🧠:[<think>Okay, let me try to tackle this problem step by step. First, part (Ⅰ) asks to determine the measure of angle ( C ) given that ( 2ccos B = 2a + b ). Hmm, triangle with sides a, b, c opposite angles A, B, C respectively. So, maybe I can use the Law of Cosines here? The Law of Cosines relates the sides of a triangle to the cosine of one of its angles. The formula is ( c^2 = a^2 + b^2 - 2abcos C ), but in this problem, we have ( 2ccos B ). Let me recall that the Law of Cosines for angle B would be ( cos B = frac{a^2 + c^2 - b^2}{2ac} ). So maybe substituting that into the given equation would help?Given equation: ( 2ccos B = 2a + b )Substitute ( cos B ):( 2c times frac{a^2 + c^2 - b^2}{2ac} = 2a + b )Simplify the left side:The 2c in the numerator and the 2ac in the denominator would cancel out to 1/a. So,( frac{a^2 + c^2 - b^2}{a} = 2a + b )Multiply both sides by a:( a^2 + c^2 - b^2 = 2a^2 + ab )Bring all terms to one side:( a^2 + c^2 - b^2 - 2a^2 - ab = 0 )Simplify:( -a^2 + c^2 - b^2 - ab = 0 )Rearranged:( c^2 = a^2 + b^2 + ab )Hmm, that's an equation relating the sides. But how does that help find angle C? Maybe use the Law of Cosines again for angle C. For angle C, the Law of Cosines is:( c^2 = a^2 + b^2 - 2abcos C )But from above, we have ( c^2 = a^2 + b^2 + ab ). So set them equal:( a^2 + b^2 - 2abcos C = a^2 + b^2 + ab )Subtract ( a^2 + b^2 ) from both sides:( -2abcos C = ab )Divide both sides by ab (assuming a and b are non-zero, which they are in a triangle):( -2cos C = 1 )Therefore,( cos C = -frac{1}{2} )What angle has cosine of -1/2? Well, in the range of 0 to 180 degrees (since it's a triangle), the angle with cosine -1/2 is 120 degrees. So angle C is 120 degrees. That seems straightforward. Let me check if I made any mistakes in substitution.Starting from the given equation:( 2ccos B = 2a + b )Used Law of Cosines for angle B, substituted, simplified, ended up with ( c^2 = a^2 + b^2 + ab ). Then used Law of Cosines for angle C, which gave ( c^2 = a^2 + b^2 - 2abcos C ). Equate the two expressions for ( c^2 ), leading to ( -2abcos C = ab ), so ( cos C = -1/2 ). Yep, that seems right. So angle C is 120 degrees. So part (Ⅰ) answer is 120°, or ( frac{2pi}{3} ) radians. Since the problem asks for the measure, probably degrees unless specified, but maybe in radians. The problem doesn't specify, but in Olympiad problems, degrees are common for angles. So 120 degrees. Got it.Now part (Ⅱ): If ( a = b ), and the area ( S ) of the triangle ( ABC ) is ( frac{sqrt{3}}{2}c ), find the perimeter of triangle ( ABC ).Alright, so given ( a = b ), meaning triangle ABC is isosceles with sides a = b. So angles opposite them, angles A and B, are equal. Wait, but angle C is already 120 degrees from part (Ⅰ). So in this triangle, angles A and B are equal, and angle C is 120°, so angles A and B must each be 30°, since 180 - 120 = 60, divided by 2 is 30. So angles A and B are 30°, angle C is 120°, sides a and b are equal. So the triangle is isosceles with sides a = b, and angle C = 120°.But we need to find the perimeter given that the area is ( frac{sqrt{3}}{2}c ). So first, let's note that in a triangle with sides a, a, c, angles 30°, 30°, 120°, the area can be expressed in terms of sides. Let's recall that the area of a triangle can be calculated using the formula ( frac{1}{2}absin C ). Wait, but here, since sides a and b are equal, and angle between them is angle C. Wait, actually, in this case, if sides a and b are equal, and angle C is between them? Wait, no. Wait, in triangle ABC, side a is opposite angle A, side b opposite angle B, side c opposite angle C. If a = b, then angles A and B are equal. But angle C is opposite side c, which is 120°. So the sides adjacent to angle C are a and b (which are equal). So angle C is between sides a and b. Therefore, the area can be calculated as ( frac{1}{2}absin C ). Since a = b, then ( frac{1}{2}a^2 sin 120° ). But angle C is 120°, so sin 120° is ( frac{sqrt{3}}{2} ). Therefore, area ( S = frac{1}{2}a^2 times frac{sqrt{3}}{2} = frac{sqrt{3}}{4}a^2 ). But the problem states that the area is ( frac{sqrt{3}}{2}c ). So equate them:( frac{sqrt{3}}{4}a^2 = frac{sqrt{3}}{2}c )Divide both sides by ( sqrt{3} ):( frac{1}{4}a^2 = frac{1}{2}c )Multiply both sides by 4:( a^2 = 2c )So ( c = frac{a^2}{2} )But we need to find the perimeter, which is ( a + a + c = 2a + c ). So if we can express c in terms of a, then we can write perimeter as ( 2a + frac{a^2}{2} ). But we need to find the numerical value. So perhaps we need another equation to relate a and c. Let's see.From part (Ⅰ), we had the relation ( c^2 = a^2 + b^2 + ab ). But since a = b, substitute:( c^2 = a^2 + a^2 + a times a = 2a^2 + a^2 = 3a^2 )Wait, wait, that seems conflicting. Wait, in part (Ⅰ), we derived that ( c^2 = a^2 + b^2 + ab ). Since here a = b, so substituting:( c^2 = a^2 + a^2 + a times a = a^2 + a^2 + a^2 = 3a^2 )Therefore, ( c = asqrt{3} ). Wait, but earlier we had from the area, ( c = frac{a^2}{2} ). So set these equal:( asqrt{3} = frac{a^2}{2} )Divide both sides by a (assuming a ≠ 0):( sqrt{3} = frac{a}{2} )Thus, ( a = 2sqrt{3} )Therefore, ( c = frac{a^2}{2} = frac{(2sqrt{3})^2}{2} = frac{4 times 3}{2} = frac{12}{2} = 6 )Hence, the sides are a = b = 2√3, and c = 6. So perimeter is ( 2a + c = 2 times 2sqrt{3} + 6 = 4sqrt{3} + 6 ). Wait, but let me check if this is correct.First, confirm the area. If a = 2√3, then area is ( frac{sqrt{3}}{4} times (2sqrt{3})^2 = frac{sqrt{3}}{4} times 12 = frac{sqrt{3}}{4} times 12 = 3sqrt{3} ). But the problem states that the area is ( frac{sqrt{3}}{2}c ). If c = 6, then area should be ( frac{sqrt{3}}{2} times 6 = 3sqrt{3} ). That matches. Good.Also, check the relation from part (Ⅰ): ( c^2 = 3a^2 ). If a = 2√3, then 3a² = 3*(4*3) = 3*12 = 36. And c = 6, so c² = 36. Correct. So that's consistent.Therefore, the perimeter is 4√3 + 6. But maybe they want it written as 6 + 4√3. Either way, it's the same. But let me see if the problem expects a simplified form or a numerical value. But since it's exact, 6 + 4√3 is fine.Wait, but let me check again the steps to be sure. When we had from part (Ⅰ), ( c^2 = 3a^2 ), which gives ( c = asqrt{3} ). From the area, we had ( c = a² / 2 ). So setting ( asqrt{3} = a² / 2 ), leading to ( a = 2sqrt{3} ). That's correct. Then c = 6. So perimeter is 2a + c = 4√3 + 6. Yes, that's correct.But let me also check using another formula for area. Since we know two sides and the included angle, area is ( frac{1}{2}absin C ). Here, a = b = 2√3, angle C = 120°, so sin 120° = √3/2. Then area is ( frac{1}{2} times 2sqrt{3} times 2sqrt{3} times frac{sqrt{3}}{2} ). Calculate that:First, ( frac{1}{2} times 2sqrt{3} times 2sqrt{3} = frac{1}{2} times 4 times 3 = 6 ). Then multiply by ( frac{sqrt{3}}{2} ): 6 × √3/2 = 3√3. Which matches the given area. So that's correct.Therefore, the perimeter is 6 + 4√3. Wait, but 4√3 is approximately 6.928, so total perimeter is approximately 12.928, but exact value is 6 + 4√3. So that's the answer.But let me check once more if I didn't mix up any formulas. Let's see. Given a = b, angle C = 120°, area is √3/2 c. Then, as we found, a = 2√3, c = 6. So perimeter is 2a + c = 4√3 + 6, which is 6 + 4√3. Yep, seems right.So part (Ⅱ) answer is ( 6 + 4sqrt{3} ). Final AnswerⅠ. boxed{120^circ}Ⅱ. boxed{6 + 4sqrt{3}}</think>Given triangle (ABC) with sides opposite angles (A), (B), and (C) being (a), (b), and (c) respectively, and the equation (2ccos B = 2a + b):(Ⅰ) Determine the measure of angle (C).1. Using the Law of Cosines for angle (B): [ cos B = frac{a^2 + c^2 - b^2}{2ac} ]2. Substituting this into the given equation: [ 2c times frac{a^2 + c^2 - b^2}{2ac} = 2a + b ]3. Simplifying: [ frac{a^2 + c^2 - b^2}{a} = 2a + b implies a^2 + c^2 - b^2 = 2a^2 + ab implies c^2 = a^2 + b^2 + ab ]4. Using the Law of Cosines for angle (C): [ c^2 = a^2 + b^2 - 2abcos C ]5. Equating the two expressions for (c^2): [ a^2 + b^2 - 2abcos C = a^2 + b^2 + ab implies -2abcos C = ab implies cos C = -frac{1}{2} ]6. Therefore, angle (C) is (120^circ).(Ⅱ) Find the perimeter of triangle (ABC) given (a = b) and area (S = frac{sqrt{3}}{2}c).1. Since (a = b), the triangle is isosceles with angles (A = B = 30^circ) and (C = 120^circ).2. The area of the triangle using sides (a) and (b) with included angle (C): [ S = frac{1}{2}a^2 sin 120^circ = frac{sqrt{3}}{4}a^2 ]3. Given (S = frac{sqrt{3}}{2}c), equating the two expressions: [ frac{sqrt{3}}{4}a^2 = frac{sqrt{3}}{2}c implies a^2 = 2c implies c = frac{a^2}{2} ]4. From part (Ⅰ), (c^2 = 3a^2): [ c = asqrt{3} ]5. Equating (c = frac{a^2}{2}) and (c = asqrt{3}): [ asqrt{3} = frac{a^2}{2} implies a = 2sqrt{3} ]6. Substituting (a = 2sqrt{3}) into (c = asqrt{3}): [ c = 2sqrt{3} times sqrt{3} = 6 ]7. The perimeter is: [ 2a + c = 2 times 2sqrt{3} + 6 = 4sqrt{3} + 6 ] Final AnswerⅠ. boxed{120^circ}Ⅱ. boxed{6 + 4sqrt{3}},<think>Okay, let's see. I have this triangle ABC with sides opposite angles A, B, and C being a, b, c respectively. The problem says that 2c cos B = 2a + b. Then part I is to find the measure of angle C. Hmm, okay. So I need to use some trigonometric identities or laws here. Probably the Law of Cosines or maybe the Law of Sines. Let me think.First, the given equation is 2c cos B = 2a + b. Let's recall that in the Law of Cosines, cos B can be expressed in terms of the sides. The Law of Cosines states that for any triangle,c² = a² + c² - 2ac cos BWait, no. Wait, the Law of Cosines for angle B would be:cos B = (a² + c² - b²)/(2ac)Wait, hold on. The Law of Cosines is:For angle B: b² = a² + c² - 2ac cos BTherefore, rearranged, cos B = (a² + c² - b²)/(2ac)So substituting this into the given equation: 2c * [(a² + c² - b²)/(2ac)] = 2a + bSimplify left side: 2c * [(a² + c² - b²)/(2ac)] = (2c / 2ac) * (a² + c² - b²) = (1/a)(a² + c² - b²)So we have (a² + c² - b²)/a = 2a + bMultiply both sides by a:a² + c² - b² = 2a² + abBring all terms to the left side:a² + c² - b² - 2a² - ab = 0Simplify:- a² + c² - b² - ab = 0Which can be written as:c² = a² + b² + abHmm, interesting. So we have c squared equals a squared plus b squared plus ab. That seems like a key equation here. Now, maybe we can relate this to angle C? Since angle C is opposite side c. Let's recall the Law of Cosines again. For angle C:c² = a² + b² - 2ab cos CBut from our previous result, c² = a² + b² + ab. So setting these equal:a² + b² - 2ab cos C = a² + b² + abSubtract a² + b² from both sides:-2ab cos C = abDivide both sides by ab (assuming a and b are non-zero, which they are in a triangle):-2 cos C = 1Therefore, cos C = -1/2So angle C must be 120 degrees, since cosine of 120° is -1/2. That's in the second quadrant. So part I answer is 120°, or 2π/3 radians. Probably degrees since it's a geometry problem. Let me check again.Wait, yes. The equation led us to cos C = -1/2, so C is 120 degrees. So that's part I done.Now part II: If a = b, and the area S of triangle ABC is (√3 / 2) c, find the perimeter of triangle ABC.Okay, so given a = b, so it's an isosceles triangle with sides a, a, c. Wait, but angle C is 120 degrees, as we found in part I. So the triangle has two sides equal (a = b), and angle C opposite side c is 120 degrees. Then, given the area is (√3 / 2) times c, find the perimeter.So let's start by using the given conditions. Since a = b, let's denote a = b = x, so sides are x, x, c. Then, perimeter is 2x + c. We need to find x and c such that the area is (√3 / 2)c.First, since a = b = x, angles opposite them are equal. So angles A and B are equal. Since angle C is 120°, the sum of angles A and B must be 60°, so each is 30°. Wait, is that correct?Wait, in triangle ABC, angles sum to 180°, so if A and B are equal, then:A + B + C = 180°But C is 120°, so A + B = 60°, and since A = B, each is 30°. Therefore, angles are 30°, 30°, 120°.So triangle is isosceles with sides a = b = x, c opposite 120°, angles at A and B are 30° each.Now, let's recall that area S = (1/2)ab sin C. Wait, but here a and b are sides adjacent to angle C. Wait, in triangle area formula, area is (1/2)*a*b*sin C, where a and b are sides forming angle C. Wait, but in this case, sides a and b are adjacent to angle C? Wait, no. Wait, angle C is opposite side c, so sides a and b are adjacent to angle C. So yes, area would be (1/2)*a*b*sin C.Given that a = b = x, and C = 120°, so sin 120° = √3 / 2. Therefore, area S = (1/2)*x*x*(√3 / 2) = (√3 / 4)x²But the problem states that area S is (√3 / 2)c. Therefore:(√3 / 4)x² = (√3 / 2)cDivide both sides by √3:(1/4)x² = (1/2)cMultiply both sides by 4:x² = 2cSo c = x² / 2Okay, so c is expressed in terms of x. Now, we also need another equation to relate x and c. Since we know the triangle sides, we can use the Law of Sines or Law of Cosines.Given that angles are 30°, 30°, 120°, sides opposite are x, x, c. Law of Sines gives:x / sin 30° = c / sin 120°sin 30° = 1/2, sin 120° = √3 / 2.Therefore:x / (1/2) = c / (√3 / 2)Simplify:2x = (2c) / √3Multiply both sides by √3:2x√3 = 2cDivide both sides by 2:x√3 = cBut from earlier, c = x² / 2. Therefore:x√3 = x² / 2Multiply both sides by 2:2x√3 = x²Divide both sides by x (assuming x ≠ 0, which it isn't):2√3 = xTherefore, x = 2√3Then, c = x² / 2 = ( (2√3)^2 ) / 2 = (4 * 3) / 2 = 12 / 2 = 6Therefore, sides a = b = 2√3, c = 6. Perimeter is 2x + c = 2*(2√3) + 6 = 4√3 + 6Wait, but let me check if this is correct. Let's verify the area. If a = b = 2√3, angle C = 120°, then area S = (1/2)*a*b*sin C = (1/2)*(2√3)*(2√3)*(√3/2)Calculate:(1/2)*(4*3)*(√3/2) = (1/2)*12*(√3/2) = 6*(√3/2) = 3√3But according to the problem, the area should be (√3 / 2)*c = (√3 / 2)*6 = 3√3. So that matches. Good.Also, check Law of Sines: a / sin A = 2√3 / sin 30° = 2√3 / 0.5 = 4√3c / sin C = 6 / sin 120° = 6 / (√3 / 2) = 6 * 2 / √3 = 12 / √3 = 4√3. So that's consistent. Good.Therefore, perimeter is 4√3 + 6. But maybe the problem expects it written as 6 + 4√3, but both are equivalent. Alternatively, they might factor 2, but probably 6 + 4√3 is acceptable.Wait, let me check if there's another way. Let me recap:From part I, angle C is 120°, so when a = b, angles at A and B are 30°, leading to an isosceles triangle with sides a = b = 2√3 and c = 6. Perimeter is 2*2√3 + 6 = 4√3 + 6. So the perimeter is 6 + 4√3. Hmm. Alternatively, they might want it written in terms of combining terms, but 6 + 4√3 is simplified as much as possible.Therefore, the answer for part II is 6 + 4√3. Let me confirm once more.Yes, if a = 2√3, b = 2√3, c = 6, then perimeter is 4√3 + 6, which is 6 + 4√3. Either form is correct, but maybe the problem expects it in a certain order. But since 6 is a constant term and 4√3 is the radical term, writing 6 + 4√3 is standard.Therefore, the final answers are:Ⅰ. 120°, so boxed{120^circ}Ⅱ. Perimeter is 6 + 4√3, so boxed{6 + 4sqrt{3}}Final AnswerⅠ. boxed{120^circ}Ⅱ. boxed{6 + 4sqrt{3}}</think>Given triangle (ABC) with sides opposite angles (A), (B), and (C) being (a), (b), and (c) respectively, and the equation (2c cos B = 2a + b):(Ⅰ) Determine the measure of angle (C).Using the Law of Cosines for angle (B):[cos B = frac{a^2 + c^2 - b^2}{2ac}]Substituting this into the given equation:[2c left( frac{a^2 + c^2 - b^2}{2ac} right) = 2a + b]Simplifying:[frac{a^2 + c^2 - b^2}{a} = 2a + b]Multiplying both sides by (a):[a^2 + c^2 - b^2 = 2a^2 + ab]Rearranging terms:[c^2 = a^2 + b^2 + ab]Using the Law of Cosines for angle (C):[c^2 = a^2 + b^2 - 2ab cos C]Equating the two expressions for (c^2):[a^2 + b^2 - 2ab cos C = a^2 + b^2 + ab]Simplifying:[-2ab cos C = ab]Solving for (cos C):[cos C = -frac{1}{2}]Thus, angle (C) is (120^circ).(Ⅱ) Find the perimeter of triangle (ABC) given (a = b) and the area (S = frac{sqrt{3}}{2}c).Given (a = b = x), the triangle is isosceles with angles (30^circ), (30^circ), and (120^circ). Using the area formula:[S = frac{1}{2}ab sin C = frac{sqrt{3}}{4}x^2]Given (S = frac{sqrt{3}}{2}c), we equate:[frac{sqrt{3}}{4}x^2 = frac{sqrt{3}}{2}c]Solving for (c):[x^2 = 2c implies c = frac{x^2}{2}]Using the Law of Sines:[frac{x}{sin 30^circ} = frac{c}{sin 120^circ}]Substituting (sin 30^circ = frac{1}{2}) and (sin 120^circ = frac{sqrt{3}}{2}):[2x = frac{2c}{sqrt{3}} implies xsqrt{3} = c]Equating the two expressions for (c):[xsqrt{3} = frac{x^2}{2} implies x = 2sqrt{3}]Thus, (c = 6). The perimeter is:[2x + c = 2(2sqrt{3}) + 6 = 4sqrt{3} + 6]Final answers:Ⅰ. (boxed{120^circ})Ⅱ. (boxed{6 + 4sqrt{3}})]

🔑:(Ⅰ) From the given information and the Law of Cosines, we have 2c times frac{a^2 + c^2 - b^2}{2ac} = 2a + b,which simplifies to a^2 + b^2 - c^2 = -ab. Now, utilizing the Law of Cosines again for angle C, we get cos C = frac{a^2 + b^2 - c^2}{2ab} = frac{-ab}{2ab} = -frac{1}{2}.Given that in triangle ABC, 0 < C < pi,we conclude that C = frac{2pi}{3},which means the measure of angle C is boxed{frac{2pi}{3}}.(Ⅱ) Because the triangle's area S = frac{1}{2}absin C = frac{sqrt{3}}{2}c, we have c = frac{1}{2}ab.Also, c^2 = a^2 + b^2 - 2abcos C = a^2 + b^2 + ab because cos C = -frac{1}{2}.Taking into consideration that a = b, we set a = b = x. From c^2 = x^2 + x^2 + x^2, we can solve for x: 3x^2 = c^2,x^2 = frac{c^2}{3},x = frac{c}{sqrt{3}}.Replacing c with ab (since c = frac{1}{2}ab and a = b = x), we get:x = frac{x^2}{sqrt{3}},so x^2 = xsqrt{3},and x = sqrt{3}, which means that a = b = sqrt{3}.Now replacing a and b with sqrt{3} in c = frac{1}{2}ab, we get:c = frac{1}{2}(sqrt{3})(sqrt{3}) = frac{3}{2}.Then, we multiply by 2 to correct the earlier simplification (c = frac{1}{2}ab initially):c = 3.Therefore, a = b = 2sqrt{3} and c = 6.The perimeter of the triangle is a + b + c = 2sqrt{3} + 2sqrt{3} + 6 = 4sqrt{3} + 6,which simplifies to boxed{6 + 4sqrt{3}}.

🧠:[<think>Okay, let's see. The problem is about the function f(x) = cos x on the interval [0, 2π]. We know that cos x has two zeros in this interval, which are at x₁ = π/2 and x₂ = 3π/2, right? Then, the equation f(x) = m, which is cos x = m, has two distinct real roots x₃ and x₄ in the same interval. So, these four numbers x₁, x₂, x₃, x₄ are arranged in ascending order to form an arithmetic sequence. We need to find the value of m.First, let's recall that cos x = m will have two solutions in [0, 2π] when m is between -1 and 1, but since the original problem says "two distinct real roots," m must be between -1 and 1, excluding the cases where m is 1 or -1, which would give only one root each. But actually, even when m is 1 or -1, there are two roots in [0, 2π], but they are not distinct. For m = 1, x = 0 and 2π, which are the same point. For m = -1, x = π. So, for m strictly between -1 and 1, we have two distinct roots. So m is in (-1, 1).The zeros of cos x are at π/2 and 3π/2. So x₁ = π/2, x₂ = 3π/2. Now, the equation cos x = m has two roots. Let's denote them as x₃ and x₄. Since cos x is even around π, the solutions will be symmetric around π. Specifically, the solutions are x = arccos m and x = 2π - arccos m. Wait, but arccos m gives a value between 0 and π. So if we take x₃ = arccos m and x₄ = 2π - arccos m. However, depending on the value of m, these solutions could be in different positions relative to π/2 and 3π/2.But the problem states that when you arrange x₁, x₂, x₃, x₄ in ascending order, they form an arithmetic sequence. So we need to order all four numbers and check if they form an arithmetic progression.Let me first list all four points:1. x₁ = π/2 ≈ 1.57082. x₂ = 3π/2 ≈ 4.71243. x₃ = arccos m4. x₄ = 2π - arccos mSince x₃ and x₄ are symmetric around π, which is approximately 3.1416. So x₃ is in [0, π] and x₄ is in [π, 2π].Now, depending on the value of m, arccos m can be in different positions. If m is positive, arccos m is in [0, π/2), and 2π - arccos m is in (3π/2, 2π]. If m is zero, arccos 0 is π/2, which is x₁. If m is negative, arccos m is in (π/2, π], and 2π - arccos m is in [π, 3π/2). But since m is such that there are two distinct roots, m cannot be 0 because then x₃ and x₄ would be π/2 and 3π/2, which are the zeros, but the problem says the zeros are x₁ and x₂, and the roots of f(x) = m are x₃ and x₄. So m cannot be zero. Wait, but if m is zero, then cos x = 0 has solutions at π/2 and 3π/2, but those are already x₁ and x₂. But the problem states that x₃ and x₄ are distinct roots of f(x) = m, which are different from x₁ and x₂. Therefore, m cannot be zero. So m must be in (-1, 0) union (0, 1).But let's check. If m is positive, then x₃ is in [0, π/2) and x₄ is in (3π/2, 2π]. If m is negative, x₃ is in (π/2, π) and x₄ is in [π, 3π/2). So depending on the sign of m, the positions of x₃ and x₄ relative to x₁ and x₂ change.So if m is positive, the order of the four points would be x₃ (in [0, π/2)), x₁ (π/2), x₂ (3π/2), x₄ (in (3π/2, 2π]). So arranged in ascending order: x₃, x₁, x₂, x₄. Similarly, if m is negative, x₃ is in (π/2, π) and x₄ is in [π, 3π/2), so the order would be x₁ (π/2), x₃, x₄, x₂ (3π/2). So ascending order: x₁, x₃, x₄, x₂.Therefore, depending on the sign of m, the order of the four numbers when sorted is different. So for m positive: x₃, x₁, x₂, x₄. For m negative: x₁, x₃, x₄, x₂.Now, these four numbers form an arithmetic sequence. Let's handle both cases.Case 1: m > 0. Then the order is x₃, x₁, x₂, x₄. So the four terms in arithmetic progression are x₃, x₁, x₂, x₄. So the common difference d must satisfy:x₁ - x₃ = x₂ - x₁ = x₄ - x₂But let's check. For an arithmetic sequence with four terms, the difference between consecutive terms is the same. So:Term2 - Term1 = Term3 - Term2 = Term4 - Term3In this case:x₁ - x₃ = x₂ - x₁ = x₄ - x₂But let's write the equations:First, x₁ - x₃ = x₂ - x₁. So:π/2 - arccos m = 3π/2 - π/2 = πTherefore:π/2 - arccos m = πWhich implies:arccos m = π/2 - π = -π/2But arccos m is in [0, π], so this is impossible. Therefore, this case cannot happen. Thus, m cannot be positive.Case 2: m < 0. Then the order is x₁, x₃, x₄, x₂. So the four terms in arithmetic progression are x₁, x₃, x₄, x₂. Therefore, the common difference d must satisfy:x₃ - x₁ = x₄ - x₃ = x₂ - x₄So let's write the equations:First, x₃ - x₁ = x₄ - x₃Which implies that x₄ = 2x₃ - x₁Second, x₄ - x₃ = x₂ - x₄Which implies that 2x₄ = x₂ + x₃So substituting x₄ from the first equation into the second:2*(2x₃ - x₁) = x₂ + x₃Simplify:4x₃ - 2x₁ = x₂ + x₃Therefore:3x₃ = x₂ + 2x₁So x₃ = (x₂ + 2x₁)/3We know x₁ = π/2, x₂ = 3π/2Therefore:x₃ = (3π/2 + 2*(π/2))/3 = (3π/2 + π)/3 = (5π/2)/3 = 5π/6Similarly, x₄ = 2x₃ - x₁ = 2*(5π/6) - π/2 = 5π/3 - π/2 = (10π/6 - 3π/6) = 7π/6Wait, but x₃ is arccos m, and x₄ is 2π - arccos m. So let's check.If x₃ = 5π/6, then arccos m = 5π/6. Therefore, m = cos(5π/6) = cos(π - π/6) = -cos(π/6) = -√3/2 ≈ -0.866.Similarly, x₄ = 2π - arccos m = 2π - 5π/6 = 7π/6, which is correct.So in this case, the four terms in order are:x₁ = π/2 ≈ 1.5708,x₃ = 5π/6 ≈ 2.6180,x₄ = 7π/6 ≈ 3.6652,x₂ = 3π/2 ≈ 4.7124.Wait, but 7π/6 is approximately 3.6652, which is less than 3π/2 ≈ 4.7124. So the order is indeed x₁, x₃, x₄, x₂. Let's check if they form an arithmetic sequence.Compute the differences:x₃ - x₁ = 5π/6 - π/2 = 5π/6 - 3π/6 = 2π/6 = π/3 ≈ 1.0472x₄ - x₃ = 7π/6 - 5π/6 = 2π/6 = π/3 ≈ 1.0472x₂ - x₄ = 3π/2 - 7π/6 = 9π/6 - 7π/6 = 2π/6 = π/3 ≈ 1.0472So yes, all differences are π/3. Therefore, it is an arithmetic sequence with common difference π/3.Therefore, the value of m is -√3/2.But let's confirm if there could be another solution. Suppose that when m is positive, even though earlier reasoning suggested it's impossible. Let's check again.If m is positive, the order is x₃, x₁, x₂, x₄. Then the differences would need to be equal:x₁ - x₃ = x₂ - x₁ = x₄ - x₂So x₁ - x₃ = x₂ - x₁ => x₃ = 2x₁ - x₂Compute that: x₁ = π/2, x₂ = 3π/2So x₃ = 2*(π/2) - 3π/2 = π - 3π/2 = -π/2But arccos m can't be negative. So x₃ can't be negative. Therefore, no solution in this case. Therefore, the only possible m is -√3/2.Hence, the answer is m = -√3/2, which in boxed form is boxed{-dfrac{sqrt{3}}{2}}.Final AnswerThe value of the real number ( m ) is boxed{-dfrac{sqrt{3}}{2}}.</think>Given the function ( f(x) = cos x ) on the interval ([0, 2pi]), the zeros are ( x_1 = frac{pi}{2} ) and ( x_2 = frac{3pi}{2} ). The equation ( cos x = m ) has two distinct roots ( x_3 ) and ( x_4 ). These four numbers form an arithmetic sequence when arranged in ascending order.For ( m ) negative, the roots ( x_3 ) and ( x_4 ) are symmetric around ( pi ). Specifically, ( x_3 = arccos m ) and ( x_4 = 2pi - arccos m ). When arranged in ascending order, the sequence is ( x_1, x_3, x_4, x_2 ).To form an arithmetic sequence, the common difference ( d ) must satisfy:[x_3 - x_1 = x_4 - x_3 = x_2 - x_4]Solving these equations:1. ( x_3 - x_1 = x_4 - x_3 )2. ( x_4 - x_3 = x_2 - x_4 )Substituting ( x_1 = frac{pi}{2} ) and ( x_2 = frac{3pi}{2} ):[x_3 = frac{x_2 + 2x_1}{3} = frac{frac{3pi}{2} + 2 cdot frac{pi}{2}}{3} = frac{5pi}{6}][x_4 = 2x_3 - x_1 = 2 cdot frac{5pi}{6} - frac{pi}{2} = frac{7pi}{6}]Thus, ( arccos m = frac{5pi}{6} ), leading to:[m = cosleft(frac{5pi}{6}right) = -frac{sqrt{3}}{2}]Verifying the arithmetic sequence:[x_3 - x_1 = frac{5pi}{6} - frac{pi}{2} = frac{pi}{3}][x_4 - x_3 = frac{7pi}{6} - frac{5pi}{6} = frac{pi}{3}][x_2 - x_4 = frac{3pi}{2} - frac{7pi}{6} = frac{pi}{3}]All differences are equal, confirming the sequence is arithmetic. Therefore, the value of ( m ) is (boxed{-dfrac{sqrt{3}}{2}}).,<think>Alright, let's try to figure this problem out step by step. The function given is f(x) = cos(x) on the interval [0, 2π]. We know that cos(x) has zeros at x = π/2 and x = 3π/2 in this interval, right? So those are x₁ and x₂. Then, the equation cos(x) = m has two distinct real roots x₃ and x₄. Since cos(x) ranges between -1 and 1, m must be between -1 and 1. But since we need two distinct real roots, m can't be equal to 1 or -1 because then the equation would have only one root (at x=0 or x=π). So m is between -1 and 1, excluding the endpoints.Now, the key part is that when we arrange these four numbers x₁, x₂, x₃, x₄ in ascending order, they form an arithmetic sequence. An arithmetic sequence means that the difference between consecutive terms is constant. Let's denote the four terms in order as a, a + d, a + 2d, a + 3d, where d is the common difference.First, let's recall the zeros of cos(x) in [0, 2π] are π/2 and 3π/2. So x₁ = π/2 and x₂ = 3π/2. The roots of cos(x) = m are x₃ and x₄. Since cos(x) is symmetric around π, the solutions to cos(x) = m will be x = arccos(m) and x = 2π - arccos(m). So x₃ = arccos(m) and x₄ = 2π - arccos(m). But depending on the value of m, arccos(m) can be in different quadrants. However, since we need two distinct roots in [0, 2π], as long as m ≠ ±1, we have two solutions.Now, we need to arrange all four numbers π/2, 3π/2, arccos(m), and 2π - arccos(m) in ascending order and have them form an arithmetic sequence. Let's think about the possible orderings.First, note that arccos(m) is in [0, π] because the range of arccos is [0, π]. Therefore, 2π - arccos(m) is in [π, 2π]. So, arccos(m) is in [0, π] and 2π - arccos(m) is in [π, 2π]. The zeros π/2 and 3π/2 are at the midpoints of the intervals [0, π] and [π, 2π], respectively.So, depending on the value of m, arccos(m) could be less than π/2, between π/2 and π, etc. Similarly, 2π - arccos(m) would be greater than π but could be less than 3π/2 or greater than 3π/2. Let's consider possible cases.Case 1: arccos(m) < π/2. Then 2π - arccos(m) > 3π/2. So the order would be arccos(m), π/2, 3π/2, 2π - arccos(m). Let's check if these four can form an arithmetic sequence.Case 2: arccos(m) > π/2. Then 2π - arccos(m) < 3π/2. So the order would be π/2, arccos(m), 3π/2, 2π - arccos(m). Let's check this case.Case 3: arccos(m) = π/2. Then 2π - arccos(m) = 3π/2. But in this case, arccos(m) = π/2 implies m = cos(π/2) = 0. But then the roots would be π/2 and 3π/2, which are the same as the zeros. But the problem states that there are two distinct zeros and two distinct roots for f(x)=m. If m=0, then the roots and zeros are the same, so we would have only two distinct points, not four. So m=0 is invalid. Therefore, arccos(m) cannot be π/2.Similarly, if arccos(m) = 3π/2, but 3π/2 is outside the range of arccos, which is [0, π], so that's impossible.So we can ignore Case 3. Let's examine Case 1 and Case 2.First, let's try Case 1: arccos(m) < π/2. Then the order is arccos(m), π/2, 3π/2, 2π - arccos(m). Let's denote them as a, a + d, a + 2d, a + 3d.So:a = arccos(m)a + d = π/2a + 2d = 3π/2a + 3d = 2π - arccos(m)From the second and third equations:a + d = π/2a + 2d = 3π/2Subtracting the first from the second: d = πThen, substituting back into a + d = π/2: a = π/2 - d = π/2 - π = -π/2But arccos(m) = a = -π/2. However, arccos(m) is in [0, π], so this is impossible. Therefore, Case 1 is invalid.So moving on to Case 2: arccos(m) > π/2. Then the order is π/2, arccos(m), 3π/2, 2π - arccos(m). Let's denote them as a, a + d, a + 2d, a + 3d.So:a = π/2a + d = arccos(m)a + 2d = 3π/2a + 3d = 2π - arccos(m)From the first and third equations:a + 2d = 3π/2But a = π/2, so π/2 + 2d = 3π/2 => 2d = π => d = π/2Then, from a + d = arccos(m):π/2 + π/2 = arccos(m) => arccos(m) = πTherefore, m = cos(π) = -1But m = -1 would give roots at arccos(-1) = π and 2π - π = π, but wait, that's the same point. Wait, cos(x) = -1 has only one root at x = π in [0, 2π]. So m = -1 gives only one root, but we need two distinct roots. Therefore, m = -1 is invalid. So this is a contradiction. Therefore, Case 2 as presented is invalid.Hmm, so maybe there's another possible ordering? Wait, perhaps my initial assumption of the ordering in Case 2 is wrong. Let me check again.If arccos(m) > π/2, then arccos(m) is in (π/2, π], so 2π - arccos(m) is in [π, 3π/2). So the zeros are at π/2 and 3π/2. So the four points would be:π/2 (zero), arccos(m) (root), 3π/2 (zero), 2π - arccos(m) (root). Wait, but 2π - arccos(m) is in [π, 3π/2), so it would come before 3π/2? Wait, no. If arccos(m) is in (π/2, π], then 2π - arccos(m) is in [π, 3π/2). So for example, if arccos(m) is slightly larger than π/2, then 2π - arccos(m) is slightly less than 3π/2. Wait, let's take an example. Let arccos(m) = 2π/3. Then 2π - arccos(m) = 2π - 2π/3 = 4π/3. So in this case, the order would be π/2, 2π/3, 4π/3, 3π/2. But 3π/2 is 4.712..., 4π/3 is about 4.188..., so 4π/3 is less than 3π/2. So the order would be π/2, arccos(m), 2π - arccos(m), 3π/2.Wait, that's different from what I thought earlier. So if arccos(m) is in (π/2, π), then 2π - arccos(m) is in (π, 3π/2). So in ascending order, the four points would be π/2, arccos(m), 3π/2, 2π - arccos(m)? Wait, no. Let me plug in actual numbers.Suppose arccos(m) = 2π/3 (which is 120 degrees), then 2π - arccos(m) = 4π/3 (240 degrees). The zeros are at π/2 (90 degrees) and 3π/2 (270 degrees). So the order would be π/2 (90), 2π/3 (120), 4π/3 (240), 3π/2 (270). Wait, but 2π/3 is approximately 2.094, and π/2 is approximately 1.570, so π/2 comes first. Then 2π/3, then 4π/3 (which is approximately 4.188), then 3π/2 (approximately 4.712). So yes, the order is π/2, arccos(m), 2π - arccos(m), 3π/2. Therefore, the four terms in order would be π/2, arccos(m), 2π - arccos(m), 3π/2. Therefore, the arithmetic sequence should be arranged as π/2, arccos(m), 2π - arccos(m), 3π/2. Let's check if these four can form an arithmetic sequence.Let me denote them as a, a + d, a + 2d, a + 3d.So:a = π/2a + d = arccos(m)a + 2d = 2π - arccos(m)a + 3d = 3π/2From the first and last equations:a = π/2a + 3d = 3π/2Subtracting, we get 3d = π => d = π/3Therefore, the common difference is π/3. Then, the four terms would be π/2, π/2 + π/3, π/2 + 2π/3, π/2 + π. Let's compute these:π/2 = 1.5708π/2 + π/3 = 1.5708 + 1.0472 ≈ 2.618π/2 + 2π/3 = 1.5708 + 2.0944 ≈ 3.6652π/2 + π = 4.7124But according to our earlier example, arccos(m) would be π/2 + π/3 = 5π/6 ≈ 2.618 (which is 150 degrees), then 2π - arccos(m) = 2π - 5π/6 = 7π/6 ≈ 3.665 (210 degrees). So the sequence would be π/2 (1.5708), 5π/6 (2.618), 7π/6 (3.665), 3π/2 (4.7124). Indeed, each term increases by π/3. So this seems to satisfy the arithmetic sequence condition.Therefore, in this case, arccos(m) = π/2 + π/3 = 5π/6. Therefore, m = cos(5π/6). What's cos(5π/6)? That's equal to -√3/2. Let's confirm:cos(5π/6) = cos(π - π/6) = -cos(π/6) = -√3/2. Yes, that's correct.Therefore, m = -√3/2. Let's check if this works.So, if m = -√3/2, then the roots of cos(x) = -√3/2 are x = 5π/6 and 7π/6. The zeros of cos(x) are π/2 and 3π/2. So arranging these four numbers in order: π/2 (≈1.5708), 5π/6 (≈2.61799), 7π/6 (≈3.66519), 3π/2 (≈4.71238). The differences between consecutive terms:5π/6 - π/2 = (5π/6 - 3π/6) = 2π/6 = π/37π/6 - 5π/6 = 2π/6 = π/33π/2 - 7π/6 = (9π/6 - 7π/6) = 2π/6 = π/3So yes, each term is π/3 apart, forming an arithmetic sequence. Therefore, this works.Therefore, the value of m is -√3/2.But wait, let's make sure there are no other possibilities. For example, could there be another ordering where the zeros and roots are arranged differently?Suppose we consider another case where the order is arccos(m), π/2, 2π - arccos(m), 3π/2. But earlier, when arccos(m) < π/2, we saw that it leads to a contradiction. Let's verify again.If arccos(m) < π/2, then 2π - arccos(m) > 3π/2. So arranging in order: arccos(m), π/2, 3π/2, 2π - arccos(m). Let's check arithmetic sequence.Set them as a, a + d, a + 2d, a + 3d.Then:a = arccos(m)a + d = π/2a + 2d = 3π/2a + 3d = 2π - arccos(m)From the second equation: d = π/2 - aFrom the third equation: a + 2d = 3π/2. Substitute d = π/2 - a:a + 2(π/2 - a) = 3π/2a + π - 2a = 3π/2- a + π = 3π/2- a = 3π/2 - π = π/2a = -π/2But arccos(m) = a = -π/2, which is impossible because arccos(m) is in [0, π]. Therefore, this case is invalid.Hence, the only valid case is when the four numbers are ordered as π/2, arccos(m), 2π - arccos(m), 3π/2, leading to m = -√3/2.Therefore, the answer should be m = -√3/2.Final AnswerThe value of the real number ( m ) is (boxed{-dfrac{sqrt{3}}{2}}).</think>Given the function ( f(x) = cos x ) on the interval ([0, 2pi]), we know the zeros are at ( x_1 = frac{pi}{2} ) and ( x_2 = frac{3pi}{2} ). The equation ( cos x = m ) has two distinct roots ( x_3 ) and ( x_4 ), which are ( arccos(m) ) and ( 2pi - arccos(m) ). These four points must form an arithmetic sequence when arranged in ascending order.1. Identify the roots and zeros: - Zeros: ( frac{pi}{2} ) and ( frac{3pi}{2} ) - Roots of ( cos x = m ): ( arccos(m) ) and ( 2pi - arccos(m) )2. Order the points: - Since ( arccos(m) in [0, pi] ) and ( 2pi - arccos(m) in [pi, 2pi] ), the valid ordering when ( arccos(m) > frac{pi}{2} ) is ( frac{pi}{2} ), ( arccos(m) ), ( 2pi - arccos(m) ), ( frac{3pi}{2} ).3. Form an arithmetic sequence: - Let the sequence be ( a, a + d, a + 2d, a + 3d ). - Set ( a = frac{pi}{2} ) and ( a + 3d = frac{3pi}{2} ). Solving gives ( 3d = pi ) so ( d = frac{pi}{3} ). - The sequence becomes ( frac{pi}{2}, frac{pi}{2} + frac{pi}{3}, frac{pi}{2} + frac{2pi}{3}, frac{pi}{2} + pi ).4. Determine ( arccos(m) ): - ( arccos(m) = frac{pi}{2} + frac{pi}{3} = frac{5pi}{6} ). - Therefore, ( m = cosleft(frac{5pi}{6}right) = -frac{sqrt{3}}{2} ).Verification shows the sequence ( frac{pi}{2}, frac{5pi}{6}, frac{7pi}{6}, frac{3pi}{2} ) is indeed an arithmetic sequence with common difference ( frac{pi}{3} ).The value of the real number ( m ) is (boxed{-dfrac{sqrt{3}}{2}}).]

🔑:The function is f(x)=cos x, where xin[0,2pi]. It has two distinct zero points x_1, x_2.therefore x_1=frac{pi}{2}, x_2=frac{3pi}{2}.The equation f(x)=m has two distinct real roots x_3, x_4. When these four numbers are arranged in ascending order, they form an arithmetic sequence.therefore x_1+x_2=x_3+x_4=2pi, hence x_1, x_2 are symmetric about the line x=pi, and x_3, x_4 are symmetric about the line x=pi.Thus, x_1, x_2 are the first and last terms of the arithmetic sequence, and x_3, x_4 are the second and third terms, respectively.therefore frac{3pi}{2}=frac{pi}{2}+3d, therefore d=frac{pi}{3},therefore x_3=x_1+d=frac{5pi}{6}, x_4=x_1+2d=frac{7pi}{6}, therefore m=cosfrac{5pi}{6}=-frac{sqrt{3}}{2},Hence, the answer is: boxed{-frac{sqrt{3}}{2}}.The given conditions are used to find x_1, x_2 by utilizing the cosine function's graph. Subsequently, x_3, x_4 are found using the definition and properties of an arithmetic sequence. Finally, the value of m=sin x_3 is determined.This problem primarily assesses knowledge of the cosine function's graph, as well as the definition and properties of an arithmetic sequence. It is considered a basic question.

🧠:[<think>Okay, let me try to figure out this problem. So, we have a square that's 90 by 90, right? It's divided by two horizontal and two vertical lines into 9 rectangles. The central rectangle has sides of 34 and 42. We need to find the total area of the four corner rectangles. Hmm, let's visualize this first.Imagine a big square. If we draw two horizontal lines and two vertical lines, we'll divide the square into a 3x3 grid, making 9 smaller rectangles. The central one has dimensions 34x42. The corners are the four rectangles at each corner of the big square. So, their areas depend on the lengths of the segments created by the dividing lines.Let me denote the vertical lines as left, middle-left, middle-right, and right. Similarly, the horizontal lines as top, middle-top, middle-bottom, and bottom. Since the square is 90 units on each side, the total length of the vertical sides should add up to 90, and the same for the horizontal sides.The central rectangle has sides 34 and 42. Let me think: in a 3x3 grid, the central rectangle is surrounded by the middle lines. So, the vertical sides of the central rectangle would be between the middle-left and middle-right vertical lines, and the horizontal sides between the middle-top and middle-bottom horizontal lines.Wait, but if the central rectangle has a width of 34 and a height of 42, then those would be the distances between the two middle vertical lines and the two middle horizontal lines, respectively. So, the sum of the widths of the left and right columns (the corner rectangles on the left and right) would be 90 minus 34. Similarly, the heights of the top and bottom rows would be 90 minus 42. But wait, actually, the vertical lines divide the square into three columns, so the total width is the sum of the left, middle, and right columns. Similarly, the horizontal lines divide it into three rows. The middle column is 34, so the left and right columns add up to 90 - 34 = 56. Similarly, the middle row is 42, so the top and bottom rows add up to 90 - 42 = 48.But each corner rectangle is in either the top or bottom row and the left or right column. So, for example, the top-left corner rectangle would have the width of the left column and the height of the top row. Similarly for the others. But we don't know the individual widths and heights of the left/right columns and top/bottom rows. We just know the total of the left and right columns is 56 and the total of the top and bottom rows is 48. Hmm, but since we need the total area of all four corner rectangles, maybe we can find that total without knowing each individual area.Let me think: the four corner rectangles are:1. Top-left: width_left * height_top2. Top-right: width_right * height_top3. Bottom-left: width_left * height_bottom4. Bottom-right: width_right * height_bottomSo, if we add all these together:Total area = width_left * height_top + width_right * height_top + width_left * height_bottom + width_right * height_bottomFactor terms:= (width_left + width_right) * (height_top + height_bottom)We already know that width_left + width_right = 56 and height_top + height_bottom = 48. So, the total area would be 56 * 48.Wait, that's interesting. So, regardless of how the widths and heights are distributed between left/right and top/bottom, the total area of the four corners is just the product of (total remaining width) and (total remaining height). Because when you distribute the left and right widths and top and bottom heights, when you cross-multiply them for the four corners, it's equivalent to multiplying the sum of left and right by the sum of top and bottom. So, 56 * 48.Let me check this logic again. If we have:Total area of corners = (width_left + width_right) * (height_top + height_bottom)But width_left + width_right = 90 - 34 = 56height_top + height_bottom = 90 - 42 = 48Therefore, 56 * 48 = ?Calculate that:56 * 48. Let's compute 50*48 + 6*48 = 2400 + 288 = 2688.So, the total area of the four corner rectangles is 2688.But let me confirm this approach. Suppose we model the square with vertical divisions a, 34, b, where a + 34 + b = 90, so a + b = 56.Similarly, horizontal divisions c, 42, d, where c + 42 + d = 90, so c + d = 48.Then, the four corner rectangles have areas:Top-left: a*cTop-right: b*cBottom-left: a*dBottom-right: b*dTotal area = a*c + b*c + a*d + b*d = c(a + b) + d(a + b) = (c + d)(a + b) = 48*56 = 2688.Yes, that's exactly the same result. Therefore, regardless of the individual values of a, b, c, d (as long as they satisfy a + b = 56 and c + d = 48), the total area of the four corners is indeed 2688.Therefore, the answer should be 2688.Final AnswerThe total area of the four corner rectangles is boxed{2688}.</think>A square of 90 by 90 is divided by two horizontal and two vertical lines into 9 rectangles. The central rectangle has sides of 34 and 42. We need to find the total area of the four corner rectangles.1. The vertical lines divide the square into three columns. The middle column has a width of 34, so the sum of the widths of the left and right columns is (90 - 34 = 56).2. The horizontal lines divide the square into three rows. The middle row has a height of 42, so the sum of the heights of the top and bottom rows is (90 - 42 = 48).The four corner rectangles are:- Top-left: width of left column (times) height of top row- Top-right: width of right column (times) height of top row- Bottom-left: width of left column (times) height of bottom row- Bottom-right: width of right column (times) height of bottom rowThe total area of these four corner rectangles is calculated as:[(text{width of left column} + text{width of right column}) times (text{height of top row} + text{height of bottom row})]Substituting the known values:[56 times 48 = 2688]Thus, the total area of the four corner rectangles is (boxed{2688}).,<think>Alright, let's see. I need to find the total area of the four corner rectangles in a 90 by 90 square that's divided into 9 smaller rectangles. The central rectangle has sides 34 and 42. Hmm, okay. So the square is divided by two horizontal and two vertical lines, making a 3x3 grid of rectangles. The middle one is 34x42, and I need the sum of the areas of the four corners.First, let me visualize this. Imagine a big square, 90 units on each side. If there are two horizontal lines, that divides the square into three rows. Similarly, two vertical lines divide it into three columns. So there are 3 rows and 3 columns, hence 9 rectangles. The central rectangle is where the middle row and middle column intersect. The central rectangle's dimensions are 34 and 42. So that must mean that the middle row has a height of 34, and the middle column has a width of 42? Wait, or is it the other way around?Wait, hold on. The central rectangle's sides are 34 and 42. Since the entire square is 90x90, the sum of the three rows (heights) must be 90, and the sum of the three columns (widths) must be 90. The central rectangle is in the middle row and middle column, so its height is the height of the middle row, and its width is the width of the middle column. Therefore, the middle row is 34 units tall, and the middle column is 42 units wide. Or is it the other way around?Wait, no. The central rectangle's sides are 34 and 42, but which one is the height and which is the width? Hmm. The problem doesn't specify, so maybe I need to consider both possibilities? Wait, but maybe the orientation is fixed. Let me think.Since the square is divided by two horizontal and two vertical lines, the vertical lines determine the column widths, and the horizontal lines determine the row heights. The central rectangle has sides of 34 and 42. So one of these is the width (determined by the vertical lines) and the other is the height (determined by the horizontal lines). So either:- The middle column is 34 wide and the middle row is 42 tall, or- The middle column is 42 wide and the middle row is 34 tall.But the total of all three columns must be 90, and the total of all three rows must be 90. So depending on which is the middle column and middle row, the remaining widths and heights will vary.Let me denote the three columns as left, middle, right. Their widths must add up to 90. Similarly, the three rows as top, middle, bottom, with heights adding up to 90.Let's suppose that the middle column is 42 and the middle row is 34. Then the remaining widths for left and right columns would be (90 - 42)/2 each? Wait, but not necessarily. The left and right columns can be different. Similarly, the top and bottom rows can be different. So unless specified, we can't assume they are equal.But the problem doesn't mention anything else about the other rectangles. So how do we proceed?Wait, maybe we can express the areas of the corner rectangles in terms of the remaining widths and heights. Let's see.Let me denote:- Columns: left (width a), middle (width 42), right (width c). So a + 42 + c = 90.- Rows: top (height x), middle (height 34), bottom (height z). So x + 34 + z = 90.Then the four corner rectangles are:- Top-left: width a, height x. Area = a*x.- Top-right: width c, height x. Area = c*x.- Bottom-left: width a, height z. Area = a*z.- Bottom-right: width c, height z. Area = c*z.So total area of the four corners is a*x + c*x + a*z + c*z = x*(a + c) + z*(a + c) = (x + z)*(a + c).But from above, we have a + c = 90 - 42 = 48, and x + z = 90 - 34 = 56. So the total area would be 48 * 56 = 2688.Wait, that seems straightforward. So regardless of the individual values of a, c, x, z, the total area of the four corners is (a + c)(x + z) = (90 - 42)(90 - 34) = 48*56.Calculating 48*56: 40*56=2240, 8*56=448, so total is 2240+448=2688.Therefore, the total area of the four corner rectangles is 2688.Wait, but let me check again. Is this possible? Because the problem didn't specify anything else, so even if the side lengths are different, the product (a + c)(x + z) is equal to the total area of the four corners. Because when you expand (a + c)(x + z) = a*x + a*z + c*x + c*z, which are exactly the four corner areas. So yes, regardless of how the remaining widths and heights are distributed, their total sum multiplied gives the total area. Therefore, since a + c = 48 and x + z = 56, the total area is 48*56=2688.That seems right. Let me verify with a different approach. The total area of the entire square is 90*90=8100. The central rectangle is 34*42=1428. The remaining area is 8100 - 1428 = 6672. But this remaining area includes the four corner rectangles plus the four edge rectangles (the ones on the sides but not corners). So to get the four corners, we need to subtract the areas of the edge rectangles.Alternatively, maybe not. Wait, but if I consider the total area:Total area = sum of all 9 rectangles.Central rectangle is 34*42=1428.The four corners, as calculated above, 2688.Then there are four edge rectangles (top-middle, middle-left, middle-right, bottom-middle) and the central one.So total area would be 2688 (corners) + 1428 (center) + [edges]. Let's see, what's the area of the edges.The edges are:Top-middle: width 42 (middle column), height x. Area=42*x.Middle-left: width a, height 34. Area=a*34.Middle-right: width c, height 34. Area=c*34.Bottom-middle: width 42, height z. Area=42*z.So total edge areas = 42x + 34a + 34c + 42z.But since a + c = 48, 34a +34c =34*48=1632.Similarly, x + z =56, so 42x +42z =42*56=2352.So total edge areas =1632 +2352=3984.Then total area: corners (2688) + edges (3984) + center (1428) = 2688 + 3984 = 6672; 6672 + 1428 = 8100, which matches the total area. So that checks out. Therefore, the calculation seems correct.Therefore, the answer is 2688.But let me check once again. Let's suppose that the middle column is 34 and the middle row is 42. Then a + c = 90 -34=56, and x + z=90 -42=48. Then total area would be 56*48=2688. Same result. So regardless of which dimension is assigned to the column or row, the total area of the corners is 2688.Therefore, the answer is 2688.Final AnswerThe total area of the four corner rectangles is boxed{2688}.</think>Given a 90 by 90 square divided into 9 rectangles by two horizontal and two vertical lines, we need to find the total area of the four corner rectangles. The central rectangle has sides 34 and 42.1. Identify the dimensions of the middle row and column: - The central rectangle has dimensions 34 and 42. These are the height of the middle row and the width of the middle column (or vice versa).2. Calculate the remaining dimensions: - The total width of the square is 90. If the middle column is 42, the sum of the widths of the left and right columns is (90 - 42 = 48). - The total height of the square is 90. If the middle row is 34, the sum of the heights of the top and bottom rows is (90 - 34 = 56).3. Determine the total area of the four corner rectangles: - The four corner rectangles are formed by the remaining widths and heights. - The total area of the four corner rectangles is the product of the remaining widths and heights: [ (90 - 42) times (90 - 34) = 48 times 56 ]4. Calculate the product: - (48 times 56 = 2688)Thus, the total area of the four corner rectangles is (boxed{2688}).]

🔑:1. 2688 .+(2 points) - the solution is correct-(0 points) - there are errors in the solution, including arithmetic errors

🧠:[<think>Okay, so I need to find the minimum number of elements that a subset X of the set {00, 01, ..., 98, 99} should have, such that in any infinite sequence of digits, there are two adjacent digits that form an element of X. Hmm, this seems related to covering all possible pairs of digits in some way. Let me try to break it down step by step.First, let me understand the problem better. The set {00, 01, ..., 98, 99} contains all two-digit numbers from 00 to 99, which is essentially all possible pairs of digits where each digit can be from 0 to 9. So there are 100 elements in total. The subset X needs to be chosen such that no matter what infinite sequence of digits someone comes up with, there will always be at least one pair of adjacent digits that is in X. The question is asking for the smallest possible size of such a subset X.This reminds me of something called a "covering set" or maybe a "hitting set" in combinatorics. A covering set is a set that intersects every set in a particular collection. In this case, the collection would be all possible infinite sequences of digits, and we need our set X to cover at least one pair from each sequence. However, the problem is not exactly the same because we're dealing with infinite sequences, which complicates things.Alternatively, maybe it's related to graph theory. If I model the digits as nodes in a graph, where each node is a digit from 0 to 9, and each edge represents a transition from one digit to another (i.e., a two-digit pair), then the entire set {00, 01, ..., 99} represents all possible edges in a directed graph with 10 nodes. The problem then becomes: what is the smallest set of edges X such that every infinite walk on this graph contains at least one edge from X. So, in other words, X is a set of edges that "cuts" all infinite walks, meaning any infinite walk must pass through X at some point.In graph theory terms, this might be equivalent to finding a feedback edge set with the additional property that every infinite path must include at least one edge from the set. However, I need to recall if such a concept exists. A feedback edge set is a set of edges whose removal makes the graph acyclic. But in this case, the graph is a complete directed graph with loops (since we have edges like 00, 11, etc.), which is highly cyclic. So if we remove X, the remaining graph should be such that it cannot support an infinite path, meaning it should be acyclic? But wait, even acyclic directed graphs can have arbitrarily long paths if they have sufficiently many edges. Wait, no. A directed acyclic graph (DAG) has a topological order, and all edges go from earlier to later in the order. So in a DAG, all paths have finite length, bounded by the number of nodes. But if the graph is infinite, but in our case, the graph is finite (10 nodes), but the walks can be infinite by revisiting nodes. Wait, actually, even a DAG with cycles can have infinite walks if there are cycles. Wait, no, DAGs don't have cycles by definition. So in a DAG, all walks must eventually terminate because you can't revisit nodes. But in our problem, the graph is a complete graph, which has all possible edges, so it's full of cycles. If we remove X, the remaining graph would still have cycles unless X is a feedback edge set that breaks all cycles. But even then, if the remaining graph is a DAG, then any infinite walk would have to eventually repeat nodes, but in a DAG, you can't have cycles. Wait, this is getting confusing.Let me try a different approach. Suppose we need to ensure that in any infinite sequence of digits, there must be some occurrence of a pair from X. So, equivalently, the complement of X (all two-digit pairs not in X) cannot appear infinitely often in any infinite sequence. Wait, but even if a pair not in X appears once, that's okay, as long as somewhere later in the sequence, a pair from X appears. But the problem states "in any infinite sequence of digits, there are two adjacent digits that form an element of X." So actually, it's not that all pairs must eventually be in X, but rather that there exists at least one pair in X somewhere in the sequence. Wait, no, the problem says "in any infinite sequence of digits, there are two adjacent digits that form an element of X." So does that mean that there must be at least one occurrence, or that there are infinitely many occurrences? The wording is a bit ambiguous. But given that it's asking for the minimal X, it's more likely that it needs that at least one occurrence exists in any infinite sequence. However, in an infinite sequence, if you have to have at least one occurrence, but the rest can be anything, that seems too easy because you could have a sequence that avoids X except once. Wait, but the problem says "in any infinite sequence of digits, there are two adjacent digits that form an element of X." So, for any infinite sequence, there exists at least one pair of adjacent digits that is in X. Wait, but every infinite sequence has a pair in X? That seems impossible unless X is the entire set, because otherwise, you could construct a sequence that never uses any pairs from X. For example, if X is missing some pair, say 00, then you could have a sequence that repeats 0 forever, so 000000..., and the pair 00 is never in X. Therefore, the problem must be interpreted as: in any infinite sequence, there are two adjacent digits that form an element of X. That is, every infinite sequence must contain at least one occurrence of a pair from X. But as my previous example shows, if X is missing even a single pair, then you can have an infinite sequence that never uses that pair, hence never has a pair from X. Therefore, the only way for X to satisfy the condition is if X is the entire set. But that contradicts the question, which asks for the minimal number of elements in X. Therefore, my interpretation must be wrong.Wait, maybe the problem is that in any infinite sequence, there are two adjacent digits that form an element of X. So, in other words, for any infinite sequence, there exists at least one pair in X. But as I saw, if X is not the entire set, then you can have an infinite sequence that never uses X. Therefore, maybe the problem is that in any infinite sequence, there are infinitely many pairs that are in X. That is, you can't have an infinite sequence that eventually stops having pairs from X. Then, in that case, X needs to be such that its complement is finite? Or maybe the complement is a finite set of pairs, so that you can only have a finite number of consecutive pairs not in X. Wait, but the problem states "in any infinite sequence of digits, there are two adjacent digits that form an element of X." The wording is ambiguous. Is it "there exist two adjacent digits" (i.e., at least one occurrence) or "for any two adjacent digits" (i.e., all pairs are in X). The former would require X to be the entire set, but the problem asks for the minimal X. The latter is impossible unless X is the entire set. So maybe the problem is misstated, or perhaps there's a different interpretation.Wait, maybe the problem is similar to the concept of unavoidable sets in formal languages. An unavoidable set of words is a set such that any infinite word must contain at least one word from the set. In this case, the words are length two, so we're dealing with an unavoidable set of bigrams. So the question reduces to: what is the smallest unavoidable set of bigrams over a 10-symbol alphabet. That seems like a known problem. Let me recall.In the context of formal languages, a set of words is unavoidable if every infinite word contains at least one word from the set. The minimal size of such a set is sought here. For a binary alphabet, the minimal unavoidable set of bigrams is 2. For example, {01, 10} is unavoidable because any infinite binary sequence must contain either 01 or 10. Wait, but actually, over a binary alphabet, if you have an infinite sequence, it's either eventually all 0s or all 1s, or it has infinitely many alternations. If it's eventually all 0s or all 1s, then it contains only 00 or 11 after some point. So if your unavoidable set is {01, 10, 00, 11}, then it's trivial. But if you want a minimal set, over the binary alphabet, the minimal unavoidable set for bigrams is 3. For example, {00, 11, 01}, but I need to check. Wait, maybe 2? If you have {00, 11}, then a sequence that alternates 010101... would avoid both 00 and 11, so that's not unavoidable. If you have {01, 10}, then a sequence of all 0s or all 1s would avoid those. So you need at least three. For example, {00, 11, 01} would be unavoidable, because if a sequence is all 0s, it has 00; all 1s has 11; otherwise, it has both 0s and 1s, so somewhere 01 or 10 must occur. But if you have 01 in the set, you can avoid 10 by having a sequence that starts with 0 and then has 1 once, then continues with 1s. So actually, maybe {00, 11, 01, 10} is unavoidable. Wait, perhaps over binary alphabet, the minimal unavoidable set of bigrams is 3. Let me check literature.But maybe this is getting too sidetracked. In our problem, the alphabet is size 10 (digits 0-9), and we need an unavoidable set of bigrams. The question is, what is the minimal size of such a set. For a binary alphabet, it's known that the minimal unavoidable set for bigrams is 3. For a larger alphabet, perhaps the minimal unavoidable set is larger? Or maybe there's a formula.Alternatively, think of it as a graph problem. Each bigram is an edge in a directed graph where the vertices are digits 0-9. Then, the problem reduces to finding the smallest set of edges X such that every infinite path in the graph contains at least one edge from X. In other words, the complement of X (the set of edges not in X) does not contain an infinite path. Therefore, the complement set must be such that it's a directed acyclic graph (DAG). Because a DAG has no infinite paths, as all paths have finite length. Therefore, if we can make the complement of X a DAG, then X is unavoidable. Therefore, the minimal size of X is equal to the number of edges in the original graph minus the maximum number of edges in a DAG on 10 vertices.So, the total number of edges in the original graph is 100 (10x10). The maximum number of edges in a DAG is obtained by arranging the vertices in a topological order and having all edges go from earlier to later in the order. For n vertices, the maximum DAG has n(n-1)/2 edges. For n=10, that's 10*9/2 = 45 edges. Therefore, the minimal X would have 100 - 45 = 55 elements. Therefore, the answer should be 55.But let me verify this reasoning. If we can model the problem as a directed graph where the complement of X must be a DAG, then X must include all edges not in some DAG. Since the largest DAG on 10 nodes has 45 edges, the minimal X would be 100 - 45 = 55. Therefore, the minimal number of elements in X is 55.But is this correct? Let me check.First, suppose that the complement of X is a DAG. Then, in any infinite path, you cannot have an infinite path entirely in the complement of X because DAGs do not have infinite paths. Therefore, any infinite path must eventually traverse an edge in X. Therefore, X is unavoidable. Therefore, if we can make the complement a DAG, then X is unavoidable. Therefore, the minimal size of X is 100 minus the maximum size of a DAG, which is 45, so 55.Conversely, suppose that X has fewer than 55 elements. Then, the complement has more than 45 edges. Since the maximum DAG has 45 edges, a graph with more than 45 edges must contain a cycle. Therefore, the complement of X would contain a cycle, which can be traversed infinitely often, creating an infinite path that does not use any edges from X. Therefore, such an X would not satisfy the condition. Therefore, 55 is indeed the minimal number.Therefore, the answer should be 55.But let me think if there's a way to have a DAG with more edges? Wait, in a directed graph, a DAG can have more edges if you consider different topological orderings. Wait, no, in a DAG, the maximum number of edges is achieved by a complete DAG where all edges go in one direction according to a topological order. For n nodes, that's n(n-1)/2 edges, which for 10 nodes is 45. Therefore, that's the maximum. So if you have a DAG with more than 45 edges, it must contain a cycle, which contradicts it being a DAG. Therefore, 45 is indeed the maximum.Therefore, the minimal size of X is 100 - 45 = 55. Therefore, the answer is 55.Final AnswerThe minimum number of elements that ( X ) should contain is (boxed{55}).</think>To solve the problem, we need to determine the minimum number of elements in a subset ( X ) of the set ({00, 01, ldots, 98, 99}) such that any infinite sequence of digits contains at least one pair of adjacent digits forming an element of ( X ).1. Graph Theory Interpretation: Model the digits as nodes in a directed graph where each node represents a digit (0-9) and each edge represents a transition between two digits (i.e., a two-digit pair). The set ({00, 01, ldots, 98, 99}) corresponds to all possible edges in this graph.2. Covering Set Requirement: The subset ( X ) must be such that any infinite sequence of digits (i.e., any infinite walk in the graph) contains at least one edge from ( X ). This means the complement of ( X ) (all edges not in ( X )) must not contain any infinite paths.3. Directed Acyclic Graph (DAG): The complement of ( X ) must be a DAG because a DAG does not contain any infinite paths. The maximum number of edges in a DAG with 10 nodes (digits 0-9) is given by the number of edges in a complete DAG, which is ( frac{10 cdot 9}{2} = 45 ).4. Minimum Size of ( X ): The total number of edges in the original graph is 100 (10 nodes each with 10 possible edges). Subtracting the maximum number of edges in a DAG from this total gives the minimum number of edges in ( X ): [ 100 - 45 = 55 ]Thus, the minimum number of elements that ( X ) should contain is (boxed{55}).,<think>Okay, let me try to figure out this problem. So, we need to find the minimum size of a subset X of the set {00, 01, ..., 98, 99} such that in any infinite sequence of digits, there are two adjacent digits that form an element of X. Hmm. So, X is like a "covering" set for all possible infinite digit sequences. Interesting.First, let me understand the problem statement again. We have the set of all two-digit numbers from 00 to 99, so that's 100 elements. We need a subset X such that no matter what infinite digit sequence someone comes up with, there must be at least one pair of adjacent digits in that sequence that is in X. The question is, what's the smallest possible size of X?So, another way to think about this is: X needs to "cover" all possible infinite sequences. That is, there's no infinite sequence that avoids all the elements of X. So, X has to be such that its complement cannot form an infinite sequence without having an element from X. So, the complement of X, let's call it Y, should not contain an infinite "Y-avoiding" sequence. Therefore, Y must be such that every infinite sequence must eventually hit X. So, the minimal X is equivalent to the maximal Y such that Y allows for an infinite sequence. Wait, maybe not exactly, but perhaps related.Alternatively, maybe this is similar to the concept of unavoidable sets in strings. An unavoidable set is a set of strings such that any infinite string must contain at least one string from the set. So, in this case, X is an unavoidable set over the alphabet of digits 0-9, and the strings in X are all of length 2. So, the question is asking for the minimal size of such an unavoidable set.Hmm. I remember that for unavoidable sets, there are certain constructions. For example, in the case of a binary alphabet, the minimal unavoidable set of certain lengths can be determined using de Bruijn sequences or other combinatorial methods. But here, the alphabet is size 10 (digits 0-9), and the strings are of length 2. So, perhaps we need to use some concepts from graph theory here?Let me think. If we model this problem as a graph, where each node represents a digit (0-9), and each edge represents a transition from one digit to another. So, the edge from node u to node v is labeled by the two-digit number uv. Then, an infinite sequence of digits corresponds to an infinite walk on this graph. The problem then reduces to finding a set X of edges such that every infinite walk must include at least one edge from X. Therefore, the complement of X (the edges not in X) must form a graph that does not contain an infinite walk, i.e., the complement graph must be such that all its walks eventually terminate or cannot continue infinitely. Wait, but in a finite graph, any infinite walk must necessarily revisit nodes, so it must contain cycles. Therefore, the complement graph Y (which is the set of edges not in X) must not contain any cycles. Otherwise, there exists an infinite walk that cycles indefinitely on a cycle in Y, thereby avoiding X.Therefore, the problem reduces to making sure that the complement graph Y is acyclic. Because if Y is acyclic, then any walk in Y must eventually reach a node with no outgoing edges in Y, thereby forcing the walk to terminate. But since we need the original graph to cover all infinite walks, by making Y acyclic, we ensure that any infinite walk must use edges from X. Therefore, the minimal X is the minimal number of edges that need to be removed from the complete graph (which has all possible edges) to make it acyclic. Wait, but in this case, the complete graph is a directed graph with 10 nodes (digits 0-9), each node has 10 outgoing edges (to all digits, including itself). So, the total number of edges is 100, which corresponds to all two-digit numbers.Therefore, the problem is equivalent to finding the minimal number of edges X such that the complement of X is a directed acyclic graph (DAG). Because if the complement is a DAG, then there are no infinite walks in the complement, so any infinite walk must use edges from X. So, the minimal size of X is equal to the total number of edges minus the maximal number of edges in a DAG on 10 nodes.But what is the maximal number of edges in a DAG on n nodes? For a directed acyclic graph, the maximum number of edges is n(n-1)/2, which is the number of edges in a complete DAG where the nodes are ordered and all edges go from earlier nodes to later nodes. However, in our case, the original graph is a complete directed graph (each node has edges to all nodes, including itself), so the complement graph Y would need to be a DAG. Therefore, the maximal number of edges in Y is the maximal number of edges a DAG can have, which is n(n-1)/2. Therefore, the minimal X would be 100 - n(n-1)/2 = 100 - 10*9/2 = 100 - 45 = 55. So, the minimal size of X is 55. Wait, but let me verify this.Wait, in a directed graph, a DAG can have up to n(n-1)/2 edges if it's a complete DAG with a topological order. However, our complement graph Y is a directed graph, which is a subgraph of the complete directed graph (digraph) with 10 nodes. So, if Y is a DAG, then it cannot have any cycles. Therefore, the maximum number of edges in Y is indeed n(n-1)/2, which for n=10 is 45. Therefore, the minimal number of edges in X is 100 - 45 = 55. Therefore, the answer should be 55. But wait, let me check if this is correct.But hold on, in a standard DAG, you can have loops? Wait, no, in a DAG, self-loops are cycles of length 1, so a DAG cannot have any self-loops. Therefore, in our case, the complement graph Y cannot have any self-loops because X must include all self-loops (since the problem states that X is a subset of {00, 01, ..., 99}, which includes all two-digit numbers, including the ones with repeated digits like 00, 11, etc.). Therefore, if we consider Y to be a DAG, then Y cannot have any cycles, including self-loops. Therefore, Y cannot have edges like 00, 11, etc., which are self-loops. Therefore, all self-loops must be in X. There are 10 self-loops (00 to 99 with both digits same). So, X must contain all 10 self-loops. Then, the remaining edges in Y are the non-loop edges. So, the original complete digraph has 100 edges (including 10 loops). If we remove the 10 loops, we have 90 edges. Then, Y is a DAG on these 90 edges (without loops). So, the maximum number of edges Y can have is n(n-1)/2 = 10*9/2 = 45. Therefore, the minimal X must contain 10 (loops) + (90 - 45) = 10 + 45 = 55. So, that's consistent with the previous calculation. Therefore, the minimal size of X is 55. Therefore, the answer is 55. But wait, let me check this again.Wait, in the problem statement, X is a subset of {00, 01, ..., 99}, which includes all two-digit numbers. So, X can include both loops (like 00) and non-loop edges. The complement Y would then be {00, ..., 99} X. If we need Y to be a DAG, then Y cannot have any cycles. However, Y is a set of two-digit numbers, which correspond to edges in the digraph. So, if Y is a DAG, then there is a topological ordering of the digits 0-9 such that all edges in Y go from earlier digits to later digits in that ordering. Since there are 10 digits, the maximum number of edges in Y is 10*9/2 = 45. Therefore, to make Y a DAG, we need to remove all edges that form cycles, which requires removing 100 - 10 - 45 = 45 edges (since 10 are loops). Wait, but 100 total edges minus 10 loops is 90. So, to make Y a DAG (without loops), we need Y to have at most 45 edges. Therefore, X must contain 90 - 45 = 45 edges plus the 10 loops, totaling 55. Therefore, X must contain at least 55 elements. Therefore, the minimal size is 55.But let me confirm this with an example. Suppose we arrange the digits in a topological order, say 0 < 1 < 2 < ... < 9. Then, in the complement graph Y, we only include edges that go from a lower digit to a higher digit. So, for example, 0→1, 0→2, ..., 0→9; 1→2, ..., 1→9; and so on, up to 8→9. There are 45 such edges. Then, all other edges (the 55 remaining) are in X. This includes all loops (10) and all edges that go from a higher digit to a lower digit (which are 45 in total: for each pair i < j, there is an edge j→i). So, X would have 10 + 45 = 55 elements. Then, any infinite sequence that would avoid X would have to only use edges from Y, which go from lower to higher digits. But since there are only 10 digits, an infinite sequence cannot keep increasing forever. Eventually, it would have to repeat a digit or go back, which would require an edge from higher to lower digit, which is in X. Therefore, such a sequence cannot be infinite. Therefore, this construction works. Thus, the minimal X has 55 elements.But wait, but in this case, the Y graph is a DAG with a topological order where edges only go from lower to higher digits. But in reality, even if you have a different topological order, the maximum number of edges remains the same. Therefore, regardless of the ordering, the maximum number of edges in a DAG is 45. Therefore, the minimal size of X is 55.But let me check if there exists a smaller X. Suppose we can find a DAG with more than 45 edges? But I don't think so. In a directed acyclic graph, the maximum number of edges is indeed n(n-1)/2 when you have a total order and all edges go in one direction. If you have a different DAG structure, you can't have more edges. For example, if you have a DAG with multiple components, each component can have a topological order, but the total number of edges can't exceed n(n-1)/2. Therefore, 45 is indeed the maximum.Therefore, the minimal size of X is 55. Therefore, the answer is 55. So, the final answer is boxed{55}.Final Answerboxed{55}</think>To solve the problem of finding the minimum number of elements in a subset ( X ) of the set ({00, 01, ldots, 98, 99}) such that any infinite sequence of digits must contain two adjacent digits forming an element of ( X ), we can model this problem using graph theory.1. Graph Representation: Consider each digit (0-9) as a node in a directed graph. Each two-digit number ( ab ) corresponds to a directed edge from node ( a ) to node ( b ). An infinite sequence of digits corresponds to an infinite walk in this graph.2. Complement Graph: The subset ( X ) must be such that its complement ( Y ) (the set of edges not in ( X )) forms a directed acyclic graph (DAG). This is because if ( Y ) were a DAG, it cannot contain any cycles, meaning any infinite walk must eventually use an edge from ( X ).3. Maximum Edges in a DAG: For a graph with ( n ) nodes (10 digits in this case), the maximum number of edges in a DAG is ( frac{n(n-1)}{2} ). For ( n = 10 ), this is ( frac{10 cdot 9}{2} = 45 ).4. Total Edges and Self-Loops: The total number of edges in the complete directed graph (including self-loops) is 100. Removing the 10 self-loops (which must be in ( X ) since they form cycles of length 1), we have 90 edges left. The maximum number of edges in a DAG (without self-loops) is 45.5. Calculating Minimum ( X ): The subset ( X ) must include all 10 self-loops and the remaining edges not in the DAG. Therefore, ( X ) contains ( 10 + (90 - 45) = 55 ) elements.Thus, the minimum number of elements that ( X ) should contain is (boxed{55}).]

🔑:[Solution] For any i, j in{0,1, cdots, 9}, X should contain i j or j i (otherwise the sequence i j i j i j cdots does not meet the requirement), there are 10+C_{10}^{2}=55 such unordered ( i, j) pairs,thus |X| geqslant 55.On the other hand, if X={i j mid 0 leqslant i leqslant j leqslant 9}, then X=55, and for any infinite sequence, let i be the smallest digit it contains, and j be the digit following i, then i j in X.Therefore, X must contain at least 55 elements.

🧠:[<think>Okay, so I need to find the probability that the complex number (m + ni)(n - mi) is real when two dice are rolled, giving numbers m and n. Hmm, let's start by understanding what this complex number is. First, I remember that when you multiply two complex numbers, the result can be real or imaginary depending on the terms. Let me expand the product (m + ni)(n - mi) to see what it simplifies to. Multiplying out the terms:- First, m * n = mn- Then, m * (-mi) = -m²i- Next, ni * n = n²i- Finally, ni * (-mi) = -m n i²So combining all these terms:mn - m²i + n²i - m n i²Wait, but i² is -1, right? So the last term becomes -m n (-1) = m n. So now the expression is:mn + mn - m²i + n²iWhich simplifies to:2mn + (n² - m²)iSo the complex number (m + ni)(n - mi) simplifies to 2mn + (n² - m²)i. For this to be a real number, the imaginary part must be zero. That means the coefficient of i, which is (n² - m²), has to be zero. Therefore, we need n² - m² = 0.So solving n² - m² = 0. This factors into (n - m)(n + m) = 0. Since n and m are numbers on a die, they are positive integers from 1 to 6. Therefore, n + m can't be zero. So the only solution is n - m = 0, which implies n = m.Therefore, the complex number is real if and only if m equals n.Now, we need to find the probability that when two dice are rolled, m equals n. When rolling two dice, there are 6 * 6 = 36 possible outcomes. The number of favorable outcomes where m = n is 6: (1,1), (2,2), ..., (6,6). Therefore, the probability is 6/36 = 1/6.Wait, but let me double-check. The problem is asking for the probability that the product is real, which we deduced happens when m = n. But is there another case where the imaginary part (n² - m²) could be zero?Wait, n² - m² = 0 implies n² = m². Since n and m are positive integers, the only solution is n = m. So no other possibilities. So yes, only when m equals n.But wait, could there be a case where n² - m² is zero even if n ≠ m? For example, if n = 3 and m = 3, then it's zero, but if n = 3 and m = -3, but since die numbers are positive, that's impossible. So only when m = n.Therefore, there are 6 favorable outcomes. So probability is 6/36 = 1/6. So the answer should be C: 1/6. But let me check again. Wait, in the expansion, did I do that correctly?Original expression: (m + ni)(n - mi). Let me expand again carefully.First, multiply m by n: mn.Then, m by -mi: -m²i.Then, ni times n: n²i.Then, ni times -mi: -m n i². Which is -m n (-1) = m n.So altogether: mn + mn + (n² - m²)i. That is, 2mn + (n² - m²)i. So the imaginary part is n² - m², which must be zero. So yes, same conclusion. So the answer is 1/6. So option C.But hold on, let me check again. Maybe there's a mistake in the multiplication. Let me use another method. The product (m + ni)(n - mi). Let's denote z1 = m + ni and z2 = n - mi. Then, their product is z1*z2. Alternatively, z2 is the conjugate of something? Let's see. If we take the conjugate of z1, which is m - ni, but z2 is n - mi. Not exactly the conjugate. Maybe it's better to just compute the product.But expanding again step by step:(m + ni)(n - mi) = m*n + m*(-mi) + ni*n + ni*(-mi)Compute each term:m*n = mnm*(-mi) = -m²ini*n = n²ini*(-mi) = -n m i² = -n m (-1) = n mSo adding all terms:mn - m²i + n²i + mnCombine like terms:Real parts: mn + mn = 2mnImaginary parts: (-m² + n²)iTherefore, the expression is 2mn + (n² - m²)i. Therefore, to be real, the coefficient of i must be zero. Therefore, n² - m² = 0, leading to n = m.Therefore, the only cases are when m = n, which are 6 outcomes. Therefore, probability 6/36 = 1/6. So answer C. Wait a second, but wait. Let me check if there is a different way that the product can be real. For example, even if the imaginary part is not zero, maybe the entire expression is real. But since the imaginary part is (n² - m²), the only way for the entire expression to be real is if that coefficient is zero. So indeed, n² - m² = 0. So the answer is C: 1/6.But wait, let me verify with an example. Suppose m = 1 and n = 1. Then the product is (1 + i)(1 - i) = 1 - i + i - i² = 1 - i + i +1 = 2. Which is real. If m = 2 and n = 3. Then the product is (2 + 3i)(3 - 2i) = 6 -4i +9i -6i² = 6 +5i +6 = 12 +5i. Which is not real. If m = 3 and n =3, then product is (3 +3i)(3 -3i) = 9 -9i +9i -9i² = 9 +9 = 18. Which is real. If m = 2 and n = 1, then product is (2 + i)(1 -2i) = 2 -4i +i -2i² = 2 -3i +2 =4 -3i. Not real. So seems like only when m = n, it's real. So 6 cases. So 1/6. So answer C.But wait, I just thought of something else. Suppose m and n are such that even though m ≠ n, (n² - m²) = 0. But since they are positive integers, n² = m² implies n = m. So no other solutions. So yeah, only when m = n. So yes, 1/6. Therefore, answer C. So the answer is C: 1/6.Final Answerboxed{C}</think>To determine the probability that the complex number ((m + ni)(n - mi)) is real when two dice are rolled, we start by expanding the product:[(m + ni)(n - mi) = mn - m^2i + n^2i - mn i^2]Since (i^2 = -1), this simplifies to:[mn - m^2i + n^2i + mn = 2mn + (n^2 - m^2)i]For the complex number to be real, the imaginary part ((n^2 - m^2)) must be zero. This gives the equation:[n^2 - m^2 = 0 implies (n - m)(n + m) = 0]Since (n) and (m) are positive integers from 1 to 6, (n + m neq 0). Therefore, the equation simplifies to (n = m).When rolling two dice, there are 36 possible outcomes. The favorable outcomes where (m = n) are (1,1), (2,2), ..., (6,6), which are 6 outcomes. Thus, the probability is:[frac{6}{36} = frac{1}{6}]The final answer is (boxed{C}).,<think>Okay, let's see. The problem is about rolling two dice, getting numbers m and n, and then figuring out the probability that the complex number (m + ni)(n - mi) is a real number. The answer choices are given, so I need to figure out which one is correct. First, I need to recall that a complex number is real if its imaginary part is zero. So, I should multiply out the given complex numbers and then check when the imaginary part cancels out. Let me start by expanding the product:(m + ni)(n - mi)Multiplying these two complex numbers. Let's do it step by step. First, multiply m by n: m*nThen, m by -mi: m*(-mi) = -m^2 iThen, ni by n: ni*n = n^2 iThen, ni by -mi: ni*(-mi) = -n m i^2But wait, i^2 is -1, so that term becomes -n m*(-1) = n mSo putting all the terms together:= mn - m^2 i + n^2 i + nmCombine like terms:Real parts: mn + nm = 2mnImaginary parts: (-m^2 i + n^2 i) = (n^2 - m^2)iSo the product is 2mn + (n^2 - m^2)iTherefore, for this complex number to be real, the imaginary part must be zero. So:n^2 - m^2 = 0Which simplifies to:n^2 = m^2Taking square roots on both sides (but since m and n are positive integers from 1 to 6, we don't have to worry about negative roots), we get:n = m or n = -mBut since both m and n are results from dice, they can't be negative. So the only possibility is n = m.Therefore, the complex number is real if and only if m equals n.So now, the problem reduces to finding the probability that when rolling two dice, both show the same number. In rolling two dice, there are 6 * 6 = 36 possible outcomes. The number of favorable outcomes where m = n is 6: (1,1), (2,2), ..., (6,6).Therefore, the probability is 6/36 = 1/6. Wait, but wait, let me check that again.Wait, hold on. The imaginary part is n^2 - m^2 = 0. So m^2 = n^2. Since m and n are positive integers from 1 to 6, that implies m = n, correct. So yes, only when they are equal. So 6 outcomes. So 6/36 is 1/6, so answer C: 1/6.But hold on, let me check the multiplication again, because sometimes when you expand complex numbers, signs can be tricky. Let me verify:(m + ni)(n - mi) = m*n + m*(-mi) + ni*n + ni*(-mi)First term: mnSecond term: -m^2 iThird term: n^2 iFourth term: -m n i^2But i^2 = -1, so fourth term becomes -m n*(-1) = m nSo total real parts: mn + mn = 2 mnTotal imaginary parts: (-m^2 + n^2) iSo indeed, the imaginary part is (n^2 - m^2) i. So setting that to zero, n^2 - m^2 = 0, so m = n. So yes, the conclusion is correct.Therefore, the probability is 6/36 = 1/6, which is answer C.But wait, hold on. Let me check if there are other possibilities. For example, maybe even if m ≠ n, but n^2 - m^2 could be zero? But in that case, m = n. Since both m and n are positive integers. So there are no other possibilities.Alternatively, could there be a case where even if m ≠ n, the imaginary part is zero? For example, if m = 1 and n = -1, but since n is a die roll, it can't be negative, so no. Hence, the only way is m = n.Therefore, the number of favorable outcomes is 6, probability 1/6. So answer C.Wait, but looking at the answer options, C is 1/6. So that's the answer.But let me think again. Wait, maybe there's a mistake in the calculation. Let me check once more.Alternative approach: compute (m + ni)(n - mi). Let's note that (m + ni)(n - mi) is equal to (m + ni)(n + (-m)i). Alternatively, perhaps there is a different way to compute this product.Alternatively, using complex numbers properties. Let me note that (m + ni)(n - mi) is equal to the product of (m + ni) and the conjugate of (mi + n), but I need to check.Wait, the conjugate of a complex number a + bi is a - bi. So if we have (mi + n), that's n + mi. Its conjugate is n - mi. So then (m + ni)(n - mi) is (m + ni) times the conjugate of (n + mi). Hmm, maybe there's another way to think about this. But maybe that's complicating things. Alternatively, perhaps think of the product as a complex number times its conjugate, but not sure. Maybe not necessary.Alternatively, since we already expanded it and found that the imaginary part is (n^2 - m^2)i, which must be zero for the number to be real. Therefore, the conclusion that m must equal n is correct.Thus, the probability is 6/36 = 1/6. So answer C.But wait, wait a second. Let me think about the problem again. Maybe the product is (m + ni)(n - mi). Let me check if there's another way to interpret the problem. For example, maybe the second term is (n - m i) instead of (n - mi). But in the problem statement, it's written as (n - mi). So m is multiplied by i. So yes, that's how it is. So the expansion is correct.Alternatively, maybe I made a mistake in signs. Let me redo the expansion:(m + ni)(n - mi) = m*n + m*(-mi) + ni*n + ni*(-mi)First term: mnSecond term: -m * m i = -m² iThird term: n * n i = n² iFourth term: -m n i². Since i² = -1, this becomes -m n (-1) = m nTherefore, real parts: mn + mn = 2 mnImaginary parts: (-m² + n²) iSo, to have the number be real, the coefficient of i must be zero: -m² + n² = 0 => n² = m² => n = m or n = -m. But since n and m are positive, n = m.Therefore, same as before. So the answer should be 1/6. Therefore, answer C.But let me check for a concrete example. Suppose m = 1, n = 1. Then the product is (1 + i)(1 - i) = 1 - i + i - i² = 1 - i² = 1 - (-1) = 2. Which is real.If m = 1, n = 2. Then (1 + 2i)(2 - i) = 1*2 + 1*(-i) + 2i*2 + 2i*(-i) = 2 - i + 4i - 2i² = 2 + 3i - 2*(-1) = 2 + 3i + 2 = 4 + 3i. Which is complex.If m = 2, n = 2. Then (2 + 2i)(2 - 2i) = 4 - 4i + 4i - 4i² = 4 - 4i² = 4 - 4*(-1) = 4 + 4 = 8. Which is real.If m = 3, n = 1. Then (3 + i)(1 - 3i) = 3*1 + 3*(-3i) + i*1 + i*(-3i) = 3 - 9i + i - 3i² = 3 - 8i -3*(-1) = 3 -8i +3 = 6 -8i. Which is complex.Wait, so in the case where m and n are different, it's complex. If they are equal, it's real. Therefore, the only cases where the product is real are when m = n, which are 6 outcomes. So 6/36=1/6. Therefore, the answer is C.But let me check if there's any case where m ≠ n but the imaginary part cancels. For example, if m = 5 and n = something else. Let's take m=5, n=5: real. If m=5, n= any other number. Let's try m=5, n=3. Then (5 +3i)(3 -5i) = 15 -25i +9i -15i² = 15 -16i +15 = 30 -16i, which is complex. If m=5, n=4: (5 +4i)(4 -5i)=20 -25i +16i -20i²=20 -9i +20=40 -9i. Still complex. So, no. Similarly, m=1, n=1 is real, m=2, n=2 is real, etc. If m and n are different, the imaginary part is non-zero. Therefore, the only way is m=n, so 6 cases. So the answer is 1/6, which is option C.Therefore, I think the correct answer is C.Final Answerboxed{C}</think>To determine the probability that the complex number ((m + ni)(n - mi)) is real when two dice are rolled, we start by expanding the product:[(m + ni)(n - mi) = mn - m^2 i + n^2 i + nm]Combining like terms, we get:- Real parts: (mn + nm = 2mn)- Imaginary parts: ((-m^2 + n^2)i)For the complex number to be real, the imaginary part must be zero:[n^2 - m^2 = 0 implies n^2 = m^2 implies n = m text{ (since } m text{ and } n text{ are positive integers)}]Thus, the complex number is real if and only if (m = n). When rolling two dice, there are 36 possible outcomes. The favorable outcomes where (m = n) are ((1,1), (2,2), ldots, (6,6)), which are 6 outcomes. Therefore, the probability is:[frac{6}{36} = frac{1}{6}]The final answer is (boxed{C}).]

🔑:AnalysisThis question tests the basic concept of complex numbers, the classical probability model, and the formula for calculating probability. It examines the ability to analyze and solve problems and is a basic question.By expanding according to the rules of polynomial multiplication and simplifying to the form a+bi (a,b in mathbb{R}), with the imaginary part being 0, we can find the relationship between m and n, count the number of elementary events that satisfy this relationship, and then calculate the probability.SolutionSince (m+ni)(n-mi)=2mn+(n^{2}-m^{2})i is a real number, it implies that n^{2}=m^{2}.Therefore, m=n can take the values 1, 2, 3, 4, 5, 6, totaling 6 possibilities.Thus, P= dfrac {6}{ C_{ 6 }^{ 1 }cdot C_{ 6 }^{ 1 }}= dfrac {1}{6},Hence, the correct option is boxed{C}.

🧠:[<think>Okay, let's see. I need to figure out which of these four statements (A, B, C, D) is correct. Let me go through each one step by step. Starting with statement A: "Two numbers with equal absolute values are equal." Hmm. Absolute value makes any number non-negative, right? So if two numbers have the same absolute value, does that mean they are equal? Wait, for example, take 3 and -3. Their absolute values are both 3, but they are not equal. So statement A is saying that if |x| = |y|, then x = y. But clearly, that's not true because x could be the negative of y. So A is incorrect. Got that.Moving on to statement B: "If a and b are opposite numbers, then a/b = -1." Opposite numbers... I think that means a = -b. So if a is the opposite of b, then a = -b. Let's check the division. If a = -b, then a/b would be (-b)/b, which simplifies to -1. Wait, but hold on. What if b is zero? Then division by zero is undefined. But the question didn't specify that a and b are non-zero. Hmm. So if a and b are opposites, then b must be -a. But if a is zero, then b is also zero. So a/b would be 0/0, which is undefined. So in that case, the statement isn't necessarily true because if both a and b are zero, the division isn't possible. However, maybe the problem assumes that a and b are non-zero? The term "opposite numbers" might imply that they are non-zero. Let me check. If a and b are opposites, then they are non-zero? Because zero's opposite is zero, but maybe opposite numbers refer to non-zero numbers. The problem is not explicit here. If we assume that a and b are non-zero, then yes, a/b = -1. But if a and b can be zero, then the statement is not always true. Since the problem didn't specify, I might need to consider this. But in typical math problems, unless stated otherwise, opposite numbers can include zero? Wait, no. Opposite numbers are numbers that are on opposite sides of zero on the number line, equidistant from zero. So for example, 5 and -5 are opposites. But zero is its own opposite. So if a and b are opposites, then either a = -b (if they are non-zero) or a = b = 0. So in the case where a = b = 0, the ratio a/b is undefined. So statement B is not always true. Therefore, B is incorrect. Hmm, so maybe statement B is incorrect because there's a case where it's undefined. So even though in the non-zero case it works, since there's a case where it's undefined, the statement is not universally correct. Therefore, B is incorrect.Moving to statement C: "If a² = b², then |a| = |b|." Let's see. If a squared equals b squared, taking the square root of both sides gives |a| = |b|. Because square root of a squared is absolute value of a. So yes, that's a mathematical truth. For example, if a is 3 and b is -3, then 9 = 9, and |3| = |-3|. So that works. Even if both are positive, or both are negative, or one is zero. So this seems to hold in all cases. Therefore, statement C is correct. Statement D: "-a is always a negative number." Hmm. Wait, if a is a negative number, then -a would be positive. For example, if a is -5, then -a is 5, which is positive. So the statement claims that -a is always negative, but that's not the case. It depends on what a is. If a is positive, -a is negative. If a is negative, -a is positive. If a is zero, -a is zero. So -a's sign depends on a's original sign. Therefore, statement D is not correct. So summarizing:- A is incorrect because numbers like 3 and -3 have equal absolute values but aren't equal.- B is incorrect because if a and b are both zero, a/b is undefined.- C is correct because a² = b² implies |a| = |b|.- D is incorrect because -a can be positive or zero depending on a.Therefore, the correct answer is statement C.</think>To determine which statement is correct:- Statement A: Incorrect. Numbers with equal absolute values can be opposites (e.g., (3) and (-3)), so they need not be equal.- Statement B: Incorrect. If (a = 0) and (b = 0) (opposites of each other), (frac{a}{b}) is undefined, not (-1).- Statement C: Correct. (a^2 = b^2) implies (sqrt{a^2} = sqrt{b^2}), so (|a| = |b|) by definition of absolute value.- Statement D: Incorrect. (-a) is negative only if (a) is positive. If (a) is negative, (-a) is positive; if (a = 0), (-a = 0).Answer: boxed{C},<think>Okay, let's see. I need to figure out which of these statements A, B, C, D is correct. Let me go through each one step by step.Starting with statement A: "Two numbers with equal absolute values are equal." Hmm. Absolute value makes any number non-negative, right? So if two numbers have the same absolute value, does that mean they are equal? Wait, no. For example, 5 and -5 both have an absolute value of 5, but they aren't equal. So statement A is not correct. So A is false.Moving on to statement B: "If a and b are opposite numbers, then a/b = -1." Opposite numbers. Let me think. If a and b are opposites, that would mean that a = -b. So then, if I divide a by b, that would be (-b)/b, which is -1. Wait, but what if b is zero? Then we would have a division by zero, which is undefined. But the question is about opposite numbers. If a and b are opposites, then if one is zero, the other must also be zero (since 0 is its own opposite). But then a/b would be 0/0, which is undefined. So does the statement consider that case? The problem says "if a and b are opposite numbers". If they are opposite numbers, can they both be zero? Yes, because 0 is the opposite of itself. So in that case, a/b is undefined. Therefore, the statement "If a and b are opposite numbers, then a/b = -1" is not always true because if a and b are both zero, the division is undefined. Therefore, statement B is not necessarily true. So B is false.Wait, but maybe opposite numbers are defined as non-zero? Let me check that. Typically, opposite numbers are numbers that add up to zero, so a = -b. If a is zero, then b is also zero. So technically, zero is the opposite of itself. So in that case, the statement B is only true if b is not zero. But since the problem doesn't specify that a and b are non-zero, then statement B can be false. So I think B is incorrect.Now statement C: "If a² = b², then |a| = |b|." Let's see. If a squared equals b squared, then taking the square root of both sides gives |a| = |b|, since square roots of squares give absolute values. Yes, that's a mathematical rule. For example, if a is 3 and b is -3, then a² is 9 and b² is 9, so |a| is 3 and |b| is 3. So in all cases, if a² equals b², their absolute values must be equal. So statement C is correct. So C is true.Then statement D: "-a is always a negative number." Hmm. If a is positive, then -a is negative. But if a is negative, then -a is positive. For example, if a is -5, then -a is 5, which is positive. Also, if a is zero, then -a is zero, which is neither positive nor negative. So -a is not always a negative number. Therefore, statement D is false.So summarizing:A: FalseB: False (due to the case where a and b are zero)C: TrueD: FalseTherefore, the correct statement is C.Final Answerboxed{C}</think>Let's analyze each statement step by step:- Statement A: "Two numbers with equal absolute values are equal." This is false because numbers with equal absolute values can be opposites. For example, (5) and (-5) have the same absolute value but are not equal.- Statement B: "If (a) and (b) are opposite numbers, then (frac{a}{b} = -1)." This is false because if (a) and (b) are both zero (since zero is its own opposite), (frac{a}{b}) is undefined. Hence, the statement does not hold in all cases.- Statement C: "If (a^2 = b^2), then (|a| = |b|)." This is true because taking the square root of both sides of (a^2 = b^2) gives (|a| = |b|), which is a fundamental property of absolute values.- Statement D: "(-a) is always a negative number." This is false because if (a) is negative, (-a) is positive. Additionally, if (a) is zero, (-a) is zero, which is neither positive nor negative.Thus, the correct statement is (boxed{C}).]

🔑:To evaluate the correctness of each statement, let's analyze them one by one:- Statement A: Two numbers with equal absolute values are equal. - Counterexample: Consider -2 and 2. We have |-2| = |2| = 2, but -2 neq 2. - Therefore, statement A is boxed{text{not correct}}.- Statement B: If a and b are opposite numbers, then frac{a}{b}=-1. - For opposite numbers, a = -b. Substituting into the fraction, we get frac{a}{b} = frac{-b}{b} = -1, assuming b neq 0 to avoid division by zero. - However, the condition b neq 0 is necessary, making statement B boxed{text{not correct}} without this clarification.- Statement C: If a^{2}=b^{2}, then |a|=|b|. - Starting from a^{2}=b^{2}, taking the square root of both sides gives |a| = |b|, since the square root of a square returns the absolute value of the original number. - This statement is boxed{text{correct}}.- Statement D: -a is always a negative number. - Counterexample: Let a = -2. Then, -a = -(-2) = 2, which is positive. - Therefore, statement D is boxed{text{not correct}}.Given the analysis above, the correct answer is boxed{C}.