervals as before. So the answer for part (1) is x ∈ [-π/3 + kπ, π/6 + kπ] for all integers k.But in the problem statement, if they are expecting a specific interval, but since none is given, the answer must be stated in terms of these intervals with k. However, in Chinese textbooks, sometimes they might expect the general solution expressed in Chinese mathematical notation, but since we are to present it here, using k as an integer is acceptable.Therefore, part (1) answer is the intervals [-π/3 + kπ, π/6 + kπ] for all integers k.Moving on to part (2): If x ∈ [-π/8, π/2], determine the range of f(x).So, we need to find the maximum and minimum values of f(x) over the interval x ∈ [-π/8, π/2].First, let's recall f(x) = sqrt(3) cosx sinx + cos²x.Alternatively, we can write this in terms of sin(2x) and cos(2x) as we did before:f(x) = (sqrt(3)/2) sin(2x) + (1 + cos(2x))/2= (sqrt(3)/2) sin(2x) + 1/2 + (1/2) cos(2x)Alternatively, this can be written as:1/2 + (sqrt(3)/2 sin(2x) + 1/2 cos(2x))Which is 1/2 + [ sin(2x) cos(π/6) + cos(2x) sin(π/6) ] because cos(π/6) = sqrt(3)/2 and sin(π/6) = 1/2.Therefore, using the sine addition formula:sin(A + B) = sinA cosB + cosA sinB.Therefore, sqrt(3)/2 sin(2x) + 1/2 cos(2x) = sin(2x + π/6).Therefore, f(x) = 1/2 + sin(2x + π/6)So, f(x) = 1/2 + sin(2x + π/6)That's a much simpler expression! So this is helpful because now we can analyze the range more easily.So, the function is 1/2 plus a sine function. The sine function varies between -1 and 1, so f(x) varies between 1/2 -1 = -1/2 and 1/2 + 1 = 3/2. But since we have a composite argument 2x + π/6, and x is restricted to [-π/8, π/2], we need to find the range over this interval.First, let's find the range of the argument 2x + π/6 when x is in [-π/8, π/2].Compute the minimum and maximum of 2x + π/6:When x = -π/8: 2*(-π/8) + π/6 = -π/4 + π/6 = (-3π/12 + 2π/12) = -π/12When x = π/2: 2*(π/2) + π/6 = π + π/6 = 7π/6Therefore, the argument 2x + π/6 ranges from -π/12 to 7π/6.Therefore, sin(2x + π/6) ranges from sin(-π/12) up to sin(7π/6), but we need to check the maximum and minimum in this interval.But actually, the sine function's maximum and minimum occur either at critical points or endpoints. So let's analyze the interval θ ∈ [-π/12, 7π/6].We can plot this interval: from -π/12 (which is -15 degrees) to 7π/6 (which is 210 degrees). So this is a little more than a full period (since 2π is 360 degrees, and 210 - (-15) = 225 degrees). So within this interval, the sine function will reach its maximum at π/2 and minimum at 3π/2, provided those points are within the interval.Check if π/2 is within [-π/12, 7π/6]. π/2 is 3π/6, which is less than 7π/6. So π/2 is within the interval. Similarly, 3π/2 is 9π/6, which is more than 7π/6, so 3π/2 is not in the interval. The minimum in the interval would then be at 7π/6, since sin(7π/6) = -1/2. Wait, let me check.Wait, sin(-π/12) = sin(-15°) = -sin(15°) ≈ -0.2588.Then, the sine function increases from -π/12 to π/2, reaching sin(π/2) = 1. Then decreases from π/2 to 7π/6. At 7π/6, sin(7π/6) = -1/2.Therefore, the maximum value of sin(θ) in θ ∈ [-π/12, 7π/6] is 1 (at θ = π/2), and the minimum is -1/2 (at θ = 7π/6).Wait, but wait, let's check if there is a lower value between θ = -π/12 and θ = 7π/6. Since between -π/12 and 7π/6, the sine function starts at sin(-π/12) ≈ -0.2588, increases to 1 at π/2, then decreases to sin(π) = 0, then to sin(7π/6) = -1/2. So the minimum is indeed -1/2 at 7π/6, and maximum 1 at π/2.Therefore, sin(2x + π/6) ranges from -1/2 to 1 in this interval.Therefore, f(x) = 1/2 + sin(2x + π/6) ranges from 1/2 + (-1/2) = 0 to 1/2 + 1 = 3/2.But wait, let me confirm if the minimum is indeed achieved at x = π/2, and the maximum at some x where 2x + π/6 = π/2.Wait, if θ = 2x + π/6, then θ = π/2 when 2x + π/6 = π/2 => 2x = π/2 - π/6 = π/3 => x = π/6.Similarly, θ = 7π/6 when x = π/2.θ = -π/12 when x = -π/8.But sin(-π/12) ≈ -0.2588, which is greater than -1/2. So the minimum of sinθ in θ ∈ [-π/12,7π/6] is -1/2 at θ =7π/6, and maximum 1 at θ=π/2.Therefore, f(x) ranges from 0 to 3/2. However, let's check if there's a critical point inside the interval that might give a higher or lower value.Wait, since f(x) is 1/2 + sin(2x + π/6), the derivative of f(x) is 2 cos(2x + π/6), which we found earlier. Critical points occur when cos(2x + π/6) = 0, i.e., when 2x + π/6 = π/2 + πk, which gives x = (π/2 - π/6)/2 + πk/2 = (π/3)/2 + πk/2 = π/6 + πk/2.So critical points in the interval x ∈ [-π/8, π/2]:Let's find k such that x = π/6 + πk/2 is within [-π/8, π/2].Try k=0: x=π/6 ≈0.523, which is less than π/2≈1.571, so it's inside.k=1: x=π/6 + π/2= (π/6 + 3π/6)=4π/6=2π/3≈2.094, which is greater than π/2, so outside.k=-1: x=π/6 - π/2= -π/3≈-1.047, which is less than -π/8≈-0.392, so outside.Therefore, the only critical point inside the interval is x=π/6.Therefore, the maximum and minimum of f(x) occur either at the endpoints x=-π/8, x=π/2, or at the critical point x=π/6.Therefore, let's compute f(x) at these three points.First, x = -π/8:Compute θ = 2*(-π/8) + π/6 = -π/4 + π/6 = (-3π/12 + 2π/12) = -π/12So f(-π/8) = 1/2 + sin(-π/12) = 1/2 - sin(π/12)Sin(π/12) is sin(15°) = (√6 - √2)/4 ≈0.2588Therefore, f(-π/8) ≈1/2 -0.2588≈0.2412. Exact value is 1/2 - (√6 - √2)/4.But let's compute it exactly:sin(π/12) = (√6 - √2)/4, so sin(-π/12)= - (√6 - √2)/4Thus f(-π/8)=1/2 - (√6 - √2)/4= [2 - (√6 - √2)] /4= [2 - √6 + √2]/4.Alternatively, leave it as 1/2 + sin(-π/12) = 1/2 - sin(π/12).But maybe we can compute exact value.But let's check at x=π/6:θ=2*(π/6) + π/6= π/3 + π/6= π/2, so sin(π/2)=1.Thus f(π/6)=1/2 +1=3/2.At x=π/2:θ=2*(π/2)+π/6= π + π/6=7π/6, sin(7π/6)= -1/2.Thus f(π/2)=1/2 + (-1/2)=0.Therefore, the function attains its maximum at x=π/6 (3/2) and minimum at x=π/2 (0). Wait, but at x=-π/8, we have f(-π/8)=1/2 - sin(π/12). Let's compute sin(π/12):sin(π/12)=sin(15°)=√( (1 - cos(30°))/2 )=√( (1 - √3/2)/2 )=√( (2 - √3)/4 )= (√(2 - √3))/2 ≈0.2588Therefore, f(-π/8)=1/2 -0.2588≈0.2412, which is greater than 0. So the minimum is at x=π/2 (0) and maximum at x=π/6 (3/2). But we should verify if there's any other extremum, but since the only critical point in the interval is x=π/6, which is a maximum, and the endpoints give 0.2412 and 0, so 0 is the minimum.Therefore, the range of f(x) over [-π/8, π/2] is [0, 3/2].But let me confirm if there's any mistake here.Wait, but when x ranges from -π/8 to π/2, θ=2x + π/6 ranges from -π/12 to 7π/6. The sine function over this interval reaches maximum at π/2 (which is inside the interval) and minimum at 7π/6 (which is the endpoint x=π/2). Therefore, sin(theta) ranges from -1/2 to 1. Hence, f(x)=1/2 + sin(theta) ranges from 0 to 3/2. The other endpoint at x=-π/8 gives sin(theta)=sin(-π/12)≈-0.2588, so f(x)=1/2 -0.2588≈0.2412, which is greater than 0, so indeed the minimum is 0 at x=π/2, and the maximum is 3/2 at x=π/6.Thus, the range of f(x) is [0, 3/2].But just to make sure, let's check if f(x) can be lower than 0 in this interval. Since sin(theta) ≥ -1/2, so 1/2 + (-1/2)=0. So the minimum is exactly 0. The maximum is 1/2 +1=3/2. So yes, the range is [0, 3/2].Therefore, the answer to part (2) is [0, 3/2].But let me verify once again.Given f(x) = 1/2 + sin(2x + π/6). The variable substitution θ = 2x + π/6, x ∈ [-π/8, π/2] gives θ ∈ [-π/12, 7π/6]. The sine function on this interval attains maximum 1 at θ=π/2 and minimum -1/2 at θ=7π/6. Therefore, f(x) ∈ [1/2 -1/2, 1/2 +1] = [0, 3/2]. Hence, the range is [0, 3/2].Therefore, the answers are:1. The function f(x) is monotonically increasing on intervals [-π/3 + kπ, π/6 + kπ] for all integers k.2. The range of f(x) over [-π/8, π/2] is [0, 3/2].Final Answer(1) The function ( f(x) ) is monotonically increasing on the intervals ( boxed{left[ -frac{pi}{3} + kpi, frac{pi}{6} + kpi right]} ) for all integers ( k ).(2) The range of ( f(x) ) when ( x in left[ -frac{pi}{8}, frac{pi}{2} right] ) is ( boxed{left[ 0, frac{3}{2} right]} ).</think>Given the vectors (overrightarrow{m} = (sqrt{3} cos x, cos x)) and (overrightarrow{n} = (sin x, cos x)), the function (f(x)) is their dot product:[f(x) = overrightarrow{m} cdot overrightarrow{n} = sqrt{3} cos x sin x + cos^2 x] (1) Finding the interval where (f(x)) is monotonically increasingFirst, we compute the derivative (f'(x)):[f(x) = sqrt{3} cos x sin x + cos^2 x = frac{sqrt{3}}{2} sin(2x) + frac{1}{2} (1 + cos(2x))]Simplifying, we get:[f(x) = frac{sqrt{3}}{2} sin(2x) + frac{1}{2} + frac{1}{2} cos(2x)]Differentiating term by term:[f'(x) = sqrt{3} cos(2x) - sin(2x)]Rewriting using trigonometric identities:[f'(x) = 2 cos(2x + frac{pi}{6})]To find where (f'(x) geq 0):[cos(2x + frac{pi}{6}) geq 0]This holds when:[-frac{pi}{2} + 2kpi leq 2x + frac{pi}{6} leq frac{pi}{2} + 2kpi]Solving for (x):[-frac{pi}{3} + kpi leq x leq frac{pi}{6} + kpi quad text{for all integers } k] (2) Determining the range of (f(x)) when (x in left[ -frac{pi}{8}, frac{pi}{2} right])Rewriting (f(x)):[f(x) = frac{1}{2} + sinleft(2x + frac{pi}{6}right)]The argument (2x + frac{pi}{6}) ranges from (-frac{pi}{12}) to (frac{7pi}{6}) as (x) varies from (-frac{pi}{8}) to (frac{pi}{2}). The sine function over this interval reaches a maximum of 1 at (frac{pi}{2}) and a minimum of (-frac{1}{2}) at (frac{7pi}{6}). Therefore:[sinleft(2x + frac{pi}{6}right) in left[-frac{1}{2}, 1right]]Thus, the range of (f(x)) is:[left[frac{1}{2} - frac{1}{2}, frac{1}{2} + 1right] = left[0, frac{3}{2}right]] Final Answers(1) The function ( f(x) ) is monotonically increasing on the intervals (boxed{left[ -frac{pi}{3} + kpi, frac{pi}{6} + kpi right]}) for all integers ( k ).(2) The range of ( f(x) ) when ( x in left[ -frac{pi}{8}, frac{pi}{2} right] ) is (boxed{left[ 0, frac{3}{2} right]}).,<think>Okay, let me try to work through this problem step by step. So, we have two vectors, m and n, each with two components. The function f(x) is their dot product. Part (1) asks for the interval where f(x) is monotonically increasing, and part (2) wants the range of f(x) when x is between -π/8 and π/2.Starting with part (1). First, I need to find the dot product of vectors m and n. The dot product is just the sum of the products of their corresponding components. So, for vectors m = (√3 cos x, cos x) and n = (sin x, cos x), the dot product f(x) should be:f(x) = (√3 cos x)(sin x) + (cos x)(cos x)Let me compute that:First component product: √3 cos x * sin xSecond component product: cos x * cos x = cos²xSo f(x) = √3 cos x sin x + cos²xHmm. Maybe I can simplify this expression using trigonometric identities. Let me recall some identities. The term √3 cos x sin x reminds me of the double-angle identity. Because sin(2x) = 2 sin x cos x, so sin x cos x = (1/2) sin 2x. But here we have √3 sin x cos x, so that would be (√3/2) sin 2x.Then, the other term is cos²x. The identity for cos²x is (1 + cos 2x)/2. So putting it all together:f(x) = √3 sin x cos x + cos²x = (√3/2) sin 2x + (1/2)(1 + cos 2x)So combining these:f(x) = (√3/2) sin 2x + (1/2) cos 2x + 1/2Hmm, that's better. Now, the first two terms, (√3/2 sin 2x + 1/2 cos 2x), look like they can be written as a single sine or cosine function. Because they are of the form A sin θ + B cos θ, which can be written as C sin(θ + φ) or C cos(θ + φ), where C = √(A² + B²) and φ is some phase shift.Let me compute that. Let me let A = √3/2 and B = 1/2. Then C = √[(√3/2)^2 + (1/2)^2] = √[(3/4) + (1/4)] = √[1] = 1.So then, A = √3/2 and B = 1/2, which corresponds to sin(2x + φ). Wait, actually, the identity is A sin θ + B cos θ = C sin(θ + φ) or something similar. Let me recall.Alternatively, we can write it as sin(2x + φ) or cos(2x - φ). Let me use the identity:A sin θ + B cos θ = C sin(θ + φ), where C = √(A² + B²) and tan φ = B/A.Wait, no, wait. Let's check:Let me recall that:A sin θ + B cos θ = C sin(θ + φ)Expanding the right side: C sin θ cos φ + C cos θ sin φComparing coefficients:A = C cos φB = C sin φTherefore, tan φ = B/A. So in our case, A is the coefficient of sin θ, which is √3/2, and B is the coefficient of cos θ, which is 1/2. So tan φ = (1/2)/(√3/2) = 1/√3, so φ = π/6.Therefore, √3/2 sin 2x + 1/2 cos 2x = sin(2x + π/6). Because C = 1, as we computed earlier.So then, f(x) = sin(2x + π/6) + 1/2Wait, let me verify this. Let me check:sin(2x + π/6) = sin 2x cos π/6 + cos 2x sin π/6Which is sin 2x*(√3/2) + cos 2x*(1/2)Which is exactly √3/2 sin 2x + 1/2 cos 2x. Perfect! So that part is correct.Therefore, f(x) = sin(2x + π/6) + 1/2So now, to find where f(x) is monotonically increasing, we need to find where its derivative f’(x) is positive.Let me compute the derivative of f(x):f’(x) = d/dx [sin(2x + π/6) + 1/2] = 2 cos(2x + π/6) * (derivative of the inside function, which is 2x + π/6, so the derivative is 2)Wait, actually, derivative of sin(u) is cos(u)*u', so here u = 2x + π/6, so u’ = 2. Therefore:f’(x) = 2 cos(2x + π/6)So f’(x) = 2 cos(2x + π/6). We need to find the intervals where f’(x) > 0, which is equivalent to cos(2x + π/6) > 0.So cos(2x + π/6) > 0. The cosine function is positive in the intervals (-π/2 + 2πk, π/2 + 2πk) for any integer k.Therefore, solving the inequality:-π/2 + 2πk < 2x + π/6 < π/2 + 2πkSubtract π/6 from all parts:-π/2 - π/6 + 2πk < 2x < π/2 - π/6 + 2πkCompute -π/2 - π/6:Convert to sixths: -3π/6 - π/6 = -4π/6 = -2π/3Similarly, π/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 + πkTherefore, for each integer k, the function f(x) is increasing on the intervals (-π/3 + πk, π/6 + πk)But the question just says "the interval where the function f(x) is monotonically increasing." However, since trigonometric functions are periodic, f(x) is increasing on infinitely many intervals. But perhaps they are asking for the general form or specific intervals within a certain domain? Wait, the problem doesn't specify a particular interval for part (1), so maybe the answer is all intervals of the form (-π/3 + πk, π/6 + πk) for integer k. But maybe we need to write it in terms of the general solution?Wait, maybe the question expects the answer in terms of the principal intervals where the function is increasing. Let me check. Since the original function f(x) is defined for all real numbers (since cosine and sine are defined everywhere), but the problem doesn't restrict x to any interval for part (1). So the answer would be that f(x) is increasing on each interval (-π/3 + πk, π/6 + πk) where k is any integer.But in the problem statement, part (1) says "the interval where the function f(x) is monotonically increasing." The wording is a bit ambiguous. Maybe in Chinese, it's more precise. But assuming that the problem is expecting the intervals in terms of a general solution, or perhaps the principal interval. Wait, but since the derivative is periodic with period π (since f’(x) = 2 cos(2x + π/6), which has period π), the function f(x) has period π as well. So the intervals of increase will repeat every π units. Therefore, the function is increasing on intervals of the form (-π/3 + πk, π/6 + πk) for integer k.But perhaps the answer should be presented as the union of all such intervals. However, usually, for such questions, they expect the general interval in terms of k. Let me check standard practice. For example, when asked where a trigonometric function is increasing, you might state the intervals with the periodicity. So, I think the answer for part (1) is that f(x) is monotonically increasing on intervals (-π/3 + kπ, π/6 + kπ) for all integers k.But let me verify this. Let's take k = 0: (-π/3, π/6). Then for k = 1: (-π/3 + π, π/6 + π) = (2π/3, 7π/6). But these are all the intervals where the function is increasing.Alternatively, maybe the problem expects a specific interval where the function is increasing, but since it's periodic, it's increasing on infinitely many intervals. So I think the answer is as above. But let me check once again.Alternatively, maybe I need to express the solution as x ∈ (-π/3 + kπ, π/6 + kπ), k ∈ Z.Yes, that's the standard way to present such solutions. So part (1) is solved.Now moving on to part (2). We need to determine the range of f(x) when x ∈ [-π/8, π/2].First, recall that f(x) = sin(2x + π/6) + 1/2.So to find the range, we can analyze the range of sin(2x + π/6) over the given interval, then add 1/2.Let me first determine the interval for 2x + π/6 when x ∈ [-π/8, π/2].Compute the lower bound: 2*(-π/8) + π/6 = -π/4 + π/6 = (-3π + 2π)/12 = (-π)/12Compute the upper bound: 2*(π/2) + π/6 = π + π/6 = 7π/6Therefore, the argument 2x + π/6 ranges from -π/12 to 7π/6.So we need to find the maximum and minimum values of sin(θ) where θ goes from -π/12 to 7π/6.The sine function reaches its maximum at π/2 and minimum at -1 at 3π/2, but depending on the interval.Let me sketch the sine curve from θ = -π/12 to θ = 7π/6.First, -π/12 is approximately -0.2618 radians, which is in the fourth quadrant. The sine of -π/12 is -sin(π/12) ≈ -0.2588.Then, as θ increases from -π/12 to 7π/6, the sine function will go from -π/12 (approx -0.2618) to 0 at θ=0, then increase to 1 at θ=π/2 (approx 1.5708), then decrease to 0 at θ=π (3.1416), then go to -1 at θ=3π/2 (approx 4.7124), and then increase back to 0 at θ=2π (6.2832), and then continue to increase until θ=7π/6 ≈ 3.6652 (wait, 7π/6 is actually about 3.6652 radians). Wait, 7π/6 is π + π/6, so it's in the third quadrant. So the sine of 7π/6 is -1/2.Wait, let's note that:θ from -π/12 to 7π/6.So in this interval, the sine function will:- Start at sin(-π/12) ≈ -0.2588- Increase to sin(π/2) = 1 at θ=π/2- Then decrease to sin(π) = 0- Then further decrease to sin(3π/2) = -1- Then increase again, but our interval stops at θ=7π/6, which is π + π/6, where sin(7π/6) = -1/2Wait, wait, 7π/6 is π + π/6, which is 210 degrees. The sine of 7π/6 is indeed -1/2.But in the interval θ ∈ [-π/12, 7π/6], the maximum of sin θ is 1 (achieved at θ=π/2), and the minimum is -1 (achieved at θ=3π/2). However, we need to check if 3π/2 is within the interval.3π/2 is approximately 4.7124. The upper limit is 7π/6 ≈ 3.6652, which is less than 3π/2 (≈4.7124). So θ=3π/2 is outside of our interval. Therefore, the minimum in the interval [-π/12, 7π/6] is the minimum between -π/12 and 7π/6.Let's check where the minimum occurs. Since θ goes from -π/12 to 7π/6, which is from about -0.2618 to 3.6652 radians. The critical points for sin θ are at θ=π/2 (1.5708) and θ=3π/2 (4.7124). But 3π/2 is outside our interval, so within our interval, the critical point is at θ=π/2 (maximum) and the endpoints. But also, we need to check if the function attains a minimum somewhere else in the interval.From θ=-π/12 to θ=7π/6:From θ=-π/12 to θ=π/2: sine increases from sin(-π/12) to 1.From θ=π/2 to θ=7π/6: sine decreases from 1 to sin(7π/6) = -1/2.Wait, so in this interval, the maximum is 1 (at θ=π/2), and the minimum is at θ=7π/6, which is -1/2. Wait, but we started at θ=-π/12, which is sin(-π/12) ≈ -0.2588. So the minimum value of sin θ over the interval is -1/2 (since -1/2 is less than -0.2588). Therefore, the range of sin θ over θ ∈ [-π/12, 7π/6] is [-1/2, 1].Wait, let's confirm. When θ increases from -π/12 to π/2, sin θ goes from sin(-π/12) ≈ -0.2588 up to 1. Then, as θ continues to 7π/6, sin θ decreases from 1 to sin(7π/6) = -1/2. So the minimum in this interval is -1/2, and the maximum is 1. Therefore, sin θ ∈ [-1/2, 1].Therefore, the range of f(x) = sin θ + 1/2 is [-1/2 + 1/2, 1 + 1/2] = [0, 3/2].But wait, is that correct? Let's verify:If sin θ ranges from -1/2 to 1, then adding 1/2 to both gives:Lower bound: -1/2 + 1/2 = 0Upper bound: 1 + 1/2 = 3/2So the range of f(x) would be [0, 3/2].But let me check if there are any other critical points. For example, between θ=-π/12 and θ=7π/6, is there a point where sin θ is lower than -1/2?Wait, 7π/6 is 210 degrees, sin(7π/6) = -1/2. θ=3π/2 is 270 degrees, which is outside the interval. So in our interval, the minimum sin θ is -1/2. However, at θ=7π/6, that's the lowest point in the interval. So that seems correct.But let's check θ=-π/12. sin(-π/12) ≈ -0.2588, which is greater than -1/2. Therefore, in the interval θ ∈ [-π/12, 7π/6], the minimum value of sin θ is -1/2, achieved at θ=7π/6, and the maximum is 1, achieved at θ=π/2.Therefore, adding 1/2, the range of f(x) is from 0 to 3/2.But let me confirm this with another approach. Let me compute f(x) at critical points and endpoints.First, critical points occur where the derivative is zero, which is when cos(2x + π/6) = 0. So 2x + π/6 = π/2 + kπ → 2x = π/2 - π/6 + kπ = (3π/6 - π/6) + kπ = (2π/6) + kπ = π/3 + kπ → x = π/6 + kπ/2.So critical points are at x = π/6 + kπ/2. Let's find all critical points within x ∈ [-π/8, π/2].Let me compute for k=0: x=π/6 ≈ 0.5236, which is within [ -π/8 ≈ -0.3927, π/2 ≈ 1.5708].For k=1: x=π/6 + π/2 = (π/6 + 3π/6) = 4π/6 = 2π/3 ≈ 2.0944, which is outside the interval.For k=-1: x=π/6 - π/2 = (π/6 - 3π/6) = -2π/6 = -π/3 ≈ -1.0472, which is outside the interval.Therefore, the only critical point within x ∈ [-π/8, π/2] is x=π/6.Additionally, we need to check the endpoints x=-π/8 and x=π/2.So let's compute f(x) at x=-π/8, x=π/6, and x=π/2.First, x=-π/8:Compute f(-π/8) = sin(2*(-π/8) + π/6) + 1/2 = sin(-π/4 + π/6) + 1/2Compute -π/4 + π/6 = (-3π/12 + 2π/12) = -π/12sin(-π/12) ≈ -0.2588, so f(-π/8) ≈ -0.2588 + 0.5 ≈ 0.2412x=π/6:f(π/6) = sin(2*(π/6) + π/6) + 1/2 = sin(π/3 + π/6) + 1/2 = sin(π/2) + 1/2 = 1 + 1/2 = 3/2x=π/2:f(π/2) = sin(2*(π/2) + π/6) + 1/2 = sin(π + π/6) + 1/2 = sin(7π/6) + 1/2 = -1/2 + 1/2 = 0Therefore, the function f(x) attains its maximum at x=π/6 with value 3/2, and its minimum at x=π/2 with value 0.But wait, when we checked earlier, the minimum of sin θ was -1/2 at θ=7π/6 (which corresponds to x=π/2), and indeed f(π/2)=0, which is the minimum. The maximum is at x=π/6, which gives 3/2.But also, we should check if there are any other local maxima or minima within the interval. However, the only critical point in the interval is x=π/6, which we already evaluated. So between x=-π/8 and x=π/6, the function is increasing (since the derivative is positive there? Wait, need to check the derivative's sign in that interval.)Wait, earlier, for part (1), we found that f(x) is increasing on intervals (-π/3 + kπ, π/6 + kπ). For k=0, the interval is (-π/3, π/6). However, our interval for part (2) includes x from -π/8 to π/2. So in the interval x ∈ [-π/8, π/6], the function is increasing (since -π/8 ≈ -0.3927, which is greater than -π/3 ≈ -1.047), so within (-π/3, π/6), which is part of the increasing interval.Then, from x=π/6 to π/2, the function is decreasing, since the derivative becomes negative. Because f’(x) = 2 cos(2x + π/6). At x=π/6, 2x + π/6 = 2*(π/6) + π/6 = π/2 + π/6 = 2π/3. cos(2π/3) = -1/2, so the derivative is 2*(-1/2) = -1 < 0. Therefore, after x=π/6, the function decreases until x=π/2.Therefore, the maximum value is at x=π/6 (3/2), and the minimum is at x=π/2 (0). But we also have the value at x=-π/8, which is approximately 0.2412. Since the function is increasing from x=-π/8 to x=π/6, the value at x=-π/8 is the minimum in the interval [-π/8, π/6], but since the function continues decreasing after π/6 down to 0 at x=π/2, the overall minimum is 0 at x=π/2.Wait, but when x=-π/8, f(x) ≈0.2412, and when x=π/2, f(x)=0. So 0 is lower than 0.2412. Therefore, the minimum is 0, and the maximum is 3/2.Therefore, the range of f(x) over x ∈ [-π/8, π/2] is [0, 3/2].But let me check another point in between. For example, let me compute f(0):f(0) = sin(0 + π/6) + 1/2 = sin(π/6) + 1/2 = 1/2 + 1/2 = 1.Which is between 0.2412 and 3/2. So that makes sense. Since the function increases from x=-π/8 to x=π/6, reaching 3/2, then decreases to 0 at x=π/2.Therefore, the range is [0, 3/2].Therefore, the answer to part (2) is [0, 3/2].But let me verify once again.Another way to see this: since the function f(x) = sin(2x + π/6) + 1/2. When x is in [-π/8, π/2], 2x + π/6 is in [-π/12, 7π/6]. In this interval, sin(theta) reaches its maximum at theta=π/2 (sin(pi/2)=1) and minimum at theta=7pi/6 (sin(7pi/6)=-1/2). Therefore, sin(theta) in [-1/2, 1], so f(x) is in [-1/2 + 1/2, 1 + 1/2] = [0, 3/2]. So this confirms the earlier conclusion.Therefore, the range is [0, 3/2].Final Answer(1) The function ( f(x) ) is monotonically increasing on the intervals ( left( -frac{pi}{3} + kpi, frac{pi}{6} + kpi right) ) for any integer ( k ).(2) The range of ( f(x) ) when ( x in left[ -frac{pi}{8}, frac{pi}{2} right] ) is (boxed{left[ 0, frac{3}{2} right]}).(boxed{left( -frac{pi}{3} + kpi, frac{pi}{6} + kpi right)}) for part (1) and (boxed{left[ 0, frac{3}{2} right]}) for part (2).Final Answer(1) (boxed{left( -frac{pi}{3} + kpi, frac{pi}{6} + kpi right)}) for all integers ( k )(2) (boxed{left[ 0, frac{3}{2} right]})</think>Given vectors (overrightarrow{m} = (sqrt{3} cos x, cos x)) and (overrightarrow{n} = (sin x, cos x)), the function (f(x) = overrightarrow{m} cdot overrightarrow{n}) is computed as follows:[f(x) = sqrt{3} cos x sin x + cos^2 x]Using trigonometric identities, this simplifies to:[f(x) = frac{sqrt{3}}{2} sin 2x + frac{1}{2} cos 2x + frac{1}{2}]This can be further simplified using a phase shift identity:[f(x) = sinleft(2x + frac{pi}{6}right) + frac{1}{2}] Part (1): Interval where (f(x)) is monotonically increasingTo determine where (f(x)) is monotonically increasing, we compute the derivative:[f'(x) = 2 cosleft(2x + frac{pi}{6}right)]We need (f'(x) > 0), which translates to (cosleft(2x + frac{pi}{6}right) > 0). Solving this inequality:[-frac{pi}{2} + 2kpi < 2x + frac{pi}{6} < frac{pi}{2} + 2kpi]Subtracting (frac{pi}{6}) and dividing by 2:[-frac{pi}{3} + kpi < x < frac{pi}{6} + kpi quad text{for all integers } k]Thus, the function (f(x)) is monotonically increasing on the intervals (left( -frac{pi}{3} + kpi, frac{pi}{6} + kpi right)) for any integer (k). Part (2): Range of (f(x)) for (x in left[ -frac{pi}{8}, frac{pi}{2} right])First, we determine the interval for (2x + frac{pi}{6}) when (x in left[ -frac{pi}{8}, frac{pi}{2} right]):[2 left( -frac{pi}{8} right) + frac{pi}{6} = -frac{pi}{12} quad text{and} quad 2 left( frac{pi}{2} right) + frac{pi}{6} = frac{7pi}{6}]Thus, (2x + frac{pi}{6}) ranges from (-frac{pi}{12}) to (frac{7pi}{6}). The sine function over this interval reaches a maximum of 1 at (frac{pi}{2}) and a minimum of (-frac{1}{2}) at (frac{7pi}{6}). Therefore, (sinleft(2x + frac{pi}{6}right)) ranges from (-frac{1}{2}) to 1. Adding (frac{1}{2}) gives the range of (f(x)) as:[left[ -frac{1}{2} + frac{1}{2}, 1 + frac{1}{2} right] = left[ 0, frac{3}{2} right]] Final Answers(1) The function (f(x)) is monotonically increasing on the intervals (boxed{left( -frac{pi}{3} + kpi, frac{pi}{6} + kpi right)}) for all integers (k).(2) The range of (f(x)) when (x in left[ -frac{pi}{8}, frac{pi}{2} right]) is (boxed{left[ 0, frac{3}{2} right]}).]

🔑:To solve the given problem, we proceed as follows: Part (1): Monotonically Increasing IntervalFirst, we express f(x) in terms of trigonometric identities:[begin{aligned}f(x) &= overrightarrow{m} cdot overrightarrow{n} &= (sqrt{3}cos x, cos x) cdot (sin x, cos x) &= sqrt{3}sin x cos x + cos^2 x &= frac{sqrt{3}}{2}sin 2x + cos^2 x &= frac{sqrt{3}}{2}sin 2x + frac{cos 2x + 1}{2} &= frac{sqrt{3}}{2}sin 2x + frac{1}{2}cos 2x + frac{1}{2} &= sin (2x + frac{pi}{6}) + frac{1}{2}end{aligned}]To find the interval where f(x) is monotonically increasing, we consider the derivative of sin (2x + frac{pi}{6}) which is positive for -frac{pi}{2} + 2kpi leqslant 2x + frac{pi}{6} leqslant frac{pi}{2} + 2kpi, where k in mathbb{Z}. Simplifying this gives:[-frac{2pi}{3} + 2kpi leqslant 2x leqslant frac{pi}{3} + 2kpi]Dividing by 2:[-frac{pi}{3} + kpi leqslant x leqslant frac{pi}{6} + kpi]Therefore, the monotonically increasing interval of f(x) is:[boxed{[-frac{pi}{3} + kpi, frac{pi}{6} + kpi]}, text{ where } k in mathbb{Z}.] Part (2): Range of f(x) for x in [-frac{pi}{8}, frac{pi}{2}]Given f(x) = sin (2x + frac{pi}{6}) + frac{1}{2}, and x in [-frac{pi}{8}, frac{pi}{2}], we find the range of f(x):[2x in [-frac{pi}{4}, pi]]Thus,[2x + frac{pi}{6} in [-frac{pi}{12}, frac{7pi}{6}]]This implies:[sin (2x + frac{pi}{6}) in [-frac{1}{2}, 1]]Adding frac{1}{2} to the range of sin (2x + frac{pi}{6}):[sin (2x + frac{pi}{6}) + frac{1}{2} in [0, frac{3}{2}]]Therefore, the range of f(x) is:[boxed{[0, frac{3}{2}]}]

🧠:[<think>Okay, let's see. I have three equations to solve here, each with different methods specified. Let me take them one by one. Starting with the first equation: ( x^2 - 2x = 2 ). The problem says to use the completing the square method. Hmm, okay. I remember that completing the square involves turning a quadratic equation into a perfect square trinomial. Let me recall the steps. First, I need to move the constant term to the right side. Wait, in this case, the equation is already set up that way. The left side is ( x^2 - 2x ), and the right side is 2. So, the next step is to take the coefficient of x, which is -2, divide it by 2 to get -1, and then square it to get 1. That number, 1, is what we need to add to both sides to complete the square. So adding 1 to both sides: left side becomes ( x^2 - 2x + 1 ), which factors into ( (x - 1)^2 ), and the right side becomes 2 + 1 = 3. So now the equation is ( (x - 1)^2 = 3 ). Then, to solve for x, we take the square root of both sides. That gives us ( x - 1 = pm sqrt{3} ). Adding 1 to both sides, we get ( x = 1 pm sqrt{3} ). Let me check if that makes sense. Plugging x = 1 + sqrt(3) into the original equation: (1 + sqrt(3))^2 - 2*(1 + sqrt(3)) = 1 + 2 sqrt(3) + 3 - 2 - 2 sqrt(3) = 4 + 2 sqrt(3) - 2 - 2 sqrt(3) = 2. Yep, that works. Similarly for x = 1 - sqrt(3), it should also give 2. So that's good. Okay, moving on to the second equation: ( 2x^2 + 3 = 3x ). This one needs to be solved using the formula method, which I assume is the quadratic formula. First, I should rearrange the equation into standard form, which is ax^2 + bx + c = 0. So subtract 3x from both sides: ( 2x^2 - 3x + 3 = 0 ). Wait, is that right? Let me check: original equation is 2x² + 3 = 3x. Subtract 3x: 2x² - 3x + 3 = 0. Yes. So then a = 2, b = -3, c = 3. Quadratic formula is x = [-b ± sqrt(b² - 4ac)] / (2a). Plugging in the values: x = [3 ± sqrt( (-3)^2 - 4*2*3 ) ] / (2*2). Let's calculate the discriminant first: (-3)^2 is 9, and 4*2*3 is 24. So discriminant is 9 - 24 = -15. Hmm, a negative discriminant. That means the solutions will be complex numbers. So sqrt(-15) is i*sqrt(15). Therefore, the solutions are x = [3 ± i sqrt(15)] / 4. Let me verify. If I substitute x = [3 + i sqrt(15)] / 4 into the original equation: 2x² + 3 = 3x. Let's compute 2x²: 2*( [3 + i sqrt(15)] / 4 )². First square the term: (9 + 6i sqrt(15) + (i^2)(15)) / 16. Since i² = -1, that becomes (9 + 6i sqrt(15) -15)/16 = (-6 + 6i sqrt(15))/16. Multiply by 2: (-12 + 12i sqrt(15))/16 = (-3 + 3i sqrt(15))/4. Then add 3: (-3 + 3i sqrt(15))/4 + 12/4 = (9 + 3i sqrt(15))/4. On the other side, 3x: 3*(3 + i sqrt(15))/4 = (9 + 3i sqrt(15))/4. So yes, both sides equal. Similarly for the other root, x = [3 - i sqrt(15)]/4. So that checks out. Third equation: ( (x + 2)^2 - 3(x + 2) = 0 ). The problem says to use an appropriate method. Let me look at this. The equation has a common factor of (x + 2). So maybe factoring would work here. Let's try factoring. Factor out (x + 2): (x + 2)[(x + 2) - 3] = 0. Simplify inside the brackets: (x + 2 - 3) = (x - 1). So the equation becomes (x + 2)(x - 1) = 0. Therefore, the solutions are x + 2 = 0 or x - 1 = 0, so x = -2 or x = 1. Let me verify these solutions. For x = -2: substitute into original equation: (-2 + 2)^2 - 3*(-2 + 2) = (0)^2 - 3*(0) = 0 - 0 = 0. Correct. For x = 1: (1 + 2)^2 - 3*(1 + 2) = 9 - 9 = 0. Also correct. So that works. Alternatively, maybe expanding the equation first and then solving. Let's try that. Expanding (x + 2)^2: x² + 4x + 4. Then subtract 3(x + 2): 3x + 6. So the equation becomes x² + 4x + 4 - 3x - 6 = 0. Combine like terms: x² + x - 2 = 0. Then factor: looking for two numbers that multiply to -2 and add to +1. Those numbers are +2 and -1. So (x + 2)(x - 1) = 0. Same result. So either way, x = -2 or x = 1. So all three equations are solved. Let me just recap the answers to make sure I didn't mix up any steps.1. Completed the square for ( x^2 - 2x = 2 ), got x = 1 ± sqrt(3). 2. Quadratic formula on ( 2x^2 + 3 = 3x ), found complex solutions: [3 ± i sqrt(15)] / 4. 3. Factored ( (x + 2)^2 - 3(x + 2) = 0 ) to get x = -2 or x = 1. Everything seems to check out. No mistakes noticed in the steps.Final Answer(1) (boxed{1 pm sqrt{3}}) (2) (boxed{dfrac{3 pm isqrt{15}}{4}}) (3) (boxed{-2}) and (boxed{1})</think> Summary of Solutions1. Equation: ( x^2 - 2x = 2 ) (using completing the square method) - Start with the equation ( x^2 - 2x = 2 ). - Complete the square by adding 1 to both sides: ( x^2 - 2x + 1 = 3 ). - Factor the left side: ( (x - 1)^2 = 3 ). - Take the square root of both sides: ( x - 1 = pm sqrt{3} ). - Solve for ( x ): ( x = 1 pm sqrt{3} ). [ boxed{1 pm sqrt{3}} ]2. Equation: ( 2x^2 + 3 = 3x ) (using the quadratic formula) - Rearrange into standard form: ( 2x^2 - 3x + 3 = 0 ). - Identify coefficients: ( a = 2 ), ( b = -3 ), ( c = 3 ). - Apply the quadratic formula: ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ). - Calculate the discriminant: ( (-3)^2 - 4(2)(3) = 9 - 24 = -15 ). - Solve for ( x ): ( x = frac{3 pm isqrt{15}}{4} ). [ boxed{dfrac{3 pm isqrt{15}}{4}} ]3. Equation: ( (x + 2)^2 - 3(x + 2) = 0 ) (using factoring) - Factor out ( (x + 2) ): ( (x + 2)((x + 2) - 3) = 0 ). - Simplify the second factor: ( (x + 2)(x - 1) = 0 ). - Solve for ( x ): ( x + 2 = 0 ) or ( x - 1 = 0 ). - Solutions: ( x = -2 ) or ( x = 1 ). [ boxed{-2} quad text{and} quad boxed{1} ],<think>Alright, let's tackle these three equations one by one. I need to solve each using the specified methods. Let me start with the first one.Problem (1): Solve ( x^2 - 2x = 2 ) using the completing the square method.Okay, completing the square. I remember the idea is to turn the quadratic equation into a perfect square trinomial. Let me recall the steps. First, move the constant term to the right side. Wait, in this case, the equation is already set up as ( x^2 - 2x = 2 ). So the constant term is already on the right. Next, we take the coefficient of x, which is -2, divide it by 2 to get -1, then square it to get 1. That's the number we need to add to both sides to complete the square.So adding 1 to both sides:Left side becomes ( x^2 - 2x + 1 ), which is a perfect square: ( (x - 1)^2 ).Right side is 2 + 1 = 3.So now we have:( (x - 1)^2 = 3 )To solve for x, take the square root of both sides:( x - 1 = pm sqrt{3} )Then add 1 to both sides:( x = 1 pm sqrt{3} )So the solutions are ( x = 1 + sqrt{3} ) and ( x = 1 - sqrt{3} ). Let me check if that makes sense. Plugging back into the original equation:For ( x = 1 + sqrt{3} ):( (1 + sqrt{3})^2 - 2(1 + sqrt{3}) )First, square term: (1 + 2sqrt{3} + 3 = 4 + 2sqrt{3})Then subtract 2 times (1 + sqrt3): ( -2 - 2sqrt{3} )So total: (4 + 2sqrt{3} - 2 - 2sqrt{3} = 2). Which matches the right side. Similarly for the other solution. Seems good.Problem (2): Solve ( 2x^2 + 3 = 3x ) using the formula method.Quadratic formula. The standard form is ( ax^2 + bx + c = 0 ). So first, rearrange the equation to standard form.Subtract 3x from both sides:( 2x^2 - 3x + 3 = 0 )Wait, hold on. Original equation: (2x^2 +3 =3x). Moving 3x to the left gives (2x^2 -3x +3=0). Correct.Now, identify a, b, c: a=2, b=-3, c=3.Quadratic formula is ( x = frac{ -b pm sqrt{b^2 -4ac} }{2a} ).Plugging in the values:( x = frac{ -(-3) pm sqrt{ (-3)^2 -4*2*3 } }{2*2} )Calculate step by step:Numerator: 3 ± sqrt(9 - 24)Which is 3 ± sqrt(-15)So sqrt(-15) is imaginary, i*sqrt(15). So solutions are complex:( x = frac{3 pm isqrt{15}}{4} )Hmm, so the equation has no real solutions, only complex ones. Let me verify if that's correct.Original equation: (2x^2 +3 =3x). Let's check if real solutions exist. The discriminant is b² -4ac = 9 - 24 = -15, which is negative. So yes, no real solutions. Correct. So the answer is complex numbers: ( frac{3}{4} pm frac{sqrt{15}}{4}i ). Alright.Problem (3): Solve ( (x + 2)^2 - 3(x + 2) = 0 ) using an appropriate method.This one looks like it can be factored. Let me see. Let’s set ( y = x + 2 ), then the equation becomes ( y^2 - 3y = 0 ). That factors as ( y(y - 3) = 0 ). So solutions are y = 0 or y = 3.Substituting back:For y = 0: x + 2 = 0 → x = -2For y = 3: x + 2 = 3 → x = 1So solutions are x = -2 and x = 1. Let me check.Plug x = -2 into original equation:( (-2 + 2)^2 - 3(-2 +2) = 0^2 -3*0 = 0 - 0 = 0. Correct.Plug x =1:( (1 +2)^2 -3(1 +2) = 9 - 9 = 0. Correct.Alternatively, could expand the original equation:( (x + 2)^2 -3(x + 2) =0 )Expanding the square: ( x^2 +4x +4 -3x -6 =0 )Combine like terms: x^2 + x -2 =0Factor: (x +2)(x -1) =0 → x = -2, x =1. Same result. Perfect. So both methods confirm the solutions. Good.Summary:1. Solutions are (1 pm sqrt{3}).2. Solutions are complex: (frac{3 pm isqrt{15}}{4}).3. Solutions are -2 and 1.Final Answer(1) (boxed{1 pm sqrt{3}})(2) (boxed{frac{3 pm isqrt{15}}{4}})(3) (boxed{-2}) and (boxed{1})</think>Problem (1): Solve ( x^2 - 2x = 2 ) using the completing the square method.1. Start with the given equation: ( x^2 - 2x = 2 ).2. Complete the square by adding 1 to both sides: [ x^2 - 2x + 1 = 2 + 1 implies (x - 1)^2 = 3 ]3. Take the square root of both sides: [ x - 1 = pm sqrt{3} ]4. Solve for ( x ): [ x = 1 pm sqrt{3} ]Answer: (boxed{1 pm sqrt{3}})Problem (2): Solve ( 2x^2 + 3 = 3x ) using the formula method.1. Rearrange the equation to standard form: [ 2x^2 - 3x + 3 = 0 ]2. Identify coefficients ( a = 2 ), ( b = -3 ), ( c = 3 ).3. Apply the quadratic formula ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ): [ x = frac{3 pm sqrt{(-3)^2 - 4 cdot 2 cdot 3}}{2 cdot 2} = frac{3 pm sqrt{9 - 24}}{4} = frac{3 pm sqrt{-15}}{4} = frac{3 pm isqrt{15}}{4} ]Answer: (boxed{frac{3 pm isqrt{15}}{4}})Problem (3): Solve ( (x + 2)^2 - 3(x + 2) = 0 ) using an appropriate method.1. Let ( y = x + 2 ), transforming the equation to: [ y^2 - 3y = 0 implies y(y - 3) = 0 ]2. Solve for ( y ): [ y = 0 quad text{or} quad y = 3 ]3. Substitute back ( y = x + 2 ): - For ( y = 0 ): ( x + 2 = 0 implies x = -2 ) - For ( y = 3 ): ( x + 2 = 3 implies x = 1 )Answer: (boxed{-2}) and (boxed{1})]

🔑: Solution:# For Equation (1) x^{2}-2x=2:1. Start by moving all terms to one side to set the equation to zero: x^{2}-2x-2=0.2. To complete the square, we need to ensure the coefficient of x^2 is 1 (which it is), and then add and subtract the square of half the coefficient of x, which is 1^2=1: [x^{2}-2x+1-1=2]3. Rearrange to form a perfect square on one side: [x^{2}-2x+1=3]4. This gives us: [(x-1)^{2}=3]5. Taking the square root of both sides: [x-1=pm sqrt{3}]6. Solving for x gives us two solutions: [x_{1}=1+sqrt{3}, quad x_{2}=1-sqrt{3}]Therefore, the solutions are x_{1}=1+sqrt{3} and x_{2}=1-sqrt{3}, which can be encapsulated as boxed{x_{1}=1+sqrt{3}, x_{2}=1-sqrt{3}}.# For Equation (2) 2x^{2}+3=3x:1. Rearrange the equation to set it to zero: 2x^{2}-3x+3=0.2. Identify coefficients a=2, b=-3, and c=3.3. Calculate the discriminant Delta = b^{2}-4ac: [Delta = (-3)^{2}-4cdot2cdot3=-15]4. Since Delta < 0, there are no real solutions.Therefore, the equation has no real solutions, encapsulated as boxed{text{No real solutions}}.# For Equation (3) (x+2)^{2}-3(x+2)=0:1. Factor by grouping: [(x+2)((x+2)-3)=0]2. Simplify inside the parentheses: [(x+2)(x-1)=0]3. Set each factor equal to zero: [x+2=0 quad text{or} quad x-1=0]4. Solve for x: [x_{1}=-2, quad x_{2}=1]Therefore, the solutions are x_{1}=-2 and x_{2}=1, which can be encapsulated as boxed{x_{1}=-2, x_{2}=1}.

🧠:[<think>Okay, let me try to figure this problem out. It's about a right triangular prism, so first, I need to recall what a right triangular prism is. A right triangular prism has two congruent triangular bases connected by rectangles. The term "right" here means that the lateral edges (the ones connecting the corresponding vertices of the two bases) are perpendicular to the bases. So, the sides are rectangles, not just parallelograms.The problem states: In a right triangular prism ABC-A₁B₁C₁, angle ACB = 90°, AC = 2 BC, A₁B is perpendicular to B₁C. We need to find the sine of the angle formed by B₁C and the lateral face A₁ABB₁.First, let me try to visualize this prism. The base triangle is ABC with a right angle at C. So, AC and BC are the legs, and AB is the hypotenuse. It's given that AC = 2 BC. Let me note that as AC = 2 BC, so maybe assign some variables here. Let me set BC = x, then AC = 2x. Since it's a right triangle at C, the length AB can be calculated using the Pythagorean theorem: AB = sqrt(AC² + BC²) = sqrt((2x)² + x²) = sqrt(4x² + x²) = sqrt(5x²) = x√5. So AB is x√5.Now, since it's a prism, the other base A₁B₁C₁ is congruent to ABC, and the lateral edges AA₁, BB₁, CC₁ are all equal in length and perpendicular to the bases. Let's denote the height of the prism (the length of the lateral edges) as h. But since it's a right prism, h is just the distance between the bases. However, the problem doesn't specify the height. Hmm, but maybe we can express the answer in terms of x or find h through the given condition that A₁B is perpendicular to B₁C.So, the key condition here is that A₁B is perpendicular to B₁C. That must relate to the geometry of the prism. Let me think.First, perhaps assign coordinates to the points. Let me place point C at the origin (0,0,0) for simplicity. Then, since angle ACB is 90°, AC and BC are along the axes. Let's set point C at (0,0,0). Then, since AC = 2 BC, let's let BC be along the x-axis and AC along the y-axis. Wait, but in a right triangle at C, AC and BC are legs. If we set point C at (0,0,0), then point B would be at (x, 0, 0) and point A would be at (0, 2x, 0). Then, point A₁ would be at (0, 2x, h), B₁ at (x, 0, h), and C₁ at (0,0,h). Wait, but is this correct? Wait, in a right triangular prism, the lateral edges are perpendicular to the base. So, if the base is in the xy-plane, then the height would be along the z-axis. So, coordinates:- C: (0, 0, 0)- B: (x, 0, 0) (since BC = x)- A: (0, 2x, 0) (since AC = 2x)- C₁: (0, 0, h)- B₁: (x, 0, h)- A₁: (0, 2x, h)Yes, that seems correct. Now, A₁B is the line segment from A₁(0, 2x, h) to B(x, 0, 0). Similarly, B₁C is the line segment from B₁(x, 0, h) to C(0, 0, 0).The problem states that A₁B is perpendicular to B₁C. So, the vectors A₁B and B₁C should be perpendicular. Let's compute these vectors.Vector A₁B is B - A₁ = (x - 0, 0 - 2x, 0 - h) = (x, -2x, -h)Vector B₁C is C - B₁ = (0 - x, 0 - 0, 0 - h) = (-x, 0, -h)Their dot product should be zero since they are perpendicular.Dot product = (x)(-x) + (-2x)(0) + (-h)(-h) = -x² + 0 + h² = h² - x²Setting this equal to zero: h² - x² = 0 => h² = x² => h = ±x. Since h is a length, we take h = x.So, the height h is equal to BC, which is x. Therefore, h = x.So now, we can fix h = x. So, the coordinates become:- C: (0, 0, 0)- B: (x, 0, 0)- A: (0, 2x, 0)- C₁: (0, 0, x)- B₁: (x, 0, x)- A₁: (0, 2x, x)Now, the problem asks for the sine of the angle formed by B₁C and the lateral face A₁ABB₁.First, let me parse this. The angle formed by a line and a plane is the angle between the line and its orthogonal projection onto the plane. The sine of this angle is equal to the sine of the angle between the line and the plane's normal vector. Alternatively, the sine of the angle θ between the line and the plane is given by sinθ = |n·v| / (|n||v|), where n is the normal vector to the plane and v is the direction vector of the line. Wait, actually, the angle between a line and a plane is complementary to the angle between the line and the normal to the plane. So, if φ is the angle between the line and the normal, then θ = 90° - φ, so sinθ = cosφ. And cosφ = |n·v| / (|n||v|). Therefore, sinθ = |n·v| / (|n||v|). But maybe I need to verify this.Alternatively, the angle between a line and a plane is defined as the angle between the line and its projection onto the plane, which is the complement of the angle between the line and the normal. So, if α is the angle between the line and the normal, then the angle between the line and the plane is 90° - α. Therefore, sin(90° - α) = cosα. Therefore, sinθ = cosα = |n·v| / (|n||v|). So yes, sinθ = |n·v| / (|n||v|). So, in order to find sinθ, where θ is the angle between B₁C and the lateral face A₁ABB₁, we need to find the normal vector to the lateral face A₁ABB₁, then take the direction vector of B₁C, compute their dot product, divide by the product of their magnitudes, and take the absolute value.Alternatively, since the lateral face A₁ABB₁ is a rectangle (since it's a right prism), the face A₁ABB₁ is the rectangle formed by the points A, B, B₁, A₁. So, this face is in the plane that includes points A, B, B₁, A₁.Wait, let's see. Points A, B, B₁, A₁. A is (0,2x,0), B is (x,0,0), B₁ is (x,0,x), A₁ is (0,2x,x). So, this face is actually a parallelogram, but since it's a right prism, it should be a rectangle. Let me check. The edges AA₁ and BB₁ are both vertical (along the z-axis), and AB is the hypotenuse of the base triangle. Since AA₁ and BB₁ are both perpendicular to the base, the face A₁ABB₁ should indeed be a rectangle. So, the lateral face A₁ABB₁ is a rectangle lying in some plane. To find the normal vector to this plane, we can take two vectors lying on the plane and compute their cross product.Let me pick vectors AB and AA₁.Vector AB is B - A = (x - 0, 0 - 2x, 0 - 0) = (x, -2x, 0)Vector AA₁ is A₁ - A = (0 - 0, 2x - 2x, x - 0) = (0, 0, x)The cross product of AB and AA₁ will give a normal vector to the plane.Compute AB × AA₁:|i   j   k ||x -2x 0||0   0  x|= i*(-2x*x - 0*0) - j*(x*x - 0*0) + k*(x*0 - (-2x)*0)= i*(-2x²) - j*(x²) + k*0= (-2x², -x², 0)So, the normal vector n is (-2x², -x², 0). We can simplify this by dividing by -x², which gives (2, 1, 0). So, the normal vector can be taken as (2, 1, 0). The direction doesn't matter for the angle calculation since we'll take the absolute value.Now, the direction vector of line B₁C is from B₁ to C: C - B₁ = (0 - x, 0 - 0, 0 - x) = (-x, 0, -x). So, the direction vector v is (-x, 0, -x). We can simplify this by dividing by -x (since direction vectors can be scaled), giving (1, 0, 1).So, now we need to compute sinθ = |n·v| / (|n||v|)First, compute n·v. n is (2, 1, 0), v is (1, 0, 1). The dot product is 2*1 + 1*0 + 0*1 = 2 + 0 + 0 = 2.Then |n·v| = |2| = 2.|n| = sqrt(2² + 1² + 0²) = sqrt(4 + 1) = sqrt(5).|v| = sqrt(1² + 0² + 1²) = sqrt(2).Therefore, sinθ = 2 / (sqrt(5) * sqrt(2)) = 2 / sqrt(10) = sqrt(10)/5 after rationalizing.Wait, 2/sqrt(10) simplifies to (2*sqrt(10))/10 = sqrt(10)/5. Yes.But wait, let me check if I did everything correctly. The normal vector was computed as (2,1,0). The direction vector of B₁C is (-x,0,-x). But when I simplified it, I divided by -x, which is a scalar, so direction vector is (1,0,1). The dot product of (2,1,0) and (1,0,1) is 2*1 + 1*0 + 0*1 = 2. Then |n| is sqrt(2² + 1² + 0²) = sqrt(5). |v| is sqrt(1 + 0 + 1) = sqrt(2). So yes, sinθ = 2/(sqrt(5)*sqrt(2)) = 2/sqrt(10) = sqrt(10)/5.But let me verify if this is indeed the correct angle. Alternatively, maybe I should visualize the line B₁C and the face A₁ABB₁.Alternatively, since the lateral face A₁ABB₁ is a rectangle, perhaps there's another way to compute the angle. The angle between the line B₁C and the face is the angle between B₁C and its projection onto the face.To compute this angle, we can find the projection of B₁C onto the face A₁ABB₁ and then find the angle between B₁C and its projection. The sine of this angle is equal to the ratio of the height (the component perpendicular to the face) to the length of B₁C.Alternatively, the angle θ between the line and the plane satisfies sinθ = length of the component of the direction vector perpendicular to the plane divided by the length of the direction vector.Which is similar to the formula we used earlier. Because the component of v perpendicular to the plane is (n·v / |n|²) * n, but since we are taking the magnitude, it would be |n·v| / |n|. Wait, maybe.Alternatively, if we have a direction vector v of the line, then the sine of the angle between the line and the plane is equal to the magnitude of the component of v perpendicular to the plane divided by |v|.But the component of v perpendicular to the plane is (v · n / |n|²) * n, so its magnitude is |v · n| / |n|. Then, sinθ = (|v · n| / |n|) / |v| = |v · n| / (|n||v|), which matches the earlier formula. So yes, that's correct.Therefore, sinθ = |n·v| / (|n||v|) = 2 / (sqrt(5)*sqrt(2)) = sqrt(10)/5 ≈ 0.632.But let me check with another approach to confirm. Let's compute the projection of B₁C onto the lateral face A₁ABB₁ and find the angle between B₁C and its projection.First, the line B₁C goes from B₁(x,0,x) to C(0,0,0). The direction vector is (-x, 0, -x) as before.The lateral face A₁ABB₁ is the rectangle containing points A, B, B₁, A₁. To find the projection of B₁C onto this face, we need to find the component of the direction vector that lies within the plane of the face.Since the normal vector of the face is (2,1,0), as computed earlier, the projection of the direction vector v onto the plane can be found by subtracting the component of v in the direction of the normal.Alternatively, the projection of v onto the plane is v minus the component of v along n.So, let's compute the component of v along n:Component of v along n = (v · n / |n|²) * nv is (-x, 0, -x), n is (2,1,0)v · n = (-x)(2) + 0(1) + (-x)(0) = -2x + 0 + 0 = -2x|n|² = (2)^2 + (1)^2 + (0)^2 = 4 + 1 = 5Therefore, component of v along n is (-2x / 5) * (2,1,0) = (-4x/5, -2x/5, 0)Therefore, the projection of v onto the plane is:v - component along n = (-x, 0, -x) - (-4x/5, -2x/5, 0) = (-x + 4x/5, 0 + 2x/5, -x - 0) = (-x/5, 2x/5, -x)So, the projection direction vector is (-x/5, 2x/5, -x). We can factor out x/5:= x/5 * (-1, 2, -5)But maybe scaling doesn't matter for the angle. Let's compute the angle between v and its projection.The angle θ between v and its projection is the angle we need, and sinθ is the ratio of the magnitude of the component of v perpendicular to the plane (which is the component along n) to the magnitude of v.Wait, but earlier we computed sinθ using the formula |n·v|/(|n||v|). Let's check if this matches.The component of v perpendicular to the plane is the part we subtracted, which had magnitude |component along n| = sqrt( (-4x/5)^2 + (-2x/5)^2 + 0^2 ) = sqrt(16x²/25 + 4x²/25) = sqrt(20x²/25) = sqrt(4x²/5) = (2x)/√5But the magnitude of v is sqrt( (-x)^2 + 0^2 + (-x)^2 ) = sqrt( x² + x² ) = sqrt(2x²) = x√2Therefore, sinθ = (2x/√5) / (x√2) ) = (2/√5)/√2 = 2/(√5√2) = 2/√10 = √10/5. So same result. Therefore, confirms that sinθ = √10/5.Therefore, the answer should be √10⁄5.But let me check again for any possible miscalculations.First, coordinates:- C at (0,0,0)- B at (x,0,0)- A at (0,2x,0)- A₁ at (0,2x,x)- B₁ at (x,0,x)- C₁ at (0,0,x)Vectors:A₁B vector is (x, -2x, -x) as h = x.B₁C vector is (-x, 0, -x)Dot product of A₁B and B₁C: (x)(-x) + (-2x)(0) + (-x)(-x) = -x² + 0 + x² = 0. Wait, that's conflicting with earlier. Wait, hold on, earlier when we set h = x, the vectors A₁B and B₁C should be perpendicular. Let's verify.Wait, if h = x, then vector A₁B is (x, -2x, -x), and vector B₁C is (-x, 0, -x). Then their dot product is x*(-x) + (-2x)*0 + (-x)*(-x) = -x² + 0 + x² = 0. So yes, that's correct. So they are indeed perpendicular.But when we calculated h, we found h = x, which matches. So that's correct.Then, for the normal vector to the lateral face A₁ABB₁, which is a rectangle. The vectors AB and AA₁ were used. AB is (x, -2x, 0) and AA₁ is (0,0,x). Cross product AB × AA₁ gives (-2x², -x², 0) as before, which simplifies to normal vector (2,1,0). Correct.The direction vector of B₁C is (-x, 0, -x). The dot product with normal vector (2,1,0):(2)(-x) + (1)(0) + (0)(-x) = -2x. Then |n·v| = |-2x| = 2x.|n| is sqrt(4 + 1) = sqrt(5), |v| is sqrt(x² + x²) = x√2.Thus sinθ = 2x / (sqrt(5) * x√2) ) = 2/(sqrt(5)sqrt(2)) = sqrt(10)/5. Correct.Therefore, after thorough calculation and verification, the sine of the angle is sqrt(10)/5.Final Answerboxed{dfrac{sqrt{10}}{5}}</think>Given a right triangular prism (ABC-A_1B_1C_1) with (angle ACB = 90^circ), (AC = 2BC), and (A_1B perp B_1C), we need to find the sine of the angle formed by (B_1C) and the lateral face (A_1ABB_1).1. Assigning Coordinates: - Place (C) at the origin ((0, 0, 0)). - Let (BC = x), then (AC = 2x). Coordinates of points: - (B = (x, 0, 0)) - (A = (0, 2x, 0)) - (A_1 = (0, 2x, h)) - (B_1 = (x, 0, h)) - (C_1 = (0, 0, h))2. Determining the Height (h): - Vectors (A_1B) and (B_1C) are perpendicular. - Vector (A_1B = (x, -2x, -h)) - Vector (B_1C = (-x, 0, -h)) - Dot product: (x(-x) + (-2x)(0) + (-h)(-h) = -x^2 + h^2 = 0 implies h = x)3. Normal Vector to the Lateral Face (A_1ABB_1): - Vectors (AB = (x, -2x, 0)) and (AA_1 = (0, 0, x)) - Cross product (AB times AA_1 = (-2x^2, -x^2, 0)) - Normal vector: ((2, 1, 0))4. Direction Vector of (B_1C): - Direction vector (B_1C = (-x, 0, -x)) simplifies to ((1, 0, 1))5. Calculating Sine of the Angle: - Dot product of normal vector ((2, 1, 0)) and direction vector ((1, 0, 1)): (2 cdot 1 + 1 cdot 0 + 0 cdot 1 = 2) - Magnitudes: (|n| = sqrt{4 + 1} = sqrt{5}), (|v| = sqrt{1 + 0 + 1} = sqrt{2}) - (sin theta = frac{|n cdot v|}{|n||v|} = frac{2}{sqrt{5} sqrt{2}} = frac{sqrt{10}}{5})[boxed{dfrac{sqrt{10}}{5}}],<think>Okay, so I need to solve this geometry problem about a right triangular prism. Let me start by understanding the problem step by step. First, the prism is named ABC-A₁B₁C₁, and it's a right triangular prism. That means the bases ABC and A₁B₁C₁ are congruent right triangles, and the lateral edges (like AA₁, BB₁, CC₁) are perpendicular to the bases. The problem states that angle ACB is 90 degrees, so triangle ABC is a right triangle with the right angle at C. Also, AC is twice BC, so AC = 2 BC. Then, it says that A₁B is perpendicular to B₁C. We need to find the sine of the angle formed by B₁C and the lateral face A₁ABB₁.Let me visualize this prism. Since it's a right triangular prism, the sides are rectangles. The base is triangle ABC with right angle at C. Let me sketch this mentally: point C is the right angle, AC is twice BC. So if I let BC = x, then AC = 2x. Then, AB can be found using the Pythagorean theorem: AB² = AC² + BC² = (2x)² + x² = 4x² + x² = 5x², so AB = x√5. Now, the prism has height equal to the length of the lateral edges, which are AA₁, BB₁, CC₁. Since it's a right prism, these edges are perpendicular to the base ABC. Therefore, the top face A₁B₁C₁ is a copy of ABC, shifted vertically (in 3D space) by the height of the prism. The problem states that A₁B is perpendicular to B₁C. Hmm. Let me parse that. A₁B and B₁C are lines in 3D space. For them to be perpendicular, their direction vectors must have a dot product of zero. So perhaps I need to assign coordinates to the points and then compute the vectors A₁B and B₁C, set their dot product to zero, and solve for the height of the prism. Once I have the height, I can find the angle between B₁C and the lateral face A₁ABB₁.Let me set up a coordinate system. Let's place point C at the origin (0, 0, 0) for simplicity. Then, since AC = 2 BC and angle C is the right angle, let's assign coordinates to points A, B, and C. Let me set BC along the x-axis and AC along the y-axis. Wait, but in the problem, angle ACB is 90 degrees, so sides AC and BC meet at C. So yes, if C is at (0,0,0), then point B is along the x-axis, point A is along the y-axis. Let me confirm: Let’s denote BC = x, so point B would be at (x, 0, 0). AC = 2x, so point A would be at (0, 2x, 0). Then, point C is at (0,0,0). The base triangle ABC is in the z=0 plane. The top face A₁B₁C₁ is then in the z=h plane, where h is the height of the prism. So coordinates:- A: (0, 2x, 0)- B: (x, 0, 0)- C: (0, 0, 0)- A₁: (0, 2x, h)- B₁: (x, 0, h)- C₁: (0, 0, h)Wait, but in a right prism, the lateral edges are perpendicular to the base, so yes, they just have the z-coordinate increased by h. So that's correct.Now, the problem says A₁B is perpendicular to B₁C. Let's find vectors for these two lines.First, vector A₁B: Point A₁ is (0, 2x, h), and point B is (x, 0, 0). So the vector from A₁ to B is (x - 0, 0 - 2x, 0 - h) = (x, -2x, -h).Vector B₁C: Point B₁ is (x, 0, h), and point C is (0, 0, 0). So the vector from B₁ to C is (0 - x, 0 - 0, 0 - h) = (-x, 0, -h).Now, these two vectors A₁B and B₁C must be perpendicular, so their dot product should be zero. Let's compute the dot product:(A₁B) • (B₁C) = (x)(-x) + (-2x)(0) + (-h)(-h) = -x² + 0 + h² = h² - x².Setting this equal to zero for perpendicularity:h² - x² = 0 ⇒ h² = x² ⇒ h = x (since height is positive, h = x).So the height of the prism h is equal to x, which is the length of BC. Therefore, h = BC = x. Therefore, in this coordinate system, h = x. So we can set x to 1 for simplicity, but maybe we can keep it as x for now. Wait, perhaps it's better to assign specific values to make calculation easier. Let me set BC = 1, so x = 1. Then AC = 2, and h = 1. Then coordinates would be:- A: (0, 2, 0)- B: (1, 0, 0)- C: (0, 0, 0)- A₁: (0, 2, 1)- B₁: (1, 0, 1)- C₁: (0, 0, 1)This might make calculations easier. Let's proceed with x = 1, so BC = 1, AC = 2, h = 1.Now, we need to find the sine of the angle formed by B₁C and the lateral face A₁ABB₁.First, let's understand what the angle between a line and a face is. The angle between a line and a plane (face) is the complement of the angle between the line and the normal to the plane. But more precisely, the angle θ between the line and the plane is defined as the angle between the line and its orthogonal projection onto the plane. Then, sin θ is equal to the sine of the angle between the line and the normal to the plane. Wait, let me recall the exact definition.Alternatively, the angle between a line and a plane is the smallest angle between the line and any line in the plane. However, the standard definition is that the angle between a line and a plane is equal to 90 degrees minus the angle between the line and the normal to the plane. So if φ is the angle between the line and the normal, then θ = 90° - φ, and sin θ = sin(90° - φ) = cos φ. So, sin θ = |cos φ|, where φ is the angle between the line and the normal.But perhaps another approach is to find the angle between the line and the plane by projecting the line onto the plane. The angle θ satisfies sin θ = opposite/hypotenuse, where opposite is the length of the component of the line's direction vector perpendicular to the plane, and hypotenuse is the magnitude of the direction vector.Since the lateral face A₁ABB₁ is a rectangle. Let's consider the plane of this face. The face A₁ABB₁ includes points A₁, A, B, B₁. Since it's a lateral face of the prism, it's a rectangle. The plane of this face can be described by the points A, B, B₁, A₁.To find the angle between line B₁C and this plane, we need to find the angle between the line and the plane. As per the formula, the sine of this angle is equal to the ratio of the length of the component of the direction vector of B₁C perpendicular to the plane divided by the magnitude of the direction vector.Alternatively, another formula: the angle θ between a line with direction vector v and a plane with normal vector n is given by sin θ = |v · n| / (|v| |n|). Wait, no. Wait, if θ is the angle between the line and the plane, then θ = 90° - φ, where φ is the angle between the line and the normal. So sin θ = sin(90° - φ) = cos φ. But cos φ is |v · n| / (|v| |n|). Therefore, sin θ = |v · n| / (|v| |n|). Wait, is that correct?Wait, let's clarify. Let’s suppose that the angle between the line and the normal is φ. Then, θ = 90° - φ. Then:sin θ = sin(90° - φ) = cos φ.But cos φ = |v · n| / (|v| |n|). Therefore, sin θ = |v · n| / (|v| |n|). However, we need to be careful. The normal vector n here is the normal to the plane. So to compute the angle between the line B₁C and the lateral face A₁ABB₁, we need the normal vector to the face A₁ABB₁.First, let's find the normal vector to the plane A₁ABB₁. Since A₁ABB₁ is a rectangle, let's find two vectors in this plane and take their cross product.Points in the plane A₁ABB₁ are A₁(0,2,1), A(0,2,0), B(1,0,0), B₁(1,0,1). So vectors in this plane can be found by subtracting coordinates. For example, vector A₁A is (0,2,0) - (0,2,1) = (0,0,-1). Vector AB is (1,0,0) - (0,2,0) = (1,-2,0). So two vectors in the plane are (0,0,-1) and (1,-2,0). The cross product of these vectors will give a normal vector.Compute cross product n = (0,0,-1) × (1,-2,0). Let's compute this:i component: (0)(0) - (-1)(-2) = 0 - 2 = -2j component: -[(0)(0) - (-1)(1)] = -[0 - (-1)] = -[1] = -1k component: (0)(-2) - (0)(1) = 0 - 0 = 0Wait, cross product formula:If vector u = (u1, u2, u3), vector v = (v1, v2, v3),then u × v = (u2v3 - u3v2, u3v1 - u1v3, u1v2 - u2v1)So for u = (0,0,-1) and v = (1,-2,0):i component: (0)(0) - (-1)(-2) = 0 - 2 = -2j component: -[(0)(0) - (-1)(1)] = -[0 - (-1)] = -[1] = -1k component: (0)(-2) - (0)(1) = 0 - 0 = 0Wait, that seems correct. So n = (-2, -1, 0). But we can also take the negative of this as the normal vector. Let me check again:Wait, cross product is u × v, so with u = A₁A = (0,0,-1) and v = AB = (1,-2,0). The cross product is:i: (0*0 - (-1)*(-2)) = 0 - 2 = -2j: - (0*0 - (-1)*1) = - (0 + 1) = -1k: (0*(-2) - 0*1) = 0 - 0 = 0Yes, so n = (-2, -1, 0). To make it a unit normal vector, we can divide by its magnitude, but maybe we don't need to since we just need the direction.Alternatively, maybe I can use a different pair of vectors in the plane to compute the normal. For instance, vector AA₁ is (0,0,1), and vector AB is (1,-2,0). Wait, but AA₁ is from A to A₁, which is (0,2,1) - (0,2,0) = (0,0,1). So vector AA₁ is (0,0,1). Then cross product AA₁ × AB would be (0,0,1) × (1,-2,0). Let's compute that:i component: (0*0 - 1*(-2)) = 0 + 2 = 2j component: - (0*0 - 1*1) = - (0 -1) = 1k component: (0*(-2) - 0*1) = 0 - 0 = 0So n = (2, 1, 0). Hmm, this is the opposite direction of the previous normal vector. Depending on the order of the cross product, the direction changes. Both are valid normal vectors, just pointing in opposite directions. Let's pick one, say n = (2,1,0). Since the problem is about the angle, the direction might not matter because we take absolute value.Now, the direction vector of line B₁C. The line B₁C goes from B₁(1,0,1) to C(0,0,0). So the direction vector is C - B₁ = (0 - 1, 0 - 0, 0 - 1) = (-1, 0, -1). Let me denote vector v = (-1, 0, -1).Now, to find the sine of the angle between line B₁C and the plane A₁ABB₁. As discussed earlier, sin θ = |v · n| / (|v| |n|), where θ is the angle between the line and the plane. Wait, but let me confirm once again.The angle between a line and a plane is defined as the complement of the angle between the line and the normal to the plane. So if φ is the angle between the line and the normal, then θ = 90° - φ. Therefore, sin θ = sin(90° - φ) = cos φ. But cos φ = |v · n| / (|v| |n|). Therefore, sin θ = |v · n| / (|v| |n|).But wait, is this correct? Let me check with an example. Suppose the line is in the plane, then the angle θ between the line and the plane is 0°, and the angle φ between the line and the normal is 90°, so cos φ = 0, so sin θ = 0, which is correct. If the line is perpendicular to the plane, then θ = 90°, φ = 0°, so cos φ = 1, so sin θ = 1, which is correct. So yes, sin θ = |v · n| / (|v| |n|). Therefore, in this case, we can compute sin θ as |v · n| divided by (|v| |n|).So let's compute v · n. The direction vector v is (-1, 0, -1), and the normal vector n is (2,1,0).Dot product: (-1)(2) + (0)(1) + (-1)(0) = -2 + 0 + 0 = -2. The absolute value is 2.|v| is the magnitude of v: sqrt((-1)^2 + 0^2 + (-1)^2) = sqrt(1 + 0 + 1) = sqrt(2).|n| is the magnitude of n: sqrt(2^2 + 1^2 + 0^2) = sqrt(4 + 1 + 0) = sqrt(5).Therefore, sin θ = |v · n| / (|v| |n|) = 2 / (sqrt(2) * sqrt(5)) ) = 2 / (sqrt(10)) ) = sqrt(10)/5 after rationalizing.Wait, 2 / sqrt(10) is equal to 2*sqrt(10)/10 = sqrt(10)/5.Therefore, sin θ = sqrt(10)/5. So the answer is sqrt(10)/5. But let me verify this.Wait, but let me check if I used the correct normal vector. Earlier, I had two different normal vectors, (-2, -1, 0) and (2,1,0). The direction of the normal vector affects the sign of the dot product, but since we take absolute value, it doesn't matter. So either way, the absolute value of the dot product would be the same. For example, if I used (-2, -1, 0) as n, then the dot product would be (-1)(-2) + 0*(-1) + (-1)(0) = 2 + 0 + 0 = 2. Absolute value is still 2. So same result.Therefore, the calculation seems correct. So sin θ = 2 / (sqrt(2)*sqrt(5)) = sqrt(10)/5.But let me verify once again if this is the correct approach. Alternatively, maybe the angle is supposed to be computed differently.Alternatively, another method to find the angle between a line and a plane is to find the angle between the line and its projection onto the plane. In this case, sin θ would be the ratio of the length of the component of the vector perpendicular to the plane over the length of the vector.So, decomposing the direction vector v into components parallel and perpendicular to the plane. The perpendicular component's length divided by the total length of v is sin θ.The perpendicular component of v to the plane is along the normal direction. So, the magnitude of the projection of v onto the normal is |v · n| / |n|. Then, sin θ = |projection| / |v| = |v · n| / (|n| |v|) * |n| / |n|? Wait, no.Wait, if we decompose v into v_parallel (component in the plane) and v_perpendicular (component perpendicular to the plane), then:v = v_parallel + v_perpendicular.The magnitude of v_perpendicular is |v · n| / |n|. Therefore, sin θ = |v_perpendicular| / |v| = |v · n| / (|n| |v|). Which is the same as before. So sin θ = |v · n| / (|n| |v|). Which is the same as before.Therefore, our previous calculation is correct, and sin θ = sqrt(10)/5. Therefore, the answer should be sqrt(10)/5.But let me check if I made a mistake in the coordinate system.Wait, in the coordinate system, I set BC = 1, so x = 1, AC = 2, h = 1. Then, the direction vector of B₁C is (-1, 0, -1). The normal vector to the lateral face A₁ABB₁ is (2,1,0). Then, the dot product is -2, absolute value 2. The magnitudes are sqrt(2) and sqrt(5). So 2/(sqrt(2)*sqrt(5)) = sqrt(10)/5. Yes, that's correct.Alternatively, if I didn't set BC = 1, but kept BC as x, then h = x. The direction vector of B₁C would be (-x, 0, -x), and normal vector would be (2x, x, 0). Then, the dot product is (-x)(2x) + 0 + (-x)(0) = -2x², absolute value 2x². The magnitude of v is sqrt( x² + 0 + x² ) = sqrt(2x²) = x√2. The magnitude of n is sqrt( (2x)^2 + x² + 0 ) = sqrt(4x² + x² ) = sqrt(5x² ) = x√5. Then, sin θ = (2x² ) / ( x√2 * x√5 ) ) = 2x² / (x²√10 ) = 2 / √10 = √10 / 5. So same result. Therefore, the answer is indeed √10 / 5.Therefore, the sine of the angle is √10⁄5. So, boxed, that's boxed{dfrac{sqrt{10}}{5}}.Final Answerboxed{dfrac{sqrt{10}}{5}}</think>Given a right triangular prism (ABC-A_1B_1C_1) with (angle ACB = 90^circ), (AC = 2BC), and (A_1B perp B_1C), we need to find the sine of the angle formed by (B_1C) and the lateral face (A_1ABB_1).1. Coordinate System Setup: - Place (C) at the origin ((0, 0, 0)). - Let (BC = x), then (AC = 2x). - Coordinates of points: - (A = (0, 2x, 0)) - (B = (x, 0, 0)) - (C = (0, 0, 0)) - (A_1 = (0, 2x, h)) - (B_1 = (x, 0, h)) - (C_1 = (0, 0, h))2. Vectors for Perpendicularity: - Vector (A_1B) is ((x, -2x, -h)). - Vector (B_1C) is ((-x, 0, -h)). - Dot product of (A_1B) and (B_1C) is (-x^2 + h^2). Setting this to zero gives (h = x).3. Normal Vector to the Lateral Face: - Vectors in the plane (A_1ABB_1): - (A_1A = (0, 0, -1)) - (AB = (1, -2, 0)) - Cross product gives normal vector (n = (-2, -1, 0)) or ((2, 1, 0)).4. Direction Vector of (B_1C): - Direction vector (v = (-1, 0, -1)).5. Calculating Sine of the Angle: - Dot product (v cdot n = -2) (absolute value is 2). - Magnitudes: (|v| = sqrt{2}), (|n| = sqrt{5}). - (sin theta = frac{|v cdot n|}{|v| |n|} = frac{2}{sqrt{2} cdot sqrt{5}} = frac{sqrt{10}}{5}).Thus, the sine of the angle formed by (B_1C) and the lateral face (A_1ABB_1) is (boxed{dfrac{sqrt{10}}{5}}).]

🔑:16. Taking C as the origin, C A, C B, C C_{1} as the x, y, z axes to establish a rectangular coordinate system, with |B C|=2, left|C C_{1}right|=a, then A(4,0,0), A_{1}(4,0, a), B(0,2,0), B_{1}(0,2, a). Since A_{1} B perp B_{1} C, we have overrightarrow{mathrm{BA}_{1}} cdot overrightarrow{mathrm{CB}_{1}}=(4,-2, a) cdot(0,2, a)=a^{2}-4=0, a=2.Let vec{n}=(x, y, z) be a normal vector to the plane A_{1} A B B_{1}, then vec{n} cdot overrightarrow{A_{1} B_{1}}=-4 x+2 y=0,vec{n} cdot overrightarrow{B B_{1}}=2 z=0, yielding vec{n}=(1,2,0).Since B_{1} C=(0,-2,-2),the sine of the angle alpha between B_{1} C and the plane A_{1} A B B_{1} issin alpha=left|vec{n} cdot overrightarrow{B_{1} C}right| /left(|vec{n}|left|overrightarrow{B C_{1}}right|right)=frac{sqrt{10}}{5} .