🧠:[<think>Alright, let's tackle this problem step by step. First, let me understand what's given and what's being asked. We have an acute triangle ABC, with sides a, b, c opposite angles A, B, C respectively. We know that sin A = 3/5 and tan(A - B) = -1/2. The questions are to find tan B and then, if b = 5, find c.Starting with part (1): Find tan B. Hmm. So we have information about angle A and the tangent of (A - B). Let me recall some trigonometric identities. The formula for tan(A - B) is [tan A - tan B] / [1 + tan A tan B]. Since tan(A - B) is given as -1/2, maybe I can set up an equation here.But first, I need to find tan A. We know sin A = 3/5. Since the triangle is acute, all angles are less than 90 degrees, so cosine of A will be positive. Using the Pythagorean identity, cos²A + sin²A = 1. Therefore, cos A = sqrt(1 - sin²A) = sqrt(1 - 9/25) = sqrt(16/25) = 4/5. So tan A = sin A / cos A = (3/5) / (4/5) = 3/4.Now, tan(A - B) = (tan A - tan B)/(1 + tan A tan B) = -1/2. Let's substitute tan A as 3/4:( (3/4) - tan B ) / (1 + (3/4) tan B ) = -1/2Let me write that equation down:(3/4 - tan B) / (1 + (3/4) tan B) = -1/2To solve for tan B, cross-multiplying:2*(3/4 - tan B) = -1*(1 + (3/4) tan B )Compute left side: 2*(3/4) - 2 tan B = 3/2 - 2 tan BRight side: -1 - (3/4) tan BSo equation becomes:3/2 - 2 tan B = -1 - (3/4) tan BLet me get all terms involving tan B on one side and constants on the other.Bring -2 tan B to the right as +2 tan B and - (3/4) tan B to the left as + (3/4) tan B. Wait, maybe it's better to collect like terms.Let's add 2 tan B to both sides and add 1 to both sides:3/2 + 1 = (2 tan B - (3/4) tan B)Left side: 3/2 + 1 = 5/2Right side: (2 - 3/4) tan B = (5/4) tan BSo 5/2 = (5/4) tan BMultiply both sides by 4/5:(5/2)*(4/5) = tan B => 4/2 = 2 = tan BWait, so tan B = 2? Let me check that again.Starting from:3/2 - 2 tan B = -1 - (3/4) tan BBring all terms to left side:3/2 - 2 tan B + 1 + (3/4) tan B = 0Combine constants: 3/2 + 1 = 5/2Combine tan terms: (-2 + 3/4) tan B = (-5/4) tan BThus:5/2 - (5/4) tan B = 0Then:5/2 = (5/4) tan BMultiply both sides by 4/5:tan B = (5/2)*(4/5) = 2Yes, that's correct. So tan B is 2. Let me just verify by plugging back into the original equation.tan(A - B) = (3/4 - 2)/(1 + (3/4)(2)) = (-5/4)/(1 + 3/2) = (-5/4)/(5/2) = (-5/4)*(2/5) = -10/20 = -1/2. Perfect, that matches the given value. So part (1) answer is 2.Now part (2): If b = 5, find c.So in triangle ABC, sides a, b, c are opposite angles A, B, C. So side b is opposite angle B, which is 5. Need to find c, which is opposite angle C.First, let me recall the Law of Sines: a/sin A = b/sin B = c/sin C = 2R, where R is the circumradius.Given that b = 5, and we know sin B. Wait, from part (1), we found tan B = 2. Since triangle is acute, angle B is less than 90 degrees, so sin B and cos B are positive.If tan B = 2, then we can construct a right triangle where the opposite side is 2 and adjacent is 1, so hypotenuse is sqrt(1 + 4) = sqrt(5). Therefore, sin B = 2/sqrt(5) and cos B = 1/sqrt(5).But maybe rationalize the denominators: sin B = 2√5 / 5, cos B = √5 / 5.Similarly, we know sin A = 3/5, and we found cos A = 4/5 earlier.So using Law of Sines: a / sin A = b / sin B.Given that sin A = 3/5, sin B = 2√5 / 5, and b = 5. Let's compute the ratio:b / sin B = 5 / (2√5 / 5) = 5 * (5 / 2√5) ) = 25 / (2√5) = rationalize denominator: 25√5 / (2*5) ) = 5√5 / 2.Therefore, the common ratio 2R = 5√5 / 2, so a = sin A * (5√5 / 2) ) = (3/5) * (5√5 / 2) ) = 3√5 / 2.Similarly, c = sin C * (5√5 / 2). So we need to find sin C.But angle C can be found since angles in triangle sum to 180 degrees. So angle C = 180 - A - B. So sin C = sin(A + B) because sin(180 - x) = sin x.So sin C = sin(A + B). Let's compute sin(A + B).We have angles A and B, and we know sin A, cos A, sin B, cos B. So sin(A + B) = sin A cos B + cos A sin B.Plugging in the values:sin A cos B + cos A sin B = (3/5)(√5/5) + (4/5)(2√5/5)Compute each term:First term: (3√5)/25Second term: (8√5)/25Add them together: (11√5)/25Therefore, sin C = 11√5 / 25Therefore, c = sin C * (5√5 / 2) ) = (11√5 / 25) * (5√5 / 2) )Simplify:The 5 in numerator and denominator cancels: 11√5 / 5 * √5 / 2Wait, let's do it step by step.Multiply numerators: 11√5 * 5√5 = 11 * 5 * (√5 * √5) = 55 * 5 = 275Denominator: 25 * 2 = 50So c = 275 / 50 = 55/10 = 11/2 = 5.5But as a fraction, 11/2. So c = 11/2.Let me check again:sin C = 11√5 / 25Multiply by 5√5 / 2:(11√5 / 25) * (5√5 / 2) = 11 * (5) * (√5 * √5) / (25 * 2) ) = 11 * 5 * 5 / (50) = (275)/50 = 5.5 = 11/2. Yep, that's correct.Therefore, the answer to part (2) is 11/2.Wait, just to make sure, is there another way to approach this? Maybe using the Law of Cosines? Let's see.Alternatively, once we have angles A and B, we can find angle C, then use the Law of Sines again. Wait, that's what we did. Alternatively, if we can find sides a and b, maybe use Law of Cosines to find c. But since we only have side b given, maybe Law of Sines is more straightforward here.Alternatively, perhaps using the ratio between sides. Since we know sides a, b, c are proportional to sin A, sin B, sin C. Since we found sin C, and we have b = 5, which is sin B = 2√5 /5.Thus, the ratio of c to b is sin C / sin B. Therefore, c = b * (sin C / sin B) = 5 * ( (11√5 /25 ) / (2√5 /5 ) ) = 5 * (11√5 /25 * 5 / 2√5 ) = 5 * (11/10 ) = 55/10 = 11/2. Same result. So that's correct.Therefore, part (2) answer is 11/2.Final Answer(1) boxed{2}(2) boxed{dfrac{11}{2}}</think>Given an acute triangle ( text{ABC} ) with sides opposite to angles ( text{A} ), ( text{B} ), and ( text{C} ) being ( a ), ( b ), and ( c ) respectively. We know ( sin text{A} = frac{3}{5} ) and ( tan (text{A} - text{B}) = -frac{1}{2} ).(1) To find ( tan text{B} ):1. Calculate ( cos text{A} ) using the Pythagorean identity: [ cos text{A} = sqrt{1 - sin^2 text{A}} = sqrt{1 - left(frac{3}{5}right)^2} = sqrt{frac{16}{25}} = frac{4}{5} ]2. Determine ( tan text{A} ): [ tan text{A} = frac{sin text{A}}{cos text{A}} = frac{frac{3}{5}}{frac{4}{5}} = frac{3}{4} ]3. Use the tangent subtraction formula: [ tan (text{A} - text{B}) = frac{tan text{A} - tan text{B}}{1 + tan text{A} tan text{B}} = -frac{1}{2} ]4. Substitute ( tan text{A} = frac{3}{4} ) and solve for ( tan text{B} ): [ frac{frac{3}{4} - tan text{B}}{1 + frac{3}{4} tan text{B}} = -frac{1}{2} ] Cross-multiplying and solving: [ 2 left( frac{3}{4} - tan text{B} right) = - left( 1 + frac{3}{4} tan text{B} right) ] Simplifying: [ frac{3}{2} - 2 tan text{B} = -1 - frac{3}{4} tan text{B} ] [ frac{5}{2} = frac{5}{4} tan text{B} ] [ tan text{B} = 2 ](2) To find ( c ) given ( b = 5 ):1. Calculate ( sin text{B} ) and ( cos text{B} ) using ( tan text{B} = 2 ): [ sin text{B} = frac{2}{sqrt{5}} = frac{2sqrt{5}}{5}, quad cos text{B} = frac{1}{sqrt{5}} = frac{sqrt{5}}{5} ]2. Use the Law of Sines: [ frac{b}{sin text{B}} = frac{c}{sin text{C}} ]3. Calculate ( sin text{C} ) using angle sum identity: [ text{C} = 180^circ - text{A} - text{B}, quad sin text{C} = sin (text{A} + text{B}) ] [ sin (text{A} + text{B}) = sin text{A} cos text{B} + cos text{A} sin text{B} ] [ sin text{C} = left( frac{3}{5} cdot frac{sqrt{5}}{5} right) + left( frac{4}{5} cdot frac{2sqrt{5}}{5} right) = frac{3sqrt{5}}{25} + frac{8sqrt{5}}{25} = frac{11sqrt{5}}{25} ]4. Use the ratio from the Law of Sines: [ c = frac{sin text{C}}{sin text{B}} cdot b = frac{frac{11sqrt{5}}{25}}{frac{2sqrt{5}}{5}} cdot 5 = frac{11}{10} cdot 5 = frac{11}{2} ]Final Answers:(1) (boxed{2})(2) (boxed{dfrac{11}{2}}),<think>Okay, let's tackle this problem step by step. It's about an acute triangle ABC with sides a, b, c opposite angles A, B, C respectively. We know that sin A = 3/5 and tan(A - B) = -1/2. We need to find tan B and then, if b = 5, find c.Starting with part (1): Find tan B.First, since we know sin A = 3/5, and triangle ABC is acute, all angles are less than 90 degrees. So angle A is acute, which means we can find cos A using the Pythagorean identity. Since sin²A + cos²A = 1, cos A = sqrt(1 - sin²A) = sqrt(1 - (9/25)) = sqrt(16/25) = 4/5. Therefore, cos A = 4/5.Then, tan A = sin A / cos A = (3/5)/(4/5) = 3/4. So tan A is 3/4.Now, we have tan(A - B) = -1/2. Let's recall the formula for tan(A - B):tan(A - B) = (tan A - tan B) / (1 + tan A tan B)We know tan(A - B) = -1/2 and tan A = 3/4. Let's plug those into the equation:-1/2 = (3/4 - tan B) / (1 + (3/4) tan B)Let me write that equation again:(3/4 - tan B) / (1 + (3/4) tan B) = -1/2Let’s solve for tan B. Let’s denote tan B as x for simplicity.So:(3/4 - x) / (1 + (3/4)x) = -1/2Cross-multiplying:2(3/4 - x) = -1(1 + (3/4)x)Simplify both sides:Left side: 2*(3/4) - 2x = 3/2 - 2xRight side: -1 - (3/4)xSo equation becomes:3/2 - 2x = -1 - (3/4)xLet’s multiply all terms by 4 to eliminate denominators:4*(3/2) - 4*2x = 4*(-1) - 4*(3/4)xWhich simplifies to:6 - 8x = -4 - 3xNow, bring all terms to one side:6 - 8x + 4 + 3x = 0Combine like terms:10 - 5x = 0So, -5x = -10Divide both sides by -5:x = 2So tan B = 2. Wait, but the problem says the triangle is acute, so angle B must be less than 90 degrees. If tan B = 2, then angle B is acute since tan is positive in acute angles (first quadrant). So that's okay.So part (1) answer is 2. Let's check if that makes sense.If tan B = 2, then angle B is arctan(2), which is around 63.43 degrees. Then angle A is arcsin(3/5). Since sin A = 3/5, angle A is approximately 36.87 degrees. Then A - B would be approximately 36.87 - 63.43 = -26.56 degrees. The tan of -26.56 degrees is approximately -0.5, which is -1/2. That matches the given condition. So that seems to check out.So part (1) is tan B = 2.Moving on to part (2): If b = 5, find c.We need to find the length of side c, opposite angle C. Given that b = 5, which is opposite angle B.Since we have some information about angles A and B, maybe we can find angle C and use the Law of Sines or Law of Cosines to find c.First, let's recall that in any triangle, the sum of the angles is 180 degrees. So angle C = 180° - angle A - angle B.We can find angle A and angle B using the information we have.We already found that tan B = 2, so angle B is arctan(2). Similarly, angle A we have sin A = 3/5, so angle A is arcsin(3/5). Alternatively, since we have tan A = 3/4, angle A is arctan(3/4).But maybe instead of working with angles, we can use trigonometric identities or the Law of Sines.Law of Sines states that a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle.We know b = 5, and we can find sin B since we know tan B = 2. Similarly, we know sin A = 3/5. Let's compute sin B.If tan B = 2, then we can imagine a right triangle where the opposite side is 2 and adjacent is 1, so hypotenuse is sqrt(1 + 4) = sqrt(5). Therefore, sin B = 2/sqrt(5) and cos B = 1/sqrt(5).Similarly, sin A is 3/5, cos A is 4/5, as we found earlier.Then, using Law of Sines:a / sin A = b / sin BGiven that a is opposite angle A, b is opposite angle B. So:a / (3/5) = 5 / (2/sqrt(5))Simplify:a = (3/5) * [5 / (2/sqrt(5))] = (3/5) * [5 * sqrt(5)/2] = (3/5)*(5 sqrt(5)/2) = 3 sqrt(5)/2So a = (3 sqrt(5))/2. But do we need a? Wait, we need c.Alternatively, maybe we can use Law of Sines again once we find angle C. But angle C = 180° - A - B. Let's compute angle C.But maybe instead of angles, we can find sin C, since sin C = sin(180° - A - B) = sin(A + B). Because sin(180° - x) = sin x.So sin C = sin(A + B). Let's compute sin(A + B).We can use the identity sin(A + B) = sin A cos B + cos A sin B.We have sin A = 3/5, cos A = 4/5.We have sin B = 2/sqrt(5), cos B = 1/sqrt(5).Therefore:sin(A + B) = (3/5)(1/sqrt(5)) + (4/5)(2/sqrt(5)) = (3)/(5 sqrt(5)) + 8/(5 sqrt(5)) = (11)/(5 sqrt(5)).Therefore, sin C = 11/(5 sqrt(5)).Alternatively, rationalizing the denominator:sin C = 11/(5 sqrt(5)) = (11 sqrt(5))/25.Now, using Law of Sines:c / sin C = b / sin BSo c = (sin C / sin B) * bPlugging in the values:sin C = 11 sqrt(5)/25, sin B = 2/sqrt(5), b = 5.Therefore:c = ( (11 sqrt(5)/25 ) / (2/sqrt(5)) ) * 5Simplify the division:(11 sqrt(5)/25 ) * (sqrt(5)/2 ) = (11 * 5 / (25 * 2)) ) = (55 / 50 ) = 11/10Then multiply by 5:11/10 * 5 = 55/10 = 11/2 = 5.5So c = 11/2. So c = 5.5, which is 11/2.Wait, let me check that calculation again step by step.c = [ (sin C) / (sin B) ] * bsin C = 11 sqrt(5)/25sin B = 2/sqrt(5)So:sin C / sin B = (11 sqrt(5)/25) / (2/sqrt(5)) = (11 sqrt(5)/25) * (sqrt(5)/2) = (11 * 5 ) / (25 * 2) = 55 / 50 = 11/10Yes, that's correct. Then multiply by b, which is 5:11/10 * 5 = 55/10 = 11/2. So c = 11/2. That is 5.5.But let's verify this with another approach to make sure.Alternatively, using Law of Cosines. Maybe we can compute the sides a, b, c with the information given.Wait, we know sides a and b? Wait, we know b = 5. We found a = (3 sqrt(5))/2. But we might need another relation.Alternatively, since we know angles A and B, we can find angle C, then use Law of Sines. Let me check angle C.We have angles A and B. Let's compute angle A and angle B in degrees to see if the triangle is valid.We had angle A: sin A = 3/5, so angle A ≈ 36.87 degrees.Angle B: tan B = 2, so angle B ≈ 63.43 degrees.So angle C ≈ 180 - 36.87 - 63.43 ≈ 79.7 degrees, which is still acute. So the triangle is indeed acute.So angle C ≈ 79.7 degrees, sin C ≈ sin 79.7 ≈ 0.9848. Let's check our previous calculation of sin C:We found sin C = 11 sqrt(5)/25. Let's compute that numerically:sqrt(5) ≈ 2.236, so 11 * 2.236 ≈ 24.596. Divided by 25: 24.596 /25 ≈ 0.9838. Which is approximately 0.984, matching the angle 79.7 degrees. So that's consistent.Therefore, sin C ≈ 0.984, sin B = 2/sqrt(5) ≈ 2/2.236 ≈ 0.894.So using Law of Sines:c = (sin C / sin B) * b ≈ (0.984 / 0.894) *5 ≈ (1.10) *5 ≈5.5, which is 11/2. So that seems correct.Alternatively, maybe use Law of Cosines on angle C, but that might be more complicated.Alternatively, since we have sides a and b, and angles A and B, maybe find side c using Law of Cosines. But we need angle C. Wait, angle C is 180 - A - B, so we can use Law of Cosines:c² = a² + b² - 2ab cos CBut we need cos C. However, we can find cos C since angle C = A + B (since A + B + C = 180, so C = 180 - (A + B), but cos(180 - (A + B)) = -cos(A + B). Wait, but angle C is in an acute triangle, so angle C < 90°, so cos C is positive.Wait, maybe we can compute cos(A + B) = cos(180 - C) = -cos C. Therefore, cos C = -cos(A + B). Hmm, perhaps.But maybe it's easier to compute cos C via sin C. Since we have sin C = 11 sqrt(5)/25, then cos C = sqrt(1 - sin²C).Compute sin²C = (121 * 5)/625 = 605/625. Therefore, cos²C = 1 - 605/625 = (625 - 605)/625 = 20/625 = 4/125. Therefore, cos C = sqrt(4/125) = 2/(5 sqrt(5)) = 2 sqrt(5)/25.Therefore, cos C = 2 sqrt(5)/25.So then, using Law of Cosines:c² = a² + b² - 2ab cos CWe have a = 3 sqrt(5)/2, b = 5.Compute a²: (9*5)/4 = 45/4.Compute b²: 25.Compute 2ab cos C:2 * (3 sqrt(5)/2) * 5 * (2 sqrt(5)/25) =Simplify step by step:First, 2 cancels with the denominator 2: 2*(3 sqrt(5)/2) = 3 sqrt(5)Then multiply by 5: 3 sqrt(5)*5 = 15 sqrt(5)Then multiply by 2 sqrt(5)/25:15 sqrt(5) * 2 sqrt(5)/25 = (30 * 5)/25 = 150/25 = 6So 2ab cos C = 6Therefore, c² = 45/4 + 25 - 6 = 45/4 + 19 = 45/4 + 76/4 = 121/4Therefore, c = sqrt(121/4) = 11/2, which matches our previous result.Therefore, c = 11/2. So part (2) answer is 11/2.Therefore, the final answers are (1) 2 and (2) 11/2.Final Answer(1) boxed{2}(2) boxed{dfrac{11}{2}}</think>Given an acute triangle (ABC) with sides (a), (b), and (c) opposite angles (A), (B), and (C) respectively, we know (sin A = frac{3}{5}) and (tan(A - B) = -frac{1}{2}).(1) To find (tan B):1. Calculate (cos A) using the Pythagorean identity: [ cos A = sqrt{1 - sin^2 A} = sqrt{1 - left(frac{3}{5}right)^2} = sqrt{frac{16}{25}} = frac{4}{5} ]2. Determine (tan A): [ tan A = frac{sin A}{cos A} = frac{frac{3}{5}}{frac{4}{5}} = frac{3}{4} ]3. Use the tangent subtraction formula (tan(A - B) = frac{tan A - tan B}{1 + tan A tan B}) and substitute the given values: [ -frac{1}{2} = frac{frac{3}{4} - tan B}{1 + frac{3}{4} tan B} ]4. Solving for (tan B): [ frac{3}{4} - tan B = -frac{1}{2} left(1 + frac{3}{4} tan Bright) ] [ 3/2 - 2tan B = -1 - frac{3}{4}tan B ] [ 10 - 5tan B = 0 implies tan B = 2 ](2) To find (c) when (b = 5):1. Calculate (sin B) and (cos B) using (tan B = 2): [ sin B = frac{2}{sqrt{5}}, quad cos B = frac{1}{sqrt{5}} ]2. Calculate (sin(A + B)) using angle addition formula: [ sin(A + B) = sin A cos B + cos A sin B = frac{3}{5} cdot frac{1}{sqrt{5}} + frac{4}{5} cdot frac{2}{sqrt{5}} = frac{11}{5sqrt{5}} = frac{11sqrt{5}}{25} ]3. Use the Law of Sines: [ frac{c}{sin C} = frac{b}{sin B} implies c = frac{sin C}{sin B} cdot b ] [ sin C = sin(A + B) = frac{11sqrt{5}}{25}, quad sin B = frac{2}{sqrt{5}} ] [ c = frac{frac{11sqrt{5}}{25}}{frac{2}{sqrt{5}}} cdot 5 = frac{11}{10} cdot 5 = frac{11}{2} ]Final answers:(1) (boxed{2})(2) (boxed{dfrac{11}{2}})]

🔑:(1) In acute triangle text{ABC}, we have sin text{A} = frac{3}{5},therefore cos text{A} = sqrt{1 - sin^2 text{A}} = frac{4}{5}, and tan text{A} = frac{sin text{A}}{cos text{A}} = frac{3}{4};Also, tan (text{A} - text{B}) = frac{tan text{A} - tan text{B}}{1 + tan text{A} tan text{B}} = frac{frac{3}{4} - tan text{B}}{1 + frac{3}{4} tan text{B}} = -frac{1}{2},therefore Solving for tan text{B}, we get tan text{B} = 2.(2) Since tan text{B} = 2, therefore frac{sin text{B}}{cos text{B}} = 2, and sin text{B} = 2cos text{B};therefore sin^2 text{B} + cos^2 text{B} = (2cos text{B})^2 + cos^2 text{B} = 5cos^2 text{B} = 1,therefore cos text{B} = sqrt{frac{1}{5}} = frac{sqrt{5}}{5}, and sin text{B} = 2cos text{B} = frac{2sqrt{5}}{5};therefore sin text{C} = sin [pi - (text{A} + text{B})] = sin (text{A} + text{B}) = sin text{A} cos text{B} + cos text{A} sin text{B} = frac{3}{5} times frac{sqrt{5}}{5} + frac{4}{5} times frac{2sqrt{5}}{5} = frac{11sqrt{5}}{25};Given that text{b} = 5, and using the sine rule frac{text{b}}{sin text{B}} = frac{text{c}}{sin text{C}},therefore text{c} = frac{text{b} cdot sin text{C}}{sin text{B}} = frac{5 times frac{11sqrt{5}}{25}}{frac{2sqrt{5}}{5}} = boxed{frac{11}{2}}.

🧠:[<think>Okay, let me look at this problem. So, there's a table with two rows and two columns. The first cell is "3+5", the second cell in the first row is "8+7". Then the second row starts with "9-5" and ends with "13-6". Hmm, what exactly is being asked here? The user hasn't specified the question, but maybe the task is to compute each of these expressions and present the results in the table format? That seems likely, since those are all arithmetic problems. Let me verify each calculation step by step to make sure I don't make any mistakes.Starting with the first cell: 3 plus 5. That's straightforward. 3 + 5 equals 8. Got that. Then the next cell in the first row is 8 plus 7. Let me add those: 8 + 7. Well, 8 plus 7 is 15. Right, because 8 plus 2 is 10, and then 5 more makes 15.Moving to the second row. The first cell there is 9 minus 5. Subtracting 5 from 9. 9 minus 5 is 4. Makes sense. Then the last cell is 13 minus 6. Let me calculate that. 13 take away 6. Since 6 plus 7 is 13, subtracting 6 gives 7. So that's 7.Wait, let me double-check all of them to be sure. Sometimes when you do simple math quickly, you might slip up. First one again: 3 + 5. Yep, 3, 4, 5, 6, 7, 8. So that's definitely 8. Second one: 8 +7. Let's see, 8 plus 7. 8 plus 7 is like 8 + 2 +5, which is 10 +5=15. Correct. Then 9-5. If I take away 5 from 9, that's 4. And 13-6. 13 minus 6. Hmm, thinking in terms of addition: 6 plus what is 13? 6 +7=13, so the answer is 7. All right, all the calculations check out.So the table with the results would look like this:| 8 | 15 || :--- | ---: || 4 | 7 |Wait, but the original table has some alignment. The first column is aligned with the colon on the left, and the second column is aligned with the colon on the right. Let me make sure I replicate that formatting. So in Markdown, the colon on the left means left-aligned, and the colon on the right means right-aligned. So the first column is left-aligned, the second column is right-aligned. So when presenting the answers, I need to keep that alignment. Let me write that out properly.Original table structure:| 3+5 | 8+7 || :--- | ---: || 9-5 | 13-6 |So replacing the expressions with their results, maintaining the alignment:| 8 | 15 || :--- | ---: || 4 | 7 |Yes, that should be correct. Each cell now contains the numerical result of the expression, and the alignment is preserved as per the original table's formatting. Let me confirm once more each calculation to ensure no errors. 3+5=8, 8+7=15, 9-5=4, 13-6=7. All correct. Alignments: first column left-aligned, second column right-aligned. Looks good. I think that's the answer needed here.</think>| 8 | 15 || :--- | ---: || 4 | 7 |boxed{begin{array}{|l|r|}hline3+5 & 8+7 hline8 & 15 hline9-5 & 13-6 hline4 & 7 hlineend{array}},<think>Okay, let me try to figure out this problem. So, there's a table here with two rows and two columns. The first cell is "3+5", then "8+7" in the top right. The second row starts with "9-5" and ends with "13-6". Hmm, I need to figure out what the question is asking. Since the user mentioned "Aufgabe 1" which is German for "Exercise 1", maybe they want me to compute the results of these arithmetic expressions? Let me check each one step by step.Starting with the first cell: 3 plus 5. That's straightforward. 3 + 5 equals 8. Got that. Then the next cell in the top row is 8 + 7. Let me add those: 8 plus 7 is 15. Okay, moving to the second row, first cell: 9 minus 5. That should be 4. And the last cell is 13 minus 6. Let me subtract: 13 minus 6 is 7. So, putting it all together, the table with the computed results would look like:| 8 | 15 || :--- | ---: || 4 | 7 |But wait, maybe there's more to it? The original table has some alignment specified. The first column is left-aligned with the colon and dashes on the left side, and the second column is right-aligned with the dashes and colon on the right. But in terms of content, if the task is just to compute each expression, then replacing them with the results as I did makes sense. Let me double-check my calculations to be sure.3 + 5: 3, 4, 5, 6, 7, 8. Yep, that's 8. 8 + 7: 8 + 7 is 15. 9 - 5: 9 minus 5 is 4. 13 - 6: 13 take away 6 equals 7. All correct. Maybe the question is to fill in the table with the answers? If that's the case, then the answers are 8, 15, 4, and 7. Alternatively, could there be another pattern or something more complex? Let me think.Looking at the original table structure:| 3+5 | 8+7 || :--- | ---: || 9-5 | 13-6 |The alignment might be a red herring, just formatting for the table. The expressions are basic arithmetic. The problem might be part of a larger set where similar tables require different operations, but based on the given information, the most straightforward answer is to compute each expression. Since the user hasn't specified a different task, like solving for variables or finding a pattern between the cells, I think calculating each one is the right approach.Another angle: Maybe the numbers in each cell relate to each other in some way? Let's see. First row: 3+5=8 and 8+7=15. The result of the first cell (8) is the first number in the second cell. Then, 8+7. Then in the second row: 9-5=4 and 13-6=7. There's no obvious connection between 4 and 13 or 7 here. Unless there's a diagonal relationship? 3+5 and 13-6: 8 and 7. 8+7 and 9-5: 15 and 4. Not sure if that's relevant. Maybe the columns or rows need to be added together? First column: 8 (from 3+5) and 4 (from 9-5), summing to 12. Second column: 15 and 7, summing to 22. But unless the problem specifies that, I don't think so.Alternatively, maybe it's a matrix or some kind of puzzle where combining the results leads to another answer. But without more context, it's hard to tell. Since the initial instruction was just "You will be given a problem. Please reason..." and the problem is this table with arithmetic expressions, the safest bet is to compute each one. Therefore, the answers are 8, 15, 4, and 7. Let me check once again to make sure I didn't make a mistake in arithmetic. 3+5=8, correct. 8+7=15, yes. 9-5=4, right. 13-6=7, definitely. I think that's all there is to it. Maybe the user wants the table filled with the results, so replacing each expression with its value. Alright, I think that's the solution.</think>The results of the arithmetic expressions in the table are computed as follows:| 3+5 = boxed{8} | 8+7 = boxed{15} || :--- | ---: || 9-5 = boxed{4} | 13-6 = boxed{7} |]

🔑:3+5=8 ; 8+7=15 ; 9-5=4 ; 13-6=7

🧠:[<think>Okay, so I need to figure out the maximum number of circles Misha can construct under these rules. Let's break down the problem step by step. First, there are 16 points named A₁ to A₁₆, each spaced 1 cm apart on a straight line. The circles Misha constructs must follow these rules:a) Circles don't intersect or touch each other.b) Each circle must contain at least one of the points A₁ to A₁₆ inside it.c) None of the points lie on the circle itself; they have to be strictly inside.d) Each circle must have a unique set of points inside it. So if one circle has, say, A₁ and A₂ inside, no other circle can have exactly A₁ and A₂. The sets have to be different.The goal is to find the maximum number of such circles. The answer is given as 31, but I need to understand why.Let me start by visualizing the setup. The points are on a line, 1 cm apart. So A₁ is at position 0, A₂ at 1 cm, A₃ at 2 cm, and so on up to A₁₆ at 15 cm. Each circle must enclose at least one point, but not have any points on the circumference. Also, circles can't overlap or touch, which means they have to be separated by some distance. And each circle's set of enclosed points must be unique.First thought: Each circle can enclose a subset of the points, but subsets have to be unique. Since there are 16 points, the total number of subsets is 2¹⁶, but most of these subsets can't be realized with non-overlapping circles. So the challenge is to select as many subsets as possible such that their corresponding circles don't overlap, and each subset is unique.But how do the circles correspond to subsets? For a circle to enclose a subset of points, those points must lie inside the circle, and the circle must not enclose any other points. But since the points are on a line, and the circles are in the plane, we can place a circle around a segment of the line containing some points.However, circles can be placed above or below the line. But since they can't intersect or touch, if two circles are on the same side, they need to be spaced apart. If they are on opposite sides, maybe they can be closer? Wait, but if you have a circle above the line and another below, but their projections on the line overlap, then the circles might intersect. Hmm.Wait, maybe the circles can be arranged in such a way that their centers are not on the line. But given that the points are on a line, perhaps the optimal way to enclose subsets is to have circles centered near the line but at different positions so that their projections on the line don't overlap. But the exact arrangement might affect whether the circles intersect.Alternatively, since the points are on a line, maybe each circle is centered around some interval on the line. For example, a circle that encloses points A_i to A_j must be large enough to contain all those points inside it, but not include any others. However, if you have two circles enclosing different intervals, they might overlap if their intervals are too close.But wait, if circles can be placed on either side of the line, maybe we can alternate them above and below to prevent overlapping. So even if two intervals are adjacent on the line, the circles can be on opposite sides and not interfere. But we have to check if that's possible.Alternatively, maybe all circles are on one side of the line, but spaced apart so that their projections on the line don't overlap. Let's think about this.Suppose we have a circle that encloses a single point. To enclose a single point, the radius of the circle needs to be less than 0.5 cm, so that it doesn't reach the adjacent points. Because the points are 1 cm apart, if the radius is 0.5 cm, the circle would touch the neighboring points, which is not allowed (rule c). So the radius must be less than 0.5 cm. Therefore, such a circle would enclose only one point and can be placed directly over that point.But if we do this for each point, we could potentially have 16 circles, each enclosing a single point. However, these circles would need to be placed such that they don't intersect. If they are all on the same side of the line, the distance between their centers (along the line) is 1 cm, but their radii are less than 0.5 cm, so the distance between centers is more than twice the radius (which is 1 cm > 2*0.5 cm). Therefore, they wouldn't intersect. Wait, but the distance between centers is 1 cm, and each radius is less than 0.5 cm. So the distance between centers is greater than the sum of the radii (which would be less than 1 cm). Therefore, the circles would not intersect. So actually, if we place all 16 circles above the line, each centered over their respective points with radius approaching 0.5 cm, they wouldn't touch or intersect. Similarly, we could place some above and some below, but the key is that they don't interfere. So in this case, we can have 16 circles each enclosing a single point. But according to rule d, each circle must contain a different set of points. If each circle contains a single different point, then all 16 are allowed. But the answer is 31, which is more than 16, so there must be a way to include more circles by enclosing multiple points.Wait, but if a circle encloses multiple points, how does that work? Let's consider a circle enclosing two adjacent points. The distance between A₁ and A₂ is 1 cm. To enclose both, the circle must have a radius greater than 0.5 cm (since the center would be between them, at 0.5 cm from each, so radius just over 0.5 cm would enclose both). But then, such a circle would have a radius of, say, 0.5 + ε cm. However, this would bring the circle close to the next point. For example, if we have a circle centered between A₁ and A₂ with radius 0.6 cm, it would enclose A₁ and A₂, but the distance from the center to A₃ is 1.5 cm, so the circle wouldn't reach A₃. Wait, no: A₁ is at 0 cm, A₂ at 1 cm. The center is at 0.5 cm. The radius is 0.6 cm, so the circle extends from 0.5 - 0.6 = -0.1 cm to 0.5 + 0.6 = 1.1 cm. So A₁ is at 0 cm, which is inside (since -0.1 to 1.1 includes 0 and 1). A₂ is at 1 cm, which is inside. A₃ is at 2 cm, which is outside. So such a circle would enclose A₁ and A₂. Similarly, a circle centered between A₂ and A₃ with radius 0.6 cm would enclose A₂ and A₃, and so on. But if we place such circles, each covering two adjacent points, how many can we fit without overlapping? Let's see. If each circle is centered between two points and has a radius of 0.6 cm, the circle spans from -0.1 to 1.1 cm for the first one, then the next would be centered at 1.5 cm, spanning from 1.5 - 0.6 = 0.9 cm to 1.5 + 0.6 = 2.1 cm. Wait, these two circles would overlap between 0.9 cm and 1.1 cm. So they would intersect. Therefore, adjacent two-point circles can't be placed next to each other on the same side.Alternatively, if we alternate sides. Place one circle above the line between A₁ and A₂, next circle below the line between A₂ and A₃, etc. However, even if they are on opposite sides, the circles might still intersect if their projections on the line overlap. Wait, the distance between the centers along the line is 1 cm (between centers at 0.5 and 1.5 cm). The vertical distance between the circles (if one is above and one is below) is the diameter in the vertical direction. But circles on opposite sides of the line would be separated vertically by twice the distance from the line. However, if the circles are just above and below the line, their vertical separation is 2h, where h is the height above/below the line. If h is greater than the radius, then the circles won't intersect. Wait, but the problem states that circles cannot touch or intersect. So even if they are on opposite sides, as long as the distance between their centers is greater than the sum of their radii.If the centers are 1 cm apart horizontally, and vertically separated by 2h, then the distance between centers is sqrt(1² + (2h)²). If the radii are r, then we need sqrt(1 + 4h²) > 2r.Suppose the circles have radius r. If we want to place circles on opposite sides such that they don't interfere, then sqrt(1 + (2h)^2) > 2r. But if the circles are very close to the line (h approaching 0), then sqrt(1 + 0) = 1 > 2r, so 2r < 1. Therefore, as long as the radius is less than 0.5 cm, then even if h approaches 0, the distance between centers is 1 cm, which is greater than 2*0.5 = 1 cm. Wait, but 1 cm is equal to 2*0.5 cm, so they would touch. Since the circles cannot touch, we need sqrt(1 + (2h)^2) > 2r. If h is zero, then we need 1 > 2r, so r < 0.5 cm. But even with h approaching zero, the vertical separation is negligible, so the centers are 1 cm apart horizontally. So if the circles have radius just under 0.5 cm, the distance between centers is 1 cm, which is just over twice the radius (which would be just under 1 cm). Therefore, they don't touch. Wait, actually, if each has radius 0.499 cm, then the distance between centers is 1 cm, which is 1 cm, and 2*0.499 is 0.998 cm. Since 1 > 0.998, the circles don't touch. Therefore, if we alternate circles above and below the line, each with radius approaching 0.5 cm, we can place a circle between every pair of adjacent points. Since there are 15 pairs between 16 points, that would give 15 circles. But we already had 16 circles for single points. But these two sets of circles might interfere with each other.Wait, for instance, if we have a circle around A₁ (radius < 0.5 cm) above the line, and then a circle between A₁ and A₂ below the line. The distance between their centers is 0.5 cm (from A₁ at 0 cm to the center between A₁ and A₂ at 0.5 cm). The vertical distance is h (say, 0.1 cm). Then the distance between centers is sqrt(0.5² + 0.1²) ≈ 0.5099 cm. The sum of radii is 0.5 (for the single-point circle) + 0.5 (for the two-point circle) = 1 cm. But the actual distance between centers is about 0.5099 cm, which is less than 1 cm, so the circles would intersect. Therefore, we can't have both a single-point circle and a two-point circle adjacent to it on opposite sides.This complicates things. So if we want to have both single-point circles and two-point circles, they need to be spaced out so that they don't interfere. Therefore, maybe we need to choose between single-point circles or two-point circles in certain regions.Alternatively, maybe the maximum number of circles isn't achieved by just single-point or two-point circles but a combination of different sizes. Let's think differently.Each circle must contain a unique subset of points. The subsets must be distinct. So if we can create circles that correspond to unique intervals or unique combinations of points, ensuring that the circles don't overlap.Another approach: Since the points are on a line, the subsets that can be enclosed by a circle are intervals. Because a circle in the plane can enclose a contiguous set of points on the line. Wait, is that necessarily true? Suppose we have a circle that encloses A₁ and A₃ but not A₂. Is that possible? Let's see. If you place a circle above the line, centered somewhere such that A₁ and A₃ are inside, but A₂ is outside. Given that A₂ is 1 cm from A₁ and 1 cm from A₃. If the circle has to enclose A₁ and A₃ but exclude A₂, the radius would have to be large enough to reach A₁ and A₃, but not reach A₂. Let's calculate. Suppose the center is somewhere along the perpendicular bisector of A₁ and A₃. The distance between A₁ and A₃ is 2 cm. The perpendicular bisector is at 1 cm from each. The center of the circle would be at (1, h) where h is the height above the line. The distance from the center to A₁ is sqrt(1² + h²), and the distance to A₂ is sqrt(0² + h²) = h. To include A₁ and A₃, the radius must be at least sqrt(1 + h²). To exclude A₂, the radius must be less than h + 0.5 cm? Wait, A₂ is at (1, 0). The distance from the center (1, h) to A₂ is h. So if the radius is greater than h, then A₂ would be inside the circle. Therefore, to exclude A₂, the radius must be less than h. But the radius also needs to be at least sqrt(1 + h²) to include A₁ and A₃. Therefore, we have sqrt(1 + h²) < h. But sqrt(1 + h²) is always greater than h for h > 0. Therefore, this is impossible. Therefore, you cannot have a circle that encloses A₁ and A₃ without enclosing A₂. Therefore, any circle enclosing two non-consecutive points must also enclose all the points in between. Therefore, the subsets of points enclosed by circles must be intervals. That's a crucial observation.Therefore, each circle must enclose a contiguous interval of points. Therefore, the possible subsets are all intervals [A_i, A_j] where 1 ≤ i ≤ j ≤ 16. The number of such intervals is 16*17/2 = 136. But due to the constraints of non-overlapping circles and unique subsets, we can't have all of them. So the problem reduces to selecting as many intervals as possible such that:1. The corresponding circles do not intersect or touch.2. Each interval is unique (which they inherently are if they are different intervals).But since the circles can't overlap, their corresponding intervals on the line must not overlap either? Wait, not necessarily. Because if the circles are placed on different sides of the line, their projections (intervals) can overlap as long as the circles themselves don't intersect. However, as we saw earlier, even if circles are on opposite sides, if their projections overlap too much, the circles might intersect.But given the earlier result that you can't have a circle enclosing non-consecutive points without including the ones in between, the intervals must be contiguous. So each circle is associated with a contiguous interval. Now, to maximize the number of circles, we need to arrange as many non-overlapping intervals as possible, considering that circles can be placed on either side of the line.Wait, but the problem allows circles to be placed anywhere in the plane, as long as they don't intersect or touch, and each contains at least one point. However, the key is that each circle must correspond to a unique interval. So even if two circles enclose the same interval but are placed on different sides, rule d) says that different circles must contain different sets of points. Therefore, two circles enclosing the same interval would have the same set of points inside, which is not allowed. Hence, each interval can be used at most once, regardless of placement.Therefore, the problem reduces to selecting a set of non-overlapping intervals (on the line), each corresponding to a unique interval of points, and arranging their corresponding circles such that the circles do not intersect or touch. However, since circles can be placed above or below the line, perhaps we can fit more circles by interleaving them on different sides.But the key constraint is that the circles cannot intersect or touch. So even if two intervals are adjacent on the line, if their corresponding circles are on opposite sides, they might not intersect. But we need to verify the distance.Suppose we have two intervals [A₁, A₂] and [A₃, A₄], placed on opposite sides of the line. The distance between the centers of these circles would be the distance between the midpoints of the intervals. For [A₁, A₂], midpoint is at 0.5 cm, and for [A₃, A₄], midpoint is at 3.5 cm. The horizontal distance is 3 cm. The vertical distance is, say, 2h if one is above and the other below. The actual distance between centers is sqrt(3² + (2h)²). If the circles have radii r₁ and r₂, then we need sqrt(9 + 4h²) > r₁ + r₂.But in order to enclose the intervals, the radii must be at least the distance from the center to the farthest point in the interval. For [A₁, A₂], the center is at 0.5 cm, so the radius must be just over 0.5 cm to include both A₁ and A₂. Similarly, for [A₃, A₄], the center is at 3.5 cm, radius just over 0.5 cm. Therefore, if h is significant, say 1 cm, then the vertical distance is 2 cm. Then the distance between centers is sqrt(3² + 2²) = sqrt(13) ≈ 3.6 cm. The sum of the radii is 0.5 + 0.5 = 1 cm. Since 3.6 > 1, they don't intersect. Therefore, circles on opposite sides of the line with non-overlapping intervals can be placed without intersecting, even if their intervals are adjacent.Wait, but in this case, the intervals [A₁, A₂] and [A₃, A₄] are separated by one point (A₃ is at 2 cm, but the interval [A₃, A₄] starts at 3 cm). Wait, actually, A₃ is at 2 cm, A₄ at 3 cm. So the midpoint of [A₃, A₄] is 2.5 cm. Wait, my mistake. Let's recast:Suppose we have two intervals: [A₁, A₂] (points at 0 cm and 1 cm) and [A₃, A₄] (points at 2 cm and 3 cm). The midpoints are 0.5 cm and 2.5 cm. The horizontal distance between centers is 2 cm. If we place one circle above the line at 0.5 cm and another below the line at 2.5 cm, the distance between centers is sqrt(2² + (2h)^2). If h is 0.5 cm (so vertical separation is 1 cm), then the distance is sqrt(4 + 1) = sqrt(5) ≈ 2.24 cm. The radii of both circles are 0.5 cm each, so the sum of radii is 1 cm. Since 2.24 > 1, the circles don't intersect. Therefore, even with a small vertical separation, circles on opposite sides with non-overlapping intervals can be placed without intersection.Therefore, the key is that intervals on the line can be arranged in such a way that their corresponding circles, placed alternately above and below the line, do not intersect. Therefore, we can model this problem similarly to interval scheduling, where each interval can be "colored" as either above or below, and intervals assigned the same color cannot overlap, but intervals on different colors can overlap as long as their circles don't intersect. However, due to the circles being on different sides, the exact constraints on overlap might be more permissive.But perhaps this is overcomplicating. Let's think recursively or in terms of combinatorics.If we can place circles on both sides of the line, then effectively, the maximum number of non-overlapping circles would be similar to twice the maximum number of non-overlapping intervals on one side. However, this might not be exact because intervals on opposite sides can be closer.But let's first consider the one-dimensional problem: what's the maximum number of non-overlapping intervals on the line, where each interval is a contiguous set of points. Since the points are discrete and spaced 1 cm apart, an interval from A_i to A_j corresponds to covering positions (i-1) cm to (j-1) cm. Wait, no. The points are at 0 cm, 1 cm, ..., 15 cm. So an interval [A_i, A_j] corresponds to covering from (i-1) cm to (j-1) cm. Wait, no: A₁ is at 0 cm, A₂ at 1 cm, ..., A₁₆ at 15 cm. So an interval from A_i to A_j covers the segment from (i-1) cm to (j-1) cm. Wait, actually, the positions are 0 to 15 cm, each 1 cm apart. So each point A_k is at (k-1) cm. Therefore, an interval [A_i, A_j] corresponds to the interval from (i-1) cm to (j-1) cm. So the length of the interval is (j - i) cm. To enclose this interval with a circle, the circle must cover from (i-1) cm to (j-1) cm. But how does the circle enclose this interval? The circle must be large enough to contain all points from A_i to A_j inside it.Alternatively, perhaps the circle is centered at the midpoint of the interval with radius equal to half the length plus some epsilon. For an interval of length L, the radius would be L/2 + ε. However, if the circle is placed close to the line, the vertical position affects the required radius. Wait, maybe a better way is to consider that the circle must enclose all points in the interval and exclude others. To enclose an interval [A_i, A_j], the circle must have all points A_i to A_j inside it and none outside. Since the points are on a straight line, the minimal circle enclosing [A_i, A_j] would have its center at the midpoint of the interval and radius just enough to cover the farthest points. However, if placed on the line, the radius would be (j - i)/2 cm. But if placed above or below the line, the radius can be slightly smaller. Wait, actually, the radius must be at least the distance from the center to the farthest point.Suppose the circle is centered at (midpoint, h) where midpoint is the center of the interval on the line, and h is the height above or below the line. The distance from the center to each endpoint of the interval is sqrt(((j - i)/2)^2 + h²). To enclose all points from A_i to A_j, the radius must be at least this distance. However, to exclude other points, the radius must be less than the distance to the next point outside the interval. For example, to exclude A_{i-1}, the radius must be less than the distance from the center to A_{i-1}, which is sqrt(( (midpoint - (i - 2)) )² + h²). Wait, this is getting complicated. Maybe a better approach is to consider that for a circle to enclose exactly the interval [A_i, A_j], it must be placed such that all points A_i to A_j are inside, and all others are outside. Given the points are on a line, the minimal enclosing circle for [A_i, A_j] is the circle with diameter from A_i to A_j. Any other circle enclosing these points would have to be at least as large. However, the circle with diameter A_iA_j would have radius (j - i)/2 cm and be centered at the midpoint. But this circle would pass through A_i and A_j, which violates rule c) that none of the points lie on the circle. Therefore, the radius must be slightly larger than (j - i)/2 cm. However, increasing the radius would cause the circle to include more points. Wait, no. If the circle is placed above the line, increasing the radius while keeping the center fixed would expand the circle in all directions, potentially including more points. Alternatively, if we increase h, the height, while adjusting the radius accordingly, maybe we can enclose only the desired points.This seems too vague. Maybe it's better to think that enclosing an interval [A_i, A_j] requires a circle that doesn't include any other points. Since the points are 1 cm apart, if the circle is placed near the line, the radius has to be just enough to include the interval but not reach the next point. For a single point A_i, the circle can have a radius approaching 0.5 cm. For two adjacent points, the circle needs a radius approaching sqrt(0.5² + h²), where h is the height, but as we saw earlier, if h is too small, the circle might include the next point. This is getting too complex. Perhaps there's a different approach.Since each circle must correspond to a unique interval, and the circles cannot overlap, the problem resembles interval scheduling where each "job" is an interval, and we need to select as many non-overlapping jobs as possible. However, in standard interval scheduling, you maximize the number of non-overlapping intervals, but here, we can use both sides of the line (above and below), so it's like two parallel lines where intervals can be scheduled on either side, as long as they don't overlap on the same side.But in our case, even if intervals are on different sides, they might still conflict if their circles are too close. However, earlier analysis suggested that if intervals are non-overlapping and placed on opposite sides, their corresponding circles can coexist without intersection. Therefore, the problem reduces to scheduling intervals on two parallel lines (above and below), where intervals on the same line must not overlap, but intervals on different lines can be adjacent or overlapping as long as their corresponding circles don't intersect.But how does the distance between intervals on different sides affect the possibility of intersection? If two intervals are adjacent on the line, say [A₁, A₂] on the top and [A₃, A₄] on the bottom, the distance between their centers is sqrt((2.5 - 0.5)^2 + (2h)^2) = sqrt(4 + 4h²). If h is large enough, this distance is greater than the sum of their radii.Wait, let's take specific examples. Suppose we have two intervals: [A₁, A₂] on top and [A₃, A₄] on bottom. The centers of these circles would be at (0.5, h) and (2.5, -h). The distance between them is sqrt((2.5 - 0.5)^2 + ( -h - h)^2) = sqrt(4 + (2h)^2). The radii of the circles: for [A₁, A₂], the radius must cover from 0 cm to 1 cm. If the center is at (0.5, h), the radius must be sqrt(0.5² + h²) to reach A₁ and A₂. Similarly, the radius for [A₃, A₄] is sqrt(0.5² + h²). The sum of the radii is 2*sqrt(0.25 + h²). The distance between centers is sqrt(4 + 4h²) = 2*sqrt(1 + h²). We need 2*sqrt(1 + h²) > 2*sqrt(0.25 + h²). Dividing both sides by 2: sqrt(1 + h²) > sqrt(0.25 + h²). Squaring both sides: 1 + h² > 0.25 + h² ⇒ 1 > 0.25, which is true. Therefore, the distance between centers is always greater than the sum of the radii, regardless of h. Therefore, circles on opposite sides can be placed adjacent to each other on the line without overlapping. Therefore, intervals on opposite sides can be adjacent or overlapping, but their corresponding circles will not intersect.Therefore, this means that we can double the number of non-overlapping intervals by placing them on both sides. However, the exact maximum number is not simply double because some intervals might be too large and interfere with others on the opposite side.Wait, but according to the above calculation, any two intervals on opposite sides can coexist regardless of their positions on the line. Because the distance between their centers will always be greater than the sum of their radii. Therefore, if we can schedule intervals on two separate lines (above and below), with no overlapping intervals on the same side, then the total number of circles would be twice the maximum number of non-overlapping intervals on one side.But what is the maximum number of non-overlapping intervals on one side? In standard interval scheduling, the maximum number is equal to the number of points if all intervals are single points, but since we can have longer intervals, it's more efficient to use longer intervals to cover more points with fewer circles. However, since we want to maximize the number of circles, we need to use as many small intervals as possible.Wait, but if we use single-point intervals, we can have 16 non-overlapping circles on one side. If we use two-point intervals, we can have 15 non-overlapping circles (each covering two points), but these would be adjacent and thus overlapping if placed on the same side. Wait, no. If you place a two-point interval between A₁ and A₂, then the next two-point interval would be between A₃ and A₄, etc. So with 16 points, you can have 8 non-overlapping two-point intervals. Similarly, four-point intervals would allow 4 non-overlapping intervals, etc. But to maximize the number, you need the smallest possible intervals. Hence, using single-point intervals allows 16 circles on one side, which is more than using two-point intervals. Therefore, on one side, the maximum number of non-overlapping intervals is 16 (each covering a single point). If we use two sides, we could have 16*2=32 circles. However, the answer is 31. So why the discrepancy?Wait, perhaps there's a constraint I missed. Let's recall rule d): Different circles contain different sets of points. So if you have a circle on the top covering A₁ and another on the bottom covering A₁, those are two different circles but both contain the same set (just A₁). Wait, no—the set is the same: {A₁}. Therefore, rule d) prohibits having two circles with the same set. Therefore, each set can be used only once, regardless of the side. Therefore, if you place a circle on the top covering A₁, you cannot place another circle on the bottom covering A₁. Because they both contain the same set {A₁}, which violates rule d).Ah, this is critical. So even if circles are on different sides, if they enclose the same set of points, they are considered duplicates and only one can be used. Therefore, each subset must be unique across all circles, regardless of their position. Therefore, the problem reduces to selecting as many unique intervals as possible such that their corresponding circles do not intersect or touch, with the additional constraint that each interval can be placed either above or below the line, but no two circles can have the same interval.Therefore, the maximum number of circles is the maximum number of non-overlapping intervals (considering both sides) where each interval is unique. However, since two circles on opposite sides can have overlapping intervals as long as their circles don't intersect, but their subsets (intervals) must be unique. Wait, no. If two circles have the same interval, even if placed on opposite sides, they have the same subset of points, violating rule d). Therefore, each interval can be used only once, on one side.Therefore, the problem becomes selecting a collection of intervals (contiguous subsets) such that:1. On each side (above and below), the intervals are non-overlapping.2. The same interval is not used on both sides.But to maximize the number of circles, we need to select as many intervals as possible, with the constraint that intervals on the same side don't overlap, and intervals on opposite sides can overlap but must be different intervals.This resembles a matching problem where we can have two sets of non-overlapping intervals (above and below) with no duplicates. The maximum number would then be the sum of the maximum number of non-overlapping intervals above and the maximum number below, minus any overlaps. But since they can overlap as long as they are different intervals, actually, it's simply the maximum number of non-overlapping intervals on one side plus the maximum number on the other side, provided all intervals are unique.However, since we can choose intervals on both sides, maybe we can get more than just twice the maximum of one side by cleverly choosing different intervals on each side. For example, if on one side we use single-point intervals, and on the other side we use two-point intervals, we might get more total circles.But let's analyze. If we use single-point intervals on one side, we can have 16 circles. On the other side, if we use two-point intervals, how many can we fit? Since two-point intervals take up 2 cm each and are spaced 1 cm apart (to avoid overlapping), but actually, on the line, two-point intervals [A₁, A₂], [A₃, A₄], etc., each take 2 points and are separated by 1 point. There are 8 such intervals in 16 points. So on the second side, we can have 8 circles. Total circles: 16 + 8 = 24. But 24 is less than 31.Alternatively, on the first side, use all single-point intervals (16), and on the second side, use all possible non-overlapping intervals of length 2 starting at odd positions: [A₂, A₃], [A₄, A₅], ..., [A₁₄, A₁₅]. That gives 7 intervals. Total 16 + 7 = 23. Still low.Alternatively, instead of single-point intervals on one side, use a mix of single and larger intervals. For example, on one side, use single-point intervals at every even position, and on the other side, use two-point intervals at odd positions. But this is getting convoluted.Alternatively, consider that the total number of possible intervals is 136, but we need to pick a subset where no two intervals overlap on the same side, and all intervals are unique. But this is too broad.Wait, perhaps there's a different approach. Since each circle must enclose at least one point, and each subset must be unique, the problem is equivalent to finding the maximum number of non-overlapping circles where each circle encloses a distinct interval of points. The key is that overlapping intervals can be placed on different sides as long as their corresponding circles don't intersect. But how to maximize this.An alternative perspective is that the problem is equivalent to covering the line with intervals (circles) such that each interval is unique, and the circles can be on either side but not overlapping. Since circles on different sides can be closer, we can model this as a graph where each node is an interval, and edges connect intervals that cannot be both selected (because their circles would intersect). Then the problem reduces to finding the maximum independent set in this graph. However, this graph is complex, and finding the maximum independent set is NP-hard, so we need a smarter approach.Another thought: since intervals on different sides can be placed without overlapping, the maximum number of circles is equal to the sum of the maximum number of intervals on each side, provided that intervals on different sides are distinct. Therefore, if we can partition all possible intervals into two groups (above and below) such that within each group, intervals are non-overlapping, then the total number would be the sum of the sizes of both groups. To maximize this sum, we need to partition the intervals into two sets of non-overlapping intervals. The maximum number of such intervals would be when we can arrange the intervals in two layers (like two colorings in interval graph coloring), where each layer is a set of non-overlapping intervals. For an interval graph, the chromatic number is equal to the maximum clique size. In our case, the maximum clique size is the maximum number of overlapping intervals at a single point. For example, at the center point, how many intervals overlap? For 16 points, the maximum number of intervals covering a single point is 16 (all intervals that include that point). Therefore, the chromatic number is 16, which is not helpful.Wait, perhaps this is not the right analogy. If we want to partition the intervals into two sets (above and below) such that within each set, intervals are non-overlapping, then the minimum number of sets needed is the chromatic number. If the chromatic number is 2, then we can have two sets, each with non-overlapping intervals, and the total number would be all intervals. But interval graphs are perfect graphs, and their chromatic number equals the maximum clique size. For interval graphs, the chromatic number is the maximum number of overlapping intervals at any point. If that maximum is, say, 8, then we need 8 colors. But in our case, with 16 points, the maximum number of intervals covering a single point is 16 (all intervals that include that point). For example, the middle point A₈ is covered by all intervals from [A₁, A₈] up to [A₈, A₁₆], which is 8 + 8 -1 = 15 intervals. Wait, no. For A₈ (position 7 cm), the number of intervals covering it is the number of intervals that start at or before A₈ and end at or after A₈. So starting points can be A₁ to A₈, and endpoints A₈ to A₁₆. Therefore, the number is 8 * 9 = 72 intervals. Wait, that's a lot. So the chromatic number would be 72, which is impractical. Therefore, this approach isn't helpful.Back to the problem. The answer is 31, which is one less than 32. 32 is 2^5, but not sure. Alternatively, 31 is the number of non-empty subsets for 5 elements, but we have 16 points. Alternatively, 31 is the maximum number of circles possible by some other construction.Wait, 16 points. If we consider all possible non-empty subsets, but we need them to be intervals. The number of non-empty intervals is 16*17/2 = 136. But due to non-overlapping constraints and unique subsets, we can't use all of them.But 31 is the answer. How?Wait, 31 is the number of non-empty intervals minus 1. 16*17/2 = 136, which is much larger. Not likely.Alternatively, 31 is the number of edges in a complete binary tree with 16 leaves? Not sure.Wait, 31 is 2⁵ - 1. Which is the number of nodes in a complete binary tree with 5 levels. Not sure how this relates.Wait, maybe the maximum number is 31 by using all possible single-point circles and all possible two-point circles alternately on both sides, but due to the constraints of rule d), some subsets can't be used. Wait, no. If we have single-point circles, they are 16. Two-point circles are 15. Total 31. But can we fit 16 + 15 = 31 circles without overlapping?Let me think. If we place all single-point circles on one side, and all two-point circles on the other side. But would they fit?Single-point circles on one side: 16 circles, each centered at the points, radius approaching 0.5 cm. Since they are spaced 1 cm apart, and radius < 0.5 cm, they don't overlap. So 16 circles on one side.Two-point circles on the other side: Each two-point circle is between two adjacent points, centered between them with radius approaching 0.5 cm. There are 15 such adjacent pairs (A₁-A₂, A₂-A₃, ..., A₁₅-A₁₆). Each of these circles has radius approaching 0.5 cm, centered at 0.5 cm, 1.5 cm, ..., 14.5 cm. The distance between centers is 1 cm, and the radius is 0.5 cm, so the circles would touch if placed on the same side. But if placed on the opposite side, as two-point circles on the bottom and single-point circles on top, then the distance between a two-point circle and a single-point circle is sqrt((0.5 cm)^2 + (2h)^2). If h is the vertical distance from the line to the circle center, then as long as h is greater than zero, even a small h, the distance between centers is sqrt(0.25 + 4h²). The sum of radii is 0.5 (for single-point) + 0.5 (for two-point) = 1 cm. Therefore, we need sqrt(0.25 + 4h²) > 1. Solving:sqrt(0.25 + 4h²) > 1Square both sides:0.25 + 4h² > 14h² > 0.75h² > 0.1875h > sqrt(0.1875) ≈ 0.433 cmSo if the vertical distance from the line to the circles is more than approximately 0.433 cm, then the circles on opposite sides don't intersect. Therefore, if we place single-point circles on top with h = 0.433 cm and two-point circles below with h = 0.433 cm, then all circles would be non-overlapping. However, the two-point circles on the bottom are spaced 1 cm apart, with radius 0.5 cm. The distance between adjacent centers is 1 cm, and each has radius 0.5 cm, so they would just touch. But the rule states circles cannot touch. Therefore, the radius must be less than 0.5 cm. If we reduce the radius slightly, say 0.49 cm, then the distance between centers (1 cm) is greater than the sum of radii (0.49 + 0.49 = 0.98 cm), so they don't touch. Therefore, by placing single-point circles on one side and two-point circles on the other side, we can have 16 + 15 = 31 circles.This matches the answer of 31. Additionally, we can check if this satisfies all the rules:a) Circles do not intersect or touch: Single-point circles are on one side, two-point circles on the other. The vertical distance ensures that single and two-point circles don't intersect. The two-point circles on their side have centers 1 cm apart with radii < 0.5 cm, so they don't touch. Similarly, single-point circles are spaced 1 cm apart with radii < 0.5 cm, so they don't touch.b) Each circle contains at least one point: Single-point circles each contain one point. Two-point circles each contain two points.c) None of the points lie on the circle: Since radii are less than 0.5 cm for single-point circles, the points are strictly inside. For two-point circles, the radius is slightly less than 0.5 cm, so the points are just inside.d) Different circles contain different sets of points: All single-point circles have unique singleton sets. All two-point circles have unique adjacent pairs. Since no two circles have the same set, this is satisfied.Therefore, the maximum number of circles is 31. It's 16 single-point circles on one side and 15 two-point circles on the other. This uses all possible single-point and adjacent two-point intervals, which are all unique and non-overlapping when placed on opposite sides with appropriate spacing. This seems to fit all the rules and maximize the number.Final Answerboxed{31}</think>Misha needs to construct the maximum number of circles according to the given rules. The key steps to determine the solution are as follows:1. Points and Intervals: There are 16 points spaced 1 cm apart. Each circle must enclose at least one point, and the circles must not intersect or touch. Each circle must contain a unique set of points.2. Single-Point Circles: Each point can be enclosed in a circle with a radius slightly less than 0.5 cm. These circles can be placed on one side of the line without overlapping. This gives 16 circles.3. Two-Point Circles: Adjacent points can be enclosed in circles centered between them with a radius slightly less than 0.5 cm. These circles can be placed on the opposite side of the line. There are 15 such adjacent pairs, giving 15 circles.4. Non-Overlapping and Unique Sets: The single-point circles and two-point circles are placed on opposite sides of the line. The vertical distance ensures they do not intersect, and each set of points enclosed by the circles is unique.By placing 16 single-point circles on one side and 15 two-point circles on the other side, Misha can construct a total of 31 circles without violating any rules.[boxed{31}],<think>Alright, let me try to figure out this problem. So, we have 16 points on a line, each 1 cm apart. Misha is constructing circles following some rules. The goal is to find the maximum number of circles she can make. The answer is supposed to be 31, but I need to understand why.First, let me restate the rules to make sure I get them:a) The circles can't intersect or touch each other. So, they have to be completely separate.b) Each circle must contain at least one of the points A1 to A16 inside it. So, no empty circles.c) None of these points can lie exactly on a circle. They have to be strictly inside or outside.d) Different circles must contain different sets of points. For example, if one circle has A1 and A2 inside, then no other circle can have exactly A1 and A2 inside. The sets of points inside each circle have to be unique.Okay, so the challenge is to arrange as many circles as possible without overlapping, each covering a unique subset of the points, with each subset being non-empty, and points can't be on the circle's boundary.Hmm. Since the points are on a line, spaced 1 cm apart, the circles can be placed either above or below the line. But they can't intersect, so their placement has to consider the distance between circles. Also, each circle must contain at least one point inside. Since points are on a straight line, the circles would have to be positioned such that their centers are near the line, but not exactly on it because if the center is on the line, the circle would be symmetric above and below. But maybe centers can be shifted to one side to include certain points.Wait, but actually, the problem doesn't specify that the circles have to be on one side of the line. They can be anywhere in the plane, as long as they don't intersect or touch, and each contains at least one point inside. However, the points are all on a straight line. So, to include a point inside the circle, the circle must be centered near that point with a radius large enough to enclose it but not so large that it intersects another circle. Also, different circles must have different subsets of points inside them.This seems related to set theory and combinatorics. The total number of non-empty subsets of 16 points is 2^16 - 1 = 65535. But obviously, we can't have that many circles because of the geometric constraints. So, the key is that each circle must correspond to a unique subset, and the circles can't overlap. But how does the geometry constrain the subsets?Wait, the problem says that different circles must contain different sets of points. So, each circle's set is unique. But the circles can't overlap. So, if two circles had overlapping regions, their subsets could potentially interfere, but since they can't intersect or touch, their subsets must be such that the regions they enclose are separate. But the points are all on a line, so maybe each circle can enclose a segment of the line, and the segments don't overlap. But circles are two-dimensional, so overlapping in the plane is different from overlapping on the line.But the problem states that the circles themselves cannot intersect or touch. So even if two circles are on opposite sides of the line, if they come too close, they might intersect. So, we need to place circles in such a way that they are all non-overlapping, each contains at least one point, and each has a unique subset of points.Wait, but how does the subset of points inside each circle relate to their positions? If two circles are on the same side of the line, their regions might overlap if they are too close. But if they're on opposite sides, maybe they can be closer without overlapping. But the main thing is that each circle must contain a unique subset of points, and they can't intersect.I need to think of a way to maximize the number of circles. The answer is 31, which is one less than 32, which is 2^5. But 16 points, so maybe there's a connection here. Wait, 31 is also 2^5 - 1. Hmm. 5 is half of 10, but not sure. Maybe there's a binary representation here.Alternatively, maybe each circle can correspond to a unique non-empty subset of points, but due to the geometric constraints, overlapping subsets can't be used. But how?Alternatively, think of the points on a line, each 1 cm apart. If we can place circles such that each circle contains exactly one point, then the maximum number would be 16. But the answer is 31, which is more than 16, so that can't be it. Wait, but 31 is less than 65535, so maybe it's about intervals. If each circle encloses an interval of points, but since the circles can't intersect, the intervals must be non-overlapping. But how many non-overlapping intervals can you have on 16 points? Well, that would be 16 choose 2, which is 120, but again, the answer is 31.Wait, perhaps the key is in rule d): different circles contain different sets of points. So even if two circles enclose the same set of points, they can't both exist. So each circle must correspond to a unique subset. However, the problem is that if two circles enclose different subsets but their regions on the line overlap, the circles might intersect in the plane. So, we need a way to represent unique subsets with circles that don't intersect.Alternatively, if you can encode each subset as a circle in such a way that no two circles intersect, which would require that the subsets are nested or something. But with 16 points, the maximum number of nested subsets is 16, but 31 is more.Wait, 31 is 2^5 -1. 5 is interesting here. Maybe each circle is associated with a binary code of length 5, which can represent 32 different subsets, minus 1 gives 31. But how does that relate to the 16 points?Alternatively, think of the line with 16 points as a binary tree structure. Each circle can represent a node in the tree, and the maximum number of nodes in a binary tree with 5 levels is 31. But I'm not sure how that connects.Alternatively, think of each circle as covering a unique interval. But intervals on a line can be arranged in a way that they are non-overlapping. The maximum number of non-overlapping intervals on 16 points would be 8 (each interval covering two points), but that's not 31. Wait, perhaps it's not intervals but individual points with different combinations, but arranged in such a way that the circles don't intersect.Wait, maybe each circle can be placed in such a way that they are in different "levels" above or below the line, each level corresponding to a different subset. But how?Alternatively, think of each circle containing a single point. But each point can be in multiple circles? But rule b) says each circle must contain at least one point, but different circles can contain overlapping points. But rule d) says different circles must contain different sets of points. So, if two circles contain the same set of points, they can't both exist. So, each circle must have a unique subset. Therefore, the problem reduces to packing as many unique subsets as possible with circles that do not intersect.But how does the geometry constrain the subsets? For instance, if a circle contains point A1 and A2, then another circle that contains A1, A2, and A3 would have to be placed in a way that doesn't intersect the first circle. But if the first circle is close to A1 and A2, the second circle would need to be larger to include A3, which might cause an overlap. Alternatively, if the second circle is placed on the other side of the line, maybe it can include A1, A2, A3 without overlapping with the first circle.This is getting a bit complicated. Let me approach it differently. Let's think of each circle as a region in the plane that contains some points. Since the points are on a straight line, maybe we can model the circles as intervals on the line, but extended into the plane.But circles in the plane can enclose multiple points along the line. For a circle to enclose certain points, those points must lie within the circle's radius. Since the points are 1 cm apart, the radius required to include a point depends on the distance from the center of the circle to the point.To minimize the radius and avoid overlapping, perhaps the circles are placed just above each point, with a radius slightly less than 0.5 cm, so that each circle contains exactly one point. But such circles would be very small and non-overlapping since the points are 1 cm apart. The distance between centers would be 1 cm, and the radius is less than 0.5 cm, so circles won't intersect. However, each circle would contain exactly one point, and since each circle's set is unique, we can have 16 such circles. But the answer is 31, so this approach is missing something.Wait, but maybe you can also have circles that contain multiple points. If a circle is larger, it can include multiple points, but you have to ensure that no two circles overlap. So, if you have a mix of small circles around single points and larger circles covering multiple points, you might be able to fit more circles.However, each circle must have a unique subset. So, for example, a circle containing A1 and A2 is different from a circle containing A1, A2, and A3, and different from a circle containing just A1. The challenge is to arrange these circles such that they don't intersect and each has a unique subset.But how to maximize the number of such circles. The answer is 31, which is 2^5 - 1. That suggests a binary structure. Maybe each circle corresponds to a binary string of length 5, but 2^5 is 32, minus 1 gives 31. How does this relate to 16 points?Alternatively, since 16 is 2^4, maybe each point can be assigned a 4-bit binary code, and the circles correspond to 5-bit codes? Not sure.Alternatively, think of each circle as a hypercube in 4 dimensions, but that seems unrelated.Wait, maybe the problem is related to the number of intervals you can have on 16 points such that no two intervals are the same. But the answer is 31, which is the number of non-empty intervals in a line with 5 points. Wait, no. For n points, the number of intervals is n(n+1)/2. For 16 points, that's 16*17/2=136. So that's not 31.Alternatively, maybe it's the number of non-empty subsets minus something. The total subsets are 2^16 -1, which is way too big.Wait, maybe the problem is similar to a binary tree where each internal node represents a circle. A full binary tree with 16 leaves has 15 internal nodes, but 15 is not 31.Alternatively, maybe it's the number of edges in a complete graph with 16 nodes, which is 120, but again, not 31.Wait, 31 is the number of non-empty proper subsets for a set of 5 elements. But we have 16 elements here. Maybe I'm overcomplicating.Alternatively, think of each circle as a bit in a binary number. 31 is 11111 in binary, which is 5 bits. Not sure.Wait, going back to the problem statement. The answer is 31. Maybe this is a known result. It's possible that the maximum number of circles is 2n - 1 where n is the number of points. But 2*16 -1 is 31. Wait, 2*16 is 32, minus 1 is 31. So perhaps that's the formula. But why?Wait, if for each point, you can have two circles: one on the left and one on the right? But 16 points would give 32 positions, minus 1 for some overlap. But how?Alternatively, each circle can be associated with a gap between two points. There are 15 gaps between 16 points. If you place a circle in each gap, but you can have two circles per gap (above and below the line), giving 30 circles. Then maybe one more somewhere, total 31. But the answer is 31, so maybe that's the reasoning.Wait, between each pair of consecutive points, there's a 1 cm gap. If you place a circle in each gap, centered between two points, with radius less than 0.5 cm, so it doesn't reach the adjacent gaps. If you place such a circle above the line, and another below the line, then for each of the 15 gaps, you can have 2 circles. That gives 30 circles. Then maybe one additional circle at the end, but there are only 16 points. The endpoints: if you place a circle to the left of A1 or to the right of A16, but those circles would only contain A1 or A16 inside. But if you place a circle to the left of A1 with radius just big enough to include A1 but not reach A2, and similarly to the right of A16. So that's two more circles. But 30 + 2 = 32. But the answer is 31. Maybe you can't have both left and right at the ends? Hmm.Wait, but if you place a circle to the left of A1, it can only contain A1, and similarly a circle to the right of A16 can only contain A16. If you do both, that's two circles. But if you place circles between the gaps, each gap can have two circles (above and below), giving 15*2=30. So total 32. But the answer is 31. Maybe there's an overlap at the ends? If you place a circle to the left of A1 and a circle to the right of A16, maybe those circles can't coexist without overlapping? But if they are placed on opposite sides of the line, they wouldn't overlap. Hmm.Alternatively, maybe the problem counts the two circles at the ends as one? No, that doesn't make sense.Wait, maybe the answer comes from considering that for n points on a line, the maximum number of non-overlapping circles each containing at least one point, with unique subsets, is 2n - 1. So, for 16 points, 2*16 -1 = 31. That could be the formula. But why?Let me think for smaller n. For example, n=1: one point. How many circles? Only one circle containing that point. According to the formula, 2*1 -1 =1, which matches.n=2: two points, spaced 1 cm apart. The formula gives 3. Let's see: we can have two circles each containing one point, placed on either side, and one circle containing both points. But the circle containing both points would have to be large enough to encompass both, but then if the two single-point circles are placed near each point, would they overlap with the big circle?If the two single-point circles are placed close to their respective points with radius less than 0.5 cm, they wouldn't reach the other point or the big circle. The big circle would have to be centered between the two points with radius slightly more than 0.5 cm to include both. If placed above the line, the big circle wouldn't interfere with the small circles below. So maybe you can have three circles. So 2*2 -1 =3, which works.Similarly, for n=3: 2*3 -1=5. Let's check. Each single point can have a small circle, two circles between the gaps (but with two circles each), and one big circle covering all three. Wait, maybe:- Three small circles, each containing one point.- Two medium circles between the gaps, each containing two points.But the medium circles would have to not overlap with the small circles. If the small circles are on one side and the medium circles on the other, they might not overlap. Then you could also have a large circle containing all three. That's 3 + 2 +1 =6. But the formula says 5. Hmm, contradiction. Wait, maybe the large circle can't be added because it would overlap with the medium circles? If the large circle is on the same side as the medium circles, it might overlap. So maybe the maximum is 5.Alternatively, arrange three small circles on one side, two medium circles on the other side. That's 5 circles. Then you can't add the large circle because it would overlap with the medium ones. So maybe 5. Which matches 2n -1=5. So perhaps the formula holds.Therefore, generalizing, for n points, the maximum number of circles is 2n -1. Thus, for 16 points, it's 31. Hence, the answer is 31.But let me check n=3 again. If you have three small circles (each around a single point) on one side, and two larger circles (each covering two adjacent points) on the other side. That's 5 circles. Then if you try to add a circle covering all three points, where would you put it? If you place it on a different side, maybe above, while the others are below. Wait, but the circle covering all three would need to have a radius large enough to cover all three points. The points are spaced 1 cm apart. The distance from the center to the farthest point would be 1 cm (if centered at the middle point). So radius 1 cm. But the small circles around each point have radius less than 0.5 cm. So the big circle's radius is 1 cm, placed above the line, while the small circles are below. The big circle would extend 1 cm above, but the small circles are 0.5 cm below. So they don't overlap. So maybe you can have the big circle as well, totaling 6. But according to the formula, it should be 5. So maybe the formula is incorrect?Alternatively, maybe the issue is that the circle covering all three points would have a different subset. If subsets must be unique, but the big circle's subset is {A1, A2, A3}, which is unique. So why can't we have it?Wait, perhaps the problem is that the circle covering all three points would overlap with the medium circles. For example, if the medium circles are covering two points each, placed on the other side. If a medium circle is between A1 and A2, placed above the line, with radius slightly more than 0.5 cm to include both A1 and A2. Then a big circle covering all three points, placed below the line, would have to be large enough to include A1, A2, A3. The distance from the center (maybe at A2) to A1 and A3 is 1 cm each. So radius 1 cm. But the medium circle above A1 and A2 has a radius of say 0.6 cm. The vertical distance between the centers is the distance above and below the line. If the circles are placed with some vertical separation, maybe they don't overlap. The vertical distance would need to be more than the sum of the radii. If the medium circle has radius 0.6 cm above, and the big circle has radius 1 cm below, the vertical distance between their centers would need to be more than 0.6 +1 =1.6 cm. If we place the circles sufficiently far apart vertically, they won't overlap. So maybe it's possible to have all 6 circles. Then the formula 2n -1 doesn't hold.Hmm, this is confusing. Maybe the answer isn't a general formula but relates to the specific arrangement for 16 points.Alternatively, think of each circle as being associated with a unique non-empty subset of points, but arranged in such a way that the circles don't overlap. To ensure non-overlapping, the subsets must be such that their corresponding circles are placed in different regions. If we can assign each subset to a unique position in the plane where the circle doesn't intersect others, then the maximum number is the number of subsets. But obviously, that's too high.But given the answer is 31, which is 2^5 -1, maybe there's a connection to binary representations where each circle is identified by a 5-bit identifier, allowing 31 unique combinations (excluding the zero). But how does this relate to the 16 points?Alternatively, think of the 16 points as being in a 4-dimensional hypercube, but that's abstract.Wait, another approach: since the circles cannot intersect, their centers must be at least some distance apart. The minimal distance between centers depends on their radii. To maximize the number of circles, we need to minimize the radii and place them as densely as possible.But each circle must contain at least one point. If we make circles as small as possible, just containing one point each, then we can have 16 circles. But since we can have different subsets, maybe we can interleave circles containing different numbers of points.Wait, but each circle must have a unique subset. If we have a circle containing A1, another containing A2, etc., that's 16 circles. Then, circles containing two points: A1 and A2, A2 and A3, etc. But each of these would need to be placed such that they don't intersect the single-point circles. If the single-point circles are on one side and the two-point circles on the other, maybe they don't intersect. Then, for two-point circles, there are 15 possible adjacent pairs. Then three-point circles, but they would need to be placed... This might escalate, but considering the non-overlapping constraint, perhaps the number is limited.But 16 +15=31, which matches the answer. So maybe the maximum number is 16 single-point circles and 15 two-point circles, totaling 31. Let's verify.If we place all single-point circles on one side of the line, each with radius less than 0.5 cm, spaced 1 cm apart. Then, their centers are 1 cm apart, radii less than 0.5 cm, so they don't intersect. Then, on the other side of the line, place circles that cover two adjacent points. Each two-point circle would need to be centered between two points, with radius slightly more than 0.5 cm to include both points. The distance between centers of these two-point circles is 1 cm (since they're between adjacent points), and their radius is about 0.5 cm. Wait, if the radius is just over 0.5 cm, then the circles between A1-A2 and A2-A3 would be centered 1 cm apart, each with radius ~0.5 cm, so they would touch or overlap. Therefore, this doesn't work.Alternatively, place the two-point circles further apart. If you skip a gap between each two-point circle, but then you can't have 15 of them. Wait, there are 15 gaps between 16 points. If you place a two-point circle in every other gap, you get 8 circles, but that's not 15.Alternatively, place the two-point circles on the other side with a larger vertical distance so that they don't interfere horizontally. But circles on the same horizontal line would still have their horizontal separation. If you have a two-point circle centered between A1 and A2, and another between A2 and A3, the distance between their centers is 1 cm. If each has a radius slightly over 0.5 cm, they would overlap. Therefore, you can't place adjacent two-point circles on the same side.Therefore, to place two-point circles without overlapping, they need to be spaced such that the distance between their centers is more than the sum of their radii. If each two-point circle has a radius of, say, 0.6 cm, then the required distance between centers is more than 1.2 cm. Since the points are 1 cm apart, the centers of the two-point circles are 1 cm apart. Therefore, even with radius 0.6 cm, they would overlap. Hence, adjacent two-point circles on the same side can't coexist.Therefore, to place two-point circles on one side, you can only place non-consecutive ones. For example, between A1-A2, then skip A2-A3, place between A3-A4, etc. This would give 8 two-point circles. But 8 is much less than 15. Therefore, this approach won't reach 31.Wait, but maybe if we alternate sides. Place a two-point circle between A1-A2 on the top, then next between A2-A3 on the bottom, then A3-A4 on top, etc. Since the vertical separation would prevent them from overlapping. The vertical distance needs to be more than the sum of their radii. If the circles have a radius of 0.6 cm, then vertical separation should be more than 1.2 cm. If we place them alternately on top and bottom with sufficient vertical spacing, they might not overlap. In that case, you could place a two-point circle in every gap, alternating sides, totaling 15 two-point circles. Then, also place 16 single-point circles on the other side. Wait, but then the total would be 15 +16=31. That matches the answer.Wait, let's see. If you have 16 single-point circles on one side (say, the top), each centered at each point with radius just under 0.5 cm. Then, on the bottom side, you place two-point circles between each pair of consecutive points. Each two-point circle is centered between two points, radius slightly over 0.5 cm. The vertical distance between the single-point circles (top) and two-point circles (bottom) would need to be sufficient so that they don't intersect. If the vertical distance is more than the sum of the radii, say single-point circles have radius 0.4 cm and two-point circles have radius 0.6 cm, then vertical distance needs to be more than 1.0 cm. If that's satisfied, they don't intersect. Similarly, the two-point circles on the bottom are spaced 1 cm apart horizontally, but since they're on the same vertical level, their horizontal separation is 1 cm, and their radii are 0.6 cm, so the distance between centers is 1 cm, and the sum of radii is 1.2 cm, which would mean they don't overlap. Wait, no. If two circles are 1 cm apart horizontally and each has radius 0.6 cm, the distance between centers is 1 cm, and the sum of radii is 1.2 cm. Since 1 cm < 1.2 cm, the circles would overlap. So even on the bottom, adjacent two-point circles would overlap. Therefore, this approach doesn't work.Therefore, placing two-point circles on the bottom would require them to be spaced such that their centers are at least 1.2 cm apart. Since the points are 1 cm apart, the centers of the two-point circles are 1 cm apart. Therefore, even on the bottom, adjacent two-point circles would overlap. So this approach fails.Alternative idea: Maybe use both sides but interleave single and double circles. For example, place a single-point circle at A1 on top, a two-point circle between A1-A2 on bottom, a single-point circle at A2 on top, etc. But then the vertical distance would need to be enough to prevent the single and two-point circles from overlapping. If the vertical distance is sufficient, maybe this works. But horizontally, the two-point circles are between the points, so their centers are offset by 0.5 cm from the single-point circles. The distance between a single-point circle at A1 (radius 0.4 cm) and a two-point circle between A1-A2 (radius 0.6 cm) would be sqrt((0.5)^2 + (vertical distance)^2). To prevent overlap, this distance must be greater than 0.4 + 0.6 = 1.0 cm. Solving sqrt(0.25 + d^2) >1.0 => 0.25 + d^2 >1 => d^2 >0.75 => d > sqrt(0.75) ≈0.866 cm. So if the vertical distance between the single-point and two-point circles is more than 0.866 cm, they don't overlap. Assuming that's possible, then this arrangement could work. Then, for 16 points, you can place 16 single-point circles on top and 15 two-point circles on the bottom, totaling 31. Since each two-point circle is between two points, offset vertically, and doesn't overlap with the single-point circles or other two-point circles. This way, all circles are non-overlapping, each contains a unique subset (single points or adjacent pairs), and all subsets are different. Thus, satisfying all the rules.Therefore, the maximum number is 31. This matches the given answer.Final Answerboxed{31}</think>Misha needs to construct the maximum number of circles according to the given rules. The key steps to determine the maximum number of circles are as follows:1. Understanding the Rules: - Circles must not intersect or touch. - Each circle must contain at least one of the 16 points. - No points can lie on the circle's boundary. - Each circle must contain a unique set of points.2. Geometric Constraints: - The points are spaced 1 cm apart on a line. - Circles must be placed such that they do not overlap or touch, even if they are on different sides of the line.3. Strategy: - Use single-point circles: Place each circle around a single point with a radius slightly less than 0.5 cm to avoid overlapping. This allows 16 circles. - Use two-point circles: Place circles between each pair of consecutive points on the opposite side of the line. These circles must have a radius slightly more than 0.5 cm to include two points but not overlap with other circles. This allows 15 circles.4. Vertical Separation: - Single-point circles are placed on one side of the line. - Two-point circles are placed on the opposite side, ensuring sufficient vertical distance to avoid overlapping with single-point circles and each other.5. Result: - The total number of circles is the sum of single-point and two-point circles: (16 + 15 = 31).Thus, the maximum number of circles Misha can construct is (boxed{31}).]

🔑:Solution: We can represent such a system of circles as a tree with 16 leaves.In such a tree, there cannot be more than 31 nodes. It is easy to construct a binary tree in which there are exactly 31 nodes.

🧠:[<think>Okay, let's see. The problem says that the terminal side of angle α is in the third quadrant. Then we need to find the value set of this expression: [sin(α/2)/|sin(α/2)|] + [cos(α/2)/|cos(α/2)|]. The options are A, B, C, D. Hmm. Alright, let's start breaking it down.First, since α is in the third quadrant. That means α is between π and 3π/2 radians, right? Because the quadrants go like this: first is 0 to π/2, second π/2 to π, third π to 3π/2, and fourth 3π/2 to 2π. So if α is in the third quadrant, then α is between π and 3π/2.Now, we have to consider α/2. Let's find out where α/2 would be. If α is between π and 3π/2, then dividing by 2, α/2 would be between π/2 and 3π/4. Wait, π/2 is 90 degrees, and 3π/4 is 135 degrees. So α/2 is in the second quadrant. That's important because the signs of sine and cosine in the second quadrant are positive for sine and negative for cosine. Because in the second quadrant, sine is positive and cosine is negative.So, sin(α/2) would be positive, and cos(α/2) would be negative. Therefore, the expression [sin(α/2)/|sin(α/2)|] would be sin(α/2) divided by its absolute value, which is 1, since it's positive. Similarly, [cos(α/2)/|cos(α/2)|] would be cos(α/2) divided by its absolute value, which is -1, because cosine is negative in the second quadrant.Therefore, adding those two together: 1 + (-1) = 0. Wait, but the answer options include A: {-2, 2}, B: {0,2}, C: {0}, D: {0,-2,2}. So according to this reasoning, the answer would be 0. But option C is {0}, but also option B and D include 0. Wait, but maybe my reasoning is incomplete?Hold on, maybe I need to check if α/2 is always in the second quadrant. Let me verify the interval again. If α is in the third quadrant, then α is between π and 3π/2. Therefore, α/2 is between π/2 and 3π/4. Yes, that's between 90 degrees and 135 degrees, which is indeed the second quadrant. So α/2 is in the second quadrant, so sine is positive, cosine is negative.So then sin(α/2)/|sin(α/2)| is 1, and cos(α/2)/|cos(α/2)| is -1, so the sum is 0. Therefore, the value should always be 0. So the value set is {0}, which is option C. But wait, the options have other choices. Maybe there's a mistake here.Wait, but maybe there's a case when sin(α/2) or cos(α/2) is zero? But in the second quadrant, sine is positive, but not zero because α/2 is between π/2 and 3π/4. So sin(α/2) is positive, not zero, and cos(α/2) is negative, not zero. Therefore, both denominators |sin(α/2)| and |cos(α/2)| are non-zero, so the expression is defined.Hence, the expression is always 1 + (-1) = 0. So the value set is {0}, which is option C. But let me check again. Maybe I messed up the quadrant?Wait, α is in the third quadrant, so α is between π and 3π/2. Then α/2 is between π/2 and 3π/4. So π/2 is 90 degrees, 3π/4 is 135 degrees. So yes, in the second quadrant. Therefore, sin positive, cos negative. So the ratios are 1 and -1. Sum is 0.Wait, but maybe the problem is not about α being in the third quadrant, but the terminal side of angle α is in the third quadrant. That's the same as α being in the third quadrant, right? Because the terminal side's quadrant defines the angle's quadrant.Alternatively, maybe angles can be more than 2π? But unless specified, we usually take α between 0 and 2π. But if α is allowed to be any angle coterminal with the third quadrant, then α could be π + 2πk to 3π/2 + 2πk for integer k. Then α/2 would be π/2 + πk to 3π/4 + πk. Depending on k, but since the trigonometric functions are periodic with period 2π, so adding 2π to α would add π to α/2. But the sine and cosine of α/2 would have the same values as when k=0. Wait, no. For example, if k=1, α would be from 3π to 7π/2. Then α/2 would be from 3π/2 to 7π/4. Then α/2 would be in different quadrants. But in that case, the original angle α's terminal side is still in the third quadrant only if k=0. Because angles are determined modulo 2π. So, if α is in the third quadrant, it's equivalent to α ∈ (π, 3π/2) + 2πk, but when you divide by 2, it's α/2 ∈ (π/2, 3π/4) + πk. So depending on k, α/2 could be in different quadrants. Wait, but the problem says "the terminal side of angle α is in the third quadrant", which usually means the principal value, i.e., α ∈ (π, 3π/2). So unless otherwise stated, we can assume α is in (π, 3π/2), so α/2 is in (π/2, 3π/4). So, as before, in the second quadrant.Therefore, sin(α/2) is positive, cos(α/2) is negative. Therefore, the expression is 1 + (-1) = 0. So the value set is {0}, which is option C. Therefore, the answer is C. But the options include A, B, C, D, with C being {0}. But let me check again. Maybe there's a case where α/2 can be in another quadrant?Wait, suppose α is exactly π. Then α is on the boundary of the third quadrant. But the problem states the terminal side is in the third quadrant, so α is strictly between π and 3π/2. So α/2 is strictly between π/2 and 3π/4. So still second quadrant. So no, even then, all α/2 are in the second quadrant. Therefore, sin(α/2) is positive, cos(α/2) is negative. Therefore, the expression is 0. So the answer should be C: {0}.But let me check if the problem is maybe different. Wait, the expression is [sin(α/2)/|sin(α/2)|] + [cos(α/2)/|cos(α/2)|]. So that's essentially sign(sin(α/2)) + sign(cos(α/2)). Since sin(α/2) is positive, its sign is +1, and cos(α/2) is negative, its sign is -1. So 1 + (-1) = 0. Therefore, regardless of the exact value of α in the third quadrant, as long as α/2 is in the second quadrant, the expression is always 0. So the value set is {0}. Therefore, answer is C.But why are there other options? Let me check if I made a mistake in determining the quadrant of α/2. Wait, suppose α is in the third quadrant. Let's take an example. Let α = 4π/3, which is in the third quadrant. Then α/2 = 2π/3, which is 120 degrees, in the second quadrant. Sin(2π/3) is √3/2, positive; cos(2π/3) is -1/2, negative. So the expression is (√3/2)/(√3/2) + (-1/2)/|-1/2| = 1 + (-1) = 0.Another example: α = 5π/4, wait, 5π/4 is in the third quadrant? Wait, 5π/4 is 225 degrees, which is in the third quadrant. Then α/2 = 5π/8, which is 112.5 degrees, still in the second quadrant. Sin(5π/8) is positive, cos(5π/8) is negative. So the expression is 1 + (-1) = 0.Another example: α = 3π/2. But 3π/2 is on the boundary between third and fourth quadrants. However, the problem states the terminal side is in the third quadrant, so α must be less than 3π/2. So let's take α approaching 3π/2 from below. Then α/2 approaches 3π/4 from below. So still in the second quadrant. So even as α approaches 3π/2, α/2 approaches 3π/4, which is 135 degrees, still in the second quadrant. So sin remains positive, cos remains negative. Therefore, expression is 0.Therefore, in all cases where α is in the third quadrant, α/2 is in the second quadrant, leading to the expression being 0. So the value set is {0}, so option C.But hold on, the answer options don't have C as a common answer. Wait, maybe the problem is written in Chinese, and the options are translated. Anyway, the options are A: {-2, 2}, B: {0,2}, C: {0}, D: {0,-2,2}. So according to my analysis, the answer is C. But maybe I need to check again.Wait, maybe I misapplied the quadrant for α/2. Let's see. Suppose α is in the third quadrant, so π < α < 3π/2. Then π/2 < α/2 < 3π/4. That's correct. So α/2 is in the second quadrant. So sin(α/2) is positive, cos(α/2) is negative. Therefore, each term in the expression is 1 and -1, so the sum is 0. Therefore, no matter what α is in the third quadrant, the expression is always 0. Therefore, the value set is {0}, so option C.But why is option A: {-2, 2} there? Maybe if the angle was in a different quadrant, but here it's third quadrant. Let's see. Suppose α is in the first quadrant. Let's test α = π/3. Then α/2 = π/6, in the first quadrant. Both sin and cos positive. Then the expression is 1 + 1 = 2. If α is in the second quadrant, say α = 2π/3. Then α/2 = π/3, still first quadrant. Wait, that's not right. Wait, α = 2π/3 is in the second quadrant, then α/2 = π/3, first quadrant. Then expression is 1 + 1 = 2. If α is in the fourth quadrant, say α = 7π/4. Then α/2 = 7π/8, which is in the fourth quadrant. Then sin(7π/8) is positive, cos(7π/8) is negative. Wait, in the fourth quadrant, sine is negative and cosine is positive. Wait, no: fourth quadrant: x positive, y negative. So cosine (x-coordinate) is positive, sine (y-coordinate) is negative. Therefore, sin(7π/8) is negative, cos(7π/8) is positive. So expression would be -1 + 1 = 0. Hmm.But the original problem is about α in the third quadrant. So maybe in different quadrants, the value set changes, but here, it's fixed.Wait, but hold on, what if α is such that α/2 is in a different quadrant? For example, if α is 5π, which is coterminal with π, which is in the third quadrant? Wait, no. 5π is equivalent to π, but π is actually on the negative x-axis, between second and third quadrants. Wait, π is the boundary between second and third quadrants. Wait, angles in the third quadrant are strictly between π and 3π/2. So if α is 5π, which is π + 2π*2, so equivalent to π, but π is not in the third quadrant. So maybe my initial reasoning holds.Alternatively, maybe the problem is not considering α in the principal range. Suppose α is in the third quadrant, but after multiple rotations. For example, α = 5π, which is equivalent to π, but that's on the negative x-axis, not in any quadrant. If α = 7π/2, which is equivalent to 7π/2 - 2π*1 = 3π/2, which is the negative y-axis, also not in a quadrant. So, if α is in the third quadrant, then it's between π and 3π/2. So α/2 is between π/2 and 3π/4, which is the second quadrant.Therefore, regardless of how many times you rotate, as long as α's terminal side is in the third quadrant, α is between π and 3π/2 when measured from standard position. Therefore, α/2 is between π/2 and 3π/4. So sin positive, cosine negative. Therefore, expression is 0.Therefore, the answer should be C: {0}. But let me check the answer options again. The options are A: {-2,2}, B: {0,2}, C: {0}, D: {0,-2,2}. So unless the problem is written incorrectly, the answer is C. But maybe I need to check if there is a different interpretation.Wait, perhaps the problem is referring to the set of possible values, not just the single value. For example, if in some cases the expression can be different. Let me think.If α is in the third quadrant, then α can be expressed as π + θ, where θ is between 0 and π/2. Therefore, α/2 = π/2 + θ/2. So θ/2 is between 0 and π/4. Therefore, α/2 is between π/2 and 3π/4. So in this interval, sin is positive and decreasing from 1 to √2/2, and cosine is negative and increasing from 0 to -√2/2. Wait, cosine is negative and going from 0 to -√2/2? Wait, no. Wait, at π/2, cosine is 0. At 3π/4, cosine is -√2/2. So cosine is decreasing from 0 to -√2/2? Wait, no, it's going from 0 to more negative. So cosine is negative in this entire interval except at π/2 where it's 0. But θ is between 0 and π/2, so θ/2 is between 0 and π/4, so α/2 is between π/2 and π/2 + π/4 = 3π/4. Therefore, cosine is negative throughout, except at π/2 where it's 0. But since α is strictly in the third quadrant, θ is between 0 and π/2, so α/2 is strictly between π/2 and 3π/4. Therefore, cos(α/2) is strictly negative, never zero. Therefore, the expression is always 1 + (-1) = 0.Therefore, there's no variation here. The expression is always 0. Therefore, the value set is {0}. So answer is C.But why is there an option A: {-2, 2}? Maybe if the angle was in a different quadrant. For example, if α was in the fourth quadrant, then α/2 could be in the second or something else. But the problem specifies the third quadrant. So maybe the options are meant to trick you. Therefore, I think the correct answer is C.But let me check another example. Let's take α = 5π/3. Wait, 5π/3 is in the fourth quadrant. No, 5π/3 is 300 degrees, which is in the fourth quadrant. But if α is 4π/3, which is 240 degrees, third quadrant. Then α/2 is 2π/3, 120 degrees, second quadrant. Then sin(2π/3) is √3/2, positive; cos(2π/3) is -1/2, negative. So expression is 1 + (-1) = 0. Same result.Another example: α = 5π/4, which is 225 degrees. Then α/2 = 5π/8, which is 112.5 degrees, second quadrant. Sin positive, cos negative. Expression is 0.So in all cases, the expression is 0. Therefore, the value set is {0}, so answer C.But the options include A, B, C, D, and C is {0}, so that must be the answer. Therefore, I think the answer is C.Final Answerboxed{C}</think>Given that the terminal side of angle ( alpha ) is in the third quadrant, we need to determine the value set of the expression ( dfrac{sin dfrac{alpha}{2}}{left|sin dfrac{alpha}{2}right|} + dfrac{cos dfrac{alpha}{2}}{left|cos dfrac{alpha}{2}right|} ).1. Quadrant Analysis: - Since ( alpha ) is in the third quadrant, ( pi < alpha < dfrac{3pi}{2} ). - Dividing by 2, we get ( dfrac{pi}{2} < dfrac{alpha}{2} < dfrac{3pi}{4} ). This places ( dfrac{alpha}{2} ) in the second quadrant.2. Signs of Trigonometric Functions: - In the second quadrant, sine is positive and cosine is negative. - Therefore, ( sin dfrac{alpha}{2} > 0 ) and ( cos dfrac{alpha}{2} < 0 ).3. Evaluating the Expression: - ( dfrac{sin dfrac{alpha}{2}}{left|sin dfrac{alpha}{2}right|} = 1 ) because the numerator and denominator are both positive. - ( dfrac{cos dfrac{alpha}{2}}{left|cos dfrac{alpha}{2}right|} = -1 ) because the numerator is negative and the denominator is positive (absolute value).4. Summing the Results: - The expression simplifies to ( 1 + (-1) = 0 ).Thus, regardless of the specific value of ( alpha ) in the third quadrant, the expression always evaluates to 0. Therefore, the value set is ( {0} ).[boxed{C}],<think>Okay, let's try to solve this problem step by step. The question is about finding the value set of the expression: [frac{sin dfrac{α}{2}}{left|sin dfrac{α}{2}right|} + frac{cos dfrac{α}{2}}{left|cos dfrac{α}{2}right|}]given that the terminal side of angle α is in the third quadrant. The options are A: {-2, 2}, B: {0, 2}, C: {0}, D: {0, -2, 2}. First, I need to recall that when an angle is in the third quadrant, its measure is between π and 3π/2 radians (or 180° and 270°). So α is in (π, 3π/2). Then, when we divide α by 2, the resulting angle α/2 will be in what quadrant? Let's compute the range for α/2.If α is between π and 3π/2, then dividing all parts by 2 gives:π/2 < α/2 < 3π/4So α/2 is between π/2 and 3π/4. Wait, π/2 is 90°, and 3π/4 is 135°, so α/2 is in the second quadrant (between 90° and 180°). That means the angle α/2 is in the second quadrant. Now, in the second quadrant, sine is positive and cosine is negative. So sin(α/2) is positive, cos(α/2) is negative. The expression given is:[frac{sin dfrac{α}{2}}{left|sin dfrac{α}{2}right|} + frac{cos dfrac{α}{2}}{left|cos dfrac{α}{2}right|}]Since sin(α/2) is positive, the absolute value of sin(α/2) is just sin(α/2). Therefore, the first term simplifies to sin(α/2)/sin(α/2) = 1. Similarly, cos(α/2) is negative, so the absolute value of cos(α/2) is -cos(α/2) (since absolute value makes it positive). Therefore, the second term is cos(α/2)/(-cos(α/2)) = -1. Therefore, the entire expression becomes 1 + (-1) = 0. Wait, but the answer choices don't have only 0. Option C is {0}, but the options also include others. Wait, but maybe I need to check if there are other possibilities. But hold on, the angle α is in the third quadrant, so α is between π and 3π/2, which makes α/2 between π/2 and 3π/4, as we said. So α/2 is strictly in the second quadrant. Therefore, both sin(α/2) and cos(α/2) will always be positive and negative respectively, right? But wait, in the second quadrant, sine is positive and cosine is negative. So regardless of the exact value of α, as long as α is in the third quadrant, α/2 is in the second quadrant, so sin(α/2) is positive, cos(α/2) is negative. Therefore, each term in the expression is 1 and -1, so the sum is 0. Therefore, the value set should only contain 0. So the answer is C: {0}. But wait, the options given include other possibilities. Let me check again. Wait, perhaps I made a mistake in determining the quadrant of α/2. Let me verify. If α is in the third quadrant, then π < α < 3π/2. Dividing by 2, we get π/2 < α/2 < 3π/4. π/2 is 90°, 3π/4 is 135°, so indeed, the angle α/2 is in the second quadrant. So sin(α/2) is positive, cos(α/2) is negative. So the first term, sin divided by its absolute value is 1, and the second term, cos divided by its absolute value is -1. So 1 + (-1) = 0. Therefore, the expression is always 0, so the value set is {0}, which is option C. But why does the problem mention "value set"? Maybe there is a case where the denominators could be zero? But sin(α/2) and cos(α/2) can't be zero in the second quadrant. Because in the second quadrant, angles between 90° and 180°, sin is positive, cos is negative, but neither is zero. For example, sin(90°) is 1, which is not zero, and cos(90°) is 0, but 90° is the boundary. Wait, but α is strictly greater than π (180°), so α/2 is strictly greater than π/2 (90°), so α/2 is in (π/2, 3π/4), so between 90° and 135°, which is entirely in the second quadrant, not including the endpoints. Therefore, sin(α/2) is positive but not zero, cos(α/2) is negative but not zero. Therefore, denominators are safe, no division by zero. Thus, the expression is always 0. Therefore, the value set should be {0}, which is option C. But looking at the options again, the original problem lists the options as:A: {-2, 2}, B: {0, 2}, C: {0}, D: {0, -2, 2}So if my reasoning is correct, the answer should be C. But why is the question phrased as "value set"? Maybe I need to check again if there's a case where α/2 could be in a different quadrant. Wait, suppose α is exactly 3π/2. Then α/2 is 3π/4, which is 135°, still in the second quadrant. If α is approaching π (180°), then α/2 approaches π/2 (90°). But since α is strictly greater than π, α/2 is strictly greater than π/2. Therefore, α/2 is always in the second quadrant, not reaching the boundaries where sine or cosine could be zero or change sign. Therefore, in all cases, the expression is 1 - 1 = 0. Therefore, the value set is {0}, so the answer is C. But let me check if there is a different interpretation. For example, sometimes with quadrants, people might consider reference angles or something else. Wait, the problem says "the terminal side of angle α is in the third quadrant". So α is an angle whose terminal side is in QIII. That means α can be written as π + θ, where θ is between 0 and π/2. So α = π + θ, θ ∈ (0, π/2). Then α/2 = π/2 + θ/2. Since θ is between 0 and π/2, θ/2 is between 0 and π/4. Therefore, α/2 is between π/2 and 3π/4, which is still QII. So same conclusion. Therefore, regardless of θ, sin(α/2) is positive, cos(α/2) is negative, leading to the sum being 0. Therefore, the answer is C. Wait, but maybe the problem is considering different branches of half-angles? For example, when you take α/2, depending on how you represent α, but α is given to be in QIII, so α is between π and 3π/2. Therefore, α/2 is uniquely determined between π/2 and 3π/4. So there's no ambiguity here. Alternatively, if the problem had not specified the terminal side, but just said α is in QIII, then sometimes angles can be coterminal, but since they mentioned terminal side is in QIII, that means α is in QIII, so the angle is between π and 3π/2. Therefore, I think the answer is C. But why do the options include other possibilities? Maybe I made a mistake. Let me think again. Wait, let's compute an example. Suppose α is 4π/3, which is in QIII (since π < 4π/3 < 3π/2). Then α/2 is 2π/3, which is 120°, in QII. Then sin(2π/3) = √3/2 (positive), cos(2π/3) = -1/2 (negative). So the expression is (√3/2)/(√3/2) + (-1/2)/(1/2) = 1 + (-1) = 0. Another example: take α = 5π/4 (Wait, 5π/4 is actually in QIII? Wait, 5π/4 is between π (≈3.14) and 3π/2 (≈4.712). Yes, 5π/4 is 225°, which is in QIII. Then α/2 is 5π/8, which is 112.5°, in QII. Then sin(5π/8) is positive, cos(5π/8) is negative. So again, the expression is 1 + (-1) = 0. Another example: take α approaching π from above, say α = π + ε, where ε is a small positive angle. Then α/2 = π/2 + ε/2, which is just slightly above π/2, so still in QII. So sin(α/2) approaches sin(π/2) = 1 (positive), cos(α/2) approaches cos(π/2) = 0 (but negative since we're slightly above π/2). So the expression is still 1 + (-1) = 0. Similarly, if α approaches 3π/2 from below, α/2 approaches 3π/4 from below, which is still in QII. So sin(3π/4) is √2/2 (positive), cos(3π/4) is -√2/2 (negative), so again the expression is 1 + (-1) = 0. Therefore, in all cases where α is in QIII, the expression equals 0. Therefore, the value set is {0}, which is option C. But wait, the answer given in the options is C, but the original problem says "value set", which typically refers to all possible values the expression can take. Since in all cases it's 0, the value set is {0}. But the options given include A: {-2, 2}, which would be if both terms were ±1 each, but in opposite signs. But in our case, since sin(α/2) is always positive and cos(α/2) is always negative, the terms are +1 and -1 always. Therefore, their sum is always 0. So {0} is the correct answer. Therefore, I think the correct answer is C. Final Answerboxed{C}</think>Given that the terminal side of angle ( alpha ) is in the third quadrant, we need to determine the value set of the expression:[frac{sin dfrac{alpha}{2}}{left|sin dfrac{alpha}{2}right|} + frac{cos dfrac{alpha}{2}}{left|cos dfrac{alpha}{2}right|}]First, we note that if ( alpha ) is in the third quadrant, then ( pi < alpha < frac{3pi}{2} ). Dividing by 2, we find the range for ( frac{alpha}{2} ):[frac{pi}{2} < frac{alpha}{2} < frac{3pi}{4}]This places ( frac{alpha}{2} ) in the second quadrant. In the second quadrant, sine is positive and cosine is negative. Therefore, ( sin frac{alpha}{2} > 0 ) and ( cos frac{alpha}{2} < 0 ).Considering the given expression, we evaluate each term separately:- The term ( frac{sin dfrac{alpha}{2}}{left|sin dfrac{alpha}{2}right|} ) simplifies to ( 1 ) because the numerator and denominator are both positive.- The term ( frac{cos dfrac{alpha}{2}}{left|cos dfrac{alpha}{2}right|} ) simplifies to ( -1 ) because the numerator is negative and the denominator is the absolute value (positive).Adding these results, we get:[1 + (-1) = 0]Thus, the value of the expression is always 0, regardless of the specific value of ( alpha ) in the third quadrant. Therefore, the value set is ({0}).The final answer is (boxed{C}).]

🔑:Analysis: This question mainly examines the concept of quadrant angles and simplification for evaluation. Given that the terminal side of α is in the third quadrant, we have 2kπ+π < α < 2kπ+ dfrac{3π}{2}, therefore kπ+ dfrac{π}{2} < dfrac{α}{2} < kπ+ dfrac{3π}{4}. When dfrac{α}{2} is in the second quadrant, dfrac{sin dfrac{α}{2}}{left|sin dfrac{α}{2}right|}+ dfrac{cos dfrac{α}{2}}{|cos dfrac{α}{2}|}=1-1=0; When dfrac{α}{2} is in the fourth quadrant, dfrac{sin dfrac{α}{2}}{left|sin dfrac{α}{2}right|}+ dfrac{cos dfrac{α}{2}}{|cos dfrac{α}{2}|}=-1+1=0, Therefore, the correct choice is boxed{C}.

🧠:[<think>Okay, so I need to find the variance of a random variable X that's uniformly distributed over the interval [α, β]. Hmm, variance... right, variance measures how spread out the values are. For a uniform distribution, since all values in the interval are equally likely, I think the variance should depend on the length of the interval. The longer the interval, the more spread out the data, so the higher the variance. That makes sense.First, I remember that the variance of a random variable is calculated as Var(X) = E[X²] - (E[X])². So I need to find the expected value of X squared and subtract the square of the expected value of X. Let me write that down: Var(X) = E[X²] - (E[X])². Now, for a uniform distribution on [α, β], the probability density function (pdf) is constant. The pdf f(x) should be 1/(β - α) for x in [α, β], and zero otherwise. Let me confirm that: yes, because the total area under the pdf must be 1, so the height is 1 divided by the length of the interval.Next, I need to find E[X], the expected value. For a uniform distribution, the mean is just the average of the endpoints, right? So E[X] should be (α + β)/2. Let me verify that by integrating x times the pdf over the interval. E[X] = ∫ from α to β of x * [1/(β - α)] dx. So that's [1/(β - α)] * [x²/2] evaluated from α to β. Plugging in, we get [1/(β - α)] * [(β²/2 - α²/2)] = [1/(β - α)] * [(β² - α²)/2]. Factoring β² - α² as (β - α)(β + α), we get [1/(β - α)] * (β - α)(β + α)/2 = (β + α)/2. Yep, that's correct. So the expected value is (α + β)/2.Now, moving on to E[X²]. This is the integral of x² times the pdf over the interval. So E[X²] = ∫ from α to β of x² * [1/(β - α)] dx. Let's compute that. Factor out the 1/(β - α), so we have [1/(β - α)] * ∫ x² dx from α to β. The integral of x² is x³/3. Evaluating from α to β gives [β³/3 - α³/3] = (β³ - α³)/3. So E[X²] = [ (β³ - α³) / 3 ] / (β - α) = (β³ - α³) / [3(β - α)].Hmm, now I need to simplify β³ - α³. Remember that β³ - α³ factors into (β - α)(β² + αβ + α²). Let me check that: (β - α)(β² + αβ + α²) = β³ + β²α + βα² - αβ² - α²β - α³ = β³ - α³. Yes, that's correct. So substituting back:E[X²] = [ (β - α)(β² + αβ + α²) ] / [3(β - α)] = (β² + αβ + α²)/3. Okay, so E[X²] is (β² + αβ + α²)/3. Now, the variance is E[X²] - (E[X])². Let's compute that.First, (E[X])² is [(α + β)/2]² = (α + β)² / 4 = (α² + 2αβ + β²)/4.Now subtract this from E[X²]:Var(X) = (β² + αβ + α²)/3 - (α² + 2αβ + β²)/4.To subtract these fractions, let's get a common denominator, which is 12.So, multiply numerator and denominator of the first term by 4: [4(β² + αβ + α²)] / 12.Multiply the second term by 3: [3(α² + 2αβ + β²)] / 12.So the variance becomes [4β² + 4αβ + 4α² - 3α² - 6αβ - 3β²] / 12.Let's combine like terms:4β² - 3β² = β²4αβ - 6αβ = -2αβ4α² - 3α² = α²So the numerator is α² - 2αβ + β². Wait, that's the same as (β - α)². Because (β - α)² = β² - 2αβ + α². So numerator is (β - α)², and denominator is 12. Therefore, Var(X) = (β - α)² / 12.So the variance of a uniform distribution on [α, β] is (β - α) squared divided by 12. Let me just check if that makes sense. If α and β are closer together, the variance is smaller, which is correct. If the interval is larger, the variance is larger. And the division by 12, that's a standard result I think. Wait, I recall that for the standard uniform distribution on [0,1], the variance is 1/12. Which fits into this formula: (1 - 0)^2 / 12 = 1/12. So that checks out. Therefore, the final answer should be (β - α)² divided by 12. So Var(X) = (β - α)² / 12.Final AnswerThe variance of the random variable ( X ) is boxed{dfrac{(beta - alpha)^2}{12}}.</think>To find the variance of a random variable ( X ) that has a uniform distribution on the interval ([alpha, beta]), we start by noting that the probability density function (pdf) is ( f(x) = frac{1}{beta - alpha} ) for ( x in [alpha, beta] ).First, we calculate the expected value ( E[X] ):[E[X] = int_{alpha}^{beta} x cdot frac{1}{beta - alpha} , dx = frac{1}{beta - alpha} left[ frac{x^2}{2} right]_{alpha}^{beta} = frac{1}{beta - alpha} left( frac{beta^2 - alpha^2}{2} right) = frac{alpha + beta}{2}]Next, we calculate the expected value ( E[X^2] ):[E[X^2] = int_{alpha}^{beta} x^2 cdot frac{1}{beta - alpha} , dx = frac{1}{beta - alpha} left[ frac{x^3}{3} right]_{alpha}^{beta} = frac{1}{beta - alpha} left( frac{beta^3 - alpha^3}{3} right)]Factoring (beta^3 - alpha^3) as ((beta - alpha)(beta^2 + alphabeta + alpha^2)), we get:[E[X^2] = frac{1}{beta - alpha} cdot frac{(beta - alpha)(beta^2 + alphabeta + alpha^2)}{3} = frac{beta^2 + alphabeta + alpha^2}{3}]The variance is then calculated as ( text{Var}(X) = E[X^2] - (E[X])^2 ):[(E[X])^2 = left( frac{alpha + beta}{2} right)^2 = frac{(alpha + beta)^2}{4} = frac{alpha^2 + 2alphabeta + beta^2}{4}][text{Var}(X) = frac{beta^2 + alphabeta + alpha^2}{3} - frac{alpha^2 + 2alphabeta + beta^2}{4}]Finding a common denominator (12) and simplifying:[text{Var}(X) = frac{4(beta^2 + alphabeta + alpha^2) - 3(alpha^2 + 2alphabeta + beta^2)}{12} = frac{alpha^2 - 2alphabeta + beta^2}{12} = frac{(beta - alpha)^2}{12}]The variance of the random variable ( X ) is (boxed{dfrac{(beta - alpha)^2}{12}}).,<think>Okay, so I need to find the variance of a random variable X that's uniformly distributed over the interval [α, β]. Hmm, let's start by recalling what a uniform distribution is. If X is uniformly distributed over [α, β], that means the probability density function (pdf) is constant over this interval and zero elsewhere. The pdf should be 1/(β - α) between α and β, right? Because the total area under the pdf has to be 1, so the height of the rectangle is 1 divided by the width (β - α). Now, variance is a measure of spread, and for continuous distributions, it's calculated as the expected value of (X - μ)², where μ is the mean or expected value of X. So, I need to first find the expected value μ, then compute E[(X - μ)²]. Let me start by finding μ. For a uniform distribution on [α, β], the mean is just the average of the endpoints, right? So μ = (α + β)/2. That seems familiar. I should verify that by integrating x times the pdf over [α, β]. Let's do that:μ = E[X] = ∫ from α to β of x * (1/(β - α)) dxIntegrating x gives (1/2)x² evaluated from α to β, multiplied by 1/(β - α):μ = (1/(β - α)) * [ (1/2)β² - (1/2)α² ) ] = (1/(2(β - α))) * (β² - α²)Factor β² - α² as (β - α)(β + α):μ = (1/(2(β - α))) * (β - α)(β + α) = (β + α)/2Yep, that checks out. So the mean is indeed (α + β)/2. Now, moving on to variance. The formula is Var(X) = E[X²] - (E[X])². So I need to compute E[X²] first. Let's do that.E[X²] = ∫ from α to β of x² * (1/(β - α)) dxIntegrating x² gives (1/3)x³ evaluated from α to β, multiplied by 1/(β - α):E[X²] = (1/(β - α)) * [ (1/3)β³ - (1/3)α³ ) ] = (1/(3(β - α))) * (β³ - α³)Again, factor β³ - α³. Remember that β³ - α³ = (β - α)(β² + αβ + α²). Let me confirm that:Yes, (β - α)(β² + αβ + α²) = β³ + β²α + βα² - αβ² - α²β - α³ = β³ - α³. The middle terms cancel out. So,E[X²] = (1/(3(β - α))) * (β - α)(β² + αβ + α²) = (β² + αβ + α²)/3Okay, so E[X²] is (β² + αβ + α²)/3. Now, Var(X) = E[X²] - (E[X])². We already have E[X] = (α + β)/2, so (E[X])² = (α + β)²/4.Let's compute that:Var(X) = (β² + αβ + α²)/3 - (α + β)²/4First, expand (α + β)^2: α² + 2αβ + β². So,Var(X) = (β² + αβ + α²)/3 - (α² + 2αβ + β²)/4To subtract these fractions, let's get a common denominator, which is 12.Multiply numerator and denominator of the first term by 4:= [4(β² + αβ + α²)]/12 - [3(α² + 2αβ + β²)]/12Now subtract the numerators:= [4β² + 4αβ + 4α² - 3α² - 6αβ - 3β²]/12Combine like terms:4β² - 3β² = β²4αβ - 6αβ = -2αβ4α² - 3α² = α²So the numerator becomes β² - 2αβ + α²Which is equal to (β - α)^2Therefore,Var(X) = (β - α)^2 / 12So the variance is (β - α)^2 divided by 12. Wait, let me check if that makes sense. If the interval is [0,1], then variance should be (1 - 0)^2 /12 = 1/12. Which I remember is the variance for the standard uniform distribution. That checks out. Another example: if the interval is [a, a + h], then the variance is h²/12. That seems right. Alternatively, I can recall that for uniform distribution, variance is (β - α)^2 /12. So, this must be the answer. But just to be thorough, let's verify once again. Alternatively, sometimes people remember the variance formula for uniform distribution as (β - α)^2 /12. So, that seems correct. So, putting it all together, after computing E[X²] and subtracting (E[X])², we end up with (β - α)^2 /12. Therefore, the variance is (β - α)^2 over 12. Final AnswerThe variance of a random variable ( X ) with a uniform distribution on ( [alpha, beta] ) is boxed{dfrac{(beta - alpha)^2}{12}}.</think>To find the variance of a random variable ( X ) that has a uniform distribution on the interval ([alpha, beta]), we start by noting the probability density function (pdf) of ( X ) is (frac{1}{beta - alpha}) for ( alpha leq x leq beta ).First, we calculate the expected value ( mu ):[mu = E[X] = int_{alpha}^{beta} x cdot frac{1}{beta - alpha} , dx]Evaluating the integral:[mu = frac{1}{beta - alpha} left[ frac{1}{2}x^2 right]_{alpha}^{beta} = frac{1}{beta - alpha} left( frac{1}{2}beta^2 - frac{1}{2}alpha^2 right) = frac{beta^2 - alpha^2}{2(beta - alpha)} = frac{(beta - alpha)(beta + alpha)}{2(beta - alpha)} = frac{alpha + beta}{2}]Next, we find ( E[X^2] ):[E[X^2] = int_{alpha}^{beta} x^2 cdot frac{1}{beta - alpha} , dx]Evaluating the integral:[E[X^2] = frac{1}{beta - alpha} left[ frac{1}{3}x^3 right]_{alpha}^{beta} = frac{1}{beta - alpha} left( frac{1}{3}beta^3 - frac{1}{3}alpha^3 right) = frac{beta^3 - alpha^3}{3(beta - alpha)}]Factoring the numerator:[beta^3 - alpha^3 = (beta - alpha)(beta^2 + alphabeta + alpha^2)]Thus,[E[X^2] = frac{(beta - alpha)(beta^2 + alphabeta + alpha^2)}{3(beta - alpha)} = frac{beta^2 + alphabeta + alpha^2}{3}]The variance is given by ( text{Var}(X) = E[X^2] - (E[X])^2 ):[text{Var}(X) = frac{beta^2 + alphabeta + alpha^2}{3} - left( frac{alpha + beta}{2} right)^2]Expanding and simplifying:[text{Var}(X) = frac{beta^2 + alphabeta + alpha^2}{3} - frac{alpha^2 + 2alphabeta + beta^2}{4}]Finding a common denominator (12):[text{Var}(X) = frac{4(beta^2 + alphabeta + alpha^2) - 3(alpha^2 + 2alphabeta + beta^2)}{12}]Simplifying the numerator:[4beta^2 + 4alphabeta + 4alpha^2 - 3alpha^2 - 6alphabeta - 3beta^2 = beta^2 - 2alphabeta + alpha^2 = (beta - alpha)^2]Thus,[text{Var}(X) = frac{(beta - alpha)^2}{12}]The variance of a random variable ( X ) with a uniform distribution on ([alpha, beta]) is boxed{dfrac{(beta - alpha)^2}{12}}.]

🔑:Solution. Using formula (2.5.15) and taking into account formulas (3.4.3) and (3.4.7), we find 204begin{aligned}D(X) & =int_{-infty}^{+infty}(x-a)^{2} p(x) d x=int_{-infty}^{alpha}(x-a)^{2} p(x) d x+int_{alpha}^{beta}(x-a)^{2} p(x) d x+ & +int_{beta}^{+infty}(x-a)^{2} p(x) d x=int_{alpha}^{beta} frac{(x-a)^{2}}{beta-alpha} d x=left.frac{1}{beta-alpha} cdot frac{(x-a)^{3}}{3}right|_{alpha} ^{beta}= & =left.frac{left(x-frac{alpha+beta}{2}right)^{3}}{3(beta-alpha)}right|_{alpha} ^{beta}=frac{left(beta-frac{alpha+beta}{2}right)^{3}-left(alpha-frac{alpha+beta}{2}right)^{3}}{3(beta-alpha)}=frac{(beta-alpha)^{2}}{12}end{aligned}Thus,D(X)=frac{(beta-alpha)^{2}}{12}

🧠:[<think>Okay, so I need to solve this system of differential equations. Let me see. The system is:x' = -4x -14yy' = 5x +13yHmm, right. These are linear differential equations with constant coefficients. I remember that for systems like this, one common method is to rewrite them in matrix form and then find the eigenvalues and eigenvectors of the coefficient matrix. That should allow me to write the general solution. Let me try that.First, let me write the system in matrix form. The equations can be represented as:d/dt [x; y] = [ -4 -14 5 13 ] [x; y]So the coefficient matrix A is:A = [ -4 -14 5 13 ]Now, to find the eigenvalues of A. Eigenvalues are found by solving the characteristic equation det(A - λI) = 0.Let me compute the determinant of (A - λI):| -4 - λ -14 || 5 13 - λ |The determinant is (-4 - λ)(13 - λ) - (-14)(5)Calculating that:First multiply (-4 - λ)(13 - λ):= (-4)(13) + (-4)(-λ) + (-λ)(13) + (-λ)(-λ)= -52 + 4λ -13λ + λ²= λ² -9λ -52Then subtract (-14)(5) which is -70, but since it's subtracted, it becomes +70.So determinant is λ² -9λ -52 +70 = λ² -9λ +18So the characteristic equation is λ² -9λ +18 = 0Let me solve this quadratic equation. The discriminant is D = 81 - 72 = 9So eigenvalues λ = [9 ± sqrt(9)] / 2 = [9 ± 3]/2Hence, λ1 = (9 +3)/2 = 12/2 = 6λ2 = (9 -3)/2 = 6/2 = 3So eigenvalues are 6 and 3. Both are real and distinct. That's good; it means we can find eigenvectors for each and the general solution will be a combination of e^(λt) times eigenvectors.Now, let's find eigenvectors for each eigenvalue.First, for λ1 = 6.We need to solve (A -6I)v = 0Compute A -6I:[ -4 -6 -14 ] = [ -10 -14 ][ 5 13 -6 ] = [5 7 ]So the matrix is:[ -10 -14 5 7 ]Let me write the equations:-10v1 -14v2 = 05v1 +7v2 = 0Note that the second equation is just the first divided by -2. So they are dependent.Let me use the second equation: 5v1 +7v2 =0. Let's solve for v1 in terms of v2:5v1 = -7v2 => v1 = (-7/5)v2So an eigenvector can be written as [v1; v2] = [ -7/5 v2; v2 ] = v2 [ -7/5; 1 ]We can choose v2 =5 to eliminate the fraction, giving the eigenvector [ -7; 5 ]So for λ1=6, an eigenvector is [-7; 5]Now for λ2=3.Compute A -3I:[ -4 -3 -14 ] = [ -7 -14 ][ 5 13 -3 ] = [5 10 ]So the matrix is:[ -7 -14 5 10 ]Again, equations:-7v1 -14v2 =05v1 +10v2 =0These equations are also dependent. Let's take the first one:-7v1 =14v2 => v1 = -2v2So eigenvectors are [v1; v2] = [ -2v2; v2 ] = v2 [ -2; 1 ]Choosing v2=1 gives the eigenvector [ -2; 1 ]So for λ2=3, an eigenvector is [ -2; 1 ]Therefore, the general solution of the system is:[ x(t); y(t) ] = C1*e^(6t)*[ -7; 5 ] + C2*e^(3t)*[ -2; 1 ]Where C1 and C2 are constants determined by initial conditions.Let me check my work to make sure I didn't make any mistakes.First, eigenvalues:Characteristic equation λ² -9λ +18=0. Solved roots at 6 and 3. Correct.Eigenvector for λ=6:Matrix A -6I: [ -10 -14; 5 7 ]Equations: -10v1 -14v2 =0 and 5v1 +7v2=0. So v1 = (-7/5)v2. So eigenvector proportional to [-7,5]. Correct.Eigenvector for λ=3:Matrix A -3I: [ -7 -14;5 10 ]Equations: -7v1 -14v2=0 => v1=-2v2. So eigenvector [-2,1]. Correct.So the general solution seems right. Hence, the solution is a linear combination of these two eigenvectors multiplied by their respective exponential terms.Alternatively, writing out the components:x(t) = -7*C1*e^(6t) -2*C2*e^(3t)y(t) = 5*C1*e^(6t) +1*C2*e^(3t)Yes, that looks correct.I think that's the solution. Let me double-check by plugging into the original equations.Suppose x(t) = -7C1 e^(6t) -2C2 e^(3t)Then x'(t) = -7*6 C1 e^(6t) -2*3 C2 e^(3t) = -42 C1 e^(6t) -6 C2 e^(3t)Now, according to the first equation: x' = -4x -14yCompute -4x -14y:-4*(-7C1 e^(6t) -2C2 e^(3t)) -14*(5C1 e^(6t) +C2 e^(3t))= 28 C1 e^(6t) +8 C2 e^(3t) -70 C1 e^(6t) -14 C2 e^(3t)= (28 -70)C1 e^(6t) + (8 -14)C2 e^(3t)= (-42 C1 e^(6t) -6 C2 e^(3t))Which matches x'(t). Good.Similarly, check y':y(t) =5C1 e^(6t) + C2 e^(3t)y'(t)=30 C1 e^(6t) +3 C2 e^(3t)According to the second equation: y' =5x +13yCompute 5x +13y:5*(-7C1 e^(6t) -2C2 e^(3t)) +13*(5C1 e^(6t) +C2 e^(3t))= -35 C1 e^(6t) -10 C2 e^(3t) +65 C1 e^(6t) +13 C2 e^(3t)= ( -35 +65 )C1 e^(6t) + ( -10 +13 )C2 e^(3t)=30 C1 e^(6t) +3 C2 e^(3t)Which matches y'(t). Perfect. So the solution satisfies both equations. Therefore, the general solution is correct.Final AnswerThe general solution is[boxed{begin{cases}x(t) = -7 C_1 e^{6t} - 2 C_2 e^{3t} y(t) = 5 C_1 e^{6t} + C_2 e^{3t}end{cases}}]where ( C_1 ) and ( C_2 ) are constants.</think>To solve the system of differential equations:[left{begin{array}{l}x' = -4x - 14y, y' = 5x + 13y,end{array}right.]we first rewrite the system in matrix form:[frac{d}{dt} begin{bmatrix} x y end{bmatrix} = begin{bmatrix} -4 & -14 5 & 13 end{bmatrix} begin{bmatrix} x y end{bmatrix}.]Next, we find the eigenvalues of the coefficient matrix ( A = begin{bmatrix} -4 & -14 5 & 13 end{bmatrix} ) by solving the characteristic equation (det(A - lambda I) = 0):[det begin{bmatrix} -4 - lambda & -14 5 & 13 - lambda end{bmatrix} = (-4 - lambda)(13 - lambda) - (-14)(5) = lambda^2 - 9lambda + 18 = 0.]The roots of the characteristic equation are:[lambda = frac{9 pm 3}{2} implies lambda_1 = 6 quad text{and} quad lambda_2 = 3.]For (lambda_1 = 6), we solve ((A - 6I)mathbf{v} = 0):[begin{bmatrix} -10 & -14 5 & 7 end{bmatrix} begin{bmatrix} v_1 v_2 end{bmatrix} = 0 implies 5v_1 + 7v_2 = 0 implies v_1 = -frac{7}{5}v_2.]Choosing (v_2 = 5), we get the eigenvector (begin{bmatrix} -7 5 end{bmatrix}).For (lambda_2 = 3), we solve ((A - 3I)mathbf{v} = 0):[begin{bmatrix} -7 & -14 5 & 10 end{bmatrix} begin{bmatrix} v_1 v_2 end{bmatrix} = 0 implies -7v_1 - 14v_2 = 0 implies v_1 = -2v_2.]Choosing (v_2 = 1), we get the eigenvector (begin{bmatrix} -2 1 end{bmatrix}).The general solution is a linear combination of the eigenvectors multiplied by their respective exponential terms:[begin{bmatrix} x(t) y(t) end{bmatrix} = C_1 e^{6t} begin{bmatrix} -7 5 end{bmatrix} + C_2 e^{3t} begin{bmatrix} -2 1 end{bmatrix}.]Thus, the solution is:[boxed{begin{cases}x(t) = -7 C_1 e^{6t} - 2 C_2 e^{3t} y(t) = 5 C_1 e^{6t} + C_2 e^{3t}end{cases}}]where ( C_1 ) and ( C_2 ) are constants.,<think>Alright, so I need to solve this system of differential equations:[begin{cases}x' = -4x -14y y' = 5x +13yend{cases}]Hmm, okay. Let me remember how to approach systems like this. I think it involves finding eigenvalues and eigenvectors, right? Because it's a linear system, and we can write it in matrix form. Let me try that first.So, writing the system as a matrix equation:[begin{pmatrix}x' y'end{pmatrix}=begin{pmatrix}-4 & -14 5 & 13end{pmatrix}begin{pmatrix}x yend{pmatrix}]Let me denote the coefficient matrix as ( A ):[A = begin{pmatrix}-4 & -14 5 & 13end{pmatrix}]To find the eigenvalues, I need to solve the characteristic equation ( det(A - lambda I) = 0 ). Let me compute that.First, subtract ( lambda ) from the diagonal entries:[A - lambda I = begin{pmatrix}-4 - lambda & -14 5 & 13 - lambdaend{pmatrix}]Now, the determinant is:[det(A - lambda I) = (-4 - lambda)(13 - lambda) - (-14)(5)]Let me compute each part step by step. First, expand the first product:[(-4 - lambda)(13 - lambda) = (-4)(13) + (-4)(-lambda) + (-lambda)(13) + (-lambda)(-lambda)][= -52 + 4lambda -13lambda + lambda^2][= lambda^2 -9lambda -52]Then, compute the second part: ( -(-14)(5) = 70 )So, the determinant equation becomes:[lambda^2 -9lambda -52 +70 = 0][lambda^2 -9lambda +18 = 0]Now, solve this quadratic equation for ( lambda ). Let me use the quadratic formula:[lambda = frac{9 pm sqrt{81 - 72}}{2} = frac{9 pm sqrt{9}}{2} = frac{9 pm 3}{2}]So, the eigenvalues are:[lambda_1 = frac{9 + 3}{2} = 6 quad text{and} quad lambda_2 = frac{9 - 3}{2} = 3]Okay, so we have two distinct real eigenvalues: 6 and 3. That should make things straightforward. Next step is to find the eigenvectors corresponding to each eigenvalue.Starting with ( lambda_1 = 6 ):We need to solve ( (A - 6I)mathbf{v} = 0 ), where ( mathbf{v} = begin{pmatrix} v_1 v_2 end{pmatrix} ).Compute ( A - 6I ):[begin{pmatrix}-4 -6 & -14 5 & 13 -6end{pmatrix}=begin{pmatrix}-10 & -14 5 & 7end{pmatrix}]So, the system is:[-10v_1 -14v_2 = 0 5v_1 +7v_2 = 0]Notice that the second equation is just the first equation divided by -2. So, they are dependent. Let's take the second equation:( 5v_1 +7v_2 = 0 )Let me solve for ( v_1 ):( 5v_1 = -7v_2 Rightarrow v_1 = -frac{7}{5}v_2 )So, the eigenvector corresponding to ( lambda = 6 ) can be written as:[mathbf{v}_1 = begin{pmatrix} -frac{7}{5} 1 end{pmatrix} quad text{or any scalar multiple.}]To avoid fractions, we can multiply both components by 5:[mathbf{v}_1 = begin{pmatrix} -7 5 end{pmatrix}]Great, so that's the eigenvector for eigenvalue 6.Now, moving on to ( lambda_2 = 3 ):Again, solve ( (A - 3I)mathbf{w} = 0 ):Compute ( A - 3I ):[begin{pmatrix}-4 -3 & -14 5 & 13 -3end{pmatrix}=begin{pmatrix}-7 & -14 5 & 10end{pmatrix}]The system is:[-7w_1 -14w_2 = 0 5w_1 +10w_2 = 0]Again, these equations are dependent. The second equation can be simplified:( 5w_1 +10w_2 = 0 Rightarrow w_1 +2w_2 = 0 Rightarrow w_1 = -2w_2 )So, the eigenvector corresponding to ( lambda = 3 ) is:[mathbf{v}_2 = begin{pmatrix} -2 1 end{pmatrix}]Perfect, now we have both eigenvalues and eigenvectors. The general solution to the system should be a combination of the solutions from each eigenvalue/eigenvector pair.Recall that for each eigenvalue ( lambda ) and eigenvector ( mathbf{v} ), the solution is ( e^{lambda t} mathbf{v} ). So, the general solution is:[begin{pmatrix}x(t) y(t)end{pmatrix}= C_1 e^{6t} begin{pmatrix} -7 5 end{pmatrix} + C_2 e^{3t} begin{pmatrix} -2 1 end{pmatrix}]Where ( C_1 ) and ( C_2 ) are constants determined by initial conditions.Let me write that out explicitly:[x(t) = -7C_1 e^{6t} -2C_2 e^{3t}][y(t) = 5C_1 e^{6t} + C_2 e^{3t}]Hmm, let me check if this makes sense. If I plug these into the original equations, they should satisfy both equations. Let me verify the first equation ( x' = -4x -14y ).Compute ( x'(t) ):[x'(t) = -7C_1 cdot 6 e^{6t} -2C_2 cdot 3 e^{3t} = -42C_1 e^{6t} -6C_2 e^{3t}]Now compute ( -4x -14y ):[-4x -14y = -4(-7C_1 e^{6t} -2C_2 e^{3t}) -14(5C_1 e^{6t} + C_2 e^{3t})][= 28C_1 e^{6t} +8C_2 e^{3t} -70C_1 e^{6t} -14C_2 e^{3t}][= (28C_1 -70C_1)e^{6t} + (8C_2 -14C_2)e^{3t}][= -42C_1 e^{6t} -6C_2 e^{3t}]Which matches ( x'(t) ). Good. Now check the second equation ( y' =5x +13y ).Compute ( y'(t) ):[y'(t) =5C_1 cdot 6 e^{6t} + C_2 cdot 3 e^{3t} =30C_1 e^{6t} +3C_2 e^{3t}]Compute (5x +13y ):[5x +13y =5(-7C_1 e^{6t} -2C_2 e^{3t}) +13(5C_1 e^{6t} + C_2 e^{3t})][= -35C_1 e^{6t} -10C_2 e^{3t} +65C_1 e^{6t} +13C_2 e^{3t}][= (-35C_1 +65C_1)e^{6t} + (-10C_2 +13C_2)e^{3t}][=30C_1 e^{6t} +3C_2 e^{3t}]Which matches ( y'(t) ). So, the solution satisfies both equations. That seems correct.Therefore, the general solution is:[x(t) = -7C_1 e^{6t} -2C_2 e^{3t}][y(t) =5C_1 e^{6t} + C_2 e^{3t}]Alternatively, since the constants ( C_1 ) and ( C_2 ) are arbitrary, we can write the solution in terms of different constants if needed, but this form should be acceptable.I think that's all. Let me recap the steps to make sure I didn't miss anything. Found the matrix, computed eigenvalues by solving the characteristic equation, found eigenvectors for each eigenvalue, then constructed the general solution using the eigenvalues and eigenvectors. Checked the solution by plugging back into the original equations, and it worked. So, this should be the correct answer.Final AnswerThe general solution is (boxed{x(t) = -7C_1 e^{6t} - 2C_2 e^{3t}}) and (boxed{y(t) = 5C_1 e^{6t} + C_2 e^{3t}}).</think>To solve the system of differential equations:[begin{cases}x' = -4x -14y y' = 5x +13yend{cases}]we first write the system in matrix form:[begin{pmatrix}x' y'end{pmatrix}=begin{pmatrix}-4 & -14 5 & 13end{pmatrix}begin{pmatrix}x yend{pmatrix}]Next, we find the eigenvalues of the coefficient matrix ( A ):[A = begin{pmatrix}-4 & -14 5 & 13end{pmatrix}]The characteristic equation is:[det(A - lambda I) = (-4 - lambda)(13 - lambda) + 70 = lambda^2 - 9lambda + 18 = 0]Solving this quadratic equation, we get the eigenvalues:[lambda_1 = 6 quad text{and} quad lambda_2 = 3]For each eigenvalue, we find the corresponding eigenvectors.For ( lambda_1 = 6 ):[(A - 6I)mathbf{v} = 0 implies begin{pmatrix}-10 & -14 5 & 7end{pmatrix}begin{pmatrix}v_1 v_2end{pmatrix}= 0]Solving this, we get the eigenvector:[mathbf{v}_1 = begin{pmatrix} -7 5 end{pmatrix}]For ( lambda_2 = 3 ):[(A - 3I)mathbf{w} = 0 implies begin{pmatrix}-7 & -14 5 & 10end{pmatrix}begin{pmatrix}w_1 w_2end{pmatrix}= 0]Solving this, we get the eigenvector:[mathbf{v}_2 = begin{pmatrix} -2 1 end{pmatrix}]The general solution of the system is a combination of the solutions from each eigenvalue/eigenvector pair:[begin{pmatrix}x(t) y(t)end{pmatrix}= C_1 e^{6t} begin{pmatrix} -7 5 end{pmatrix} + C_2 e^{3t} begin{pmatrix} -2 1 end{pmatrix}]Expressing this explicitly, we have:[x(t) = -7C_1 e^{6t} -2C_2 e^{3t}][y(t) = 5C_1 e^{6t} + C_2 e^{3t}]Thus, the general solution is:[boxed{x(t) = -7C_1 e^{6t} - 2C_2 e^{3t}}]and[boxed{y(t) = 5C_1 e^{6t} + C_2 e^{3t}}]]

🔑:Solution. Immediately note that the unknown functions are denoted by different letters, and we will consider the variable t as the argument.The essence of the method is to transition from the given system to another, equivalent system, in which differential groups or complete differentials can be separated. For example, if we perform the specified operations (multiplications by corresponding numbers and subsequent addition of the results) on the equations of the given system, we will sequentially obtain:begin{aligned}& Rightarrow 5 x^{prime}+7 y^{prime}=15 x+21 y Leftrightarrow frac{d}{d t}(5 x+7 y)=3(5 x+7 y)end{aligned}This was the goal of our transformation. We will transform the last equality and integrate: int frac{d(5 x+7 y)}{5 x+7 y}=3 int d t, ln |5 x+7 y|=3 t+ln C_{1}, 5 x+7 y=C_{1} e^{3 t}. We have obtained one relationship between the unknown functions, which we will rewrite as: 7 y=C_{1} e^{3 t}-5 x.We will form a new system by appending the last equality to the first equation of the original system:left{begin{array}{l|l}7 y=C_{1} e^{3 t}-5 x & 2 -14 y=x^{prime}+4 x . & 1end{array}+right.By adding (as indicated in the diagram) these equations, we obtain a linear (non-homogeneous) first-order equation: x^{prime}-6 x=-2 C_{1} e^{3 t}, which can be solved using the substitution method: x=u v, x^{prime}=u^{prime} v+u v^{prime}. We obtain sequentiallyu^{prime} v+u v^{prime}-6 u v=-2 C_{1} e^{3 t} Rightarrowleft{begin{array} { r l } { v ^ { prime } } & { = 6 v } { u ^ { prime } v } & { = - 2 C _ { 1 } e ^ { 3 t } }end{array} Rightarrow left{begin{array}{rl}v & =e^{6 t} u & =frac{2}{3} C_{1} e^{-3 t}+C_{2}end{array}right.right.x=u v=frac{2}{3} C_{1} e^{3 t}+C_{2} e^{6 t} - this is the first unknown function of the system. The second unknown function y=y(t) is found from the relationship 7 y= =C_{1} e^{3 t}-5 x.text { Answer. }left{begin{array}{l}x=frac{2}{3} C_{1} e^{3 t}+C_{2} e^{6 t} y=-frac{1}{3} C_{1} e^{-3 t}-frac{5}{7} C_{2} e^{6 t}end{array}right.