🧠:[<think>Okay, so I have this problem about an acute triangle ABC. The angles are A, B, C opposite sides a, b, c respectively. The first part asks me to find angle A given the equation 2 sin A (c cos B + b cos C) = √3 a. Hmm, let's start with part (1).First, I remember there are some trigonometric identities and laws related to triangles, like the Law of Cosines and the Law of Sines. Maybe those can help here. The equation given has terms like c cos B and b cos C. Wait, in the Law of Cosines, we have expressions like c² = a² + b² - 2ab cos C, but here we have c cos B. Maybe those terms can be related to projections?Let me think. In a triangle, the length of a side multiplied by the cosine of an adjacent angle gives the projection of that side onto another side. For example, c cos B would be the projection of side AB onto side BC, or something like that? Wait, maybe it's the projection of side c (which is AB) onto side a (which is BC). Hmm, not sure. Alternatively, maybe there's a formula that relates c cos B + b cos C.Wait, actually, in a triangle, the sum of two sides each multiplied by the cosine of the opposite angle... Is there a formula for that? Let me recall. Wait, maybe this relates to the formula for the length of a median? Or maybe something else.Alternatively, I can use the Law of Cosines to express cos B and cos C. Let's try that. Let's write cos B and cos C using the Law of Cosines.From Law of Cosines:cos B = (a² + c² - b²)/(2ac)cos C = (a² + b² - c²)/(2ab)So, substituting these into the expression c cos B + b cos C:c * [(a² + c² - b²)/(2ac)] + b * [(a² + b² - c²)/(2ab)]Simplify each term:First term: [c(a² + c² - b²)] / (2ac) = (a² + c² - b²)/(2a)Second term: [b(a² + b² - c²)] / (2ab) = (a² + b² - c²)/(2a)So adding these together:[ (a² + c² - b²) + (a² + b² - c²) ] / (2a)Simplify numerator:a² + c² - b² + a² + b² - c² = 2a²So total expression is 2a² / (2a) = aSo c cos B + b cos C = aWow, that's neat. So the original equation is 2 sin A * a = √3 aTherefore, divide both sides by a (assuming a ≠ 0, which it isn't in a triangle):2 sin A = √3Thus sin A = √3 / 2So angle A is either 60 degrees or 120 degrees. But since the triangle is acute, all angles are less than 90 degrees, so A must be 60 degrees.So part (1) answer is 60 degrees, or π/3 radians.Okay, that wasn't too bad. For part (2), given c = 2√2 and a = 3.First, part (i): Find sin(2C + A). Since we already found A = 60 degrees, so angle A is 60 degrees. So we need to find angle C first, then compute 2C + A, and find its sine.But to find angle C, we might need to use the Law of Sines or Law of Cosines. Since we know sides a and c, and angle A, maybe Law of Sines is the way to go.Law of Sines: a / sin A = c / sin CGiven a = 3, angle A = 60°, c = 2√2.So 3 / sin 60° = 2√2 / sin CCompute sin 60° = √3 / 2, so 3 / (√3 / 2) = 2√2 / sin CSimplify left side: 3 * 2 / √3 = 6 / √3 = 2√3Therefore, 2√3 = 2√2 / sin CDivide both sides by 2: √3 = √2 / sin CThen sin C = √2 / √3 = √(2/3) = √6 / 3So sin C = √6 / 3. Since the triangle is acute, angle C is less than 90°, so cos C is positive.Compute cos C = sqrt(1 - sin²C) = sqrt(1 - (6/9)) = sqrt(3/9) = sqrt(1/3) = 1/√3 = √3 / 3.Now, we need to find sin(2C + A). Since A is 60°, so 2C + 60°. Let's compute 2C + A = 2C + 60°.First, let's find angle C. From sin C = √6 / 3, angle C is arcsin(√6 / 3). Let me compute that in degrees. But maybe we don't need the exact angle, but to express sin(2C + 60°) in terms of sin and cos.Alternatively, use trigonometric identities. Let's express sin(2C + A) as sin(2C + 60°). Using the sine addition formula:sin(2C + 60°) = sin 2C cos 60° + cos 2C sin 60°We can compute sin 2C and cos 2C using double-angle formulas.First, compute sin 2C = 2 sin C cos C = 2*(√6/3)*(√3/3) = 2*(√18)/9 = 2*(3√2)/9 = 6√2/9 = 2√2/3.Compute cos 2C = 1 - 2 sin²C = 1 - 2*(6/9) = 1 - 12/9 = 1 - 4/3 = -1/3.Alternatively, cos 2C = 2 cos²C - 1 = 2*(1/3) - 1 = 2/3 - 1 = -1/3. Same result.So sin(2C + 60°) = sin 2C cos 60° + cos 2C sin 60°Plug in the values:= (2√2/3)(1/2) + (-1/3)(√3/2)Simplify:= (√2/3) - (√3)/6To combine these, they need a common denominator. Let's write √2/3 as 2√2/6.So 2√2/6 - √3/6 = (2√2 - √3)/6So sin(2C + A) = (2√2 - √3)/6.Wait, is that correct? Let me check the calculations again.Wait, step by step:sin(2C + 60°) = sin2C cos60 + cos2C sin60sin2C = 2 sinC cosC = 2*(√6/3)*(√3/3) = 2*(√18)/9 = 2*(3√2)/9 = 6√2/9 = 2√2/3. Correct.cos60 = 1/2, so sin2C * cos60 = (2√2/3)*(1/2) = √2/3.cos2C = -1/3, sin60 = √3/2, so cos2C * sin60 = (-1/3)*(√3/2) = -√3/6.Adding these together: √2/3 - √3/6. To combine the terms, convert √2/3 to 2√2/6:2√2/6 - √3/6 = (2√2 - √3)/6. Yes, that's correct.So part (i) answer is (2√2 - √3)/6.Now part (ii): Find the area of triangle ABC.We can use the formula (1/2)ab sin C, but need to figure out which sides and angles we know.We know sides a = 3 (opposite angle A), c = 2√2 (opposite angle C). We also know angle A = 60°, angle C we can compute (since we found sin C = √6/3, but maybe we can use other methods.Alternatively, use Heron's formula, but we need to know all three sides. We know a and c, but we need b.Alternatively, use the formula (1/2) * a * b * sin C. Wait, but we don't know b. Alternatively, use (1/2) * a * c * sin B. But we need angle B.Alternatively, use the Law of Sines to find side b. Let's try that.From Law of Sines: a / sin A = b / sin B = c / sin C.We know a = 3, angle A = 60°, c = 2√2, sin C = √6/3. Let's compute the common ratio.First, compute a / sin A = 3 / (√3/2) = 3 * 2 / √3 = 6 / √3 = 2√3. So the common ratio is 2√3.Therefore, c = 2√2, so c / sin C = 2√3. Let's check: c / sin C = 2√2 / (√6/3) = 2√2 * 3 / √6 = 6√2 / √6 = 6√2 / (√6) = 6 / √3 = 2√3. Correct.Therefore, side b is 2√3 * sin B. But we need to find sin B. Since the triangle angles sum to 180°, angle B = 180° - A - C = 60° - C. Wait, no: angle A is 60°, angle B = 180° - 60° - C = 120° - C. Hmm, but maybe we can use Law of Cosines.Alternatively, since we know sides a, c, and angles A and C, maybe we can compute side b via Law of Cosines.Law of Cosines: b² = a² + c² - 2ac cos B. But we don't know angle B. Alternatively, use Law of Sines to find angle B.But maybe another approach: since we know angles A and C, and sides a and c, we can find side b using Law of Sines.Wait, since a / sin A = b / sin B = c / sin C, and we know a, sin A, c, sin C, so we can find angle B via angle sum.Alternatively, compute angle B as 180° - A - C, then compute sin B.We have angle C: since sin C = √6/3, angle C is arcsin(√6/3). Let's compute angle C in degrees, but maybe we don't need to. Instead, angle B = 120° - C (since A = 60°). Then sin B = sin(120° - C). Let's compute sin B.Alternatively, use sin B = sin(180° - A - C) = sin(A + C) because sin(180° - x) = sin x. Wait, no. Wait, 180° - A - C = B, so sin B = sin(180° - A - C) = sin(A + C). Wait, sin(180° - x) = sin x, so sin(180° - (A + C)) = sin(B). Wait, I think I confused the identity. Let's rephrase.Actually, angle B = 180° - A - C, so angle B is equal to 180° - (A + C). Therefore, sin B = sin(A + C) only if A + C = 180° - B, but that's not the case. Wait, no. Wait, if angle B = 180° - A - C, then sin B = sin(180° - A - C) = sin(A + C) since sin(180° - x) = sin x. Wait, that's not right. Wait, sin(180° - x) = sin x. So if angle B = 180° - (A + C), then sin B = sin(A + C). Wait, but A + C = 180° - B, so sin(A + C) = sin(180° - B) = sin B. Wait, that's circular. Let me think again.Wait, angle B = 180° - A - C. So sin B = sin(180° - A - C) = sin(A + C). Wait, no. Wait, sin(180° - x) = sin x. So sin(180° - (A + C)) = sin(A + C). Therefore, sin B = sin(A + C). Wait, but angle B is 180° - A - C, so sin B = sin(A + C). Hmm, but A + C = 180° - B, so sin(A + C) = sin(180° - B) = sin B. So that gives sin B = sin B, which is a tautology. Hmm, maybe this approach isn't helpful.Alternatively, express sin B as sin(180° - A - C) = sin(A + C) but that's the same thing. Maybe using the sine addition formula. Let's do that.sin(A + C) = sin A cos C + cos A sin CWe know sin A = √3/2, cos A = 1/2 (since angle A is 60°). sin C = √6/3, cos C = √3/3.So sin B = sin(A + C) = (√3/2)(√3/3) + (1/2)(√6/3)Compute each term:First term: (√3 * √3)/(2*3) = 3/(6) = 1/2Second term: (1 * √6)/(2*3) = √6 / 6So sin B = 1/2 + √6 /6 = (3 + √6)/6Therefore, sin B = (3 + √6)/6Then, since a / sin A = 2√3 (from earlier), so side b = 2√3 * sin B = 2√3 * (3 + √6)/6Simplify:2√3 * (3 + √6)/6 = (√3)(3 + √6)/3 = √3(3 + √6)/3= (3√3 + √3 * √6)/3 = (3√3 + √18)/3 = (3√3 + 3√2)/3 = √3 + √2So side b is √3 + √2Now, to find the area, we can use formula (1/2)ab sin C or (1/2)bc sin A, etc.Let's use (1/2)ab sin C. We have sides a = 3, b = √3 + √2, and sin C = √6/3.So area = (1/2) * 3 * (√3 + √2) * (√6 / 3)Simplify:(1/2) * 3 * (√3 + √2) * √6 / 3 = (1/2) * (√3 + √2) * √6Multiply √6 with (√3 + √2):√6 * √3 = √18 = 3√2√6 * √2 = √12 = 2√3So total is 3√2 + 2√3Therefore, area = (1/2)(3√2 + 2√3) = (3√2 + 2√3)/2Alternatively, check with another formula. Let's confirm using another method to be safe.Alternatively, use Heron's formula. We have sides a = 3, b = √3 + √2, c = 2√2.Compute semi-perimeter s = (a + b + c)/2 = (3 + √3 + √2 + 2√2)/2 = (3 + √3 + 3√2)/2. This seems messy, but maybe possible.But Heron's formula would require computing sqrt[s(s - a)(s - b)(s - c)], which might be complicated. Let's see.Alternatively, use formula (1/2)ac sin B. We have a = 3, c = 2√2, sin B = (3 + √6)/6Compute area = (1/2)*3*2√2*(3 + √6)/6 = (3*2√2 / 2)*(3 + √6)/6 = (3√2)*(3 + √6)/6Simplify:(3√2)(3 + √6) /6 = [9√2 + 3√12]/6 = [9√2 + 3*2√3]/6 = [9√2 + 6√3]/6 = (3√2 + 2√3)/2Same result. So that's good. Therefore, the area is (3√2 + 2√3)/2.Alternatively, maybe they want it written differently? Let me check again.Wait, maybe using (1/2)ab sin C:a = 3, b = √3 + √2, sin C = √6/3So (1/2)*3*(√3 + √2)*(√6/3) = (1/2)*(√3 + √2)*√6As before, same steps. So yes, same answer.Alternatively, perhaps there's a simpler form? Let me compute (√3 + √2)*√6:√3*√6 = √18 = 3√2√2*√6 = √12 = 2√3Thus, (√3 + √2)*√6 = 3√2 + 2√3, so area is (3√2 + 2√3)/2. That seems as simplified as it can get.Therefore, part (ii) answer is (3√2 + 2√3)/2.So summarizing:(1) Angle A is 60°, or π/3 radians.(2)(i) sin(2C + A) = (2√2 - √3)/6(2)(ii) Area = (3√2 + 2√3)/2Final Answer(1) boxed{dfrac{pi}{3}} (2)(i) boxed{dfrac{2sqrt{2} - sqrt{3}}{6}} (2)(ii) boxed{dfrac{3sqrt{2} + 2sqrt{3}}{2}}</think>In acute triangle (triangle ABC), we are given the equation (2sin Aleft(ccos B + bcos Cright) = sqrt{3}a).(1) To find the value of angle (A):- Using the Law of Cosines, we express (ccos B) and (bcos C): [ ccos B = frac{a^2 + c^2 - b^2}{2a}, quad bcos C = frac{a^2 + b^2 - c^2}{2a} ]- Adding these expressions, we get: [ ccos B + bcos C = frac{a^2 + c^2 - b^2 + a^2 + b^2 - c^2}{2a} = frac{2a^2}{2a} = a ]- Substituting back into the given equation: [ 2sin A cdot a = sqrt{3}a implies 2sin A = sqrt{3} implies sin A = frac{sqrt{3}}{2} ]- Since (triangle ABC) is acute, (A = 60^circ) or (frac{pi}{3}) radians.(2) Given (c = 2sqrt{2}) and (a = 3):(i) Find (sin(2C + A)):- Using the Law of Sines: [ frac{a}{sin A} = frac{c}{sin C} implies frac{3}{sin 60^circ} = frac{2sqrt{2}}{sin C} implies sin C = frac{sqrt{6}}{3} ]- (cos C = sqrt{1 - sin^2 C} = frac{sqrt{3}}{3})- Using double-angle identities: [ sin 2C = 2sin C cos C = frac{2sqrt{2}}{3}, quad cos 2C = -frac{1}{3} ]- Using the sine addition formula: [ sin(2C + 60^circ) = sin 2C cos 60^circ + cos 2C sin 60^circ = frac{sqrt{2}}{3} - frac{sqrt{3}}{6} = frac{2sqrt{2} - sqrt{3}}{6} ](ii) Find the area of (triangle ABC):- Using the Law of Sines to find (b): [ frac{a}{sin A} = 2sqrt{3} implies b = 2sqrt{3} cdot sin B ]- (sin B = sin(120^circ - C) = frac{3 + sqrt{6}}{6})- Calculating (b): [ b = sqrt{3} + sqrt{2} ]- Using the area formula (frac{1}{2}ab sin C): [ text{Area} = frac{1}{2} cdot 3 cdot (sqrt{3} + sqrt{2}) cdot frac{sqrt{6}}{3} = frac{3sqrt{2} + 2sqrt{3}}{2} ]Final Answers:(1) (boxed{dfrac{pi}{3}}) (2)(i) (boxed{dfrac{2sqrt{2} - sqrt{3}}{6}}) (2)(ii) (boxed{dfrac{3sqrt{2} + 2sqrt{3}}{2}}),<think>Okay, so I need to solve this problem about an acute triangle ABC. Let me start by understanding what's given and what needs to be found.The problem says: In acute triangle ABC, with angles A, B, C opposite sides a, b, c respectively, it's known that 2 sin A (c cos B + b cos C) = √3 a.Part (1) asks for the value of angle A. Then part (2) gives c = 2√2 and a = 3, and asks for sin(2C + A) and the area of the triangle.Let me tackle part (1) first.First, let's recall some trigonometric identities and laws related to triangles. The Law of Cosines might be helpful here. Also, perhaps the projection formula, which states that in any triangle, the length of a side is equal to the sum of the projections of the other two sides onto it. Wait, the expression c cos B + b cos C reminds me of that.In a triangle, the length of side a is equal to b cos C + c cos B. That's one of the projection formulas. Let me confirm that. Yes, in any triangle, a = b cos C + c cos B. Similarly, b = a cos C + c cos A, and c = a cos B + b cos A.So, given that, the expression inside the parentheses in the given equation is c cos B + b cos C. Wait, but according to the projection formula, that's equal to a. So c cos B + b cos C = a.Therefore, substituting into the given equation:2 sin A (a) = √3 aSo, 2 sin A * a = √3 aSince a is a positive length, we can divide both sides by a:2 sin A = √3Therefore, sin A = √3 / 2So, sin A = √3 / 2. Then, angle A is either 60 degrees or 120 degrees. However, since the triangle is acute, all angles must be less than 90 degrees. Therefore, angle A must be 60 degrees.Therefore, the answer to part (1) is 60 degrees, or π/3 radians.Let me check that again. The key step was recognizing that c cos B + b cos C is equal to a. So substituting that into the equation simplifies it directly. Then solving for sin A gives √3 / 2. Since the triangle is acute, angle A can't be 120 degrees, so it's 60 degrees. That seems solid.Moving on to part (2), given c = 2√2 and a = 3.First, (i) asks for sin(2C + A). Since we already found A = 60 degrees in part (1), so A is 60°, so angle A is π/3 radians.We need to find sin(2C + A). To compute this, we need to find the value of angle C first. Then compute 2C + A, and find its sine.But to find angle C, we need more information. Since we know sides a and c, and angle A, perhaps we can use the Law of Sines.Law of Sines states that a / sin A = b / sin B = c / sin C.Given that a = 3, c = 2√2, and angle A = 60°, let's compute sin C.From the Law of Sines:a / sin A = c / sin CSo, 3 / sin 60° = 2√2 / sin CWe know sin 60° = √3 / 2, so:3 / (√3 / 2) = 2√2 / sin CSimplify the left side:3 * 2 / √3 = 6 / √3 = 2√3 (rationalizing the denominator: 6 / √3 = 2√3)Therefore, 2√3 = 2√2 / sin CDivide both sides by 2:√3 = √2 / sin CTherefore, sin C = √2 / √3Simplify that:sin C = √(2/3) = √6 / 3So sin C = √6 / 3.Since the triangle is acute, angle C must be less than 90°, so this is valid because sin C is positive and less than 1, and arcsin(√6/3) is an acute angle.Now, let's find angle C.Compute angle C: sin C = √6 / 3 ≈ 0.816. Let me compute this value. Since √6 ≈ 2.449, so √6 / 3 ≈ 0.816. The arcsin of 0.816 is approximately 54.7 degrees. But maybe we can express it exactly?Alternatively, perhaps use the Law of Cosines to find angle C. Let me check.But since we have two sides and the included angle? Wait, actually, we have sides a and c, and angle A. Wait, perhaps we can find another side or angle.Alternatively, since we have sides a and c, and angle A, maybe use the Law of Cosines to find another side.Wait, let's see. We have a = 3, c = 2√2, angle A = 60 degrees.Law of Cosines says that a² = b² + c² - 2bc cos A.But we can use that to find side b.Wait, but maybe first check the Law of Sines result. We found sin C = √6 / 3. Let's confirm that with the Law of Cosines.Alternatively, let's compute side b first.From the Law of Sines:a / sin A = b / sin B = c / sin CWe have a = 3, sin A = √3 / 2, c = 2√2, sin C = √6 / 3.So, 3 / (√3 / 2) = 2√2 / (√6 / 3)Compute 3 / (√3 / 2) = 3 * 2 / √3 = 6 / √3 = 2√3.And 2√2 / (√6 / 3) = 2√2 * 3 / √6 = 6√2 / √6 = 6√2 / (√2 * √3) ) = 6 / √3 = 2√3. So both ratios are equal, which checks out. Therefore, sin C = √6 / 3 is correct.Now, angle C is arcsin(√6 / 3). Let's compute that. Since the triangle is acute, angle C is acute, so angle C = arcsin(√6 / 3). Maybe we can relate this angle with some known triangle?Alternatively, maybe we can compute angle B first.Since angles in a triangle sum to 180°, angle B = 180° - A - C = 180° - 60° - C = 120° - C.But we might not need angle B for part (i). Wait, part (i) is sin(2C + A). Since A is 60°, 2C + A = 2C + 60°. So we need to find sin(2C + 60°).Alternatively, perhaps express 2C + A in terms of angles we know.But maybe we can compute angle C numerically first?Given that sin C = √6 / 3 ≈ 0.816, so angle C ≈ 54.7356 degrees. Then 2C + A ≈ 2*54.7356 + 60 ≈ 109.4712 + 60 ≈ 169.4712 degrees. Then sin(169.4712°) ≈ sin(180° - 10.5288°) ≈ sin(10.5288°) ≈ 0.183. But this is approximate. However, maybe there's an exact value.Alternatively, perhaps use trigonometric identities.We need to compute sin(2C + A). Let's write this as sin(2C + 60°). Let's express this as sin(2C + 60°). Using the sine addition formula:sin(2C + 60°) = sin 2C cos 60° + cos 2C sin 60°We can compute sin 2C and cos 2C if we know sin C and cos C.We already know sin C = √6 / 3. Therefore, cos C = sqrt(1 - (√6 / 3)^2) = sqrt(1 - 6/9) = sqrt(1 - 2/3) = sqrt(1/3) = 1/√3 = √3 / 3.Therefore:sin 2C = 2 sin C cos C = 2*(√6 / 3)*(√3 / 3) = 2*(√18)/9 = 2*(3√2)/9 = 6√2/9 = 2√2/3.cos 2C = 1 - 2 sin² C = 1 - 2*(6/9) = 1 - 12/9 = 1 - 4/3 = -1/3.Alternatively, cos 2C can be calculated as cos² C - sin² C = (1/3 - 6/9) = (1/3 - 2/3) = -1/3. Same result.Therefore, sin(2C + 60°) = sin 2C * cos 60° + cos 2C * sin 60°Plug in the values:sin 2C = 2√2 / 3cos 60° = 1/2cos 2C = -1/3sin 60° = √3 / 2Therefore:sin(2C + 60°) = (2√2 / 3)(1/2) + (-1/3)(√3 / 2)Simplify:= (√2 / 3) - (√3 / 6)To combine these terms, let's get a common denominator of 6:= (2√2 / 6) - (√3 / 6)= (2√2 - √3) / 6Therefore, sin(2C + A) = (2√2 - √3)/6.Wait, but let me double-check the calculation.First, sin 2C = 2 sin C cos C = 2*(√6/3)*(√3/3) = 2*(√18)/9 = 2*(3√2)/9 = 6√2/9 = 2√2/3. Correct.cos 2C = 1 - 2 sin² C = 1 - 2*(6/9) = 1 - 12/9 = -1/3. Correct.Then:sin(2C + 60°) = sin2C cos60 + cos2C sin60= (2√2/3)(1/2) + (-1/3)(√3/2)= (√2/3) - (√3)/6Yes, that's (2√2 - √3)/6 when combining over denominator 6. So that's the exact value.But let me check if that's simplified correctly. Let me see:√2/3 = 2√2/6, so 2√2/6 - √3/6 = (2√2 - √3)/6. Correct.Therefore, sin(2C + A) = (2√2 - √3)/6.Is this the simplest form? I think so. Alternatively, maybe rationalize or something, but I don't think it can be simplified further.Therefore, part (i) answer is (2√2 - √3)/6.Now, part (ii) asks for the area of triangle ABC.To find the area, we can use several formulas. The most straightforward might be (1/2)ab sin C, but we need to know sides and angles. Alternatively, since we have sides a and c, and angles A and C, perhaps use (1/2)*a*c*sin B.Alternatively, use Heron's formula if we can find all sides. Let me see.First, let's see if we can find all sides. We know a = 3, c = 2√2. Let's find side b.From the Law of Sines: a / sin A = b / sin B = c / sin C.We know a = 3, sin A = √3 / 2.Therefore, 3 / (√3 / 2) = 2√3. So the common ratio is 2√3.Therefore, c / sin C = 2√3. Since c = 2√2, then sin C = c / (2√3) = (2√2) / (2√3) = √2 / √3 = √6 / 3. Which matches our previous result.Similarly, to find side b: b = sin B * 2√3.But angle B = 180° - A - C = 120° - C. So sin B = sin(120° - C).Alternatively, since we can compute angle B once we know angle C.But perhaps a better approach is to use area = (1/2)ab sin C, where a and b are two sides, and C is the included angle. Wait, but in our case, sides a and c are given, but angle B is between them. Wait, actually, sides a and c are not adjacent. Wait, sides a, b, c are opposite angles A, B, C respectively.So, sides a and c are opposite angles A and C. Therefore, they are not adjacent sides unless angle B is between them. Wait, no. In triangle ABC, side a is opposite angle A, so side a is BC, side b is AC, and side c is AB.Therefore, if we have sides a = BC = 3, c = AB = 2√2. So sides BC and AB are 3 and 2√2 respectively. The angle between them would be angle B. Wait, angle B is at point B, between sides AB and BC. So, angle B is between sides c and a.Therefore, the area can be calculated as (1/2)*AB*BC*sin B = (1/2)*c*a*sin B.But we need sin B. Since we can find angle B.Alternatively, we can use another formula. Let's see. If we can find side b, then we can use Heron's formula.Alternatively, from Law of Sines, we can compute side b.We have:a / sin A = 2√3, so b = 2√3 * sin B.But angle B = 180° - A - C = 120° - C. So sin B = sin(120° - C).Alternatively, perhaps compute angle C first. Since we have sin C = √6 / 3, and angle C is acute, we can compute cos C = √3 / 3 as before.Then, angle B = 180° - 60° - C = 120° - C.So, sin B = sin(120° - C) = sin 120° cos C - cos 120° sin C.Compute that:sin 120° = √3 / 2, cos 120° = -1/2.Therefore,sin B = (√3 / 2)(√3 / 3) - (-1/2)(√6 / 3)Simplify:= ( (√3 * √3) / 6 ) + ( (1 * √6) / 6 )= (3 / 6 ) + (√6 / 6 )= (1/2) + (√6 / 6 )Therefore, sin B = 1/2 + √6 / 6 = (3 + √6)/6Therefore, sin B = (3 + √6)/6Thus, side b = 2√3 * sin B = 2√3 * (3 + √6)/6Simplify:= (2√3)(3 + √6)/6= (√3)(3 + √6)/3= [3√3 + √3 * √6]/3= [3√3 + √18]/3= [3√3 + 3√2]/3= √3 + √2So side b = √3 + √2.Therefore, now we have all sides: a = 3, b = √3 + √2, c = 2√2.Now, we can compute the area using Heron's formula.First, compute the semi-perimeter, s = (a + b + c)/2s = (3 + √3 + √2 + 2√2)/2 = (3 + √3 + 3√2)/2. Wait, hold on:Wait, c = 2√2, so adding a, b, c:3 + (√3 + √2) + 2√2 = 3 + √3 + √2 + 2√2 = 3 + √3 + 3√2Therefore, s = (3 + √3 + 3√2)/2But Heron's formula would require calculating sqrt[s(s - a)(s - b)(s - c)], which might be complicated with radicals. Perhaps another method is better.Alternatively, use area = (1/2)ab sin C.Take sides a and b with angle C between them. Wait, but sides a and b are BC and AC, respectively. The angle between them would be angle at C? Wait, no. Let me clarify.Wait, sides a, b, c are opposite angles A, B, C. So, side a is BC, side b is AC, side c is AB.Therefore, angle A is at vertex A, between sides b and c.Angle B is at vertex B, between sides a and c.Angle C is at vertex C, between sides a and b.Therefore, if we take sides a and b (lengths 3 and √3 + √2), the angle between them is angle C.Therefore, area = (1/2)*a*b*sin C.We know sin C = √6 / 3, so:Area = (1/2)*3*(√3 + √2)*(√6 / 3)Simplify:= (1/2)*(√3 + √2)*(√6)Multiply √6 into the parentheses:= (1/2)*(√3*√6 + √2*√6)Simplify each term:√3*√6 = √18 = 3√2√2*√6 = √12 = 2√3Therefore:Area = (1/2)*(3√2 + 2√3) = (3√2 + 2√3)/2Alternatively, 3√2/2 + √3But let me check the calculation again.Area = (1/2)*a*b*sin C = (1/2)*3*(√3 + √2)*(√6/3)The 3 in the numerator and denominator cancels:= (1/2)*(√3 + √2)*√6= (1/2)*(√3*√6 + √2*√6)= (1/2)*(√18 + √12)= (1/2)*(3√2 + 2√3)= (3√2 + 2√3)/2Yes, that's correct.Alternatively, maybe use another formula. Let's confirm using (1/2)*a*c*sin B.Given a = 3, c = 2√2, sin B = (3 + √6)/6.Therefore, area = (1/2)*3*2√2*sin B = 3√2*sin B= 3√2*(3 + √6)/6= (3√2)(3 + √6)/6= (3√2*3 + 3√2*√6)/6= (9√2 + 3√12)/6Simplify:= (9√2 + 3*2√3)/6= (9√2 + 6√3)/6= (3√2 + 2√3)/2Same result. So that's correct.Therefore, the area is (3√2 + 2√3)/2.Alternatively, factor out 1/2:= (3√2)/2 + (2√3)/2 = (3√2)/2 + √3But perhaps the first form is better.Alternatively, rationalizing or other forms, but this seems as simplified as possible.Therefore, the area is (3√2 + 2√3)/2.Let me check if this is the simplest form. Since there are no like terms, yes. Alternatively, factor out √3 or √2, but I don't think that's necessary.Therefore, the answer to part (ii) is (3√2 + 2√3)/2.But let me see if there's another way. Maybe using coordinates.Suppose we place the triangle in coordinate system. Let me try that.Let’s place vertex A at the origin (0,0). Since angle A is 60°, and sides AB and AC meet at angle A. Let’s let side AB be along the x-axis. So vertex B is at (c, 0) = (2√2, 0). Vertex C is somewhere in the plane.We need to find coordinates of C. Since side AC has length b = √3 + √2, and angle at A is 60°, so the coordinates of C can be (b cos 60°, b sin 60°) = ( (√3 + √2)*1/2, (√3 + √2)*√3/2 )But then side BC has length a = 3.So the distance from B(2√2, 0) to C( (√3 + √2)/2, (√3 + √2)√3/2 ) should be 3.Let me compute that distance.Compute coordinates of C:x = (√3 + √2)/2y = (√3 + √2)*√3/2 = (3 + √6)/2Coordinates of B: (2√2, 0)Distance BC:sqrt[ ( ( (√3 + √2)/2 - 2√2 )^2 + ( (3 + √6)/2 - 0 )^2 ) ]Simplify the x-component:( (√3 + √2)/2 - 2√2 ) = (√3 + √2 - 4√2)/2 = (√3 - 3√2)/2The y-component:(3 + √6)/2Therefore, the distance squared is:[ (√3 - 3√2)/2 ]^2 + [ (3 + √6)/2 ]^2Compute first term squared:( (√3)^2 - 2*√3*3√2 + (3√2)^2 ) / 4= (3 - 6√6 + 9*2)/4= (3 - 6√6 + 18)/4= (21 - 6√6)/4Second term squared:(9 + 6√6 + 6)/4 = (15 + 6√6)/4Add both terms:(21 - 6√6 + 15 + 6√6)/4 = (36)/4 = 9Therefore, distance BC = sqrt(9) = 3, which matches a = 3. Therefore, coordinates are correct.Therefore, coordinates of C are ( (√3 + √2)/2, (3 + √6)/2 )Therefore, the area can be calculated using the coordinates. Since we placed A at (0,0), B at (2√2, 0), and C at ( (√3 + √2)/2, (3 + √6)/2 )The area can be computed via the shoelace formula:Area = (1/2)| (x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)) |Plug in:x_A = 0, y_A = 0x_B = 2√2, y_B = 0x_C = (√3 + √2)/2, y_C = (3 + √6)/2Thus,Area = (1/2)| 0*(0 - (3 + √6)/2) + 2√2*((3 + √6)/2 - 0) + (√3 + √2)/2*(0 - 0) |Simplify:= (1/2)| 0 + 2√2*( (3 + √6)/2 ) + 0 |= (1/2)| √2*(3 + √6) | = (1/2)(√2*(3 + √6)) = (3√2 + √12)/2 = (3√2 + 2√3)/2Same result as before. Therefore, the area is indeed (3√2 + 2√3)/2.Therefore, part (ii) answer is (3√2 + 2√3)/2.So summarizing:(1) Angle A is 60°, or π/3 radians.(2)(i) sin(2C + A) = (2√2 - √3)/6(2)(ii) Area = (3√2 + 2√3)/2I need to make sure all steps are correct and there are no miscalculations.Reviewing part (1):Used the projection formula correctly, substituted into the equation, solved for sin A = √3 / 2, concluded A = 60° since triangle is acute. Correct.Part (2)(i):Used Law of Sines to find sin C = √6 / 3, found cos C = √3 / 3, applied sine addition formula for sin(2C + 60°), calculated sin 2C and cos 2C, substituted into formula, simplified to (2√2 - √3)/6. All steps check out.Part (2)(ii):Used Law of Sines to find sin B, found side b, then used area formula (1/2)ab sin C, calculated step-by-step, verified via coordinates and shoelace formula. All steps lead to the same result, so confident in the answer.Yes, all steps seem correct. Therefore, the answers are as above.Final Answer(1) boxed{60^circ}(2)(i) boxed{dfrac{2sqrt{2} - sqrt{3}}{6}}(2)(ii) boxed{dfrac{3sqrt{2} + 2sqrt{3}}{2}}</think>Given an acute triangle (triangle ABC) with angles (A), (B), and (C) opposite sides (a), (b), and (c) respectively, we know that (2sin A(ccos B + bcos C) = sqrt{3}a).(1) To find the value of angle (A):- Using the projection formula, (ccos B + bcos C = a).- Substituting this into the given equation: (2sin A cdot a = sqrt{3}a).- Simplifying, we get (2sin A = sqrt{3}), leading to (sin A = frac{sqrt{3}}{2}).- Since the triangle is acute, (A = 60^circ).(2) Given (c = 2sqrt{2}) and (a = 3):(i) To find (sin(2C + A)):- Using the Law of Sines: (frac{a}{sin A} = frac{c}{sin C}).- Solving for (sin C), we get (sin C = frac{sqrt{6}}{3}).- Calculating (cos C = frac{sqrt{3}}{3}).- Using trigonometric identities: (sin 2C = frac{2sqrt{2}}{3}) and (cos 2C = -frac{1}{3}).- Applying the sine addition formula: (sin(2C + 60^circ) = sin 2C cos 60^circ + cos 2C sin 60^circ).- Simplifying, we get (sin(2C + A) = frac{2sqrt{2} - sqrt{3}}{6}).(ii) To find the area of (triangle ABC):- Using the Law of Sines, we find side (b = sqrt{3} + sqrt{2}).- Using the area formula: (text{Area} = frac{1}{2}ab sin C).- Substituting values and simplifying, we get the area as (frac{3sqrt{2} + 2sqrt{3}}{2}).Final Answers:(1) (boxed{60^circ})(2)(i) (boxed{dfrac{2sqrt{2} - sqrt{3}}{6}})(2)(ii) (boxed{dfrac{3sqrt{2} + 2sqrt{3}}{2}})]

🔑: Solution:# Part (1): Finding the Value of Angle AGiven the equation 2sin Aleft(ccos B+bcos Cright)=sqrt{3}a, we apply the Law of Sines and trigonometric identities to simplify it as follows:[2sin Aleft(sin Ccos B+sin Bcos Cright)=sqrt{3}sin A implies 2sin Asin(B+C)=sqrt{3}sin A]Since B+C=pi-A in a triangle, and sin(pi-A)=sin A, we have:[2sin^2 A = sqrt{3}sin A]Given that sin A > 0 (since A is acute), dividing both sides by sin A yields:[2sin A = sqrt{3} implies sin A = frac{sqrt{3}}{2}]Thus, A = frac{pi}{3}.# Part (2)(i): Finding sin(2C+A)Given c=2sqrt{2}, a=3, and A=frac{pi}{3}, by the Law of Sines:[frac{a}{sin A} = frac{c}{sin C} implies frac{3}{frac{sqrt{3}}{2}} = frac{2sqrt{2}}{sin C}]Solving for sin C gives:[sin C = frac{sqrt{6}}{3}]Since C is acute, cos C = sqrt{1-sin^2 C} = frac{sqrt{3}}{3}. Therefore:[sin 2C = 2sin C cos C = 2 times frac{sqrt{6}}{3} times frac{sqrt{3}}{3} = frac{2sqrt{2}}{3}][cos 2C = cos^2 C - sin^2 C = frac{1}{3} - frac{2}{3} = -frac{1}{3}]Thus, sin(2C+A) = sin(2C+frac{pi}{3}) can be calculated as:[sin(2C+A) = sin 2Ccos frac{pi}{3} + cos 2Csin frac{pi}{3} = frac{2sqrt{2}}{3} times frac{1}{2} + left(-frac{1}{3}right) times frac{sqrt{3}}{2} = frac{2sqrt{2}-sqrt{3}}{6}]# Part (2)(ii): Finding the Area of Triangle triangle ABCGiven c=2sqrt{2}, a=3, and A=frac{pi}{3}, by the Law of Cosines:[a^2 = b^2 + c^2 - 2bccos A implies 9 = b^2 + 8 - 2 times 2sqrt{2}b times frac{1}{2}]Solving for b gives b = sqrt{3} + sqrt{2} (negative solution is rejected). Therefore, the area S of triangle ABC is:[S = frac{1}{2}bcsin A = frac{1}{2} times (sqrt{3}+sqrt{2}) times 2sqrt{2} times frac{sqrt{3}}{2} = frac{3sqrt{2}+2sqrt{3}}{2}]Final Answers:- For angle A: boxed{A = frac{pi}{3}}- For sin(2C+A): boxed{sin(2C+A) = frac{2sqrt{2}-sqrt{3}}{6}}- For the area of triangle ABC: boxed{S = frac{3sqrt{2}+2sqrt{3}}{2}}

🧠:[<think>Okay, so I need to figure out by what factor the surface area of a regular octahedron is greater than the surface area of a cube inscribed in it, where the cube's vertices lie on the edges of the octahedron. Hmm, let's start by visualizing both shapes. A regular octahedron has eight triangular faces, each of which is an equilateral triangle. A cube, of course, has six square faces. When a cube is inscribed in an octahedron, their centers probably coincide, and each vertex of the cube touches an edge of the octahedron. First, maybe I should assign coordinates to both shapes to make it easier. Let me recall that a regular octahedron can be thought of as two square pyramids glued together at their square bases. If I place the octahedron with its vertices along the coordinate axes, the coordinates might be something like (±1, 0, 0), (0, ±1, 0), and (0, 0, ±1). Wait, no, actually, a regular octahedron has six vertices, two along each axis. Let me confirm: yes, if the edge length is √2, then the vertices would be at (±1, 0, 0), (0, ±1, 0), and (0, 0, ±1). But the edge length between two adjacent vertices (like from (1,0,0) to (0,1,0)) would be √[(1-0)^2 + (0-1)^2 + (0-0)^2] = √2. So the edge length is √2 for that octahedron.Alternatively, maybe it's better to start with an octahedron with edge length 1, but I need to check. Wait, actually, if the edge length is 1, then the distance between two adjacent vertices would be 1, so maybe the coordinates would be different. Let me think. For a regular octahedron with edge length 'a', the coordinates can be (0, 0, ±h), and (±h, 0, 0), (0, ±h, 0), but I need to make sure the distance between adjacent vertices is 'a'. So the distance between (h,0,0) and (0,h,0) would be sqrt[(h)^2 + (-h)^2] = sqrt(2h²) = h√2. So if we set h√2 = a, then h = a/√2. So, if the edge length is a, then the vertices are at (0, 0, ±a/√2), (±a/√2, 0, 0), and (0, ±a/√2, 0). Hmm, maybe that's getting too complicated. Alternatively, maybe standard coordinates for a regular octahedron are (±1, 0, 0), (0, ±1, 0), (0, 0, ±1), which gives edge lengths of √2 between adjacent vertices, so edge length √2. Let me check that. For example, the distance between (1,0,0) and (0,1,0) is sqrt[(1-0)^2 + (0-1)^2 + (0-0)^2] = sqrt(1 + 1) = sqrt(2). So yes, if we take those coordinates, the edge length is sqrt(2). So in that case, the octahedron has edge length sqrt(2). But maybe I can just work with that.Alternatively, perhaps it's easier to scale the octahedron so that it has edge length 1. Let me see. If the edge length is 1, then the coordinates would be scaled accordingly. Let me think. Let me denote the edge length as 'a'. So the distance between two vertices like (h,0,0) and (0,h,0) should be 'a'. So sqrt(h² + h²) = a => h√2 = a => h = a / √2. So then the coordinates would be (a/√2, 0, 0), etc., and (0, 0, a/√2). But maybe this is complicating things. Alternatively, let's work with the octahedron with vertices at (±1,0,0), (0,±1,0), (0,0,±1). Then its edge length is sqrt(2). Let's see if that's okay.Now, if I inscribe a cube inside this octahedron such that the cube's vertices lie on the edges of the octahedron. How would that cube be positioned? Let's think. The octahedron has edges between (±1,0,0) and (0,±1,0), (0,0,±1), etc. So each edge of the octahedron is between two vertices where one coordinate is ±1 and the others are 0. For example, an edge from (1,0,0) to (0,1,0). The cube inscribed in the octahedron would have its vertices lying on these edges. So, each edge of the octahedron can have at most one vertex of the cube, since the cube has 12 edges, but the octahedron has 12 edges as well. Wait, a regular octahedron has 12 edges? Let me confirm: each of the 8 triangular faces has 3 edges, but each edge is shared by two faces, so total edges are (8*3)/2 = 12. Yes. So the octahedron has 12 edges, same as the cube. So maybe each edge of the octahedron has a vertex of the cube? But the cube only has 8 vertices. Wait, the cube has 8 vertices. So maybe each vertex of the cube is placed on a different edge of the octahedron, but since the octahedron has 12 edges, there are more edges than needed. So perhaps each vertex of the cube is placed on an edge of the octahedron, but not all edges of the octahedron have a cube vertex.Wait, let's think. Let me visualize the octahedron and the cube inside. The octahedron has two pyramids glued at the base. If I put a cube inside, the cube must be oriented such that its edges are aligned along the octahedron's edges. Wait, the cube's edges are not aligned with the octahedron's edges, but the cube's vertices lie on the octahedron's edges. So each vertex of the cube is somewhere along an edge of the octahedron.Let me consider one octant of the octahedron. In each octant, there's a triangular face. Wait, maybe it's better to parameterize the cube's position. Suppose the cube is centered at the origin, just like the octahedron. Then, the cube's vertices will have coordinates (±k, ±k, ±k) for some k. But these points need to lie on the edges of the octahedron. The edges of the octahedron are between the points like (1,0,0) and (0,1,0). So the parametric equation of such an edge can be represented as (1 - t, t, 0), where t ranges from 0 to 1. Similarly, other edges can be parameterized similarly. So, the cube's vertex lying on this edge would have coordinates (1 - t, t, 0), and similarly for other edges. But the cube's vertex in this edge must also satisfy the cube's coordinates. Since the cube is symmetric, all its vertices should be equidistant from the origin, right? So, if one vertex is (a, a, a), then all vertices would be (±a, ±a, ±a). But wait, in the octahedron's edges, the coordinates can't have three non-zero components. For example, the edge between (1,0,0) and (0,1,0) is in the plane z=0, so any point on that edge has z=0. So if the cube's vertex is on that edge, its z-coordinate must be 0. But if the cube has a vertex at (a, a, 0), then there must be other vertices at (a, a, 0), (a, -a, 0), (-a, a, 0), etc., but wait, that would imply the cube has vertices in the same plane, which is not possible. Wait, maybe my approach is wrong.Alternatively, perhaps the cube is oriented such that its space diagonal is aligned with the octahedron's axis. Wait, the octahedron has three axes: x, y, z. The cube's space diagonal is along one of these? Hmm, maybe not. Let me think again. If the cube is inscribed in the octahedron with vertices on edges, then each vertex of the cube lies on a different edge of the octahedron.Since the octahedron has 12 edges, and the cube has 8 vertices, so 8 edges of the octahedron will have a cube vertex each. Each edge of the octahedron can have at most one cube vertex. So, how are these edges selected? Probably, each pair of opposite edges of the octahedron will have one cube vertex. Wait, the octahedron has edges in three different directions. Let me see: in the octahedron with vertices at (±1,0,0), (0,±1,0), (0,0,±1), the edges are along the directions between these points. For example, edges between (1,0,0) and (0,1,0), which lies in the xy-plane; edges between (1,0,0) and (0,0,1), which is in the xz-plane; similarly others. So each edge is in a different plane. But maybe the cube's vertices lie on edges that are in different octants. Alternatively, let's parameterize the cube. Let me consider an edge of the octahedron, say between (1,0,0) and (0,1,0). A point on this edge can be written as (1 - t, t, 0), where t is between 0 and 1. Similarly, a point on the edge between (1,0,0) and (0,0,1) can be written as (1 - s, 0, s), with s between 0 and 1. If these points are vertices of the cube, then they must satisfy the cube's structure, i.e., all edges of the cube must be equal. But this seems complicated. Maybe there's a better approach. Let's consider the cube inscribed in the octahedron. Since both are regular polyhedrons, there might be a known scaling factor between them. Let me recall that a cube can be inscribed in a regular octahedron such that the edges of the cube are parallel to the edges of the octahedron. Wait, but how?Alternatively, let's think about the dual relationship between the cube and the octahedron. The octahedron is the dual of the cube, meaning that the vertices of one correspond to the centers of the faces of the other. But in this problem, the cube is inscribed in the octahedron with vertices on the edges, not on the faces. So it's a different relationship.Perhaps using coordinates is the way to go. Let me assign coordinates to the octahedron. Let's suppose the octahedron is the standard one with vertices at (±1, 0, 0), (0, ±1, 0), (0, 0, ±1). Then, its edges are of length √2, as we saw earlier. Now, let's consider the cube inscribed within it. Let's assume that the cube is axis-aligned, so its edges are parallel to the coordinate axes. Wait, but if the cube is axis-aligned, then its vertices would be at (±k, ±k, ±k). However, these points need to lie on the edges of the octahedron. Let's see. The edges of the octahedron are between two vertices where one coordinate is ±1 and the others are zero. For example, take the edge from (1, 0, 0) to (0, 1, 0). A point on this edge can be parametrized as (1 - t, t, 0) for t between 0 and 1. Similarly, other edges can be parameterized similarly. But a cube vertex at (k, k, k) would lie on which edge? Let me check.Wait, if the cube is axis-aligned, then its vertices are at (±k, ±k, ±k). But these points would not lie on the edges of the octahedron. For example, the point (k, k, k) would have all three coordinates non-zero, but all edges of the octahedron have two coordinates non-zero at most. So this suggests that the cube is not axis-aligned. Therefore, the cube must be rotated somehow inside the octahedron.Alternatively, maybe the cube is oriented such that each of its vertices lies on a different edge of the octahedron. Let's consider one vertex of the cube. Let's pick an edge of the octahedron, say the edge from (1, 0, 0) to (0, 1, 0). Let a vertex of the cube lie on this edge at some point (1 - t, t, 0). Similarly, another vertex of the cube would lie on the edge from (1, 0, 0) to (0, 0, 1) at (1 - s, 0, s), and another on the edge from (1, 0, 0) to (0, -1, 0), etc. Wait, but this might get complicated. Maybe symmetry can help here.Since both the cube and the octahedron are regular and symmetric, the cube must be placed such that all its vertices are equidistant from the origin, and each lies on an edge of the octahedron. Let's suppose that all the cube's vertices are at a distance 'd' from the origin. Since the cube is inscribed in the octahedron, the maximum distance from the origin to any vertex of the cube must be equal to the distance from the origin to the faces of the octahedron. Wait, but actually, the octahedron's vertices are at distance 1 from the origin (if we take the standard octahedron with vertices at (±1, 0, 0), etc.), and its edges are between these vertices. So the edges themselves are at a distance from the origin. Wait, maybe the cube's vertices lie along the edges of the octahedron, somewhere between the octahedron's vertices and the origin.Let me think again. Let's consider an edge of the octahedron, say from (1, 0, 0) to (0, 1, 0). A point on this edge can be written as (1 - t, t, 0), where t is between 0 and 1. Similarly, a cube vertex lying on this edge would have coordinates (1 - t, t, 0). Since the cube is symmetric, the same parameter 't' should be used for all edges. That is, each edge of the octahedron that hosts a cube vertex will have the cube vertex at the same fraction 't' along the edge. For example, on the edge from (1, 0, 0) to (0, 1, 0), the cube vertex is at (1 - t, t, 0), on the edge from (0, 1, 0) to (0, 0, 1) it's at (0, 1 - t, t), etc. Then, these points must form a cube. To form a cube, the distance between adjacent vertices should be equal. Let's pick two adjacent vertices of the cube. For example, take the vertex on the edge from (1,0,0) to (0,1,0), which is (1 - t, t, 0), and the vertex on the edge from (0,1,0) to (0,0,1), which is (0, 1 - t, t). The distance between these two points should be equal to the edge length of the cube. Let's compute that distance.Distance squared between (1 - t, t, 0) and (0, 1 - t, t) is:[(1 - t) - 0]^2 + [t - (1 - t)]^2 + [0 - t]^2= (1 - t)^2 + (2t - 1)^2 + t^2Let me compute each term:First term: (1 - t)^2 = 1 - 2t + t^2Second term: (2t - 1)^2 = 4t^2 -4t + 1Third term: t^2Summing them up:1 -2t + t^2 + 4t^2 -4t + 1 + t^2= (1 + 1) + (-2t -4t) + (t^2 + 4t^2 + t^2)= 2 -6t + 6t^2So the distance is sqrt(6t^2 -6t + 2)Similarly, let's compute the distance between two other adjacent vertices of the cube. For example, take the vertex on the edge from (1,0,0) to (0,1,0) which is (1 - t, t, 0) and the vertex on the edge from (1,0,0) to (0,0,1) which is (1 - s, 0, s). Wait, but if all parameters are the same 't', then on the edge from (1,0,0) to (0,0,1), the point would be (1 - t, 0, t). So the distance between (1 - t, t, 0) and (1 - t, 0, t) is sqrt[( (1 - t) - (1 - t))^2 + (t - 0)^2 + (0 - t)^2] = sqrt[0 + t^2 + t^2] = sqrt(2t^2) = t√2.But if the cube is regular, these two distances should be equal. That is, the edge length between vertices on adjacent edges of the octahedron should be the same. Therefore, we have:sqrt(6t^2 -6t + 2) = t√2Squaring both sides:6t^2 -6t + 2 = 2t^2Subtract 2t^2 from both sides:4t^2 -6t + 2 = 0Divide equation by 2:2t^2 -3t + 1 = 0Factor:(2t -1)(t -1) = 0So t = 1/2 or t = 1t = 1 would put the cube's vertices at the octahedron's vertices, but then the cube would collapse because the edge length would be zero (since two adjacent vertices would be at the same point). So t = 1/2 is the valid solution.Therefore, t = 1/2. So each cube vertex is halfway along the edges of the octahedron.Therefore, the coordinates of the cube's vertices are at the midpoints of the octahedron's edges. Wait, but hold on. For example, on the edge from (1,0,0) to (0,1,0), the midpoint is (0.5, 0.5, 0). Similarly, on the edge from (0,1,0) to (0,0,1), the midpoint is (0, 0.5, 0.5), and on the edge from (1,0,0) to (0,0,1), the midpoint is (0.5, 0, 0.5). So these points (0.5, 0.5, 0), (0, 0.5, 0.5), and (0.5, 0, 0.5) are vertices of the cube. Wait, but does this form a cube?Let me check the distances. The distance between (0.5, 0.5, 0) and (0, 0.5, 0.5) is sqrt[(0.5)^2 + 0 + ( -0.5)^2] = sqrt(0.25 + 0.25) = sqrt(0.5) = √(1/2) = (√2)/2 ≈ 0.707. Similarly, the distance from (0.5, 0.5, 0) to (0.5, 0, 0.5) is the same. So all edges of the cube would have length √2/2. But a cube with edge length √2/2 has a space diagonal of (√2/2)*√3 = √6/2. Let's check the distance from (0.5, 0.5, 0) to (-0.5, -0.5, 0), which would be sqrt[(1)^2 + (1)^2 + 0] = √2. But that's a face diagonal, not a space diagonal. Wait, but hold on, the cube in question here—wait, actually, the points we have are (0.5, 0.5, 0), (0, 0.5, 0.5), (0.5, 0, 0.5), and their negatives. Let's list all eight vertices of the cube:1. (0.5, 0.5, 0)2. (0.5, -0.5, 0)3. (-0.5, 0.5, 0)4. (-0.5, -0.5, 0)5. (0.5, 0, 0.5)6. (0.5, 0, -0.5)7. (-0.5, 0, 0.5)8. (-0.5, 0, -0.5)Wait, no, that doesn't seem right. Wait, actually, each edge of the octahedron has a midpoint, and there are 12 edges, but the cube only has 8 vertices, so we need to select 8 of these midpoints. Wait, maybe not. Let me think again.Wait, actually, each vertex of the cube lies on a different edge of the octahedron. Since the octahedron has 12 edges, we have 12 possible edges, but the cube has 8 vertices, so 8 edges are used. Each edge of the octahedron can host at most one vertex of the cube. But when we take t = 1/2, the midpoints of the edges of the octahedron, which are 12 points. However, the cube only needs 8 vertices. So perhaps the cube is formed by selecting 8 of these midpoints. Wait, but how?Wait, let me consider that in the standard octahedron with vertices at (±1,0,0), (0,±1,0), (0,0,±1), the edges can be categorized based on their direction. For example, edges in the xy-plane, edges in the xz-plane, and edges in the yz-plane. Each pair of opposite edges (like the edge from (1,0,0) to (0,1,0) and the edge from (-1,0,0) to (0,-1,0)) would have midpoints that are symmetric.If the cube is inscribed such that each vertex is the midpoint of an edge of the octahedron, then the cube would have vertices at these midpoints. Let me check if these midpoints form a cube. Let's take the midpoint of the edge from (1,0,0) to (0,1,0): (0.5, 0.5, 0). Similarly, the midpoint of the edge from (0,1,0) to (0,0,1): (0, 0.5, 0.5). The midpoint of the edge from (0,0,1) to (1,0,0): (0.5, 0, 0.5). If these three points are vertices of the cube, then the cube's edges would be the distances between these points. Let's compute the distance between (0.5, 0.5, 0) and (0, 0.5, 0.5): sqrt[(0.5)^2 + 0 + (-0.5)^2] = sqrt(0.25 + 0.25) = sqrt(0.5) = √2/2. Similarly, the distance between (0.5, 0.5, 0) and (0.5, 0, 0.5) is also √2/2. So these edges are of equal length. Now, to check if all edges are equal, let's consider another vertex. Take the midpoint of the edge from (-1,0,0) to (0,-1,0): (-0.5, -0.5, 0). The distance from (-0.5, -0.5, 0) to (-0.5, 0, -0.5) (midpoint of another edge) would be sqrt[0 + ( -0.5 - 0)^2 + (0 - (-0.5))^2] = sqrt[0.25 + 0.25] = √0.5 again. So all edges of the cube are of length √2/2. Therefore, the cube inscribed with vertices at the midpoints of the octahedron's edges is indeed a cube with edge length √2/2.Wait, but hold on. The cube has 12 edges, each of length √2/2. Then, the surface area of the cube would be 6*(edge length)^2 = 6*( (√2/2)^2 ) = 6*(2/4) = 6*(1/2) = 3. So the surface area of the cube is 3.Now, let's compute the surface area of the octahedron. The regular octahedron with vertices at (±1,0,0), (0,±1,0), (0,0,±1) has 8 triangular faces, each face being an equilateral triangle. The edge length of the octahedron is the distance between two adjacent vertices, which is sqrt[(1-0)^2 + (0 - 1)^2 + (0 - 0)^2] = sqrt(2), as we saw before. So each face is an equilateral triangle with side length sqrt(2). The area of one such triangle is (sqrt(3)/4)*(sqrt(2))^2 = (sqrt(3)/4)*2 = sqrt(3)/2. Therefore, the total surface area of the octahedron is 8*(sqrt(3)/2) = 4*sqrt(3).Therefore, the surface area of the octahedron is 4√3, and the surface area of the cube is 3. Therefore, the ratio of the octahedron's surface area to the cube's surface area is (4√3)/3 ≈ 4*1.732/3 ≈ 6.928/3 ≈ 2.309. But the problem states that the cube is inscribed in the octahedron such that its vertices lie on the edges of the octahedron. In our case, the cube is inscribed with vertices at the midpoints, so is this the correct configuration?Wait, but in the problem statement, it just says "a cube is inscribed in a regular octahedron such that its vertices lie on the edges of the octahedron". There could be different ways to inscribe a cube, but the one with vertices at midpoints is a standard one. However, perhaps the cube is scaled differently. Wait, in our calculation, we assumed t = 1/2, which led to the cube with edge length √2/2. But is there another possible cube that can be inscribed with a different scaling?Wait, when we solved the equation earlier, we found that t = 1/2 is the only valid solution. So maybe that is the only possible cube inscribed in the octahedron with vertices on its edges. Therefore, the surface area ratio is 4√3 / 3. But let's confirm the surface areas again.First, octahedron: edge length sqrt(2). Each face is an equilateral triangle with side length sqrt(2). Area per face is (sqrt(3)/4)*(sqrt(2))² = (sqrt(3)/4)*2 = sqrt(3)/2. Total surface area: 8 faces * sqrt(3)/2 = 4*sqrt(3). Correct.Cube: edge length sqrt(2)/2. Surface area: 6*(sqrt(2)/2)^2 = 6*(2/4) = 6*(1/2) = 3. Correct.So the ratio is 4√3 / 3. However, the problem asks "By what factor is the surface area of the octahedron greater than the surface area of the inscribed cube?" So the factor is 4√3 / 3. But let's check if we need to rationalize or present it differently. Alternatively, maybe there's a simplification.But wait, maybe the octahedron and cube are scaled differently. In my analysis, I assumed the octahedron has vertices at (±1, 0, 0), etc., but maybe in the problem, the octahedron and cube are of different sizes. Wait, no, the problem states "a cube is inscribed in a regular octahedron", so their sizes are related by the inscribing. Therefore, the ratio should be purely based on their geometric relationship, so the factor is 4√3 / 3.But let me verify with another approach to ensure this is correct.Alternative approach: Let's consider the octahedron and cube in terms of edge lengths. Let’s denote the edge length of the octahedron as 'a', and the edge length of the cube as 'b'. We need to find the ratio of their surface areas.For a regular octahedron with edge length 'a', the surface area is 2√3 a². For a cube with edge length 'b', the surface area is 6b². So the ratio is (2√3 a²) / (6b²) = (√3/3)(a² / b²). So we need to find a² / b².In the previous coordinate system, the octahedron had edge length sqrt(2) (distance between (1,0,0) and (0,1,0)), and the cube had edge length sqrt(2)/2. So a = sqrt(2), b = sqrt(2)/2, so a² / b² = (2) / ( (2)/4 ) = 2 / (0.5) = 4. So the ratio is √3 /3 * 4 = (4√3)/3, which matches our previous result. Therefore, the surface area of the octahedron is (4√3)/3 times that of the cube. So the answer should be 4√3 / 3, which can be written as (4/3)√3, but typically rationalized or simplified. However, 4√3 /3 is a standard form.Wait, but hold on. Let me check the formula for the surface area of a regular octahedron. The formula for the surface area of a regular octahedron is 2√3 * a², where 'a' is the edge length. Let me confirm that. Yes, because a regular octahedron has 8 equilateral triangular faces. Each face has area (√3/4)a², so total surface area is 8*(√3/4)a² = 2√3 a². So in our case, when the edge length a = sqrt(2), the surface area is 2√3*(sqrt(2))² = 2√3*2 = 4√3. Correct. And the cube with edge length b = sqrt(2)/2 has surface area 6*(sqrt(2)/2)^2 = 6*(0.5) = 3. So the ratio is 4√3 / 3.Therefore, the factor is 4√3 divided by 3. So the answer should be 4√3/3. Therefore, the surface area of the octahedron is greater than that of the cube by a factor of 4√3/3.But let me check again if this is correct. Another way to think is scaling. If the cube is inscribed in the octahedron, then perhaps there's a scaling factor between them. If the cube is scaled by a factor 'k' relative to the octahedron, then the surface areas would be scaled by k². But in our case, the cube is not a scaled version of the octahedron, since they are different shapes. So the ratio is not a simple square factor but involves the geometry of both shapes.Alternatively, maybe the problem expects a different approach. For example, using the fact that the cube inscribed in the octahedron touches the midpoints of the octahedron's edges. Since the octahedron's edges are divided in a 1:1 ratio by the cube's vertices, the cube is formed by connecting these midpoints. Then, the edge length of the cube can be found by the distance between two adjacent midpoints, which we computed as √2/2. Therefore, the cube's edge length is half the edge length of the octahedron's edge multiplied by √2. Wait, the octahedron's edge length is sqrt(2), so half of that is sqrt(2)/2, but our cube's edge length is sqrt(2)/2. Wait, actually, in the standard octahedron with edge length a = sqrt(2), the cube's edge length is sqrt(2)/2, so the cube is scaled down by a factor of 1/2 in terms of edge length compared to the octahedron's edge length. But since surface area scales with the square, the surface area of the cube would be (1/2)^2 = 1/4 of the octahedron's surface area if they were similar, but they are not similar. However, in our case, the surface area of the octahedron is 4√3, and the cube is 3. Therefore, the ratio is 4√3 /3. Alternatively, since the problem doesn't specify the edge length of the octahedron, we can assume a general regular octahedron and find the ratio in terms of their edge lengths. Let me try that.Let’s denote the edge length of the regular octahedron as 'a'. Then, its surface area is 2√3 a². Now, we need to find the edge length 'b' of the inscribed cube. As before, if we place the octahedron in a coordinate system with vertices at (±a/√2, 0, 0), (0, ±a/√2, 0), (0, 0, ±a/√2). Wait, this might be necessary because earlier we considered a specific octahedron with edge length sqrt(2), but perhaps scaling is needed here.Wait, let's clarify. If we have a regular octahedron with edge length 'a', then the distance between two adjacent vertices is 'a'. The coordinates of the octahedron can be defined as follows: if we set the vertices along the axes at (±h, 0, 0), (0, ±h, 0), (0, 0, ±h), then the edge length between (h,0,0) and (0,h,0) is sqrt[(h)^2 + (h)^2] = h√2. Therefore, to have edge length 'a', we need h√2 = a => h = a / √2. Therefore, the coordinates are (±a/√2, 0, 0), etc. Now, the cube inscribed in this octahedron will have its vertices on the edges of the octahedron. As before, each edge of the octahedron is from (±a/√2, 0, 0) to another vertex. Let's take the edge from (a/√2, 0, 0) to (0, a/√2, 0). A point on this edge can be parameterized as (a/√2 - t, t, 0), where t ranges from 0 to a/√2. Similarly, the cube's vertex on this edge would be at a fraction 'k' along the edge. To maintain symmetry, all cube vertices would be placed at the same fraction 'k' along their respective edges. Following the earlier reasoning, we can set up equations for the edge length of the cube. Let’s denote t = k * a/√2, so the coordinates of the cube vertex on the edge from (a/√2, 0, 0) to (0, a/√2, 0) is (a/√2 - t, t, 0) = (a/√2 - k*a/√2, k*a/√2, 0) = (a/√2(1 - k), a/√2 k, 0). Similarly, the cube vertex on the edge from (0, a/√2, 0) to (0, 0, a/√2) is (0, a/√2(1 - k), a/√2 k). Now, the distance between these two points should be equal to the edge length 'b' of the cube. Let's compute this distance:Δx = [0 - a/√2(1 - k)] = -a/√2(1 - k)Δy = [a/√2(1 - k) - a/√2 k] = a/√2(1 - k - k) = a/√2(1 - 2k)Δz = [a/√2 k - 0] = a/√2 kSo, the squared distance is:[ -a/√2(1 - k) ]² + [ a/√2(1 - 2k) ]² + [ a/√2 k ]²= (a²/2)(1 - k)^2 + (a²/2)(1 - 2k)^2 + (a²/2)k²Factor out a²/2:(a²/2)[ (1 - 2k + k²) + (1 - 4k + 4k²) + k² ]= (a²/2)[1 - 2k + k² + 1 - 4k + 4k² + k²]= (a²/2)[2 -6k +6k²]Therefore, the squared distance is (a²/2)(6k² -6k +2). This distance should equal the squared edge length of the cube, which is b². Additionally, we can consider another edge of the cube, say from the vertex (a/√2(1 -k), a/√2 k, 0) to (a/√2(1 -k), 0, a/√2 k). The distance here is:Δx = 0Δy = [0 - a/√2 k] = -a/√2 kΔz = [a/√2 k - 0] = a/√2 kSquared distance = 0 + (a²/2 k²) + (a²/2 k²) = a²/2 (2k²) = a² k²Therefore, this squared distance is a² k².Since both distances are edges of the cube, they must be equal:(a²/2)(6k² -6k +2) = a² k²Divide both sides by a²:(1/2)(6k² -6k +2) = k²Multiply both sides by 2:6k² -6k +2 = 2k²Subtract 2k² from both sides:4k² -6k +2 =0Divide by 2:2k² -3k +1 =0Which factors as:(2k -1)(k -1) =0Therefore, k=1/2 or k=1. Again, k=1 is invalid as before, so k=1/2.Therefore, the cube's edge length b is sqrt(a² k²) = a k = a*(1/2). Wait, no, wait. The second distance we computed was sqrt(a² k²) = a k. But this edge length must also equal the first edge length. Wait, from the first calculation:squared edge length = (a²/2)(6k² -6k +2). Plugging k=1/2:= (a²/2)(6*(1/4) -6*(1/2) +2) = (a²/2)(1.5 -3 +2) = (a²/2)(0.5) = a²/4Therefore, edge length b = a/2. But from the second edge calculation, we had edge length = a k = a*(1/2) = a/2. So both give the same result. Therefore, the edge length of the cube is a/2. Wait, but this contradicts our previous result where the cube's edge length was sqrt(2)/2 when the octahedron's edge length was sqrt(2). Let me check.If the octahedron has edge length a, then the cube inscribed has edge length b = a/2. Therefore, the surface area of the octahedron is 2√3 a², and the surface area of the cube is 6b² = 6*(a²/4) = (3/2)a². Therefore, the ratio is (2√3 a²) / ( (3/2)a² ) = (2√3)/(3/2) = (4√3)/3. Therefore, the ratio is indeed 4√3/3, regardless of the edge length 'a' of the octahedron. This confirms our previous result. Therefore, the surface area of the regular octahedron is greater than that of the inscribed cube by a factor of 4√3/3.Final AnswerThe surface area of the octahedron is greater than that of the cube by a factor of boxed{dfrac{4sqrt{3}}{3}}.</think>To determine the factor by which the surface area of a regular octahedron is greater than the surface area of an inscribed cube, we start by considering the geometric relationship between the two polyhedrons.1. Coordinates and Edge Lengths: - A regular octahedron with vertices at ((pm 1, 0, 0)), ((0, pm 1, 0)), and ((0, 0, pm 1)) has edge length (sqrt{2}). - The inscribed cube's vertices lie on the edges of the octahedron. By symmetry, each vertex of the cube is at the midpoint of an edge of the octahedron.2. Cube's Edge Length: - The midpoints of the octahedron's edges are at coordinates like ((pm 0.5, pm 0.5, 0)), ((0, pm 0.5, pm 0.5)), and ((pm 0.5, 0, pm 0.5)). - The edge length of the cube is the distance between two adjacent midpoints, which is (sqrt{(0.5)^2 + (0.5)^2} = frac{sqrt{2}}{2}).3. Surface Areas: - Octahedron: The surface area of a regular octahedron with edge length (a) is (2sqrt{3}a^2). For (a = sqrt{2}), the surface area is (2sqrt{3}(sqrt{2})^2 = 4sqrt{3}). - Cube: The surface area of a cube with edge length (b) is (6b^2). For (b = frac{sqrt{2}}{2}), the surface area is (6left(frac{sqrt{2}}{2}right)^2 = 3).4. Ratio Calculation: - The ratio of the surface area of the octahedron to the surface area of the cube is (frac{4sqrt{3}}{3}).Thus, the surface area of the octahedron is greater than the surface area of the inscribed cube by a factor of (boxed{dfrac{4sqrt{3}}{3}}).,<think>Okay, let me try to figure out this problem. So, we have a cube inscribed in a regular octahedron, with the cube's vertices lying on the edges of the octahedron. We need to find the factor by which the surface area of the octahedron is greater than that of the cube. Hmm, alright.First, I should recall some properties of a regular octahedron. A regular octahedron has eight triangular faces, each of which is an equilateral triangle. It has six vertices and twelve edges. Each edge is of equal length. Now, a cube inscribed in this octahedron... so the cube is sitting inside the octahedron, touching its edges. The cube's vertices are on the edges of the octahedron. That means each vertex of the cube is somewhere along an edge of the octahedron, right?I need to relate the surface areas. To do that, I probably need to find the edge lengths of both the octahedron and the cube, and then compute their surface areas. The ratio of the surface areas will give the factor. But how to relate their edge lengths?Let me visualize this. A regular octahedron can be thought of as two square pyramids glued together at their bases. If I inscribe a cube in it, the cube would be oriented such that its edges are aligned along the octahedron's edges? Maybe not exactly, but the cube's vertices are on the octahedron's edges. So each edge of the octahedron might have a vertex of the cube somewhere along it.Wait, but an octahedron has twelve edges, and a cube has twelve edges as well. But no, the cube's vertices are on the octahedron's edges, not necessarily that each octahedron edge has a cube vertex. Wait, a cube has eight vertices. The octahedron has twelve edges. So each cube vertex is on a different edge of the octahedron? So eight edges of the octahedron will have a cube vertex each, and the remaining four edges won't? Hmm, not sure. Maybe each edge of the octahedron is intersected by a cube vertex? But eight vertices can't cover twelve edges. Hmm.Alternatively, maybe each edge of the cube lies along a space diagonal of the octahedron? Wait, but the octahedron's space diagonals... Maybe it's better to assign coordinates to both shapes and work it out algebraically.Yes, that might be a good approach. Let me place the regular octahedron in a coordinate system. A regular octahedron can be centered at the origin with vertices at (±1, 0, 0), (0, ±1, 0), and (0, 0, ±1). Wait, no, actually, that's a different octahedron. Wait, let me confirm.A regular octahedron can be represented with vertices at (±1, 0, 0), (0, ±1, 0), and (0, 0, ±1). That's correct. Each of these vertices is connected to four others. So the edges of the octahedron are between these points. For example, (1, 0, 0) is connected to (0, 1, 0), (0, -1, 0), (0, 0, 1), and (0, 0, -1). Each edge is between a pair of vertices where one coordinate is ±1 and the other two are 0.Now, if we inscribe a cube in this octahedron such that its vertices lie on the edges of the octahedron, we need to find the coordinates of the cube's vertices. Let's consider the cube. A cube has eight vertices, each with coordinates (±a, ±a, ±a) for some scaling factor a. But in this case, the cube is inscribed in the octahedron, so the cube's vertices must lie on the edges of the octahedron.But the octahedron's edges are between the points like (1,0,0) and (0,1,0). So, parametrizing one edge: from (1,0,0) to (0,1,0), we can write points along this edge as (1 - t, t, 0) where t ranges from 0 to 1. Similarly, other edges can be parametrized similarly.A cube inscribed in the octahedron would have its vertices on these edges. Let's consider one vertex of the cube. Suppose it's on the edge from (1,0,0) to (0,1,0). Let's denote this vertex as (1 - t, t, 0). Similarly, another vertex of the cube might be on the edge from (1,0,0) to (0,0,1), which would be (1 - s, 0, s). Wait, but the cube has vertices that are connected by edges of the cube, so these points must be connected in such a way that the cube's edges are orthogonal.Alternatively, perhaps the cube is oriented such that its edges are parallel to the coordinate axes. Wait, but in that case, the cube's edges would not lie on the octahedron's edges. Wait, maybe the cube is oriented differently.Alternatively, maybe the cube is dual to the octahedron. Wait, but the dual of a cube is an octahedron, yes. But the dual would have vertices at the centers of the cube's faces. But in this problem, the cube is inscribed in the octahedron with vertices on the edges, so it's a different relationship.Let me think. If the cube is inscribed in the octahedron, then each vertex of the cube is on an edge of the octahedron. Since the octahedron has edges along its axes, perhaps the cube's vertices are located at points where their coordinates are (k, k, k) but normalized? Wait, no.Wait, let me take a step back. Let's parameterize the octahedron. Let's assume the regular octahedron has edge length L. Then, the surface area of the octahedron is 2√3 L². Because each face is an equilateral triangle with area (√3/4)L², and there are 8 faces, so total surface area is 8*(√3/4)L² = 2√3 L². Wait, actually, no. Wait, the regular octahedron with edge length L has 8 equilateral triangular faces. Each face has area (√3/4)L², so total surface area is 8*(√3/4)L² = 2√3 L². Yes, that's correct.Now, the cube inscribed in the octahedron. Let's denote the edge length of the cube as a. The surface area of the cube is 6a². We need to find the ratio (2√3 L²)/(6a²) = (√3/3)(L/a)². So we need to find the ratio L/a.But how to relate L and a? To do this, we need to express the edge length of the octahedron in terms of the cube's edge length. Or vice versa.Assuming the octahedron is given, and the cube is inscribed in it. Let's set up a coordinate system. Let me consider the octahedron with vertices at (±1, 0, 0), (0, ±1, 0), (0, 0, ±1). So this octahedron has edge length √2, because the distance between (1,0,0) and (0,1,0) is √[(1-0)^2 + (0 - 1)^2 + (0 - 0)^2] = √2. So the edge length L is √2.But perhaps scaling the octahedron to have edge length L, but for simplicity, let's work with this octahedron with edge length √2. Then, we can find the cube inscribed in it.So, in this coordinate system, the edges of the octahedron are between the points like (1,0,0) and (0,1,0), etc. Let's consider the cube inscribed in the octahedron. Let's say that each vertex of the cube lies on an edge of the octahedron. Since the octahedron has 12 edges and the cube has 8 vertices, each vertex of the cube is on a different edge of the octahedron. Each edge of the octahedron can have at most one vertex of the cube, since the cube only has 8 vertices and the octahedron has 12 edges. So 8 edges will have a cube vertex each, and 4 edges won't.Alternatively, maybe each edge of the octahedron is intersected by a cube edge? Hmm, not sure. Let me think of parametrizing the cube's vertices.Suppose a vertex of the cube is on the edge from (1,0,0) to (0,1,0). Let's parametrize this edge as (1 - t, t, 0) where t ∈ [0,1]. Similarly, another vertex could be on the edge from (1,0,0) to (0,0,1), parametrized as (1 - s, 0, s), and so on.But the cube's vertices must satisfy the cube's structure. That is, each vertex must differ by ±a in coordinates, but in this case, since they lie on the octahedron's edges, their coordinates are constrained.Alternatively, maybe all the cube's vertices are at the midpoints of the octahedron's edges? Let's check. The midpoints of the octahedron's edges would be points like (0.5, 0.5, 0), (0.5, 0, 0.5), etc. If these are the cube's vertices, then connecting them would form a cube. Let's see. For instance, (0.5, 0.5, 0), (0.5, 0, 0.5), (0, 0.5, 0.5), etc. Wait, but these points would form an octahedron as well? Wait, no. Let me see. If you take all midpoints of the octahedron's edges, there are 12 edges, so 12 midpoints, which would form a cuboctahedron. But we only have 8 vertices for the cube. So that approach might not work.Wait, perhaps the cube is scaled such that its vertices lie along the edges of the octahedron. Let me think. Suppose the cube is oriented with its space diagonal aligned with the coordinate axes. Wait, the octahedron's vertices are along the coordinate axes. If we inscribe a cube, maybe the cube is oriented such that its edges are at 45 degrees relative to the octahedron's edges?Alternatively, maybe the cube is dual to the octahedron in some way. But the dual of an octahedron is a cube, but that's when the dual is formed by placing vertices at the centers of the original's faces. However, in this problem, the cube's vertices are on the edges of the octahedron, which is different.Hmm, perhaps it's easier to use coordinates. Let me assume the octahedron has vertices at (±1, 0, 0), (0, ±1, 0), (0, 0, ±1). Let me consider a cube inscribed in it. Each vertex of the cube is on an edge of the octahedron. Let's pick one edge of the octahedron, say from (1,0,0) to (0,1,0). Let's place a vertex of the cube here. Let me parameterize this point as (1 - t, t, 0) where t is between 0 and 1.Similarly, another vertex of the cube could be on the edge from (1,0,0) to (0,0,1), which would be (1 - s, 0, s), and another on the edge from (1,0,0) to (0,-1,0), which would be (1 - u, -u, 0), etc. But this seems messy. Maybe there's a symmetry here.Since the cube is regular and the octahedron is regular, the cube must be symmetric with respect to the octahedron's symmetry. So the parameter t should be the same for all edges where the cube's vertices lie.Wait, if all the cube's vertices are equally placed along the octahedron's edges, then t would be the same for all. Let's assume that. So each vertex of the cube is a fraction t along the edge from the octahedron's vertex towards the adjacent vertex.For example, take the edge from (1,0,0) to (0,1,0). The cube's vertex on this edge is at (1 - t, t, 0). Similarly, on the edge from (1,0,0) to (0,0,1), the cube's vertex is at (1 - t, 0, t). Similarly, on the edge from (1,0,0) to (0,-1,0), it's at (1 - t, -t, 0), and so on for all edges.Wait, but the cube has eight vertices. The octahedron has twelve edges. So if we put a cube vertex on each of eight edges, but which edges?Wait, perhaps each vertex of the cube lies on a different edge of the octahedron. But the cube has eight vertices, so we need to choose eight edges of the octahedron. The octahedron has twelve edges, so we need to pick eight of them. But how?Alternatively, maybe each edge of the octahedron that is connected to a positive axis vertex (like (1,0,0)) has two cube vertices: one on each of the four edges connected to (1,0,0). But that would be four edges, each contributing a vertex. Similarly for (-1,0,0), (0,1,0), etc. Wait, but that would be too many. Each vertex of the cube is on an edge of the octahedron. Maybe the cube has one vertex on each of the twelve edges? No, cube only has eight vertices.Wait, perhaps each edge of the octahedron that is adjacent to a positive axis vertex (like (1,0,0)) has one cube vertex. Since the octahedron has three edges per vertex (for each of the six vertices), but we need to pick eight edges. Maybe this is getting too confusing.Let me try another approach. Let's consider that the cube is inscribed in the octahedron such that each vertex of the cube lies on an edge of the octahedron. Let's assign coordinates to the cube's vertices such that they lie on the octahedron's edges. Let's assume the octahedron is the standard one with vertices at (±1,0,0), (0,±1,0), (0,0,±1).Suppose a vertex of the cube is on the edge between (1,0,0) and (0,1,0). Let's parameterize this vertex as (1 - t, t, 0). Similarly, another vertex on the edge between (1,0,0) and (0,0,1) would be (1 - t, 0, t). Another vertex on the edge between (1,0,0) and (0,-1,0) would be (1 - t, -t, 0). Similarly, vertices on edges connected to (-1,0,0) would be (-1 + t, t, 0), (-1 + t, 0, t), etc. Wait, but this might not form a cube.Alternatively, let's consider that all the cube's vertices are of the form (±a, ±a, ±a), but scaled such that they lie on the edges of the octahedron. Wait, the octahedron's edges are between points like (1,0,0) and (0,1,0). So any point on that edge has coordinates (1 - t, t, 0). If the cube's vertex is at (a, a, a), then that point would have to lie on some edge of the octahedron.But (a, a, a) is not on any edge of the octahedron unless two coordinates are zero. Wait, no, the edges of the octahedron are along lines where two coordinates vary and one is zero, or one coordinate varies and the other two are zero? Wait, no. The edges of the octahedron connect vertices that differ in two coordinates. For example, (1,0,0) is connected to (0,1,0), (0,0,1), (0,-1,0), and (0,0,-1). So each edge is along a line where one coordinate decreases from 1 to 0, and another increases from 0 to 1, while the third remains 0. So parametrizing such an edge as (1 - t, t, 0), t ∈ [0,1].Therefore, a point on this edge has coordinates (1 - t, t, 0). Similarly, a point on the edge from (1,0,0) to (0,0,1) is (1 - t, 0, t). So if the cube's vertices are on these edges, then the coordinates of the cube's vertices would be of the form (1 - t, t, 0), (1 - t, 0, t), (1 - t, -t, 0), (1 - t, 0, -t), and similar for the edges connected to (-1,0,0), (0,1,0), (0,-1,0), etc.But the cube has eight vertices. Let's see. If we take one vertex on each of the edges connected to (1,0,0), that would be four vertices: (1 - t, t, 0), (1 - t, 0, t), (1 - t, -t, 0), (1 - t, 0, -t). Similarly, for (-1,0,0), we would have (-1 + t, t, 0), (-1 + t, 0, t), (-1 + t, -t, 0), (-1 + t, 0, -t). That gives eight vertices in total, which is the number of vertices for a cube. So maybe this is the cube.So, the cube's vertices are:1. (1 - t, t, 0)2. (1 - t, 0, t)3. (1 - t, -t, 0)4. (1 - t, 0, -t)5. (-1 + t, t, 0)6. (-1 + t, 0, t)7. (-1 + t, -t, 0)8. (-1 + t, 0, -t)Now, let's check if these points form a cube. For that, the distances between adjacent vertices should be equal. Let's compute the distance between vertex 1 and vertex 2: points (1 - t, t, 0) and (1 - t, 0, t). The distance squared is [(1 - t - (1 - t))^2 + (t - 0)^2 + (0 - t)^2] = [0 + t² + t²] = 2t². So the distance is t√2.Similarly, distance between vertex 1 and vertex 5: (1 - t, t, 0) and (-1 + t, t, 0). Distance squared is [(1 - t - (-1 + t))^2 + (t - t)^2 + (0 - 0)^2] = [(2 - 2t)^2 + 0 + 0] = (2 - 2t)^2. So distance is |2 - 2t|.But in a cube, all edges should have the same length. So, the edges between vertices like 1 and 2 should be equal to edges between vertices like 1 and 5. So, set t√2 = |2 - 2t|.Since t is between 0 and 1 (because the points are on the edges from (1,0,0) towards adjacent vertices), 2 - 2t is positive. Therefore, t√2 = 2 - 2t.Solving for t:t√2 + 2t = 2t(√2 + 2) = 2t = 2 / (√2 + 2)Multiply numerator and denominator by (√2 - 2):t = [2(√2 - 2)] / [(√2 + 2)(√2 - 2)] = [2√2 - 4] / [2 - 4] = [2√2 - 4] / (-2) = (4 - 2√2)/2 = 2 - √2So t = 2 - √2 ≈ 0.5858. Which is between 0 and 1, as required.Therefore, the edge length of the cube is t√2 = (2 - √2)√2 = 2√2 - 2.Wait, let's verify:t = 2 - √2 ≈ 0.5858t√2 = (2 - √2)√2 = 2√2 - (√2 * √2) = 2√2 - 2Which is approximately 2*1.414 - 2 ≈ 2.828 - 2 = 0.828, which seems reasonable.Alternatively, we can compute the edge length as |2 - 2t|. Let's check with t = 2 - √2:2 - 2t = 2 - 2*(2 - √2) = 2 - 4 + 2√2 = -2 + 2√2. The absolute value is 2√2 - 2, which matches the previous result.So, the edge length of the cube is 2√2 - 2. Let's compute this numerically: 2*1.414 - 2 ≈ 2.828 - 2 = 0.828. Hmm.But let's check if this is correct. If the cube has edge length a = 2√2 - 2, then the surface area of the cube is 6a².But first, let's compute the edge length of the octahedron. The octahedron in this coordinate system has vertices at distance 1 from the origin. Wait, but the edge length of the octahedron here is the distance between (1,0,0) and (0,1,0), which is √[(1-0)^2 + (0 - 1)^2 + (0 - 0)^2] = √2. So the octahedron's edge length is √2.But in the problem statement, the octahedron is regular. So, given that, we can compute the surface area of the octahedron as 2√3 L², where L is edge length. So here, L = √2, so surface area is 2√3*(√2)^2 = 2√3*2 = 4√3.The cube's edge length is a = 2√2 - 2. Let's compute a²: (2√2 - 2)^2 = (2√2)^2 - 2*2√2*2 + 2² = 8 - 8√2 + 4 = 12 - 8√2.Therefore, the surface area of the cube is 6*(12 - 8√2) = 72 - 48√2.Wait, but the surface area ratio would be 4√3 / (72 - 48√2). Hmm, that seems complicated. Maybe I made a mistake here. Let's check again.Wait, no. The octahedron in the coordinate system I chose has edge length √2. If we want a general octahedron with edge length L, then the coordinates would be scaled accordingly. But since we're looking for a ratio, scaling factors might cancel out.But let's confirm.Alternatively, maybe we should work with a unit octahedron. Let's assume the octahedron has edge length 1. Then, the surface area would be 2√3. Then, the cube inscribed in it would have edge length a, and we need to find the ratio 2√3 / (6a²) = (√3)/(3a²). So we need to find a in terms of the octahedron's edge length.But in the previous calculation, we considered the octahedron with edge length √2 (distance between (1,0,0) and (0,1,0) is √2). If we want the octahedron to have edge length L = 1, then we need to scale down the coordinates by a factor of √2. So the vertices would be at (±1/√2, 0, 0), (0, ±1/√2, 0), (0, 0, ±1/√2). Then, the edge length between these points is 1.In that case, the coordinates of the octahedron are scaled by 1/√2. Therefore, the parameter t in the previous calculation would also be scaled by 1/√2. Wait, this seems messy. Maybe it's better to recast the problem with the octahedron of edge length 1.Let me try that. Let the octahedron have edge length L = 1. Let me determine the coordinates.The regular octahedron with edge length L can be constructed by placing its vertices at (±k, 0, 0), (0, ±k, 0), (0, 0, ±k), where k is chosen so that the distance between any two adjacent vertices is L = 1.The distance between (k, 0, 0) and (0, k, 0) is √[(k)^2 + (-k)^2] = √(2k²) = k√2. We want this to equal 1, so k√2 = 1 => k = 1/√2.Therefore, the coordinates of the octahedron are (±1/√2, 0, 0), (0, ±1/√2, 0), (0, 0, ±1/√2). Now, let's find the cube inscribed in this octahedron.Following the previous approach, each vertex of the cube is on an edge of the octahedron. Let's take the edge from (1/√2, 0, 0) to (0, 1/√2, 0). A point on this edge is (1/√2 - t, t, 0), where t ranges from 0 to 1/√2.Similarly, a point on the edge from (1/√2, 0, 0) to (0, 0, 1/√2) is (1/√2 - t, 0, t).Following the same logic as before, the cube's edge length would be determined by the parameter t such that the distance between adjacent vertices is equal. Let's denote the parameter here as t (since the coordinates are scaled by 1/√2 compared to the previous case). Then, the cube's edge length would be t√2, and the distance between vertices connected through the octahedron's edges would be 2k - 2t = 2(1/√2) - 2t = √2 - 2t.Setting these equal: t√2 = √2 - 2t.Solving for t:t√2 + 2t = √2t(√2 + 2) = √2t = √2 / (√2 + 2)Multiply numerator and denominator by (√2 - 2):t = √2(√2 - 2) / [(√2 + 2)(√2 - 2)] = (2 - 2√2) / (2 - 4) = (2 - 2√2)/(-2) = (-2 + 2√2)/2 = (-1 + √2)So t = √2 - 1 ≈ 0.4142.Therefore, the edge length of the cube is t√2 = (√2 - 1)√2 = 2 - √2 ≈ 0.5858.So the cube has edge length a = 2 - √2.Now, the surface area of the octahedron is 2√3 L² = 2√3*(1)^2 = 2√3.The surface area of the cube is 6a² = 6*(2 - √2)^2.Let's compute (2 - √2)^2:(2 - √2)^2 = 4 - 4√2 + 2 = 6 - 4√2.Therefore, the surface area of the cube is 6*(6 - 4√2) = 36 - 24√2.Therefore, the ratio of the octahedron's surface area to the cube's surface area is:(2√3) / (36 - 24√2).Simplify this expression:Factor numerator and denominator:Numerator: 2√3Denominator: 12*(3 - 2√2)So ratio = (2√3) / [12(3 - 2√2)] = (√3)/(6(3 - 2√2))To rationalize the denominator:Multiply numerator and denominator by (3 + 2√2):= [√3*(3 + 2√2)] / [6*(3 - 2√2)(3 + 2√2)]Compute denominator: (3)^2 - (2√2)^2 = 9 - 8 = 1Thus, denominator becomes 6*1 = 6So ratio = [√3*(3 + 2√2)] / 6Simplify:= [3√3 + 2√3*√2]/6= [3√3 + 2√6]/6But the question asks for the factor by which the surface area of the octahedron is greater than the cube's. So we have this ratio, but perhaps it can be simplified further or expressed in a different form.Alternatively, maybe there's a mistake in the calculation. Let me check.Wait, wait. If the edge length of the octahedron is 1, then surface area is 2√3*1² = 2√3. The cube's surface area is 6*(2 - √2)^2 = 6*(4 - 4√2 + 2) = 6*(6 - 4√2) = 36 - 24√2. Then, the ratio is 2√3 / (36 - 24√2). Which is the same as √3 / (18 - 12√2). Then rationalizing:Multiply numerator and denominator by (18 + 12√2):Numerator: √3*(18 + 12√2)Denominator: (18)^2 - (12√2)^2 = 324 - 288 = 36Thus, ratio becomes [√3*(18 + 12√2)] / 36 = [18√3 + 12√6]/36 = [3√3 + 2√6]/6Hmm, same as before. So the ratio is (3√3 + 2√6)/6. But this seems complicated. Maybe there's a simpler way to express this.Alternatively, let me compute the numerical value to check:3√3 ≈ 3*1.732 ≈ 5.1962√6 ≈ 2*2.449 ≈ 4.898Total numerator ≈ 5.196 + 4.898 ≈ 10.094Divide by 6: ≈ 10.094 / 6 ≈ 1.682So the ratio is approximately 1.682. Let me check with the approximate surface areas:Octahedron surface area: 2√3 ≈ 3.464Cube surface area: 36 - 24√2 ≈ 36 - 24*1.414 ≈ 36 - 33.936 ≈ 2.064Ratio: 3.464 / 2.064 ≈ 1.679, which is close to our previous calculation. So approximately 1.68.But the problem asks for an exact value. The expression (3√3 + 2√6)/6 is exact, but can we simplify it more?Alternatively, maybe we made a mistake in the parametrization. Let me double-check the cube edge length.Wait, the cube's edge length was found to be a = 2 - √2 when the octahedron has edge length 1. Then the cube's surface area is 6*(2 - √2)^2 = 6*(6 - 4√2) = 36 - 24√2. So the ratio is 2√3 / (36 - 24√2). Let's factor 12 from the denominator:= 2√3 / [12*(3 - 2√2)] = √3 / [6*(3 - 2√2)]Then, rationalizing:Multiply numerator and denominator by (3 + 2√2):= [√3*(3 + 2√2)] / [6*(9 - 8)] = [√3*(3 + 2√2)] / [6*1] = [3√3 + 2√6]/6Yes, same result. Therefore, the exact ratio is (3√3 + 2√6)/6.But maybe this can be written as (sqrt(3)/3 + sqrt(6)/3)/2, but that doesn't seem simpler. Alternatively, factor sqrt(3)/6:= sqrt(3)/6*(3 + 2*sqrt(2)) = sqrt(3)/6*(3 + 2*sqrt(2)).Alternatively, perhaps there's a different approach to get a simpler expression. Let me think again.Alternative approach: Let's consider the octahedron and cube in terms of their geometric duals or other properties.A cube inscribed in an octahedron... wait, the dual of a cube is an octahedron. So if the cube is inscribed in the octahedron, maybe there's a dual relationship here. But in the dual relationship, the vertices of one correspond to the faces of the other. However, in this problem, the cube's vertices are on the edges of the octahedron, so it's a different relationship.Alternatively, perhaps using linear algebra. The regular octahedron can be represented as the set of points (x, y, z) where |x| + |y| + |z| = k for some constant k. In contrast, the cube is the set of points where max(|x|, |y|, |z|) = a. Maybe we can relate these two equations.But the octahedron's edges are where two coordinates vary while the third is zero. The cube inscribed in it would have its vertices where the edges of the octahedron intersect the cube's faces. Wait, maybe not.Alternatively, if we consider the octahedron |x| + |y| + |z| = 1 and the cube max(|x|, |y|, |z|) = a. The intersection of these would be where the cube's vertices lie on the octahedron's edges. For a cube vertex at (a, a, a), but this point must satisfy |a| + |a| + |a| = 1 => 3a = 1 => a = 1/3. But this point (1/3, 1/3, 1/3) is not on an edge of the octahedron, since edges of the octahedron are where two coordinates are non-zero and the third is zero. Therefore, this approach is incorrect.Wait, but the cube's vertices are on the edges of the octahedron. Each edge of the octahedron is a line segment between two vertices of the octahedron. For example, between (1,0,0) and (0,1,0). So any point on this edge can be written as (1 - t, t, 0) for t ∈ [0,1]. Similarly for other edges.Therefore, the cube's vertices must lie on such edges. Let's assume that the cube is such that each of its vertices is at a point (1 - t, t, 0) on an edge of the octahedron. As we did before.But as we calculated earlier, the edge length of the cube is a = 2 - √2 when the octahedron has edge length 1. Therefore, the ratio of surface areas is 2√3 / [6*(2 - √2)^2] = 2√3 / [6*(6 - 4√2)] = 2√3 / (36 - 24√2) = √3 / (18 - 12√2). Rationalizing gives us (3√3 + 2√6)/6.But is this the simplest form? Alternatively, perhaps the answer is supposed to be 3, but that contradicts our approximate calculation of ~1.68. So maybe not. Alternatively, maybe there's a miscalculation.Wait, let's check once again:If the octahedron has edge length 1, its surface area is 2√3*(1)^2 = 2√3.The cube's edge length is a = 2 - √2 ≈ 0.5858. Then, the cube's surface area is 6*(2 - √2)^2.Compute (2 - √2)^2:= 4 - 4√2 + 2 = 6 - 4√2.Thus, cube's surface area is 6*(6 - 4√2) = 36 - 24√2 ≈ 36 - 33.94 ≈ 2.06.Therefore, the ratio is 2√3 / (36 - 24√2) ≈ 3.464 / 2.06 ≈ 1.68. So the surface area of the octahedron is approximately 1.68 times that of the cube.But the problem states "by what factor is the surface area of the octahedron greater than the surface area of the inscribed cube?" So the answer should be this ratio. But we need to present it in exact form.We have the ratio as (3√3 + 2√6)/6. Let's check if this can be simplified.Factor numerator:3√3 + 2√6 = 3√3 + 2√(2*3) = 3√3 + 2√2√3 = √3(3 + 2√2)Therefore, ratio = √3(3 + 2√2)/6 = (3 + 2√2)√3 / 6 = (3√3 + 2√6)/6.Yes, that's the same as before. So this is the simplified exact form.But perhaps the problem expects a rationalized form or another expression. Alternatively, maybe there's a different approach where the ratio comes out more neatly.Wait, let me consider scaling factors.Suppose we have a regular octahedron and inscribe a cube in it. The octahedron and cube are both regular polyhedra, so their sizes are related by a scaling factor. The edge length of the cube can be related to the edge length of the octahedron via some ratio, and since surface area scales with the square of the scaling factor, the ratio of surface areas would be the square of the reciprocal of that scaling factor.Wait, if the cube is inscribed in the octahedron, then the scaling factor from cube to octahedron is such that the octahedron is larger by some factor. Let me call the edge length of the octahedron L and the edge length of the cube a. Then, the surface area ratio is (Surface area of octahedron)/(Surface area of cube) = (2√3 L²)/(6a²) = (√3/3)(L/a)².So we need to find (L/a)² * √3/3. Therefore, if we can find L/a, then we can compute the ratio.From our previous calculation, with the octahedron of edge length L = 1, we found a = 2 - √2. Therefore, L/a = 1/(2 - √2). Rationalizing:1/(2 - √2) = (2 + √2)/[(2 - √2)(2 + √2)] = (2 + √2)/(4 - 2) = (2 + √2)/2 = 1 + (√2)/2.Therefore, (L/a)² = [1 + (√2)/2]^2 = 1 + √2 + ( (√2)/2 )² = 1 + √2 + (2)/4 = 1 + √2 + 0.5 = 1.5 + √2 ≈ 1.5 + 1.414 ≈ 2.914.Then, multiplying by √3/3:(√3/3)*(1.5 + √2) ≈ (1.732/3)*(1.5 + 1.414) ≈ 0.577*(2.914) ≈ 1.68, which matches our previous result.But this still doesn't give us a neat exact expression. However, the exact ratio is (√3/3)*( (2 + √2)/2 )².Wait, let's compute it step by step.First, (L/a) = 1/(2 - √2) = (2 + √2)/2.Therefore, (L/a)² = (2 + √2)^2 / 4 = (4 + 4√2 + 2)/4 = (6 + 4√2)/4 = (3 + 2√2)/2.Therefore, the surface area ratio is (√3/3)*(3 + 2√2)/2 = (√3*(3 + 2√2))/6 = (3√3 + 2√6)/6.Which is the same as before. So there's no further simplification, it seems.Therefore, the exact factor by which the surface area of the octahedron is greater than that of the cube is (3√3 + 2√6)/6.But let me check if this is equivalent to another expression. Suppose we factor out √3:= √3*(3 + 2√2)/6 = (3 + 2√2)/(2√3).But this is not necessarily simpler.Alternatively, rationalizing:(3 + 2√2)/(2√3) = [ (3 + 2√2) / (2√3) ] * (√3/√3) = [ (3√3 + 2√6 ) / 6 ].Which brings us back to the same expression.Therefore, I think this is as simplified as it gets. So the answer is (3√3 + 2√6)/6. However, I need to check if this can be written as √3 + √6)/2 or something similar, but testing numerically:(3√3 + 2√6)/6 ≈ (5.196 + 4.899)/6 ≈ 10.095/6 ≈ 1.682.(√3 + √6)/2 ≈ (1.732 + 2.449)/2 ≈ 4.181/2 ≈ 2.09, which is different. So that's not it.Alternatively, perhaps the answer is 3. But that doesn't match the approximate value. Therefore, the exact value is (3√3 + 2√6)/6.But let me see the original problem statement again: "A cube is inscribed in a regular octahedron such that its vertices lie on the edges of the octahedron. By what factor is the surface area of the octahedron greater than the surface area of the inscribed cube?"Given that the answer is likely expected to be in a simplified radical form, perhaps expressed as (3√3 + 2√6)/6, but maybe there's a different approach leading to a different expression.Alternatively, perhaps I made a mistake in assuming the cube's edge length. Let me check again.We found that for the octahedron with edge length L = 1, the cube's edge length a = 2 - √2. Therefore, a ≈ 0.5858.But let's check the distance between two adjacent vertices of the cube. For example, between (1 - t, t, 0) and (1 - t, 0, t), where t = √2 - 1 ≈ 0.4142.So (1 - t) ≈ 1 - 0.4142 ≈ 0.5858.So the coordinates are (0.5858, 0.4142, 0) and (0.5858, 0, 0.4142). The distance between them is sqrt[(0.5858 - 0.5858)^2 + (0.4142 - 0)^2 + (0 - 0.4142)^2] = sqrt[0 + 0.1716 + 0.1716] = sqrt[0.3432] ≈ 0.586, which matches a ≈ 0.5858. So that's correct.Therefore, the cube's edge length is indeed a = 2 - √2 when L = 1.Thus, the surface area ratio is (3√3 + 2√6)/6.But let me check if there's an alternative expression. Let's compute (3√3 + 2√6)/6:= (3√3)/6 + (2√6)/6 = (√3)/2 + (√6)/3 ≈ 0.866 + 0.816 ≈ 1.682, which matches our approximation.Therefore, I think this is the correct answer.Final AnswerThe surface area of the octahedron is greater than that of the cube by a factor of boxed{dfrac{3sqrt{3} + 2sqrt{6}}{6}}.</think>To solve the problem of finding the factor by which the surface area of a regular octahedron is greater than that of an inscribed cube, we start by considering the geometric relationship between the two polyhedra.1. Coordinates and Edge Lengths: - A regular octahedron with edge length ( L ) can be represented with vertices at ( (pm 1, 0, 0) ), ( (0, pm 1, 0) ), and ( (0, 0, pm 1) ). The edge length of this octahedron is ( sqrt{2} ). - The inscribed cube's vertices lie on the edges of the octahedron. We parameterize these vertices as points along the edges of the octahedron. For example, a vertex on the edge from ( (1, 0, 0) ) to ( (0, 1, 0) ) can be represented as ( (1 - t, t, 0) ).2. Determining the Parameter ( t ): - By solving for the parameter ( t ) such that the distances between adjacent vertices of the cube are equal, we find ( t = 2 - sqrt{2} ).3. Edge Length of the Cube: - The edge length ( a ) of the cube is calculated as ( tsqrt{2} = (2 - sqrt{2})sqrt{2} = 2sqrt{2} - 2 ). However, simplifying for a unit octahedron (edge length 1), the cube's edge length is ( 2 - sqrt{2} ).4. Surface Areas: - The surface area of the octahedron with edge length 1 is ( 2sqrt{3} ). - The surface area of the cube with edge length ( 2 - sqrt{2} ) is ( 6(2 - sqrt{2})^2 = 6(6 - 4sqrt{2}) ).5. Ratio Calculation: - The ratio of the surface areas is given by ( frac{2sqrt{3}}{6(6 - 4sqrt{2})} ). Rationalizing the denominator and simplifying, we find the exact ratio to be ( frac{3sqrt{3} + 2sqrt{6}}{6} ).Thus, the factor by which the surface area of the octahedron is greater than the surface area of the inscribed cube is:[boxed{dfrac{3sqrt{3} + 2sqrt{6}}{6}}]]

🔑:Solution.Square ABCD is the common base of the regular pyramids SABCD and S_1ABCD, which together form the given regular octahedron, O is the center of this square, MNPQM_1N_1P_1Q_1 is the cube mentioned in the problemFig. 11.36Fig. 11.37K is the center of the square MNPQ - the intersection point of SO and the plane of the square MNPQ (Fig. 11.36). Let AD=a, MQ=2b. Then MP=2bsqrt{2}, AC=asqrt{2}, SO=AO=frac{asqrt{2}}{2}, SK=SO-KO=frac{asqrt{2}}{2}-b. triangle SMP sim triangle SAC. From the similarity, it follows: frac{MP}{AC}=frac{SK}{SO}; quad frac{2bsqrt{2}}{asqrt{2}}=frac{frac{asqrt{2}}{2}-b}{frac{asqrt{2}}{2}}; 2bsqrt{2}=asqrt{2}-2b; 2b=frac{asqrt{2}}{sqrt{2+1}}=asqrt{2}(sqrt{2-1}). The surface area of the octahedron S_1=8S_{triangle ASD}=8 cdot frac{a^2sqrt{3}}{4}=2a^2sqrt{3}. The surface area of the cube S_2=6S_{MNPQ}=6 cdot (2b)^2= =6(asqrt{2}(sqrt{2}-1))^2=12a^2(sqrt{2}-1)^2=12a^2(3-2sqrt{2})Then frac{S_1}{S_2}=frac{2a^2sqrt{3}}{12a^2(3-2sqrt{2})}=frac{sqrt{3}(3+2sqrt{2})}{6}.Answer: frac{sqrt{3}(3+2sqrt{2})}{6}.11.140. Find the volume of a regular triangular pyramid (Fig. 11.37), where the plane angle at the vertex is 90^circ, and the distance between a lateral edge and the opposite side is d.## Solution.We will assume that B is the vertex of the pyramid, and triangle ASC is the base. SD is the height of the right isosceles triangle ASC, dropped to the hypotenuse AC. Then SD is the common perpendicular to the skew lines BS and AC, and its length is the distance between them, SD=d. Therefore, AC=2SD=2d, S_{triangle ASC}=frac{1}{2}AC cdot SD=d^2, BS=AS=SDsqrt{2}. The volume of the pyramid V=frac{1}{3}S_{triangle ASC} cdot BS=frac{d^3}{3}sqrt{2}.Answer: frac{d^3}{3}sqrt{2}.

🧠:[<think>Okay, let's see. There are three employees who got a total bonus of 2970 rubles. The problem gives relationships between their bonuses. The second one received 1/3 of what the first got plus 180 rubles. The third got 1/3 of the second's amount plus 130 rubles. I need to figure out how much each person received. Hmm, algebra time.Let me denote the first employee's bonus as, let's say, x rubles. Then the second one would be (1/3)x + 180. The third one would be 1/3 of the second's amount plus 130. So first, let me write that out step by step.First employee: xSecond employee: (1/3)x + 180Third employee: (1/3)*(second employee) + 130So substituting the second into the third, the third employee's bonus would be (1/3)*[(1/3)x + 180] + 130Now, the total is x + [(1/3)x + 180] + [(1/3)*(1/3x + 180) + 130] = 2970Let me compute each part step by step.First, let's expand the third employee's bonus:(1/3)*(1/3x + 180) = (1/9)x + 60Then add 130: (1/9)x + 60 + 130 = (1/9)x + 190So now, total bonuses:x (first) + (1/3x + 180) (second) + (1/9x + 190) (third) = 2970Combine like terms:x + 1/3x + 1/9x + 180 + 190 = 2970First, combine the x terms. Let's convert them to ninths to add:x = 9/9x1/3x = 3/9x1/9x = 1/9xSo total x terms: 9/9x + 3/9x + 1/9x = 13/9xThen the constants: 180 + 190 = 370Therefore, the equation becomes:13/9x + 370 = 2970Subtract 370 from both sides:13/9x = 2970 - 3702970 - 370 is 2600? Let me check: 2970 - 300 = 2670, then +30 more is 2700, so 2970 - 370 = 2600. Wait, no. 2970 - 300 = 2670, then subtract 70 more: 2670 -70 = 2600. Yes, correct.So 13/9x = 2600To solve for x, multiply both sides by 9/13:x = 2600 * (9/13)Compute 2600 divided by 13 first. Let's see, 13*200=2600. So 2600/13=200Thus, x=200*9=1800So the first employee received 1800 rubles.Now, the second employee is (1/3)x + 180 = (1 would have transitive triples.. Thus, the answer is 3.Final Answerboxed{3}</think>To determine the maximum size of set ( S ) such that the sum of any two distinct elements is not divisible by 17, we need to ensure that the sum of any two distinct elements does not equate to 0 modulo 17. This requires that for any two elements ( a ) and ( b ), ( a + b notequiv 0 mod 17 ).1. Key Insight: For any two elements ( a ) and ( b ), ( a + b notequiv 0 mod 17 ). This implies that for each element ( a ), there cannot be an element. So, but, if x=0.6. So, but, if x=0.6. So, but, if x=0.6. So, but, the equation is set to 1. But, the equation is set to 1. 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Let me call the first employee's bonus "A", the second one's "B", and the third one's "C". So, according to the problem, A + B + C = 2970 rubles. That's the total.Now, the second part: the second employee received 1/3 of what the first received plus 180 rubles. So, translating that into an equation, B = (1/3)A + 180. Got that.Then, the third employee received 1/3 of the second's amount plus 130 rubles. So, C = (1/3)B + 130. Alright.So, we have three equations:1. A + B + C = 29702. B = (1/3)A + 1803. C = (1/3)B + 130Since we have expressions for B and C in terms of A, maybe we can substitute those into the first equation and solve for A first. Let's try that.Starting with equation 2: B is (1/3)A + 180. So, substitute that into equation 3 for B. So, C = (1/3)( (1/3)A + 180 ) + 130. Let me compute that.First, (1/3) of (1/3)A is (1/9)A. Then, (1/3) of 180 is 60. So, C = (1/9)A + 60 + 130. Combining the constants, 60 + 130 is 190. So, C = (1/9)A + 190.So now, we have B and C in terms of A. Let's plug both into the first equation.Original equation 1: A + B + C = 2970.Substituting B and C:A + [ (1/3)A + 180 ] + [ (1/9)A + 190 ] = 2970.Now, let's combine like terms. Let's first handle all the terms with A.A is the same as (9/9)A. Then (1/3)A is (3/9)A, and (1/9)A is (1/9)A. Adding those together:(9/9 + 3/9 + 1/9)A = (13/9)A.Then, the constants are 180 + 190 = 370.So, the equation becomes:(13/9)A + 370 = 2970.Now, subtract 370 from both sides:(13/9)A = 2970 - 370.2970 minus 370. Let's compute that. 2970 - 300 is 2670, then add back 30 to get 2670 + 30 = 2700. Wait, no. Wait, 2970 - 370: 2970 - 300 is 2670, then subtract 70 more: 2670 - 70 = 2600. Wait, no, that's not right. Wait, 2970 - 370. 2970 - 300 is 2670, then subtract 70: 2670 - 70 = 2600. Yeah, so (13/9)A = 2600.Then, to solve for A, multiply both sides by (9/13):A = 2600 * (9/13).Let me compute that. 2600 divided by 13. Let's see, 13 times 200 is 2600, so 2600 /13 = 200. Then, 200 *9 = 1800. So, A is 1800 rubles.Okay, so the first employee received 1800 rubles.Now, let's find B. From equation 2: B = (1/3)A + 180.Plugging in A = 1800:B = (1/3)*1800 + 180. Compute that: 1800 divided by 3 is 600. So, 600 + 180 = 780. So, B is 780 rubles.Now, let's find C. From equation 3: C = (1/3)B + 130.Plugging in B = 780:C = (1/3)*780 + 130. 780 divided by 3 is 260. So, 260 + 130 = 390. So, C is 390 rubles.Let me check if these add up to 2970.A = 1800, B = 780, C = 390. 1800 + 780 = 2580. 2580 + 390 = 2970. Perfect, that matches the total. So, the calculations seem correct.But let me just double-check the equations to make sure I didn't make any substitution errors.Second employee: B = (1/3)A + 180. So, (1/3)*1800 is 600, plus 180 is 780. Correct.Third employee: C = (1/3)B + 130. (1/3)*780 is 260, plus 130 is 390. Correct.Total: 1800 + 780 + 390 is 2970. Perfect.So, the answer is:First employee: 1800 rubles,Second employee: 780 rubles,Third employee: 390 rubles.Final AnswerThe first employee received boxed{1800} rubles, the second received boxed{780} rubles, and the third received boxed{390} rubles.</think>Let ( A ), ( B ), and ( C ) represent the bonuses received by the first, second, and third employees respectively. We know the following:1. The total bonus is 2970 rubles: [ A + B + C = 2970 ]2. The second employee's bonus is given by: [ B = frac{1}{3}A + 180 ]3. The third employee's bonus is given by: [ C = frac{1}{3}B + 130 ]First, substitute the expression for ( B ) into the expression for ( C ):[C = frac{1}{3}left(frac{1}{3}A + 180right) + 130][C = frac{1}{9}A + 60 + 130][C = frac{1}{9}A + 190]Now substitute ( B ) and ( C ) into the total bonus equation:[A + left(frac{1}{3}A + 180right) + left(frac{1}{9}A + 190right) = 2970]Combine the terms involving ( A ):[A + frac{1}{3}A + frac{1}{9}A + 180 + 190 = 2970][left(frac{9}{9}A + frac{3}{9}A + frac{1}{9}Aright) + 370 = 2970][frac{13}{9}A + 370 = 2970]Solve for ( A ):[frac{13}{9}A = 2970 - 370][frac{13}{9}A = 2600][A = 2600 times frac{9}{13}][A = 1800]Next, find ( B ) using the value of ( A ):[B = frac{1}{3} times 1800 + 180][B = 600 + 180][B = 780]Finally, find ( C ) using the value of ( B ):[C = frac{1}{3} times 780 + 130][C = 260 + 130][C = 390]The first employee received (boxed{1800}) rubles, the second received (boxed{780}) rubles, and the third received (boxed{390}) rubles.]

🔑:## Solution.Let x rubles be the bonus of the first employee. Then frac{1}{3} x+180 rubles for the second; frac{1}{3}left(frac{1}{3} x+180right)+130 rubles for the third. According to the condition, x+frac{1}{3} x+180+frac{1}{3}left(frac{1}{3} x+180right)+130=2970, from which x=1800 rubles - the bonus of the first employee; frac{1}{3} cdot 1800+180=780 rubles - the bonus of the second; frac{1}{3} cdot 780+130=390 rubles - the third.Answer: 1800, 780, and 390 rubles.

🧠:[<think>Okay, let's try to solve this hyperbola problem. Hmm, hyperbola equation is given as x²/a² - y²/b² = 1. Focus at F. Then, we need to draw a line l through F perpendicular to the x-axis. Since it's perpendicular to the x-axis, that's a vertical line. So line l is vertical, passing through F. Then, this line intersects the two asymptotes at points A and B. Then, P is one of the intersection points of line l with the hyperbola. O is the origin. Then, we have vector OP = mOA + nOB, and m*n = 2/9. We need to find the eccentricity of the hyperbola. The options are given.First, let's recall some hyperbola basics. The standard hyperbola equation is x²/a² - y²/b² = 1. The foci are at (±c, 0), where c² = a² + b². Eccentricity e = c/a. So we need to find e, which would be c/a. The problem gives us some conditions involving points A, B, P, and vectors. Let's break this down step by step.First, identify the coordinates of focus F. Since the hyperbola is centered at the origin and opens left and right, the foci are on the x-axis. The coordinates of F would be (c, 0). But wait, the problem just says "a focus at F", so since there are two foci, but since the line l is drawn through F and is vertical, and intersects the asymptotes at A and B, maybe we need to consider which focus. But since the hyperbola is symmetric, maybe it doesn't matter. But let's just assume F is (c, 0). Then, the vertical line through F is x = c. So line l is x = c.Now, find points A and B where line l intersects the asymptotes. The asymptotes of the hyperbola are y = ±(b/a)x. So substituting x = c into the asymptotes equations:For asymptote y = (b/a)x, when x = c, y = (b/a)c. So point A is (c, (b/a)c).For asymptote y = -(b/a)x, when x = c, y = -(b/a)c. So point B is (c, -(b/a)c).So points A and B are (c, bc/a) and (c, -bc/a).Now, point P is where line l intersects the hyperbola. So substitute x = c into the hyperbola equation:(c)²/a² - y²/b² = 1=> c²/a² - y²/b² = 1We can solve for y:y²/b² = c²/a² - 1Multiply both sides by b²:y² = b²(c²/a² - 1)But c² = a² + b², so substitute that in:y² = b²((a² + b²)/a² - 1) = b²( (a² + b² - a²)/a² ) = b²(b²/a²) = b⁴/a²Therefore, y = ±b²/aTherefore, the points P are (c, b²/a) and (c, -b²/a). Since the problem says "Let P be one of the points where l intersects the hyperbola," so P can be either of these. Let's pick one, say (c, b²/a). The other one would be similar, just with a negative sign, but since we're dealing with vectors and scalars m and n, maybe the sign won't matter, or it would just affect the sign of m and n. But let's proceed with (c, b²/a).Now, we need to express vector OP as a linear combination of vectors OA and OB. So:OP = m OA + n OBFirst, let's write down the coordinates of points O, A, B, P.O is the origin: (0, 0)A is (c, bc/a)B is (c, -bc/a)P is (c, b²/a)So vector OA is (c, bc/a), vector OB is (c, -bc/a), and vector OP is (c, b²/a).So, the equation is:(c, b²/a) = m*(c, bc/a) + n*(c, -bc/a)Let's write the components separately.For the x-component:c = m*c + n*cFor the y-component:b²/a = m*(bc/a) + n*(-bc/a)Simplify x-component equation:c = c*(m + n)Divide both sides by c (since c ≠ 0):1 = m + nSo, m + n = 1. (Equation 1)Now, y-component equation:b²/a = (mbc/a) - (nbc/a)Factor out (bc/a):b²/a = (bc/a)*(m - n)Multiply both sides by a/(bc):(b²/a) * (a/(bc)) = m - nSimplify left side:(b²/a)*(a/(bc)) = (b² * a)/(a * b c) )= b²/(b c) = b/cSo:b/c = m - nTherefore, m - n = b/c. (Equation 2)Now, we have two equations:1) m + n = 12) m - n = b/cWe can solve for m and n. Let's add equations 1 and 2:( m + n ) + ( m - n ) = 1 + b/c => 2m = 1 + b/c => m = (1 + b/c)/2Similarly, subtract equation 2 from equation 1:( m + n ) - ( m - n ) = 1 - (b/c ) => 2n = 1 - b/c => n = (1 - b/c)/2Now, the problem states that m*n = 2/9. So let's compute m*n:m * n = [(1 + b/c)/2] * [(1 - b/c)/2] = [1 - (b/c)^2]/4Set this equal to 2/9:[1 - (b/c)^2]/4 = 2/9Multiply both sides by 4:1 - (b/c)^2 = 8/9Therefore:(b/c)^2 = 1 - 8/9 = 1/9Take square root:b/c = 1/3So, b/c = 1/3. Therefore, c = 3b.But wait, we know that for hyperbola, c² = a² + b². Since c = 3b, substitute into that equation:(3b)² = a² + b² => 9b² = a² + b² => 8b² = a² => a² = 8b²Therefore, a = sqrt(8) b = 2*sqrt(2) bThen, the eccentricity e = c/a = 3b / (2*sqrt(2) b ) = 3/(2*sqrt(2)) = 3*sqrt(2)/4Which is option A: 3√2/4Wait, but let me verify once again. Let's check steps.Starting from m*n = 2/9. So m*n = [1 - (b/c)^2]/4 = 2/9Thus, 1 - (b/c)^2 = 8/9 => (b/c)^2 = 1/9 => b/c = 1/3 => c = 3bThen, c² = a² + b² => (3b)^2 = a² + b² => 9b² = a² + b² => a² = 8b² => a = 2√2 bThus, e = c/a = 3b / (2√2 b ) = 3/(2√2) = 3√2/4, which is option A.But wait, the answer options have option A as 3√2/4, which matches. The answer should be A. However, let me check if there is any mistake in coordinates.Wait, when we found point P, substituting x = c into hyperbola equation gives y² = (c²/a² - 1)*b². But c² = a² + b², so substituting:y² = ((a² + b²)/a² - 1)*b² = (1 + b²/a² - 1)*b² = (b²/a²)*b² = b^4/a², so y = ±b²/a. So point P is (c, ±b²/a). Then vector OP is (c, b²/a). Then OA is (c, bc/a) and OB is (c, -bc/a). Then decomposition is correct.So the components lead us to m + n = 1 and m - n = b/c. Then solving gives m = (1 + b/c)/2, n = (1 - b/c)/2. Then m*n = (1 - (b/c)^2)/4. So equate to 2/9. Then (1 - (b/c)^2)/4 = 2/9 => (1 - (b/c)^2) = 8/9 => (b/c)^2 = 1/9 => b/c = 1/3. Then c = 3b. Then, using c² = a² + b² => 9b² = a² + b² => a² = 8b² => a = 2√2 b. Then e = c/a = 3b/(2√2 b) = 3/(2√2) = 3√2/4. Which is option A. So the answer should be A.But let me check again if there is a mistake in the direction. The line l is vertical through focus F, so x = c. Asymptotes are y = ±(b/a)x. So when x = c, y = ±(b/a)c. So points A and B are (c, (b/a)c) and (c, -(b/a)c). Then OP is (c, b²/a). So the vectors OA is (c, bc/a), OB is (c, -bc/a), OP is (c, b²/a). Then, yes, the components:For x: c = m*c + n*c => 1 = m + nFor y: b²/a = m*(bc/a) + n*(-bc/a) => b²/a = (m - n)*(bc/a) => b²/a = (m - n)*(bc/a) => divide both sides by (bc/a):(b²/a) / (bc/a) ) = (m - n) => (b²/a)*(a/(bc)) = (m - n) => b/c = (m - n). So correct.Thus, the calculations are correct. Therefore, the answer should be option A: 3√2/4.Wait, but the answer options given are (A) 3√2/4, (B) 9/8, (C) 3√5/5, (D) 3√2/2. So according to our calculation, answer is A. But wait, just to be thorough, let me check if the problem allows for another possibility. For example, if point P is (c, -b²/a), then OP would be (c, -b²/a). Then the y-component would be -b²/a = m*(bc/a) + n*(-bc/a). So same process: -b²/a = (m - n)*(bc/a). Then, -b²/a = (m - n)*(bc/a) => divide both sides by (bc/a):(-b²/a) / (bc/a) ) = m - n => (-b/c) = m - n. Then, we would have:m + n = 1m - n = -b/cSo solving:Add: 2m = 1 - b/c => m = (1 - b/c)/2Subtract: 2n = 1 + b/c => n = (1 + b/c)/2Then m*n = [(1 - b/c)/2] * [(1 + b/c)/2] = [1 - (b/c)^2]/4 same as before. So m*n still is [1 - (b/c)^2]/4. So even if P is the other point, the product m*n is same. So same result. Therefore, regardless of which P we take, the result is the same. Therefore, the answer is indeed A.But just to cross-verify, let's take numbers. Let's suppose a = 2√2, b = 1. Then c = sqrt(a² + b²) = sqrt(8 + 1) = 3. Then e = c/a = 3/(2√2) = 3√2/4, which is option A. Then points:Asymptotes: y = ±(b/a)x = ±(1/(2√2))x.Points A and B on x = 3: y = ±(1/(2√2))*3 = ±3/(2√2) = ±3√2/4. So A is (3, 3√2/4), B is (3, -3√2/4).Point P: (3, b²/a) = (3, 1/(2√2)) = (3, √2/4). Then OP = (3, √2/4). OA is (3, 3√2/4), OB is (3, -3√2/4). Then, express OP as m*OA + n*OB:So:3 = 3m + 3n => 1 = m + n√2/4 = m*(3√2/4) + n*(-3√2/4)Divide second equation by √2/4:1 = 3m - 3n => 3m - 3n = 1 => m - n = 1/3So we have:m + n = 1m - n = 1/3Solving:Adding: 2m = 4/3 => m = 2/3Then n = 1 - 2/3 = 1/3Therefore, m = 2/3, n = 1/3. Then m*n = (2/3)(1/3) = 2/9, which matches the given condition. So yes, it's correct. Hence, the eccentricity is 3√2/4, option A. Therefore, the answer is A.Final Answerboxed{A}</think>Given the hyperbola (frac{x^2}{a^2} - frac{y^2}{b^2} = 1) with a focus at (F), a vertical line (l) through (F) intersects the asymptotes at points (A) and (B). Point (P) is where line (l) intersects the hyperbola. We need to find the eccentricity of the hyperbola given that (overrightarrow{OP} = m overrightarrow{OA} + n overrightarrow{OB}) and (mn = frac{2}{9}).1. Identify coordinates of points: - Focus (F) is ((c, 0)) where (c = sqrt{a^2 + b^2}). - Vertical line (l) through (F) is (x = c). - Asymptotes are (y = pm frac{b}{a}x). Intersecting with (x = c), points (A) and (B) are ((c, frac{bc}{a})) and ((c, -frac{bc}{a})). - Point (P) on the hyperbola when (x = c) is ((c, frac{b^2}{a})).2. Vector decomposition: - Vectors (overrightarrow{OA} = (c, frac{bc}{a})), (overrightarrow{OB} = (c, -frac{bc}{a})), and (overrightarrow{OP} = (c, frac{b^2}{a})). - Solving (overrightarrow{OP} = m overrightarrow{OA} + n overrightarrow{OB}): - (c = mc + nc Rightarrow m + n = 1) - (frac{b^2}{a} = m frac{bc}{a} - n frac{bc}{a} Rightarrow frac{b^2}{a} = frac{bc}{a}(m - n) Rightarrow frac{b}{c} = m - n)3. Solve for (m) and (n): - (m + n = 1) - (m - n = frac{b}{c}) - Solving these, (m = frac{1 + frac{b}{c}}{2}) and (n = frac{1 - frac{b}{c}}{2}) - Given (mn = frac{2}{9}), substituting (m) and (n) gives (left(1 - left(frac{b}{c}right)^2right)/4 = frac{2}{9}) - Solving, (left(frac{b}{c}right)^2 = frac{1}{9} Rightarrow frac{b}{c} = frac{1}{3} Rightarrow c = 3b)4. Calculate eccentricity: - Using (c^2 = a^2 + b^2), substitute (c = 3b): - (9b^2 = a^2 + b^2 Rightarrow a^2 = 8b^2 Rightarrow a = 2sqrt{2}b) - Eccentricity (e = frac{c}{a} = frac{3b}{2sqrt{2}b} = frac{3}{2sqrt{2}} = frac{3sqrt{2}}{4})Thus, the eccentricity of the hyperbola is (boxed{A}).,<think>Okay, let's see. I need to find the eccentricity of the hyperbola given some conditions. The hyperbola equation is x²/a² - y²/b² = 1, with a and b positive. A focus F is on the hyperbola. Then, a line l is drawn through F perpendicular to the x-axis, meaning it's a vertical line. This line intersects the two asymptotes at points A and B. Then, point P is one of the intersections of line l with the hyperbola. The origin is O. The problem states that the vector OP can be expressed as mOA + nOB, where m and n are real numbers such that mn = 2/9. We need to find the eccentricity e of the hyperbola.First, recall that the eccentricity e of a hyperbola is given by e = c/a, where c is the distance from the center to each focus, and c² = a² + b². So, ultimately, we need to relate the given condition mn = 2/9 to the parameters a and b, which will allow us to compute e.Let's start by identifying coordinates of key points. The hyperbola is centered at the origin (0,0). The foci are located at (±c, 0), where c² = a² + b². The problem mentions a focus F. Since the hyperbola is symmetric, it might not matter whether F is (c,0) or (-c,0), but since the line l is drawn through F perpendicular to the x-axis (vertical line), the line l will have equation x = c or x = -c. However, the asymptotes of the hyperbola are y = (b/a)x and y = -(b/a)x. If we take F as (c, 0), then the vertical line x = c will intersect the asymptotes at points A and B.Wait, but let me confirm: if we take a vertical line through F, which is on the x-axis, then x = c (assuming F is (c,0)) intersects the asymptotes y = (b/a)x and y = -(b/a)x. So substituting x = c into the asymptote equations gives y = (b/a)c and y = -(b/a)c. Therefore, points A and B would be (c, (b/a)c) and (c, -(b/a)c). But wait, the line l is vertical through F, which is (c,0). So the two points of intersection with the asymptotes are (c, (b/a)c) and (c, -(b/a)c). So these are points A and B.Then, the line x = c intersects the hyperbola at point P. Let's find the coordinates of P. Substitute x = c into the hyperbola equation:(c)²/a² - y²/b² = 1 => c²/a² - y²/b² = 1. Then, solving for y:y²/b² = c²/a² - 1 = (c² - a²)/a². But since c² = a² + b², this becomes (a² + b² - a²)/a² = b²/a². Therefore, y²/b² = b²/a² => y² = b⁴/a² => y = ±b²/a. So the points where line x = c intersects the hyperbola are (c, b²/a) and (c, -b²/a). Therefore, point P can be either of these. Let's pick one, say (c, b²/a). The other would be similar.Now, we need to express vector OP as a linear combination of vectors OA and OB. Let's first find vectors OA and OB.Point A is (c, (b/a)c) and point B is (c, -(b/a)c). Therefore, vectors OA and OB are the same as the coordinates of A and B, since they are position vectors from the origin. So OA = (c, (b/a)c) and OB = (c, -(b/a)c). Then, vector OP is (c, b²/a). The problem states that OP = m OA + n OB. So:(c, b²/a) = m (c, (b/a)c) + n (c, -(b/a)c)Breaking this into components:For the x-component: c = m*c + n*c => c = c(m + n) => dividing both sides by c (since c ≠ 0), we get 1 = m + n.For the y-component: b²/a = m*(b/a)c + n*(-(b/a)c) => b²/a = (b c/a)(m - n)Let me write that as:b²/a = (b c /a)(m - n)Multiply both sides by a:b² = b c (m - n)Divide both sides by b (assuming b ≠ 0):b = c (m - n)So we have two equations:1) m + n = 12) m - n = b / cWe need to solve for m and n. Let's denote equation 1 and equation 2:From 1: m + n = 1From 2: m - n = b/cAdding equations 1 and 2: 2m = 1 + b/c => m = (1 + b/c)/2Subtracting equation 2 from equation 1: 2n = 1 - b/c => n = (1 - b/c)/2Therefore, m = (1 + b/c)/2 and n = (1 - b/c)/2.The problem states that mn = 2/9. So let's compute mn:mn = [(1 + b/c)/2] * [(1 - b/c)/2] = [1 - (b/c)^2]/4 = [1 - b²/c²]/4Set this equal to 2/9:[1 - b²/c²]/4 = 2/9Multiply both sides by 4:1 - b²/c² = 8/9Then,b²/c² = 1 - 8/9 = 1/9 => b/c = 1/3 (since b and c are positive)Therefore, b/c = 1/3 => c = 3b.But we also know that for hyperbola, c² = a² + b². Since c = 3b, then:(3b)² = a² + b² => 9b² = a² + b² => a² = 8b² => a = 2√2 bBut we need the eccentricity e = c/a. c = 3b and a = 2√2 b, so:e = 3b / (2√2 b) = 3/(2√2) = 3√2/4.Wait, that's one of the answer choices, option A: 3√2/4. So is that the answer?Wait, but let me check again my steps to see if there's any mistake.First, the equation of the hyperbola: x²/a² - y²/b² = 1. The foci are at (±c, 0) with c² = a² + b². The line through focus F is x = c. The asymptotes are y = ±(b/a)x. The points A and B are (c, bc/a) and (c, -bc/a). Then, the intersection point P with the hyperbola is (c, y) where y = ±b²/a. So we took P as (c, b²/a). Then, vectors OA and OB are (c, bc/a) and (c, -bc/a). Then, OP is (c, b²/a). Expressed as m OA + n OB, so:c = m c + n c => m + n = 1b²/a = m (bc/a) + n (-bc/a) => (b c/a)(m - n) = b²/a => multiply both sides by a: b c (m - n) = b² => c (m - n) = b => m - n = b/cTherefore, m + n = 1, m - n = b/c. Then, solving:m = (1 + b/c)/2, n = (1 - b/c)/2mn = [(1 + b/c)(1 - b/c)]/4 = (1 - (b/c)^2)/4 = 2/9Therefore,1 - (b/c)^2 = 8/9 => (b/c)^2 = 1/9 => b/c = 1/3 => c = 3bThen, c² = a² + b² => (3b)^2 = a² + b² => 9b² = a² + b² => a² = 8b² => a = 2√2 bTherefore, eccentricity e = c/a = 3b/(2√2 b) = 3/(2√2) = (3√2)/4, which is option A.But wait, the options given include (A) 3√2/4, which is what we got. But let me check the answer choices again:(A) 3√2/4(B) 9/8(C) 3√5/5(D) 3√2/2So, our answer is (A). However, let me verify once more. Let's compute e = c/a. We found that c = 3b and a = 2√2 b. Therefore, e = 3b / (2√2 b) = 3/(2√2) = 3√2/4. Yes, that's correct. Rationalizing the denominator: 3/(2√2) = 3√2/4. So the answer is indeed (A). Therefore, the correct answer should be option A.But wait, the problem statement says "the hyperbola x²/a² - y²/b² =1 (a>0, b>0) have a focus at F". Wait, hyperbola's foci are at (±c,0), so F is (c,0). Then, the vertical line through F is x = c. The asymptotes are y = ±(b/a)x. So when x = c, y = ±(b/a)c. Therefore, points A and B are (c, (b/a)c) and (c, -(b/a)c). Then, point P is (c, b²/a). Then, vectors OA and OB are (c, (b/a)c) and (c, -(b/a)c). Then, vector OP is (c, b²/a). Expressing OP as m OA + n OB. So component-wise:x-component: c = m c + n c => m + n =1y-component: b²/a = m*(b/a)c + n*(-(b/a)c) = (b c/a)(m -n)Thus, b²/a = (b c/a)(m -n) => divide both sides by (b/a):b = c (m -n)Hence, m -n = b/cTherefore, solving m + n =1 and m -n = b/c, we get m = (1 + b/c)/2 and n = (1 - b/c)/2. Then, mn = [ (1 + b/c)(1 - b/c) ] /4 = (1 - (b/c)^2)/4 = 2/9.Hence, (1 - (b²/c²)) = 8/9 => (b/c)^2 =1/9 => b/c =1/3 => c=3b.Then, c² =a² +b² => 9b² =a² +b² => a²=8b² => a=2√2 b.Thus, e =c/a=3b/(2√2 b)=3/(2√2)=3√2/4. Which is answer choice A. So seems correct.But wait, just to be thorough, maybe the line l is through F and perpendicular to the x-axis. Since the x-axis is horizontal, a line perpendicular to it is vertical, so yes, x = c. So all steps are correct.Alternatively, if I had chosen the focus at (-c,0), the vertical line would be x = -c. Then points A and B would be (-c, -b/a*(-c)) = (-c, bc/a) and (-c, -bc/a). Similarly, the intersection with hyperbola would be (-c, ±b²/a). Then, vectors OA and OB would be (-c, bc/a) and (-c, -bc/a). Then, vector OP would be (-c, b²/a) or (-c, -b²/a). Then, same steps: OP = m OA + n OB. For x-component: -c = m*(-c) + n*(-c) => -c = -c(m +n) => m +n =1. For y-component: ±b²/a = m*(bc/a) + n*(-bc/a) => same as before: b²/a = bc/a (m -n) => same equation. So same result. So regardless of which focus we take, the result is the same.Therefore, the answer should be option A: 3√2/4. Therefore, I think that's correct.Final Answerboxed{A}</think>Given the hyperbola (frac{x^2}{a^2} - frac{y^2}{b^2} = 1) with foci at ((pm c, 0)) where (c^2 = a^2 + b^2), we consider a vertical line (l) through the focus (F(c, 0)). This line intersects the asymptotes (y = pm frac{b}{a}x) at points (A) and (B), and the hyperbola at point (P). We need to find the eccentricity (e) given that (overrightarrow{OP} = m overrightarrow{OA} + n overrightarrow{OB}) with (mn = frac{2}{9}).1. Coordinates of Points: - Points (A) and (B) on the asymptotes are ((c, frac{bc}{a})) and ((c, -frac{bc}{a})). - Point (P) on the hyperbola is ((c, frac{b^2}{a})).2. Vectors: - (overrightarrow{OA} = (c, frac{bc}{a})) - (overrightarrow{OB} = (c, -frac{bc}{a})) - (overrightarrow{OP} = (c, frac{b^2}{a}))3. Linear Combination: - For the x-component: (c = mc + nc implies m + n = 1) - For the y-component: (frac{b^2}{a} = m frac{bc}{a} - n frac{bc}{a} implies frac{b^2}{a} = frac{bc}{a}(m - n)) - Simplifying: (b = c(m - n))4. Solving for (m) and (n): - From (m + n = 1) and (m - n = frac{b}{c}), we get (m = frac{1 + frac{b}{c}}{2}) and (n = frac{1 - frac{b}{c}}{2}). - Product (mn = frac{(1 + frac{b}{c})(1 - frac{b}{c})}{4} = frac{1 - frac{b^2}{c^2}}{4} = frac{2}{9}). - Solving (frac{1 - frac{b^2}{c^2}}{4} = frac{2}{9}) gives (frac{b^2}{c^2} = frac{1}{9} implies frac{b}{c} = frac{1}{3}).5. Eccentricity Calculation: - (c = 3b) and (c^2 = a^2 + b^2 implies 9b^2 = a^2 + b^2 implies a^2 = 8b^2 implies a = 2sqrt{2}b). - Eccentricity (e = frac{c}{a} = frac{3b}{2sqrt{2}b} = frac{3}{2sqrt{2}} = frac{3sqrt{2}}{4}).Thus, the eccentricity of the hyperbola is (boxed{A}).]

🔑:4. A.Substitute the coordinates of points Aleft(c, frac{b c}{a}right), Bleft(c,-frac{b c}{a}right) intooverrightarrow{O P}=m overrightarrow{O A}+n overrightarrow{O B} text {, }to get point Pleft((m+n) c,(m-n) frac{b c}{a}right).Substitute the coordinates of point P into the hyperbola equation to getbegin{array}{l}frac{(m+n)^{2} c^{2}}{a^{2}}-frac{(m-n)^{2} b^{2} c^{2}}{a^{2} b^{2}}=1 Rightarrow 4 m n c^{2}=a^{2} Rightarrow e=frac{c}{a}=frac{1}{sqrt{4 m n}}=sqrt{frac{9}{8}}=frac{3 sqrt{2}}{4} .end{array}

🧠:[<think>Okay, let's see. I need to find the limit as x approaches 0 of [sin(bx) - sin(ax)] divided by ln[tan(π/4 + a x)]. Hmm. Alright, so first, maybe I should recall some trigonometric identities or limit properties that can help simplify this expression. Let me think.The numerator is sin(bx) - sin(ax). I remember that there's a sine subtraction formula: sin A - sin B = 2 cos[(A+B)/2] sin[(A - B)/2]. Maybe I can apply that here. Let's try:sin(bx) - sin(ax) = 2 cos[(bx + ax)/2] * sin[(bx - ax)/2] = 2 cos[(a + b)x / 2] * sin[(b - a)x / 2]Okay, so the numerator becomes 2 cos[(a + b)x / 2] * sin[(b - a)x / 2]. That might be easier to handle when x approaches 0 because I know that as x approaches 0, sin(something) is approximately equal to that something. Similarly, cos(something) approaches 1. So maybe I can use Taylor series expansions here.For the numerator: as x → 0, cos[(a + b)x / 2] ≈ 1 - [(a + b)^2 x^2]/8 + ... but since x is approaching 0, maybe the higher order terms can be neglected. And sin[(b - a)x / 2] ≈ (b - a)x / 2 - [(b - a)^3 x^3]/48 + ... So up to the first order, the numerator is approximately 2 * 1 * [(b - a)x / 2] = (b - a)x. So the numerator behaves like (b - a)x when x is near 0.Now the denominator is ln[tan(π/4 + a x)]. Let me think about that. Let's analyze the argument of the tangent first. π/4 is 45 degrees, and adding a x to it. When x is approaching 0, this is a small angle around π/4. So tan(π/4 + a x). Hmm. Remember that tan(π/4) is 1. Maybe I can use the tangent addition formula or expand tan(π/4 + a x) using a Taylor series around x = 0.Alternatively, since x is approaching 0, maybe I can set θ = a x, so θ approaches 0. Then tan(π/4 + θ). Let's recall that tan(π/4 + θ) = [tan(π/4) + tan θ] / [1 - tan(π/4) tan θ] = [1 + tan θ] / [1 - tan θ]. So tan(π/4 + θ) = (1 + tan θ)/(1 - tan θ). Then ln of that is ln[(1 + tan θ)/(1 - tan θ)] = ln(1 + tan θ) - ln(1 - tan θ). Hmm. Alternatively, maybe I can write this as 2 artanh(tan θ)? Because artanh(z) = (1/2) ln[(1 + z)/(1 - z)]. So yes, ln[(1 + z)/(1 - z)] = 2 artanh(z). So here, z is tan θ. Therefore, the denominator is 2 artanh(tan θ) where θ = a x.But I'm not sure if that helps. Alternatively, maybe expand tan(π/4 + a x) using a Taylor series. Let's try that.First, let me recall the expansion of tan(π/4 + ε) where ε is small. Let's set ε = a x. So tan(π/4 + ε) = [tan(π/4) + tan ε] / [1 - tan(π/4) tan ε] = (1 + tan ε) / (1 - tan ε). Now, tan ε ≈ ε + ε^3/3 + ... So substituting in, we have:(1 + ε + ε^3/3 + ...) / (1 - ε - ε^3/3 - ...) ≈ [1 + ε] / [1 - ε] when ε is small, neglecting higher order terms. Then, using the expansion 1/(1 - ε) ≈ 1 + ε + ε^2 + ε^3 + ..., so multiplying numerator and denominator:(1 + ε) * (1 + ε + ε^2 + ...) ≈ 1 + 2ε + 2ε^2 + ... So perhaps tan(π/4 + ε) ≈ 1 + 2ε + 2ε^2 + ... But maybe there's a better way to get the expansion.Alternatively, let's compute the Taylor series of tan(π/4 + ε) around ε = 0. Let f(ε) = tan(π/4 + ε). Then f(0) = tan(π/4) = 1. The first derivative f’(ε) = sec^2(π/4 + ε), so f’(0) = sec^2(π/4) = 1 / cos^2(π/4) = 1 / (√2/2)^2 = 1 / (1/2) = 2. The second derivative f''(ε) = 2 sec^2(π/4 + ε) tan(π/4 + ε), so f''(0) = 2 * 2 * 1 = 4. So the expansion up to second order is f(ε) ≈ f(0) + f’(0) ε + (1/2) f''(0) ε^2 = 1 + 2ε + 2ε^2 + ... So tan(π/4 + ε) ≈ 1 + 2ε + 2ε^2.Therefore, tan(π/4 + a x) ≈ 1 + 2a x + 2(a x)^2. So ln[tan(π/4 + a x)] ≈ ln(1 + 2a x + 2a^2 x^2). Now, using the expansion ln(1 + y) ≈ y - y^2/2 + y^3/3 - ..., where y = 2a x + 2a^2 x^2. So substituting in:ln(1 + 2a x + 2a^2 x^2) ≈ (2a x + 2a^2 x^2) - (2a x + 2a^2 x^2)^2 / 2 + ... Let's compute up to the first order term. The first term is 2a x + 2a^2 x^2. The next term is (2a x)^2 / 2 = 4a^2 x^2 / 2 = 2a^2 x^2. So subtracting that, the expansion becomes approximately 2a x + 2a^2 x^2 - 2a^2 x^2 = 2a x. Therefore, ln[tan(π/4 + a x)] ≈ 2a x when x is near 0. So the denominator behaves like 2a x as x approaches 0.Therefore, putting numerator and denominator together, the limit becomes:lim_{x→0} [ (b - a)x ] / [ 2a x ] = lim_{x→0} (b - a)/(2a) = (b - a)/(2a).Wait, but let me check again. The numerator was approximated as (b - a)x, and the denominator as 2a x. So the x terms cancel out, leaving (b - a)/(2a). So is that the answer?Wait, but let me verify if my approximations are correct. Maybe I made a mistake in expanding the denominator.Starting again with the denominator:tan(π/4 + a x) ≈ 1 + 2a x + 2a^2 x^2. Then ln(1 + 2a x + 2a^2 x^2). Let me let y = 2a x + 2a^2 x^2. Then ln(1 + y) ≈ y - y^2/2 + y^3/3 - ... So up to the first order, it's y - y^2/2. Let's compute y = 2a x + 2a^2 x^2. Then y^2 = (2a x)^2 + 2*(2a x)*(2a^2 x^2) + (2a^2 x^2)^2 = 4a^2 x^2 + 8a^3 x^3 + 4a^4 x^4. So y^2 / 2 = 2a^2 x^2 + 4a^3 x^3 + 2a^4 x^4. Then ln(1 + y) ≈ y - y^2/2 = (2a x + 2a^2 x^2) - (2a^2 x^2 + 4a^3 x^3 + ...) = 2a x - 2a^2 x^2 - 4a^3 x^3 + ... So up to the first order, it's 2a x, and the next term is -2a^2 x^2. But when x approaches 0, the dominant term is 2a x. So in the denominator, the leading term is 2a x, so the denominator is approximately 2a x. Therefore, when we take the ratio [ (b - a)x ] / [2a x], the x cancels, so we get (b - a)/(2a).But let me check if this is correct. Alternatively, perhaps using L'Hospital's Rule because both numerator and denominator approach 0 as x approaches 0. Let's verify:When x → 0, numerator: sin(bx) - sin(ax) → 0 - 0 = 0.Denominator: ln[tan(π/4 + a x)] → ln(tan π/4) = ln(1) = 0. So indeed, 0/0 indeterminate form. Therefore, L'Hospital's Rule applies. So take derivatives of numerator and denominator.So applying L'Hospital's Rule once:lim_{x→0} [ d/dx (sin b x - sin a x) ] / [ d/dx ln(tan(π/4 + a x)) ]Compute numerator derivative:d/dx [sin b x - sin a x] = b cos b x - a cos a x. At x = 0, this is b cos 0 - a cos 0 = b*1 - a*1 = b - a.Denominator derivative:d/dx [ln(tan(π/4 + a x))] = [1 / tan(π/4 + a x)] * sec^2(π/4 + a x) * a. Because derivative of ln(u) is (1/u) * u', and derivative of tan(u) is sec^2(u) * u', so here u = π/4 + a x, so u' = a. Therefore, derivative is [1 / tan(u)] * sec^2(u) * a.But let's compute this at x = 0. Then u = π/4, so tan(u) = 1, sec^2(u) = 1 + tan^2(u) = 1 + 1 = 2. Therefore, the derivative is [1 / 1] * 2 * a = 2a.Therefore, after applying L'Hospital's Rule once, the limit becomes (b - a)/(2a). So this confirms the previous result. So the answer should be (b - a)/(2a). Let me check again if this is correct.Alternatively, perhaps I can use the standard limit lim_{x→0} sin(kx)/x = k. Also, for the denominator, we can write ln(tan(π/4 + a x)) ≈ ln(1 + 2a x) ≈ 2a x when x is near 0. Wait, but how?Wait, from the previous analysis, tan(π/4 + a x) ≈ 1 + 2a x, so ln(1 + 2a x) ≈ 2a x - (2a x)^2 / 2 + ... ≈ 2a x - 2a^2 x^2. So as x approaches 0, the dominant term is 2a x, so the denominator is approximately 2a x. Then the numerator is sin(bx) - sin(ax). If we approximate sin(bx) ≈ bx - (bx)^3 / 6 and sin(ax) ≈ ax - (ax)^3 / 6, then sin(bx) - sin(ax) ≈ (bx - ax) - [ (b^3 x^3 - a^3 x^3 ) / 6 ] ≈ (b - a)x - (b^3 - a^3)x^3 / 6. So the leading term is (b - a)x, so the numerator is ~ (b - a)x, denominator ~2a x. So the ratio is ~ (b - a)/(2a), same as before.Therefore, regardless of the method, whether using trigonometric identities and Taylor expansions or L'Hospital's Rule, we get the same result.Therefore, the limit is (b - a)/(2a). Let me just write that in the box.Final AnswerThe limit is boxed{dfrac{b - a}{2a}}.</think>To calculate the limit of the function:lim _{x rightarrow 0} frac{sin b x - sin a x}{ln left(tanleft(frac{pi}{4} + a xright)right)}we start by simplifying the numerator and the denominator using trigonometric identities and Taylor series expansions.Numerator Analysis:Using the sine subtraction formula:sin(bx) - sin(ax) = 2 cosleft(frac{(b+a)x}{2}right) sinleft(frac{(b-a)x}{2}right)For small ( x ), we approximate:cosleft(frac{(a+b)x}{2}right) approx 1 quad text{and} quad sinleft(frac{(b-a)x}{2}right) approx frac{(b-a)x}{2}Thus, the numerator approximates to:2 cdot 1 cdot frac{(b-a)x}{2} = (b-a)xDenominator Analysis:Using the tangent addition formula for small ( theta = ax ):tanleft(frac{pi}{4} + axright) = frac{1 + tan(ax)}{1 - tan(ax)} approx frac{1 + ax}{1 - ax} approx 1 + 2axTaking the natural logarithm:lnleft(tanleft(frac{pi}{4} + axright)right) approx ln(1 + 2ax) approx 2axCombining Results:The limit becomes:lim _{x rightarrow 0} frac{(b-a)x}{2ax} = lim _{x rightarrow 0} frac{b-a}{2a} = frac{b-a}{2a}Verification using L'Hospital's Rule:Both the numerator and denominator approach 0 as ( x to 0 ), so applying L'Hospital's Rule:- Derivative of the numerator: ( b cos(bx) - a cos(ax) rightarrow b - a )- Derivative of the denominator: ( frac{2a}{tan(pi/4 + ax) cos^2(pi/4 + ax)} rightarrow 2a )Thus, the limit is:frac{b - a}{2a}Final Answer:The limit is boxed{dfrac{b - a}{2a}}.,<think>Okay, so I need to find the limit as x approaches 0 of [sin(bx) - sin(ax)] divided by ln(tan(π/4 + ax)). Hmm, let me think. First, when dealing with limits as x approaches 0, especially involving trigonometric functions, I remember that Taylor series expansions or L'Hôpital's Rule can be helpful. Let me check if this limit is in an indeterminate form like 0/0 or ∞/∞, which would make L'Hôpital applicable. Let's evaluate the numerator and denominator separately as x approaches 0.Starting with the numerator: sin(bx) - sin(ax). When x is near 0, sin(bx) ≈ bx - (bx)^3/6 and similarly sin(ax) ≈ ax - (ax)^3/6. So subtracting them, sin(bx) - sin(ax) ≈ (bx - ax) - [(bx)^3 - (ax)^3]/6. The leading term here would be (b - a)x. So as x approaches 0, the numerator approaches 0.Now the denominator: ln(tan(π/4 + ax)). Let's analyze the inside of the natural log first. When x is near 0, π/4 + ax is a small angle added to π/4. Let's recall that tan(π/4) = 1. So tan(π/4 + ax) can be expanded using the tangent addition formula. The formula for tan(A + B) is [tan A + tan B]/[1 - tan A tan B]. Here, A is π/4 and B is ax. Since tan(π/4) = 1, so tan(π/4 + ax) = [1 + tan(ax)] / [1 - tan(ax)]. Alternatively, maybe using a Taylor expansion for tan(π/4 + ax) around x=0. Let's note that when x is near 0, ax is small. So tan(π/4 + ax) can be approximated. Let's recall that the derivative of tan(u) at u=π/4 is sec^2(π/4) = 2. So, expanding tan(π/4 + ax) around x=0 using first-order Taylor series: tan(π/4) + (ax) * sec^2(π/4) + higher terms. tan(π/4) is 1, so this becomes 1 + 2ax + ... So tan(π/4 + ax) ≈ 1 + 2ax. Therefore, ln(tan(π/4 + ax)) ≈ ln(1 + 2ax). And for small y, ln(1 + y) ≈ y - y^2/2 + y^3/3 - ... So ln(1 + 2ax) ≈ 2ax - (2ax)^2/2 + ... Therefore, the denominator is approximately 2ax as x approaches 0. So the denominator approaches 0 as x approaches 0.Therefore, the entire expression is of the form 0/0 as x approaches 0, which means L'Hôpital's Rule can be applied. Alternatively, since both numerator and denominator can be approximated by their Taylor series, maybe expanding both up to the first non-zero term would work.Let me first try using L'Hôpital's Rule. So, take the derivative of the numerator and the derivative of the denominator.Numerator derivative: d/dx [sin(bx) - sin(ax)] = b cos(bx) - a cos(ax).Denominator derivative: d/dx [ln(tan(π/4 + ax))]. Let's compute this step by step. Let’s set u = tan(π/4 + ax), so the derivative is (1/u) * d/dx [tan(π/4 + ax)]. Then, derivative of tan(π/4 + ax) is sec^2(π/4 + ax) * a. So putting together, the derivative of the denominator is [1/tan(π/4 + ax)] * a sec^2(π/4 + ax).Simplify that: [1/tan(π/4 + ax)] * a sec^2(π/4 + ax) = a * sec^2(π/4 + ax)/tan(π/4 + ax). But sec^2(theta) is 1 + tan^2(theta). However, maybe there's a simpler way. Since 1/tan(theta) = cot(theta), so this becomes a * sec^2(theta) * cot(theta) where theta = π/4 + ax. Then, sec(theta) is 1/cos(theta), so sec^2(theta) = 1/cos^2(theta), and cot(theta) = cos(theta)/sin(theta). So multiplying together: a * [1/cos^2(theta)] * [cos(theta)/sin(theta)] = a / [cos(theta) sin(theta)].Alternatively, note that tan(theta) = sin(theta)/cos(theta), so 1/tan(theta) = cos(theta)/sin(theta). So, putting it all together, the derivative of the denominator is a * sec^2(theta) * cos(theta)/sin(theta) = a * [1/cos^2(theta)] * [cos(theta)/sin(theta)] = a / [cos(theta) sin(theta)].But theta is π/4 + ax. So, as x approaches 0, theta approaches π/4. Therefore, cos(theta) approaches cos(π/4) = √2/2, and sin(theta) approaches sin(π/4) = √2/2. Therefore, the denominator derivative as x approaches 0 becomes a / [(√2/2)(√2/2)] = a / (2/4) = a / (1/2) = 2a. Wait, that seems important.Wait, but we are taking the derivative of the denominator, which is [ln(tan(π/4 + ax))]', which evaluated at x approaching 0, but the derivative itself is a function of x. Wait, no. When applying L'Hôpital's Rule, we need to compute the limit of the derivatives as x approaches 0, not evaluate the derivatives at x=0. Hmm, maybe I need to re-examine.Wait, actually, L'Hôpital's Rule says that if the original limit is 0/0 or ∞/∞, then the limit is equal to the limit of the ratio of their derivatives, provided that latter limit exists. So, after differentiating numerator and denominator, we need to compute the limit as x approaches 0 of [b cos(bx) - a cos(ax)] divided by [a / (cos(π/4 + ax) sin(π/4 + ax))].Wait, maybe I miscalculated the derivative of the denominator. Let me redo that.Starting again, denominator is ln(tan(π/4 + ax)). Let’s denote theta = π/4 + ax. Then, d/dx [ln(tan theta)] = (1/tan theta) * sec^2 theta * d theta/dx. Since theta = π/4 + ax, d theta/dx = a. So, derivative is (1/tan theta) * sec^2 theta * a. But 1/tan theta = cot theta, so this becomes a cot theta sec^2 theta. But cot theta sec^2 theta = (cos theta / sin theta) * (1 / cos^2 theta) ) = 1/(sin theta cos theta). Therefore, derivative is a / (sin theta cos theta). So yes, the derivative is a divided by [sin(theta) cos(theta)], where theta = π/4 + ax.So the derivative of the denominator is a / [sin(π/4 + ax) cos(π/4 + ax)].Now, as x approaches 0, theta approaches π/4. Therefore, sin(theta) cos(theta) approaches sin(π/4) cos(π/4) = (√2/2)(√2/2) = 1/2. Therefore, the denominator derivative approaches a / (1/2) = 2a.Meanwhile, the numerator derivative is b cos(bx) - a cos(ax). As x approaches 0, cos(bx) approaches 1 and cos(ax) approaches 1. Therefore, the numerator derivative approaches b - a.Therefore, applying L'Hôpital's Rule once, the limit becomes (b - a) / (2a). Wait, is that correct? Let me verify.So after applying L’Hospital once, the limit becomes:lim_{x->0} [b cos(bx) - a cos(ax)] / [a / (sin(π/4 + ax) cos(π/4 + ax))]But as x approaches 0, the denominator of this expression is [a / (sin(π/4) cos(π/4))] = a / ( (√2/2)(√2/2) ) = a / (0.5) = 2a. The numerator becomes b*1 - a*1 = b - a. Therefore, the entire limit is (b - a)/(2a).Wait, but let me confirm if we can actually apply L’Hospital once and directly get this result, because sometimes if the derivatives still result in 0/0, we need to apply L’Hospital multiple times. But in this case, after differentiating once, the numerator becomes b cos(bx) - a cos(ax). When x approaches 0, that becomes b - a. The denominator becomes a / [sin(theta) cos(theta)] with theta approaching π/4, which is finite (2a). Therefore, the limit is (b - a)/(2a).Alternatively, maybe I can approach this using Taylor series expansions instead of L’Hospital's Rule, to cross-verify.Starting with the numerator: sin(bx) - sin(ax). Using Taylor series around x=0:sin(bx) = bx - (b^3 x^3)/6 + ... sin(ax) = ax - (a^3 x^3)/6 + ... Subtracting: (bx - ax) - [ (b^3 - a^3)x^3 ]/6 + ... Which is (b - a)x - [(b^3 - a^3)/6]x^3 + ... So the leading term is (b - a)x.For the denominator: ln(tan(π/4 + ax)).First, expand tan(π/4 + ax). Using the identity tan(A + B) = (tan A + tan B)/(1 - tan A tan B). Let A = π/4, so tan A = 1, and B = ax, so tan B ≈ ax + (ax)^3/3 + ... Therefore, tan(π/4 + ax) ≈ [1 + ax + (ax)^3/3] / [1 - ax - (ax)^3/3]. But since ax is small, we can approximate this as [1 + ax] / [1 - ax] up to the first order. Multiplying numerator and denominator by 1 + ax (to rationalize?), or maybe expanding as a Taylor series.Alternatively, [1 + ax] / [1 - ax] ≈ (1 + ax)(1 + ax + (ax)^2 + ...) ≈ 1 + 2ax + 2(ax)^2 + ... for small ax. But maybe a better approach is to use tan(π/4 + ax) = [1 + tan(ax)] / [1 - tan(ax)]. Since tan(ax) ≈ ax + (ax)^3/3 + ..., so [1 + ax + ...] / [1 - ax - ...] ≈ (1 + ax)/(1 - ax) ≈ 1 + 2ax + 2(ax)^2 + ... when expanded as a series for small ax.Alternatively, maybe use the expansion tan(π/4 + ax) ≈ 1 + 2ax. Wait, let me check. Let’s consider theta = π/4 + ax. Then theta ≈ π/4 when x is near 0. So tan(theta) can be expanded around theta = π/4. Let’s let h = theta - π/4 = ax. Then, tan(pi/4 + h) = tan(pi/4) + h * sec^2(pi/4) + (h^2)/2 * d^2/dtheta^2 tan(theta) |_{theta=pi/4} + ... tan(pi/4 + h) = 1 + h*(2) + (h^2)/2*(2*2) + ... because the derivative of tan(theta) is sec^2(theta), which at theta=pi/4 is 2, and the second derivative is 2 sec^2(theta) tan(theta), which at theta=pi/4 is 2*2*1=4. So:tan(pi/4 + h) ≈ 1 + 2h + 2h^2 + ... Therefore, with h = ax, we get 1 + 2ax + 2a^2x^2 + ... So up to first order, tan(pi/4 + ax) ≈ 1 + 2ax.Therefore, ln(tan(pi/4 + ax)) ≈ ln(1 + 2ax) ≈ 2ax - (2ax)^2/2 + (2ax)^3/3 - ... So the leading term is 2ax. Therefore, the denominator is approximately 2ax.Therefore, the numerator is approximately (b - a)x, the denominator is approximately 2ax. So the ratio is (b - a)x / (2ax) = (b - a)/(2a). Which matches the result from L’Hospital's Rule. So this confirms that the limit is (b - a)/(2a).Therefore, the answer should be (b - a)/(2a). Let me just check if there's any possible mistake here. For example, could there be a miscalculation in the derivative of the denominator?Wait, when I applied L’Hospital, I took the derivative of the numerator, which is correct: b cos(bx) - a cos(ax). Then the derivative of the denominator is correct as well: a / [sin(pi/4 + ax) cos(pi/4 + ax)]. Then, evaluating as x approaches 0, sin(pi/4 + ax) cos(pi/4 + ax) approaches sin(pi/4) cos(pi/4) = (sqrt(2)/2)(sqrt(2)/2) = 1/2. Therefore, denominator derivative approaches a / (1/2) = 2a. Then, the numerator approaches b - a. Therefore, the ratio is (b - a)/(2a). So that seems correct.Alternatively, suppose a=0. Wait, but if a=0, then the original problem would have denominator ln(tan(pi/4 + 0)) = ln(tan(pi/4)) = ln(1) = 0. So denominator is 0, and numerator becomes sin(bx) - sin(0) = sin(bx). So the expression becomes sin(bx)/0, which would approach sign(sin(bx)/0) * infinity. But since a=0 is probably not allowed here. Similarly, if a=0, the original problem is not defined? Wait, if a=0, the denominator is ln(1) = 0, and numerator is sin(bx) - 0 = sin(bx). So the limit becomes lim_{x->0} sin(bx)/0, which is undefined (approaches infinity or negative infinity depending on the side). But since the problem didn't specify any constraints on a and b, other than being constants, perhaps a is non-zero. So we can assume a ≠ 0. Similarly, if a=0, the limit does not exist (infinite). Therefore, the answer (b - a)/(2a) is valid when a ≠ 0.Another check: suppose a = b. Then the numerator becomes sin(ax) - sin(ax) = 0, so the entire expression is 0 divided by something approaching 0. Wait, but if a = b, the numerator is 0 for all x, so the limit is 0 / something, which would be 0. But according to our formula (b - a)/(2a) when a = b, gives 0/(2a) = 0, which is correct. So that's consistent.Another test case: suppose a = 1, b = 2. Then the limit should be (2 - 1)/(2*1) = 1/2. Let's verify numerically. Take x = 0.001. Compute numerator: sin(2*0.001) - sin(1*0.001) ≈ 0.002 - 0.00099999983 ≈ 0.00100000017. Denominator: ln(tan(pi/4 + 0.001)) ≈ ln(tan(pi/4 + 0.001)). tan(pi/4 + 0.001) ≈ 1 + 2*0.001 = 1.002, so ln(1.002) ≈ 0.0019980027. Then, the ratio ≈ 0.00100000017 / 0.0019980027 ≈ 0.5005. Which is approximately 0.5, which is 1/2. So that checks out.Another test case: a = 2, b = 3. The limit should be (3 - 2)/(2*2) = 1/4. Let x = 0.0001. Numerator: sin(0.0003) - sin(0.0002) ≈ 0.0003 - 0.00019999998 ≈ 0.00010000002. Denominator: ln(tan(pi/4 + 0.0002)) ≈ ln(1 + 2*0.0002) ≈ ln(1.0004) ≈ 0.000399968. Ratio ≈ 0.00010000002 / 0.000399968 ≈ 0.250008, which is approximately 1/4. So that works too.Therefore, after multiple checks, it seems that the limit is indeed (b - a)/(2a). Therefore, the final answer is (b - a)/(2a), which in boxed form is boxed{dfrac{b - a}{2a}}.Final Answerboxed{dfrac{b - a}{2a}}</think>To calculate the limit:lim _{x rightarrow 0} frac{sin b x - sin a x}{ln left(tanleft(frac{pi}{4} + a xright)right)}we start by recognizing that both the numerator and the denominator approach 0 as ( x ) approaches 0, forming an indeterminate form ( 0/0 ). This allows us to apply L'Hôpital's Rule.Step 1: Differentiate the numerator and the denominatorThe numerator is ( sin(bx) - sin(ax) ). Its derivative is:b cos(bx) - a cos(ax)The denominator is ( lnleft(tanleft(frac{pi}{4} + axright)right) ). Let ( theta = frac{pi}{4} + ax ). The derivative of the denominator is:frac{d}{dx} ln(tan(theta)) = frac{1}{tan(theta)} cdot sec^2(theta) cdot a = frac{a}{sin(theta) cos(theta)}Step 2: Evaluate the derivatives at ( x to 0 )As ( x to 0 ), ( theta to frac{pi}{4} ). Therefore, ( sin(theta) cos(theta) to sinleft(frac{pi}{4}right) cosleft(frac{pi}{4}right) = frac{sqrt{2}}{2} cdot frac{sqrt{2}}{2} = frac{1}{2} ). Thus, the derivative of the denominator approaches:frac{a}{frac{1}{2}} = 2aThe derivative of the numerator as ( x to 0 ) is:b cos(0) - a cos(0) = b - aStep 3: Apply L'Hôpital's RuleThe limit becomes:lim_{x to 0} frac{b cos(bx) - a cos(ax)}{frac{a}{sinleft(frac{pi}{4} + axright) cosleft(frac{pi}{4} + axright)}} = frac{b - a}{2a}Thus, the limit of the given function is:boxed{dfrac{b - a}{2a}}]

🔑:## Solutionbegin{aligned}& lim _{x rightarrow 0} frac{sin b x-sin a x}{ln left(tanleft(frac{pi}{4}+a xright)right)}=lim _{x rightarrow 0} frac{2 sin frac{b x-a x}{2} cos frac{b x+a x}{2}}{ln left(frac{tan frac{pi}{4}+tan a x}{1-tan frac{pi}{4} cdot tan a x}right)}= & =lim _{x rightarrow 0} frac{2 sin frac{b x-a x}{2} cos frac{b x+a x}{2}}{ln left(frac{1+tan a x}{1-tan a x}right)}=lim _{x rightarrow 0} frac{2 sin frac{b x-a x}{2} cos frac{b x+a x}{2}}{ln left(frac{1-tan a x+2 tan a x}{1-tan a x}right)}= & =lim _{x rightarrow 0} frac{2 sin frac{x(b-a)}{2} cos frac{x(b+a)}{2}}{ln left(1+frac{2 tan a x}{1-tan a x}right)}=end{aligned}Using the substitution of equivalent infinitesimals:begin{aligned}& sin frac{x(b-a)}{2} sim frac{x(b-a)}{2} quad, quad x rightarrow 0left(frac{x(b-a)}{2} rightarrow 0right) & ln left(1+frac{2 tan a x}{1-tan a x}right) sim frac{2 tan a x}{1-tan a x}, quad x rightarrow 0left(frac{2 tan a x}{1-tan a x} rightarrow 0right)end{aligned}We get:=lim _{x rightarrow 0} frac{2 frac{x(b-a)}{2} cos frac{x(b+a)}{2}}{frac{2 tan a x}{1-tan a x}}=lim _{x rightarrow 0} frac{x(b-a)(1-tan a x) cos frac{x(b+a)}{2}}{2 tan a x}=Using the substitution of equivalent infinitesimals:tan a x sim a x, as x rightarrow 0(a x rightarrow 0)We get:=lim _{x rightarrow 0} frac{x(b-a)(1-tan a x) cos frac{x(b+a)}{2}}{2 a x}=begin{aligned}& =lim _{x rightarrow 0} frac{(b-a)(1-tan a x) cos frac{x(b+a)}{2}}{2 a}= & =frac{(b-a)(1-tan(a cdot 0)) cos frac{0 cdot(b+a)}{2}}{2 a}= & =frac{(b-a)(1-0) cos 0}{2 a}=frac{b-a}{2 a}end{aligned}## Problem Kuznetsov Limits 16-28

🧠:[<think>Okay, so I have this system of equations to solve. The first equation is x minus y equals pi over 3, and the second one is tangent x minus tangent y equals square root of 3. Alright, let's see how to approach this. First, maybe I can express one variable in terms of the other using the first equation. Since x - y = π/3, that means x = y + π/3. So if I substitute this into the second equation, I can have an equation with just one variable, which should be easier to solve. Let me try that.So the second equation is tan x - tan y = √3. Substituting x with y + π/3 gives tan(y + π/3) - tan y = √3. Hmm, okay. Now I need to simplify tan(y + π/3). I remember the tangent addition formula: tan(a + b) = (tan a + tan b) / (1 - tan a tan b). Let me apply that here.Let me denote a as y and b as π/3. Then tan(y + π/3) = (tan y + tan(π/3)) / (1 - tan y tan(π/3)). I know that tan(π/3) is √3. So substituting that in, we get (tan y + √3) / (1 - tan y * √3). So now, the equation becomes [(tan y + √3) / (1 - tan y * √3)] - tan y = √3. Let me write this down step by step:[(tan y + √3)/(1 - √3 tan y)] - tan y = √3To solve this equation for tan y, let me first combine the terms into a single fraction. Let's find a common denominator. The first term is over (1 - √3 tan y) and the second term is just tan y, which can be written as tan y * (1 - √3 tan y)/(1 - √3 tan y). So:[(tan y + √3) - tan y (1 - √3 tan y)] / (1 - √3 tan y) = √3Expanding the numerator:tan y + √3 - tan y + √3 (tan y)^2Simplify numerator:tan y - tan y cancels out, leaving √3 + √3 (tan y)^2So numerator is √3 (1 + (tan y)^2)Denominator is (1 - √3 tan y)So now the equation is [√3 (1 + (tan y)^2)] / (1 - √3 tan y) = √3Multiply both sides by (1 - √3 tan y):√3 (1 + (tan y)^2) = √3 (1 - √3 tan y)Divide both sides by √3:1 + (tan y)^2 = 1 - √3 tan ySubtract 1 from both sides:(tan y)^2 = -√3 tan yBring all terms to one side:(tan y)^2 + √3 tan y = 0Factor out tan y:tan y (tan y + √3) = 0So the solutions are tan y = 0 or tan y = -√3Now, let's consider each case.Case 1: tan y = 0If tan y = 0, then y is an angle where tangent is 0, which occurs at y = kπ, where k is an integer. Then x = y + π/3 = kπ + π/3. So the solutions in this case are (x, y) = (kπ + π/3, kπ). Let's check if these satisfy the original second equation.Compute tan x - tan y. Since tan y = 0, it's tan x - 0 = tan x. x is kπ + π/3, so tan x = tan(kπ + π/3). The tangent function has a period of π, so tan(kπ + π/3) = tan(π/3) = √3. So tan x - tan y = √3 - 0 = √3, which matches the second equation. So these solutions are valid.Case 2: tan y = -√3If tan y = -√3, then y is an angle where tangent is -√3. The principal solution is y = -π/3 + kπ, since tan(-π/3) = -√3. Therefore, the general solution is y = -π/3 + kπ. Then x = y + π/3 = (-π/3 + kπ) + π/3 = kπ. So x = kπ, y = -π/3 + kπ. Let's check this in the second equation.Compute tan x - tan y. x is kπ, so tan x = tan(kπ) = 0. y is -π/3 + kπ, so tan y = tan(-π/3 + kπ) = tan(-π/3) = -√3 (since tangent has period π). Therefore, tan x - tan y = 0 - (-√3) = √3, which matches the second equation. So these solutions are also valid.Therefore, the solutions are all pairs (x, y) where either:1. x = kπ + π/3 and y = kπ, for any integer k, or2. x = kπ and y = -π/3 + kπ, for any integer k.But since k is an integer, we can express these solutions in different forms. For example, in the second case, y = -π/3 + kπ can also be written as y = 2π/3 + (k-1)π, so maybe there's a way to combine these, but probably not necessary. The question just asks for the angles that satisfy the system, so we can present both families of solutions.Alternatively, if we need to express all solutions without overlapping, but since k is any integer, both forms represent all solutions. So the answer is:x = kπ + π/3 and y = kπ, or x = kπ and y = kπ - π/3 for any integer k.But let me verify with specific values. Let's take k=0 first.Case 1: x = π/3, y = 0. Then tan(π/3) - tan(0) = √3 - 0 = √3. Correct.Case 2: x = 0, y = -π/3. Then tan(0) - tan(-π/3) = 0 - (-√3) = √3. Correct.Take k=1.Case 1: x = π + π/3 = 4π/3, y = π. tan(4π/3) - tan(π) = tan(π/3) - 0 = √3. Correct.Case 2: x = π, y = π - π/3 = 2π/3. tan(π) - tan(2π/3) = 0 - (-√3) = √3. Correct.Similarly for negative k. So these solutions are valid.Therefore, the general solutions are the two families as above. Since the problem says "angles" that satisfy the system, we can write the solutions in terms of x and y with the parameter k.Hence, all solutions are given by:x = π/3 + kπ, y = kπ,andx = kπ, y = -π/3 + kπ,for any integer k.But maybe we can combine these into a single expression? Let's see.Alternatively, note that if x = kπ + π/3, then y = kπ. Similarly, if x = kπ, then y = kπ - π/3. But these are two separate families. So unless there's a way to express them as one, but I think they are distinct. So the answer would be both sets.Alternatively, since k is an integer, we can write the solutions as:For any integer k,x = kπ + π/3 and y = kπ,orx = kπ and y = kπ - π/3.So these are the two sets of solutions.The problem doesn't specify a particular interval, like [0, 2π), so the answer should be in terms of all angles differing by integer multiples of π. Therefore, the solutions are as above.Final AnswerThe solutions are all angles such that ( x = frac{pi}{3} + kpi ) and ( y = kpi ), or ( x = kpi ) and ( y = -frac{pi}{3} + kpi ) for any integer ( k ). Thus, the angles are (boxed{x = frac{pi}{3} + kpi}) and (boxed{y = kpi}) or (boxed{x = kpi}) and (boxed{y = -frac{pi}{3} + kpi}) where ( k ) is any integer.</think>Given the system of equations:begin{aligned}x - y & = frac{pi}{3} tan x - tan y & = sqrt{3}end{aligned}We start by expressing ( x ) in terms of ( y ) using the first equation: ( x = y + frac{pi}{3} ). Substituting this into the second equation gives:tan(y + frac{pi}{3}) - tan y = sqrt{3}Using the tangent addition formula, we have:tan(y + frac{pi}{3}) = frac{tan y + tan frac{pi}{3}}{1 - tan y tan frac{pi}{3}} = frac{tan y + sqrt{3}}{1 - sqrt{3} tan y}Substituting this back into the equation, we get:frac{tan y + sqrt{3}}{1 - sqrt{3} tan y} - tan y = sqrt{3}Combining the terms into a single fraction:frac{sqrt{3}(1 + tan^2 y)}{1 - sqrt{3} tan y} = sqrt{3}Simplifying, we find:1 + tan^2 y = 1 - sqrt{3} tan yThis reduces to:tan^2 y + sqrt{3} tan y = 0Factoring out (tan y):tan y (tan y + sqrt{3}) = 0This gives two cases:1. (tan y = 0)2. (tan y = -sqrt{3})Case 1: (tan y = 0)- ( y = kpi ) for any integer ( k )- ( x = y + frac{pi}{3} = kpi + frac{pi}{3} )Case 2: (tan y = -sqrt{3})- ( y = -frac{pi}{3} + kpi ) for any integer ( k )- ( x = y + frac{pi}{3} = kpi )Thus, the solutions are:1. ( x = frac{pi}{3} + kpi ) and ( y = kpi )2. ( x = kpi ) and ( y = -frac{pi}{3} + kpi ) Final AnswerThe solutions are all angles such that ( x = frac{pi}{3} + kpi ) and ( y = kpi ), or ( x = kpi ) and ( y = -frac{pi}{3} + kpi ) for any integer ( k ). Thus, the angles are (boxed{x = frac{pi}{3} + kpi}) and (boxed{y = kpi}) or (boxed{x = kpi}) and (boxed{y = -frac{pi}{3} + kpi}) where ( k ) is any integer.,<think>Okay, so I need to solve this system of equations where the first equation is x minus y equals pi over 3, and the second one is tangent x minus tangent y equals square root of 3. Hmm, let me think. I remember that tangent has some identities, maybe I can use those. Let me write down the equations again to make it clear.First equation: x - y = π/3. So that means x is y plus π/3. Maybe I can substitute that into the second equation. The second equation is tan x - tan y = √3. If I substitute x as y + π/3, then the equation becomes tan(y + π/3) - tan y = √3. That might be a good approach. Let me try expanding tan(y + π/3) using the tangent addition formula. The formula is tan(a + b) = (tan a + tan b)/(1 - tan a tan b). So applying that here:tan(y + π/3) = [tan y + tan(π/3)] / [1 - tan y tan(π/3)]I know that tan(π/3) is √3, so substituting that in:= [tan y + √3] / [1 - tan y * √3]So now, substituting back into the second equation:[ (tan y + √3) / (1 - √3 tan y) ) ] - tan y = √3Let me write that more clearly:( (tan y + √3) / (1 - √3 tan y) ) - tan y = √3Hmm, this looks a bit complicated, but maybe I can simplify it. Let's let t = tan y to make it easier. Let t = tan y. Then the equation becomes:[ (t + √3) / (1 - √3 t) ) ] - t = √3So now, let me combine these terms. Let's write it as:( t + √3 ) / (1 - √3 t ) - t = √3To combine these, I need a common denominator. The first term has denominator (1 - √3 t), and the second term is just t, so I can write the second term as t*(1 - √3 t)/(1 - √3 t). So:[ (t + √3) - t(1 - √3 t) ] / (1 - √3 t) = √3Let me compute the numerator:(t + √3) - t(1 - √3 t) = t + √3 - t + √3 t^2 = √3 + √3 t^2So numerator is √3(1 + t^2). Therefore, the equation becomes:√3(1 + t^2) / (1 - √3 t) = √3We can divide both sides by √3, assuming √3 ≠ 0, which it isn't. So:(1 + t^2)/(1 - √3 t) = 1Multiply both sides by (1 - √3 t):1 + t^2 = 1 - √3 tSubtract 1 from both sides:t^2 = -√3 tBring all terms to one side:t^2 + √3 t = 0Factor out t:t(t + √3) = 0So solutions are t = 0 or t = -√3.But t = tan y, so tan y = 0 or tan y = -√3.Now, let's consider each case.Case 1: tan y = 0. Then y is an angle where tangent is 0. That occurs at y = kπ, where k is an integer. Then x = y + π/3, so x = kπ + π/3.Case 2: tan y = -√3. Tangent is -√3 at angles where y = -π/3 + kπ, because tan(-π/3) = -√3, and tangent has a period of π. So y = -π/3 + kπ. Then x = y + π/3 = (-π/3 + kπ) + π/3 = kπ. So x = kπ.Now, let's check if these solutions satisfy the original equations. Because sometimes when we manipulate equations, especially involving trigonometric identities, we might introduce extraneous solutions or lose some solutions. Let me verify both cases.First, check Case 1: y = kπ, x = kπ + π/3.Compute tan x - tan y. Since tan y = tan(kπ) = 0. tan x = tan(kπ + π/3). Since tan has period π, tan(kπ + π/3) = tan(π/3) = √3. So tan x - tan y = √3 - 0 = √3, which matches the second equation. So this is a valid solution.Case 2: y = -π/3 + kπ, x = kπ.Compute tan x - tan y. tan x = tan(kπ) = 0. tan y = tan(-π/3 + kπ). Since tan is periodic with period π, tan(-π/3 + kπ) = tan(-π/3) = -√3. So tan x - tan y = 0 - (-√3) = √3, which also matches the second equation. So this is also a valid solution.Therefore, both cases are valid. So the general solutions are:For Case 1: x = kπ + π/3, y = kπ, where k is an integer.For Case 2: x = kπ, y = -π/3 + kπ, where k is an integer.But the problem says "Which angles satisfy the system of equations". Depending on the context, if they are looking for all solutions, then these are the general solutions. However, sometimes problems like this expect solutions within a specific interval, like [0, 2π), but since no interval is specified here, we should present the general solution.But let me check if there's another way to write these solutions. Let's see.Case 1: y = kπ, so x = kπ + π/3.Case 2: y = -π/3 + kπ, x = kπ.Alternatively, for Case 2, we can write y = (k - 1/3)π, but that's the same as y = -π/3 + kπ. So both forms are equivalent.Alternatively, if we consider k to be any integer, these solutions can be written as:For Case 1: x = π/3 + kπ, y = 0 + kπ.Wait, but x and y are related by x = y + π/3, so if y is kπ, then x is kπ + π/3. So they differ by π/3. Similarly, in Case 2, y = -π/3 + kπ, so x = y + π/3 = -π/3 + kπ + π/3 = kπ. So x is kπ, y is kπ - π/3. Which is the same as y = (k - 1)π + 2π/3. But that might complicate things.Alternatively, maybe writing both solutions in terms of k and m. But I think the way it is now is acceptable.So the solutions are all pairs (x, y) such that either:1. x = π/3 + kπ and y = kπ for some integer k, or2. x = kπ and y = -π/3 + kπ for some integer k.Alternatively, combining both cases, we can write x = nπ + π/3 and y = nπ, or x = nπ and y = nπ - π/3, where n is an integer.Wait, let me check that. Let's take Case 1: x = kπ + π/3, y = kπ.If we let n = k, then x = nπ + π/3, y = nπ.Similarly, Case 2: x = kπ, y = -π/3 + kπ. Let n = k - 1, so k = n + 1. Then x = (n + 1)π, y = -π/3 + (n + 1)π = nπ + 2π/3. Wait, that might not be helpful. Alternatively, if we take n = k, then x = nπ, y = -π/3 + nπ. Which is equivalent to y = nπ - π/3.So depending on how you index them, both cases can be written as:Either x = nπ + π/3 and y = nπ, or x = nπ and y = nπ - π/3, for integer n.But perhaps the problem expects the solutions in terms of specific expressions. Let me check if there's a way to combine these solutions.Alternatively, we can note that both solutions can be represented by shifting the integer multiples of π. But since the equations are periodic with period π, the solutions are going to repeat every π. So the general solution is as we found.Therefore, the answer is that the angles satisfy:For any integer k,x = π/3 + kπ and y = kπ,ORx = kπ and y = -π/3 + kπ.I think that's the correct conclusion. Let me just verify with specific values. Let's take k=0.Case 1: x = π/3, y = 0.Check x - y = π/3 - 0 = π/3, correct.tan(π/3) - tan(0) = √3 - 0 = √3, correct.Case 2: x = 0, y = -π/3.Check x - y = 0 - (-π/3) = π/3, correct.tan(0) - tan(-π/3) = 0 - (-√3) = √3, correct.Another example, k=1.Case 1: x = π/3 + π = 4π/3, y = π.Check x - y = 4π/3 - π = π/3, correct.tan(4π/3) - tan(π) = tan(π + π/3) - 0 = tan(π/3) = √3, correct.Case 2: x = π, y = -π/3 + π = 2π/3.Check x - y = π - 2π/3 = π/3, correct.tan(π) - tan(2π/3) = 0 - tan(π - π/3) = 0 - (-√3) = √3, correct.So all these examples check out. Therefore, the solutions are correct.Final AnswerThe solutions are all angles where ( x = frac{pi}{3} + kpi ) and ( y = kpi ), or ( x = kpi ) and ( y = -frac{pi}{3} + kpi ) for any integer ( k ). Thus, the angles can be written as:[boxed{x = frac{pi}{3} + kpi text{ and } y = kpi} quad text{or} quad boxed{x = kpi text{ and } y = -frac{pi}{3} + kpi}]where ( k ) is any integer.</think>Given the system of equations:[begin{aligned}x - y & = frac{pi}{3} tan x - tan y & = sqrt{3}end{aligned}]We start by substituting ( x = y + frac{pi}{3} ) from the first equation into the second equation. Using the tangent addition formula, we have:[tan(y + frac{pi}{3}) = frac{tan y + tan frac{pi}{3}}{1 - tan y tan frac{pi}{3}} = frac{tan y + sqrt{3}}{1 - sqrt{3} tan y}]Substituting this into the second equation gives:[frac{tan y + sqrt{3}}{1 - sqrt{3} tan y} - tan y = sqrt{3}]Letting ( t = tan y ), we rewrite the equation as:[frac{t + sqrt{3}}{1 - sqrt{3} t} - t = sqrt{3}]Combining the terms over a common denominator and simplifying, we get:[frac{sqrt{3}(1 + t^2)}{1 - sqrt{3} t} = sqrt{3}]Dividing both sides by (sqrt{3}) and solving for ( t ):[1 + t^2 = 1 - sqrt{3} t implies t^2 + sqrt{3} t = 0 implies t(t + sqrt{3}) = 0]This gives ( t = 0 ) or ( t = -sqrt{3} ). Therefore, (tan y = 0 ) or (tan y = -sqrt{3} ).For (tan y = 0 ), ( y = kpi ) (where ( k ) is an integer), and substituting back, ( x = kpi + frac{pi}{3} ).For (tan y = -sqrt{3} ), ( y = -frac{pi}{3} + kpi ), and substituting back, ( x = kpi ).Verifying both cases, we find they satisfy the original equations. Thus, the solutions are:[boxed{x = frac{pi}{3} + kpi text{ and } y = kpi} quad text{or} quad boxed{x = kpi text{ and } y = -frac{pi}{3} + kpi}]where ( k ) is any integer.]

🔑:The equation can be written asfrac{sin (x-y)}{cos x cos y}=sqrt{3}Sincesin (x-y)=sin frac{pi}{3}=frac{sqrt{3}}{2}it follows that:2 cos x cos y=cos (x+y)+cos (x-y)=1However,cos (x-y)=cos frac{pi}{3}=frac{1}{2}thuscos (x+y)=frac{1}{2} text {, i.e., } x+y= pm frac{pi}{3}+2 k piTherefore, we have two systems of equations:I. x-y=frac{pi}{3} and x+y=frac{pi}{3}+2 k pi. From this, x=frac{pi}{3}+k pi and y=k pi, where k is any positive or negative integer, or zero.Jenő Tornai (Kegyesrendi g. VI. o. Veszprém).