🧠:[<think>Okay, let's see. I need to solve this problem about an arithmetic sequence and a quadratic inequality. Hmm, part (1) asks for the general formula of the arithmetic sequence {a_n}, which has the first term a and common difference b. But before that, there's an inequality given: ax² - 3x + 2 > 0, and its solution set is (-∞,1) ∪ (b, +∞). So, I probably need to find the values of a and b first using this inequality, right?Alright, let's start with the inequality. The quadratic inequality ax² - 3x + 2 > 0 has solutions x < 1 or x > b. So, that means the quadratic opens upwards (since the solution is outside the roots) and the roots are at x = 1 and x = b. Wait, but in a quadratic inequality, the solution set being (-∞,1) ∪ (b, +∞) implies that the quadratic is positive outside the interval between its two roots. So, the quadratic must open upwards, which means the coefficient of x², which is a, must be positive. So, a > 0.Also, the roots of the quadratic equation ax² - 3x + 2 = 0 are x = 1 and x = b. Let's recall that for a quadratic equation ax² + bx + c = 0, the sum of the roots is -b/a and the product is c/a. Wait, but here, the quadratic is ax² - 3x + 2, so comparing to standard form, the coefficient of x is -3, and the constant term is 2. So, the sum of the roots should be -(-3)/a = 3/a, and the product of the roots is 2/a.But the roots are given as 1 and b. So, sum of roots is 1 + b = 3/a, and product is 1 * b = 2/a.Therefore, we have two equations:1 + b = 3/a ...(1)b = 2/a ...(2)So, substituting equation (2) into equation (1):1 + (2/a) = 3/aMultiply both sides by a to eliminate denominators:a*1 + 2 = 3So, a + 2 = 3Subtract 2 from both sides:a = 1Then, from equation (2), b = 2/a = 2/1 = 2.So, a = 1 and b = 2. Therefore, the arithmetic sequence {a_n} has first term a = 1 and common difference b = 2.Therefore, the general formula for the arithmetic sequence is:a_n = a + (n - 1)d = 1 + (n - 1)*2 = 1 + 2n - 2 = 2n -1.Wait, let me check that. First term is a = 1, so when n = 1, a_1 = 1 + (1 - 1)*2 = 1, which is correct. Then a_2 = 1 + (2 - 1)*2 = 3, a_3 = 5, etc. So yes, the nth term is 2n -1. So part (1) is done.Now part (2): Let the sequence {b_n} satisfy [something?] Wait, the original problem says "the sequence {b_n} satisfy = 1/(a_n a_{n+1})", find the sum S_n of the first n terms.Wait, maybe the problem says: "Let the sequence {b_n} satisfy b_n = 1/(a_n a_{n+1}), find the sum S_n..." Yes, that makes sense. So, the terms of {b_n} are defined as 1 divided by the product of consecutive terms of the arithmetic sequence {a_n}.Given that {a_n} is an arithmetic sequence with a_n = 2n -1, so a_n = 2n -1, then a_{n+1} = 2(n+1) -1 = 2n + 2 -1 = 2n +1. Therefore, the product a_n * a_{n+1} = (2n -1)(2n +1). That's a difference of squares: (2n)^2 -1^2 = 4n² -1. Wait, but (2n -1)(2n +1) = 4n² -1. Therefore, b_n = 1/(4n² -1).So, b_n = 1/((2n -1)(2n +1)) = 1/(4n² -1). To find the sum S_n = sum_{k=1}^n b_k.This looks like a telescoping series. To compute the sum, we can use partial fractions.Let me decompose 1/(4n² -1) into partial fractions. The denominator factors as (2n -1)(2n +1). So, we can write:1/((2n -1)(2n +1)) = A/(2n -1) + B/(2n +1)Multiply both sides by (2n -1)(2n +1):1 = A(2n +1) + B(2n -1)This must hold for all n. Let's solve for A and B.Expand the right side:1 = (2A + 2B)n + (A - B)Set coefficients equal:For the coefficient of n: 2A + 2B = 0For the constant term: A - B = 1So, from the first equation: 2A + 2B = 0 => A + B = 0 => B = -ASubstitute into the second equation: A - (-A) = 1 => 2A = 1 => A = 1/2Then B = -1/2Therefore, the partial fractions decomposition is:1/((2n -1)(2n +1)) = (1/2)/(2n -1) - (1/2)/(2n +1)Therefore, b_n = 1/2 [1/(2n -1) - 1/(2n +1)]So, the sum S_n = sum_{k=1}^n [1/2 (1/(2k -1) - 1/(2k +1))] = 1/2 sum_{k=1}^n [1/(2k -1) - 1/(2k +1)]This is a telescoping series. Let's write out the terms:For k = 1: 1/2 [1/1 - 1/3]For k = 2: 1/2 [1/3 - 1/5]For k = 3: 1/2 [1/5 - 1/7]...For k = n: 1/2 [1/(2n -1) - 1/(2n +1)]When we add all these up, the intermediate terms cancel out. Let's see:Total sum = 1/2 [ (1/1 - 1/3) + (1/3 - 1/5) + (1/5 - 1/7) + ... + (1/(2n -1) - 1/(2n +1)) ]Each negative term cancels with the positive term in the next fraction:-1/3 cancels with +1/3, -1/5 cancels with +1/5, etc., up to -1/(2n -1) would cancel with the next term, but since we stop at k = n, the last term is -1/(2n +1).Therefore, the sum telescopes to:1/2 [1 - 1/(2n +1)] = 1/2 [ (2n +1 -1)/(2n +1) ) ] = 1/2 [ (2n)/(2n +1) ) ] = (1/2)(2n)/(2n +1) = n/(2n +1)So, the sum S_n = n/(2n +1)Wait, let me verify with small n. For n =1:S_1 = b_1 = 1/(1*3) = 1/3. According to the formula, 1/(2*1 +1) = 1/3. Correct.For n =2:S_2 = 1/3 + 1/(3*5) = 1/3 + 1/15 = 5/15 + 1/15 = 6/15 = 2/5. According to the formula, 2/(2*2 +1) = 2/5. Correct.For n =3:S_3 = 1/3 + 1/15 + 1/(5*7) = 1/3 + 1/15 + 1/35. Let's compute:1/3 = 35/105, 1/15 = 7/105, 1/35 = 3/105. Total: 35 +7 +3 = 45/105 = 9/21 = 3/7. Formula gives 3/(2*3 +1) = 3/7. Correct. So, the formula seems to hold.Therefore, the sum S_n is n/(2n +1).So, summarizing:1) The general formula for {a_n} is a_n = 2n -12) The sum S_n = n/(2n +1)Final Answer(1) The general term of the sequence is boxed{a_n = 2n - 1}(2) The sum of the first ( n ) terms of the sequence ( {b_n} ) is boxed{S_n = dfrac{n}{2n + 1}}</think>Given an arithmetic sequence ({a_n}) with the first term (a) and common difference (b), and the solution set of the inequality (ax^2 - 3x + 2 > 0) is ((-infty, 1) cup (b, +infty)):1. Finding the general formula for the sequence ({a_n}): - The quadratic inequality (ax^2 - 3x + 2 > 0) has solutions (x in (-infty, 1) cup (b, +infty)), indicating the roots of the quadratic equation (ax^2 - 3x + 2 = 0) are (1) and (b). - Using Vieta's formulas, the sum of the roots is (frac{3}{a}) and the product of the roots is (frac{2}{a}). - Solving the equations (1 + b = frac{3}{a}) and (1 cdot b = frac{2}{a}), we find (a = 1) and (b = 2). - The arithmetic sequence ({a_n}) has the first term (a = 1) and common difference (b = 2). The general term is: [ a_n = 1 + (n - 1) cdot 2 = 2n - 1 ]2. Finding the sum of the first (n) terms of the sequence ({b_n}): - The sequence ({b_n}) is defined as (b_n = frac{1}{a_n a_{n+1}}). - Substituting (a_n = 2n - 1) and (a_{n+1} = 2n + 1), we get: [ b_n = frac{1}{(2n - 1)(2n + 1)} ] - Using partial fractions decomposition: [ frac{1}{(2n - 1)(2n + 1)} = frac{1}{2} left( frac{1}{2n - 1} - frac{1}{2n + 1} right) ] - The sum (S_n) is a telescoping series: [ S_n = sum_{k=1}^n b_k = frac{1}{2} sum_{k=1}^n left( frac{1}{2k - 1} - frac{1}{2k + 1} right) ] - This telescopes to: [ S_n = frac{1}{2} left( 1 - frac{1}{2n + 1} right) = frac{1}{2} left( frac{2n}{2n + 1} right) = frac{n}{2n + 1} ] Final Answer(1) The general term of the sequence is boxed{a_n = 2n - 1}(2) The sum of the first ( n ) terms of the sequence ( {b_n} ) is boxed{S_n = dfrac{n}{2n + 1}},<think>Okay, let's see. I need to solve this problem about an arithmetic sequence and a quadratic inequality. Let's start with part (1). The problem says:Given an arithmetic sequence {a_n} with the first term a and common difference b. The solution set of the inequality ax² - 3x + 2 > 0 is (-∞,1) ∪ (b, +∞). Then, I need to find the general formula for the sequence {a_n}.First, let's recall that an arithmetic sequence has the general term a_n = a + (n-1)b. So, once we know a and b, we can write the formula. Therefore, part (1) reduces to finding the values of a and b.The key here is the quadratic inequality. The solution set is given as (-∞,1) ∪ (b, +∞). That tells me that the quadratic ax² - 3x + 2 is positive outside the interval [1, b]. So, the quadratic opens upwards (since it's positive outside the interval) and has roots at x=1 and x=b. Also, since the solution is two intervals, the quadratic must have two distinct real roots. So, discriminant must be positive. Let's check that later.First, let's note that for a quadratic equation ax² + bx + c = 0, the sum of roots is -b/a and the product is c/a. Wait, here the quadratic is ax² - 3x + 2. So, comparing to the standard form:Standard quadratic: Ax² + Bx + C = 0Our quadratic: a x² - 3 x + 2 = 0Therefore, A = a, B = -3, C = 2.Sum of roots = -B/A = 3/aProduct of roots = C/A = 2/aBut the roots are given as 1 and b. So, sum of roots is 1 + b, and product is 1 * b = b.Therefore, we can set up equations:1 + b = 3/aandb = 2/aSo, substituting the second equation into the first:1 + (2/a) = 3/aWait, that gives 1 = 3/a - 2/a = (3 - 2)/a = 1/aTherefore, 1 = 1/a ⇒ a = 1.Then, from the product of roots, b = 2/a = 2/1 = 2.So, a = 1 and b = 2.Therefore, the arithmetic sequence has first term a = 1 and common difference b = 2. So, the general term is:a_n = a + (n - 1)b = 1 + (n - 1)*2 = 1 + 2n - 2 = 2n -1.Wait, let's check that:a_1 = 1 + (1-1)*2 = 1 + 0 = 1a_2 = 1 + (2-1)*2 = 1 + 2 = 3a_3 = 1 + 3*2 = 7? Wait, 1 + (3-1)*2 = 1 + 4 = 5. Wait, that's 5. Wait, 2n -1:n=1: 2*1 -1 =1n=2: 4 -1=3n=3:6-1=5. Yes, that's correct. So the general term is 2n -1. So, that's part (1) done.But wait, let me verify the quadratic inequality with a=1. The quadratic is 1*x² -3x +2 >0, which is x² -3x +2 >0. Let's factor that: (x-1)(x-2) >0. So, the roots are 1 and 2. The quadratic opens upwards (coefficient positive), so the inequality is satisfied when x <1 or x>2. Therefore, the solution set is (-∞,1) ∪ (2, ∞), which matches the given solution set (-∞,1) ∪ (b, ∞). Therefore, b must be 2, which is consistent with our previous result. So, a=1, b=2. Therefore, the arithmetic sequence is 1, 3, 5, 7,... with general term a_n=2n-1.So, part (1) is solved.Now part (2):Let the sequence {b_n} satisfy = 1/(a_n a_{n+1}). Find the sum of the first n terms of {b_n}, S_n.First, let's parse the problem. The sequence {b_n} is defined as b_n = 1/(a_n a_{n+1}). Then, we need to find S_n = sum_{k=1}^n b_k.Given that {a_n} is an arithmetic sequence with a_n = 2n -1, so a_{n+1} = 2(n+1) -1 = 2n +2 -1 = 2n +1.Therefore, b_n = 1/( (2n -1)(2n +1) )So, we need to compute the sum S_n = sum_{k=1}^n [1/( (2k -1)(2k +1) ) ]This looks like a telescoping series. To compute such a sum, we can use partial fractions.Let's decompose 1/( (2k -1)(2k +1) ) into partial fractions.Let me set:1/( (2k -1)(2k +1) ) = A/(2k -1) + B/(2k +1)Multiply both sides by (2k -1)(2k +1):1 = A(2k +1) + B(2k -1)Now, let's solve for A and B.Expanding the right side:1 = (2A + 2B)k + (A - B)This must hold for all k, so the coefficients of k and the constants must match.Coefficient of k: 2A + 2B = 0Constant term: A - B =1So, from the first equation: 2A + 2B =0 ⇒ A = -BSubstitute into the second equation: (-B) - B =1 ⇒ -2B =1 ⇒ B= -1/2Then A = -B = 1/2Therefore, the partial fraction decomposition is:1/( (2k -1)(2k +1) ) = (1/2)/(2k -1) - (1/2)/(2k +1)So, 1/( (2k -1)(2k +1) ) = (1/2)[1/(2k -1) - 1/(2k +1)]Therefore, each term b_k is (1/2)[1/(2k -1) - 1/(2k +1)]So, the sum S_n = sum_{k=1}^n (1/2)[1/(2k -1) -1/(2k +1) ]This is a telescoping sum. Let's write out the terms:For k=1: (1/2)[1/1 -1/3]For k=2: (1/2)[1/3 -1/5]For k=3: (1/2)[1/5 -1/7]...For k=n: (1/2)[1/(2n -1) -1/(2n +1)]When we add all these terms, most of the intermediate terms cancel out. Specifically, the -1/3 from the first term cancels with the +1/3 from the second term, the -1/5 from the second term cancels with the +1/5 from the third term, and so on, up to the -1/(2n -1) from the (n-1)-th term cancels with the +1/(2n -1) from the n-th term. What's left is the first term of the first expression and the last term of the last expression.Therefore, S_n = (1/2)[1/1 -1/(2n +1)]Simplifying this:S_n = (1/2)[1 - 1/(2n +1)] = (1/2)[ (2n +1 -1)/(2n +1) ) ] = (1/2)[ (2n)/(2n +1) ) ] = (1/2)*(2n)/(2n +1) = n/(2n +1)Therefore, the sum S_n is n/(2n +1)Let me check with n=1:b_1 = 1/(1*3) =1/3. S_1=1/3. According to the formula, 1/(2*1 +1)=1/3. Correct.n=2:b_1 +b_2 =1/3 +1/(3*5)=1/3 +1/15=5/15 +1/15=6/15=2/5. Formula: 2/(2*2 +1)=2/5. Correct.n=3:S_3=2/5 +1/(5*7)=2/5 +1/35=14/35 +1/35=15/35=3/7. Formula:3/(2*3 +1)=3/7. Correct.So the formula works. Therefore, the sum S_n is n/(2n +1).So, summarizing:(1) The general term of the arithmetic sequence is a_n =2n -1.(2) The sum S_n =n/(2n +1)Final Answer(1) The general formula for the sequence is boxed{a_n = 2n - 1}. (2) The sum of the first ( n ) terms of the sequence ( {b_n} ) is boxed{S_n = dfrac{n}{2n + 1}}.</think>Given an arithmetic sequence ({a_n}) with the first term (a) and common difference (b), and the solution set of the inequality (ax^2 - 3x + 2 > 0) is ((-infty, 1) cup (b, +infty)):(1) To find the general formula for the sequence ({a_n}):- The quadratic inequality (ax^2 - 3x + 2 > 0) has roots at (x = 1) and (x = b). - The sum of the roots is (1 + b = frac{3}{a}) and the product of the roots is (1 cdot b = frac{2}{a}).- Solving these equations, we find (a = 1) and (b = 2).- The arithmetic sequence has the first term (a = 1) and common difference (b = 2). The general term is (a_n = 1 + (n - 1) cdot 2 = 2n - 1).Thus, the general formula for the sequence is (boxed{a_n = 2n - 1}).(2) To find the sum of the first (n) terms of the sequence ({b_n}) where (b_n = frac{1}{a_n a_{n+1}}):- The sequence ({b_n}) is given by (b_n = frac{1}{(2n - 1)(2n + 1)}).- Using partial fractions, we decompose (b_n) as (frac{1}{2} left( frac{1}{2n - 1} - frac{1}{2n + 1} right)).- The sum (S_n) is a telescoping series: [ S_n = sum_{k=1}^n frac{1}{2} left( frac{1}{2k - 1} - frac{1}{2k + 1} right) ]- Most terms cancel out, leaving: [ S_n = frac{1}{2} left( 1 - frac{1}{2n + 1} right) = frac{1}{2} left( frac{2n}{2n + 1} right) = frac{n}{2n + 1} ]Thus, the sum of the first (n) terms of the sequence ({b_n}) is (boxed{S_n = dfrac{n}{2n + 1}}).]

🔑:Solution: (1) Since the solution set of the inequality ax^{2}-3x+2 > 0 is (-infty,1)cup(b,+infty) therefore x=1, x=b are the roots of the equation ax^{2}-3x+2=0 therefore begin{cases} 1+b= frac {3}{a} b= frac {2}{a}end{cases} Solving this, we get b=2, a=1 therefore The sequence {a_n} is an arithmetic sequence with the first term 1 and common difference 2 By the general formula of an arithmetic sequence, we have a_n=1+2(n-1)=2n-1 (2) Since b_n= frac {1}{a_{n}a_{n+1}}= frac {1}{(2n-1)(2n+1)}= frac {1}{2}( frac {1}{2n-1}- frac {1}{2n+1}) S_{n}=b_{1}+b_{2}+ldots+b_{n} = frac {1}{2}(1- frac {1}{3}+ frac {1}{3}- frac {1}{5}+ldots+ frac {1}{2n-1}- frac {1}{2n+1}) = frac {1}{2}(1- frac {1}{2n+1})= frac {1}{2}times frac {2n}{2n+1}= frac {n}{2n+1}Therefore, the answers are: (1) The general formula for the sequence {a_n} is boxed{a_n=2n-1} (2) The sum of the first n terms of the sequence {b_n}, S_{n}, is boxed{frac {n}{2n+1}}

🧠:[<think>Okay, so I need to figure out the possible length of the third stick in this triangle problem. Let's see... The problem says that three wooden sticks are connected end to end to form a triangle. Two of the sticks are 3 cm and 9 cm long, and we need to determine which of the options A through D could be the length of the third stick. The options are 3 cm, 6 cm, 10 cm, or 12 cm. Hmm, triangles have some rules about their side lengths, right? The triangle inequality theorem. Let me recall that.The triangle inequality theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the remaining side. So, for any triangle with sides a, b, and c, the following must hold:1. a + b > c2. a + c > b3. b + c > aSo, applying this to the problem here. Let's denote the third side as x. The given sides are 3 cm and 9 cm. So, we need to check the inequalities for each possible x value given in the options. Let's list them again:A: 3 cmB: 6 cmC: 10 cmD: 12 cmFirst, let's check each option one by one.Starting with option A: 3 cm.So, the sides would be 3, 3, and 9. Let's check the triangle inequalities.1. 3 + 3 > 9? That would be 6 > 9, which is false. So, this combination doesn't satisfy the triangle inequality. Therefore, a triangle cannot be formed with sides 3, 3, and 9. So, option A is invalid.Next, option B: 6 cm.So, sides are 3, 6, and 9. Let's check the inequalities.1. 3 + 6 > 9? That's 9 > 9. Wait, the inequality says the sum must be greater than the third side. 9 is not greater than 9. It's equal. So, equality here would mean that the three sides form a degenerate triangle, which is essentially a straight line. But a triangle must have a positive area. Therefore, a degenerate triangle isn't considered a valid triangle in this context. Therefore, 3, 6, 9 would not form a valid triangle. So, option B is also invalid.Moving on to option C: 10 cm.Sides are 3, 9, and 10. Check the inequalities.1. 3 + 9 > 10? 12 > 10, which is true.2. 3 + 10 > 9? 13 > 9, which is true.3. 9 + 10 > 3? 19 > 3, which is true.All three inequalities hold here. Therefore, a triangle with sides 3, 9, and 10 cm is valid. So, option C is a possible answer.Option D: 12 cm.Let's check the inequalities again. The sides would be 3, 9, and 12.1. 3 + 9 > 12? 12 > 12? Again, this is equality. So, similar to option B, this would result in a degenerate triangle. So, 3, 9, 12 can't form a valid triangle. Therefore, option D is invalid.So, the only valid option here is C: 10 cm.Wait a second, let me just double-check my reasoning for each option because sometimes I might make a mistake in the inequalities.Starting with option A: 3, 3, 9. The sum of the two smaller sides is 3 + 3 = 6, which is less than 9, so that's invalid. Correct.Option B: 3, 6, 9. The sum of 3 and 6 is 9, which is equal to the third side, so degenerate. Correct.Option C: 3, 9, 10. The sum of 3 and 9 is 12, which is greater than 10. The sum of 3 and 10 is 13, which is greater than 9. The sum of 9 and 10 is 19, which is greater than 3. All correct. So, valid.Option D: 3, 9, 12. Sum of 3 and 9 is 12, which is equal to the third side, so degenerate. Correct.Therefore, the only possible answer is C: 10 cm. So, I think that's the right choice.Final Answerboxed{C}</think>To determine the possible length of the third stick in a triangle with sides 3 cm and 9 cm, we apply the triangle inequality theorem, which states that the sum of the lengths of any two sides must be greater than the length of the remaining side.1. Option A: 3 cm - Sides: 3 cm, 3 cm, 9 cm - Check inequalities: - (3 + 3 = 6) (not greater than 9) - Invalid triangle.2. Option B: 6 cm - Sides: 3 cm, 6 cm, 9 cm - Check inequalities: - (3 + 6 = 9) (equal to 9, forming a degenerate triangle) - Invalid triangle.3. Option C: 10 cm - Sides: 3 cm, 9 cm, 10 cm - Check inequalities: - (3 + 9 = 12 > 10) - (3 + 10 = 13 > 9) - (9 + 10 = 19 > 3) - All inequalities hold. Valid triangle.4. Option D: 12 cm - Sides: 3 cm, 9 cm, 12 cm - Check inequalities: - (3 + 9 = 12) (equal to 12, forming a degenerate triangle) - Invalid triangle.The only valid length for the third stick is 10 cm.[boxed{C}],<think>Okay, so I have this problem here where three wooden sticks are connected end to end to form a triangle. Two of the sticks are 3 cm and 9 cm long, and I need to figure out which of the options A to D could be the length of the third stick. Let me think about how to approach this.First, I remember that for any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side. This is called the triangle inequality theorem. So, if I have three sides a, b, and c, then:a + b > c a + c > b b + c > aRight? So, applying this theorem here. Let's denote the third side as x. Then the three sides would be 3 cm, 9 cm, and x cm.So, according to the triangle inequality, the following inequalities must hold:1. 3 + 9 > x 2. 3 + x > 9 3. 9 + x > 3Let me compute each of these.First inequality: 3 + 9 > x That's 12 > x, which is the same as x < 12.Second inequality: 3 + x > 9 Subtracting 3 from both sides gives x > 6.Third inequality: 9 + x > 3 Subtracting 9 from both sides gives x > -6. But since lengths can't be negative, this inequality doesn't really add any new information because x has to be a positive number anyway. So, effectively, the important inequalities are x > 6 and x < 12. So the third stick must be longer than 6 cm and shorter than 12 cm.Now, looking at the answer choices: A is 3 cm, B is 6 cm, C is 10 cm, D is 12 cm.So, let's check each one.A: 3 cm. If the third stick is 3 cm, then the sides are 3, 3, 9. Let's check the triangle inequalities.3 + 3 > 9? 6 > 9? No, that's not true. So, 3 cm can't work. So A is out.B: 6 cm. So sides are 3, 6, 9. Check the inequalities.3 + 6 > 9? 9 > 9? No, that's equal. The inequality requires the sum to be greater than the third side, not equal. So, 6 cm doesn't work either. B is out.C: 10 cm. Then sides are 3, 9, 10.Check inequalities:3 + 9 > 10? 12 > 10? Yes.3 + 10 > 9? 13 > 9? Yes.9 + 10 > 3? 19 > 3? Yes. All three inequalities are satisfied. So 10 cm is a valid length. So C works.D: 12 cm. Sides 3, 9, 12.Check inequalities:3 + 9 > 12? 12 > 12? No, again equality here. So that's invalid. So D is out.Therefore, the only possible answer is C: 10 cm.Wait, let me double-check. The problem says "the length of the third stick can be ( )", so they might be asking which is a possible length. So since 10 cm satisfies all the inequalities, that's the right answer.But hold on, let me confirm once more. For the sides 3, 9, 10:3 + 9 = 12, which is greater than 10. 3 + 10 = 13, which is greater than 9. 9 + 10 = 19, which is greater than 3. All good.For 6 cm: 3 + 6 = 9, which is equal to the third side, so it's a degenerate triangle, which technically isn't a triangle because it would just be a straight line. Similarly, 12 cm gives 3 + 9 = 12, again a degenerate triangle.So yeah, only 10 cm works. So answer is C.Final Answerboxed{C}</think>To determine the possible length of the third stick forming a triangle with sticks of lengths 3 cm and 9 cm, we use the triangle inequality theorem. The theorem states that the sum of the lengths of any two sides must be greater than the length of the remaining side. Denote the third side as ( x ).The inequalities to check are:1. ( 3 + 9 > x ) which simplifies to ( 12 > x )2. ( 3 + x > 9 ) which simplifies to ( x > 6 )3. ( 9 + x > 3 ) which simplifies to ( x > -6 ) (this is always true since lengths are positive)Combining the relevant inequalities, we get ( 6 < x < 12 ).Checking the given options:- A: 3 cm does not satisfy ( x > 6 )- B: 6 cm does not satisfy ( x > 6 ) (since it is equal)- C: 10 cm satisfies ( 6 < 10 < 12 )- D: 12 cm does not satisfy ( x < 12 ) (since it is equal)Thus, the only valid length for the third stick is 10 cm.[boxed{C}]]

🔑:To solve this problem, we start by letting the length of the third stick be a cm. According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. This gives us two inequalities:1. 3 + 9 > a2. a + 3 > 93. a + 9 > 3From these, we can simplify to find the range of possible values for a:- From the first inequality, we have 12 > a.- From the second and third inequalities, considering the largest and smallest given sides, we get the most restrictive condition by subtracting the smallest side from the largest side, which gives us 9 - 3 < a.Combining these inequalities gives us 6 < a < 12. This means the length of the third stick must be greater than 6cm but less than 12cm. Looking at the options given:A: 3cm is not greater than 6cm.B: 6cm is not strictly greater than 6cm.C: 10cm fits within the range 6 < a < 12.D: 12cm is not less than 12cm.Therefore, the correct answer is boxed{C}.

🧠:[<think>Okay, so I have this problem here: Given the expression ((sqrt{x} - frac{2}{x^2})^n), the sum of the binomial coefficients of all terms in its expansion is 1024. I need to find three things: (1) the value of (n), (2) the constant term in the expansion, and (3) the number of rational terms in the expansion. Let me start by trying to understand each part step by step.First, part (1): Find the value of (n). The problem says that the sum of the binomial coefficients of all terms in the expansion is 1024. Hmm, I remember that for any binomial expression ((a + b)^n), the sum of the binomial coefficients is obtained by setting (a = 1) and (b = 1), so the sum is ((1 + 1)^n = 2^n). But in this case, our binomial is ((sqrt{x} - frac{2}{x^2})^n). Wait, so how does that work? Is it the same idea?Wait, actually, the sum of the coefficients is found by setting the variables to 1. So if we substitute (x = 1) into the expression, we get ((sqrt{1} - frac{2}{1^2})^n = (1 - 2)^n = (-1)^n). But the problem states that the sum of the binomial coefficients is 1024. However, (-1)^n can be either 1 or -1, depending on whether (n) is even or odd. But 1024 is a positive number, specifically (2^{10}), so maybe there's a misunderstanding here.Wait, perhaps I'm missing something. The problem says "the sum of the binomial coefficients of all terms in its expansion". In the standard binomial expansion ((a + b)^n), the coefficients are the binomial coefficients (binom{n}{k}), so the sum is indeed (2^n). However, in this case, the expression is ((sqrt{x} - frac{2}{x^2})^n), which when expanded, each term has a coefficient multiplied by some power of (x). So does the problem mean the sum of just the coefficients, ignoring the variables? For example, if a term is ( binom{n}{k} (sqrt{x})^{n - k} (-frac{2}{x^2})^k ), then the coefficient would be ( binom{n}{k} (-2)^k ). So the sum of all coefficients would be the sum over (k) from 0 to (n) of ( binom{n}{k} (-2)^k ). Alternatively, perhaps the question is just asking for the sum of the binomial coefficients (binom{n}{k}), ignoring the other constants. But the wording says "the sum of the binomial coefficients of all terms in its expansion", which is a bit ambiguous. Wait, in standard terminology, the binomial coefficients are just the (binom{n}{k}) terms. So maybe the problem is asking for the sum of (binom{n}{k}) for all terms, which is indeed (2^n). But in the given problem, the sum is 1024, which is (2^{10}). Therefore, (n = 10). Is that possible?Wait, but if we just consider the sum of the binomial coefficients, regardless of the other constants, then yes, the sum is (2^n = 1024), which would imply (n = 10). But in reality, the expansion here has coefficients that include ((-2)^k), so maybe the problem is referring to the sum of all coefficients in the expansion, considering the constants as well. That would be substituting (x = 1) into the expression, which gives ((sqrt{1} - frac{2}{1^2})^n = (-1)^n), as I thought earlier. But (-1)^n can't be 1024. So there must be a misunderstanding here.Wait, let's re-examine the problem statement: "the sum of the binomial coefficients of all terms in its expansion is 1024". The term "binomial coefficients" usually refers to the (binom{n}{k}) terms. However, in the expansion, each term is (binom{n}{k} (sqrt{x})^{n - k} left(-frac{2}{x^2}right)^k), so the coefficients would actually be (binom{n}{k} (-2)^k). Therefore, if the problem is referring to the sum of the coefficients (including the constants like (-2^k)), then the sum would be obtained by setting (x = 1), which gives ((sqrt{1} - frac{2}{1^2})^n = (-1)^n). But if they are referring to the sum of the binomial coefficients (binom{n}{k}) without considering the constants, then the sum is (2^n). But the problem says "the sum of the binomial coefficients of all terms in its expansion", so the binomial coefficients are the (binom{n}{k}), so sum is (2^n). Therefore, 1024 is (2^{10}), so (n = 10). That seems more plausible. So maybe part (1) is straightforward: (n = 10). Let me check with substituting (x = 1):If we substitute (x = 1) into the expression ((sqrt{x} - frac{2}{x^2})^n), we get ((1 - 2)^n = (-1)^n). If the problem had said "the sum of all coefficients", that would be ((-1)^n), but since they specify "binomial coefficients", which are (binom{n}{k}), the sum is (2^n). Therefore, answer to part (1) is (n = 10). Let me confirm this with an example. If n = 1, the expansion is (sqrt{x} - 2/x^2), coefficients are 1 and -2. But binomial coefficients are 1 and 1, sum is 2, which is 2^1. So even if there are other constants multiplied, the binomial coefficients are still 1 and 1. So, in that case, the sum is 2, not 1 - 2 = -1. Therefore, the problem is referring to the sum of the binomial coefficients (the (binom{n}{k})), which is 2^n = 1024, so n=10. Okay, that makes sense.So part (1): n=10.Moving on to part (2): Find the constant term in the expansion. So we need to find the term where the exponent of x is 0. Let's first write the general term in the expansion. The expansion is (sum_{k=0}^{10} binom{10}{k} (sqrt{x})^{10 - k} left(-frac{2}{x^2}right)^k).Let's simplify the exponents. The exponent of x in each term is given by ((sqrt{x})^{10 - k}) which is (x^{(10 - k)/2}), and (left(-frac{2}{x^2}right)^k) which is ((-2)^k x^{-2k}). So combining these, the exponent of x is (frac{10 - k}{2} - 2k = frac{10 - k - 4k}{2} = frac{10 -5k}{2}). We need this exponent to be 0 for the constant term. Therefore:(frac{10 -5k}{2} = 0)Solving for k:10 -5k =05k =10k=2So the constant term corresponds to k=2. Let's compute that term:Term = (binom{10}{2} (sqrt{x})^{10 - 2} left(-frac{2}{x^2}right)^2)Compute each part:(binom{10}{2} = 45)((sqrt{x})^8 = x^{4})(left(-frac{2}{x^2}right)^2 = left(-2right)^2 x^{-4} = 4x^{-4})Multiply them together:45 * x^4 * 4x^{-4} = 45 * 4 * x^{4 -4} = 180 * x^0 = 180So the constant term is 180. Let me verify this. When k=2, exponent is (10 - 5*2)/2 = 0, yes. Coefficient is 45 * (-2)^2 = 45*4=180. Correct. So part (2) answer is 180.Part (3): Find the number of rational terms in the expansion. So we need to determine how many terms in the expansion have exponents of x that are integers, since rational terms would be terms where x is raised to an integer power. Because x is a variable, and if the exponent is an integer, the term is a rational function (if exponent is positive, it's a polynomial term; if negative, it's a rational term with denominator). So the question is, how many terms have exponents of x that are integers.From the general term, as before, exponent of x is (frac{10 -5k}{2}). So we need (frac{10 -5k}{2}) to be an integer. Let's denote this exponent as m:( m = frac{10 -5k}{2} )We need m ∈ ℤ. Therefore, 10 -5k must be even. Let's write that:10 -5k ≡ 0 mod 210 is even, 5k is 5 times k. Since 5 is odd, 5k ≡ k mod 2. Therefore:10 -5k ≡ 0 -k mod 2 ⇒ -k ≡ 0 mod 2 ⇒ k ≡ 0 mod 2Thus, k must be even. Therefore, for k even, the exponent is integer, hence the term is rational. Since k ranges from 0 to 10 inclusive, the number of even k is 6 (k=0,2,4,6,8,10). Therefore, the number of rational terms is 6.Wait, let me confirm. If k is even, then m = (10 -5k)/2 must be integer. Let's check for k=0: (10 -0)/2 =5, integer. k=2: (10 -10)/2=0, integer. k=4: (10 -20)/2= -5, integer. k=6: (10 -30)/2=-10, integer. k=8: (10 -40)/2=-15, integer. k=10: (10 -50)/2=-20, integer. So all even k's result in integer exponents. For odd k's: Let's try k=1: (10 -5)/2=5/2=2.5, not integer. k=3: (10 -15)/2=-2.5, not integer. Similarly, k=5: (10 -25)/2=-7.5, etc. So indeed, only even k's give integer exponents, so number of rational terms is 6. Therefore, part (3) answer is 6.Wait, but the problem says "the number of rational terms". Is there a possibility that some terms might have the same exponent? For example, different k's leading to same exponent. Let's check:For k=0: exponent 5k=2: exponent 0k=4: exponent -5k=6: exponent -10k=8: exponent -15k=10: exponent -20So each even k gives a distinct exponent. Therefore, all these terms are distinct. So total 6 terms. Therefore, the answer is 6. So part (3) answer is 6.But let me think again. A term is considered rational if the exponent of x is an integer. Since x is a variable, and if the exponent is an integer, then the term is a rational function (if exponent is positive or negative integer). So yes, each term with integer exponent is a rational term, so 6 terms. So that's correct.Therefore, to summarize:1) n=102) Constant term=1803) Number of rational terms=6I think that's all. Let me just recheck part (2) once more. For k=2:Term is C(10,2)*(sqrt(x))^8*(-2/x^2)^2 = 45*x^4*(4/x^4) = 45*4=180. Yes, correct.So all answers are confirmed.Final Answer(1) boxed{10}(2) boxed{180}(3) boxed{6}</think>Given the expression ((sqrt{x} - frac{2}{x^2})^n), we need to solve three parts:1. Find the value of (n): - The sum of the binomial coefficients in the expansion is given as 1024. - The sum of the binomial coefficients in any binomial expansion ((a + b)^n) is (2^n). - Therefore, (2^n = 1024), which implies (n = 10) because (2^{10} = 1024). - Answer: (boxed{10})2. Find the constant term in the expansion: - The general term in the expansion is (binom{10}{k} (sqrt{x})^{10 - k} left(-frac{2}{x^2}right)^k). - The exponent of (x) in the term is (frac{10 - 5k}{2}). For the term to be constant, this exponent must be 0. - Solving (frac{10 - 5k}{2} = 0) gives (k = 2). - Substituting (k = 2) into the term: (binom{10}{2} (sqrt{x})^8 left(-frac{2}{x^2}right)^2 = 45 cdot x^4 cdot 4x^{-4} = 180). - Answer: (boxed{180})3. Find the number of rational terms in the expansion: - A term is rational if the exponent of (x) is an integer. - The exponent (frac{10 - 5k}{2}) must be an integer, implying (10 - 5k) must be even. - This requires (k) to be even. The even values of (k) from 0 to 10 are 0, 2, 4, 6, 8, 10, giving 6 terms. - Answer: (boxed{6}),<think>Okay, so I need to solve this problem about the binomial expansion of (√x - 2/x²)^n. There are three parts: finding n, the constant term, and the number of rational terms. Let me start with the first part.(1) Find the value of n.The problem says the sum of the binomial coefficients of all terms in its expansion is 1024. Hmm. Wait, the sum of the binomial coefficients in a binomial expansion (a + b)^n is obtained by setting a = 1 and b = 1, right? Because then each term becomes just the binomial coefficient. So if we have (1 + 1)^n = 2^n, which is the sum of the coefficients. But in this case, the binomial is (√x - 2/x²)^n. Wait, but to get the sum of the coefficients, we should set x = 1, right? Because substituting x = 1 would make each term just the coefficient times (1)^something, which is 1. So the sum of the coefficients is (√1 - 2/1²)^n = (1 - 2)^n = (-1)^n. But the problem states that this sum is 1024. Wait, (-1)^n = 1024? But 1024 is positive, and (-1)^n can only be 1 or -1. That can't be right. Did I do something wrong here?Wait, maybe the question is different. It says "the sum of the binomial coefficients of all terms in its expansion is 1024". Wait, binomial coefficients are just the coefficients without the constants from the terms. Wait, but in the expansion of (a + b)^n, the binomial coefficients are the C(n,k). But in this case, each term is C(n,k)*(√x)^(n-k)*(-2/x²)^k. So the coefficients would include the constants from each term, like C(n,k)*(-2)^k. So maybe the problem is referring to the sum of all those coefficients? So if we set x = 1, then we get the sum of all terms, which would be the sum of coefficients times (1)^something, which is just the sum of the coefficients. So that would be (√1 - 2/1²)^n = (1 - 2)^n = (-1)^n. But that's supposed to be 1024. But (-1)^n can't be 1024. Hmm, this is conflicting. Maybe I misunderstood the question.Wait, perhaps "the sum of the binomial coefficients of all terms" refers to the sum of the binomial coefficients without considering the other constants. But in the expansion of (a + b)^n, the coefficients are C(n,k). But in this case, the expansion is (√x - 2/x²)^n, so each term is C(n,k)*(√x)^(n - k)*(-2/x²)^k. So the coefficients would be C(n,k)*(-2)^k. So if we want the sum of the coefficients, that would be the sum of C(n,k)*(-2)^k for k from 0 to n. But that is equal to (1 - 2)^n = (-1)^n, which again is either 1 or -1. But the problem states that the sum is 1024. That doesn't add up. Maybe there's a mistake here.Wait, but maybe the problem is translated from Chinese, and the term "binomial coefficients" is meant to refer to the coefficients of the terms in the expansion, including the constants. So in that case, the sum would be obtained by setting x = 1, which gives (-1)^n. But then how does (-1)^n equal 1024? That's impossible because 1024 is 2^10. So maybe the problem is not referring to setting x = 1? Wait, perhaps the original question is different. Wait, let me check again.Original problem: Given (√x - 2/x²)^n, the sum of the binomial coefficients of all terms in its expansion is 1024.Wait, perhaps "binomial coefficients" here actually refer to the coefficients without the variables, meaning just the numerical coefficients. So for example, in the expansion, each term is of the form C(n,k)*(√x)^(n - k)*(-2/x²)^k. So the coefficient would be C(n,k)*(-2)^k. So the sum of all coefficients would be the sum over k of C(n,k)*(-2)^k. Which is equal to (1 - 2)^n = (-1)^n. So again, that would be 1 or -1. But the problem says 1024. That's not possible. Therefore, I must be misunderstanding the problem.Wait, maybe the problem is not in English, and there's a mistranslation. Alternatively, maybe "binomial coefficients" here refers to the coefficients in the expansion when written in terms of x, so considering the exponents. But even then, the coefficients are still C(n,k)*(-2)^k. So unless there's another interpretation. Alternatively, maybe the coefficients of the terms when expressed as polynomials, but even so, the sum would still be (-1)^n. Hmm.Wait, maybe the question is not about the expansion of (√x - 2/x²)^n, but about (sqrt(x) - 2/x²)^n, but perhaps the user made a typo? Let me check again.The problem is written as: (sqrt(x) - 2/x²)^n. So it's (x^{1/2} - 2x^{-2})^n. Then, the sum of the binomial coefficients of all terms in its expansion is 1024.Wait, maybe "binomial coefficients" here refer to the multinomial coefficients? But no, it's a binomial. Wait, perhaps the user is referring to the coefficients of the terms, considering the numerical factors. For example, each term is C(n,k) * (-2)^k * x^{(n - k)/2 - 2k}. So the coefficient is C(n,k)*(-2)^k. So the sum of all coefficients would be sum_{k=0}^n C(n,k)*(-2)^k = (1 - 2)^n = (-1)^n. But that equals 1024? That's not possible. Therefore, maybe there is a miscalculation here.Wait, unless the problem is referring to the sum of the absolute values of the coefficients? Then, sum_{k=0}^n C(n,k)*2^k = (1 + 2)^n = 3^n. If that were the case, 3^n = 1024. But 3^6 = 729, 3^7 = 2187, so that's not 1024. So that's not possible either. Hmm.Alternatively, maybe the problem is referring to the sum of the coefficients of the terms with integer exponents? But that would complicate. Wait, let's check the problem statement again.Original problem:Given (√x - 2/x²)^n, the sum of the binomial coefficients of all terms in its expansion is 1024.(1) Find n.Wait, perhaps in the original problem, the binomial coefficients refer to the coefficients when expressed in terms of a certain variable substitution. For example, if we let y = √x, then 1/x² = y^{-4}, so the expression becomes (y - 2y^{-4})^n. Then, the binomial expansion would have terms C(n,k)*y^{n - k}*(-2)^k*y^{-4k} = C(n,k)*(-2)^k*y^{n - 5k}. Then, the coefficients here are C(n,k)*(-2)^k. The sum of these coefficients would be when y = 1, which gives (1 - 2)^n = (-1)^n. Again, same problem. So I don't see how that sum can be 1024. Therefore, maybe the problem is mistyped? Or perhaps the original binomial is different.Wait, maybe the original binomial is (sqrt(x) + 2/x²)^n instead of minus? Then, the sum of coefficients would be (1 + 2)^n = 3^n. If 3^n = 1024, but 3^6 = 729, 3^7 = 2187, so that's not possible either. Hmm.Alternatively, if the binomial was (sqrt(x) + 1/x²)^n, then the sum would be (1 + 1)^n = 2^n. Then 2^n = 1024, which is 2^10 = 1024. Therefore, n = 10. So maybe there was a mistranslation or typo in the original problem. If the binomial is (sqrt(x) + 1/x²)^n, then the sum of coefficients is 2^n = 1024, so n=10.But in the problem as given, it's (sqrt(x) - 2/x²)^n. So unless the problem says "sum of the absolute values of the coefficients", which would be different. Alternatively, maybe the user meant "sum of the coefficients is 1024 in absolute value", but then (-1)^n = ±1024. But 1024 is 2^10, which is positive. So if (-1)^n = 1024, that's impossible. So I must be missing something.Wait, maybe the term "binomial coefficients" here refers to the coefficients C(n,k) without the other constants. So the sum of C(n,k) from k=0 to n is 2^n. But the problem says the sum is 1024. Therefore, 2^n = 1024, which implies n = 10. So maybe the problem is using "binomial coefficients" in the sense of just the combinatorial coefficients C(n,k), not multiplied by any other constants. So in the expansion, each term has a coefficient C(n,k)*(-2)^k, but if we ignore the (-2)^k part and just sum C(n,k), that sum is 2^n. If the problem is referring to that, then 2^n = 1024, so n=10. But is that the standard definition? Typically, binomial coefficients are C(n,k), but in the expansion, the coefficients include the constants from the terms. So I'm confused.Alternatively, perhaps the problem is stated correctly, but there's a different approach. Let's think again.If we have the expression (a + b)^n, the sum of the coefficients is obtained by setting a = 1 and b = 1. So in this case, a = sqrt(x) and b = -2/x². So setting x = 1 gives (1 - 2)^n = (-1)^n. If this is supposed to be 1024, then (-1)^n = 1024. But 1024 is positive, and (-1)^n is either 1 or -1. So this is impossible. Therefore, the problem must have a typo or mistranslation. Alternatively, perhaps the coefficients are being considered differently.Wait, maybe the original problem is in Chinese, and "binomial coefficients" was mistranslated. Maybe it refers to the coefficients of the terms when expanded, which include the numerical factors. So the sum of those coefficients is 1024. But as we saw, that sum is (-1)^n. So unless n is even, and (-1)^n = 1, but 1 ≠ 1024. So this is a problem. Therefore, unless the problem is incorrect, or I'm misinterpreting it.Alternatively, maybe the user made a mistake in writing the problem. Wait, perhaps the binomial is (sqrt(x) + 2/x²)^n instead of minus. Then the sum would be (1 + 2)^n = 3^n. But 3^6=729, 3^7=2187. Not 1024. Alternatively, if the coefficient is (sqrt(x) - 1/x²)^n, then the sum is (1 - 1)^n = 0. Hmm.Alternatively, perhaps the original problem had (sqrt(x) + 1/x)^n, then the sum is (1 + 1)^n = 2^n = 1024, so n=10. But that's not the given problem.Alternatively, maybe the binomial is (sqrt(x) - 2/x)^n. Then the sum would be (1 - 2)^n = (-1)^n. Still, same problem.Alternatively, maybe the problem refers to the expansion in terms of x, but with different exponents. Wait, but even so, the sum of coefficients is obtained by x=1.Wait, unless the problem is not referring to substituting x=1, but something else. Wait, perhaps the term "sum of the binomial coefficients" is meant to refer to the coefficients of the terms in the expansion when written in the form of a polynomial, but considering the exponents as variables. But that's not standard terminology.Alternatively, maybe the problem is referring to the sum of the coefficients of the terms that are binomial coefficients in some other sense. But I can't see how.Alternatively, maybe the problem is correct, and there is a trick here. Let's think. If the sum of the coefficients is 1024, which is 2^10. So maybe n=10. But then (-1)^10=1, which is not 1024. Wait, so even if n=10, the sum would be 1, not 1024. Therefore, there's a contradiction here.Wait, perhaps the problem is not substituting x=1, but x= some other value. Wait, but how would we get 1024? If we set x to a value such that the expression (sqrt(x) - 2/x²) equals 2, then 2^n=1024 implies n=10. Let's see: Suppose (sqrt(x) - 2/x²) = 2. Then we can solve for x. But is this necessary? The problem states that the sum of the coefficients is 1024. The standard way is to substitute x=1. But in this case, that gives -1^n. If that's supposed to be 1024, then there is an inconsistency.Alternatively, maybe there's a misunderstanding in the problem's phrasing. Maybe it's not the sum of coefficients, but the sum of the absolute values of the coefficients. Then, sum_{k=0}^n C(n,k)*2^k = (1 + 2)^n = 3^n. Then 3^n=1024. But 3^6=729, 3^7=2187, so n is not integer. Therefore, that's impossible. So maybe the problem is incorrect, or there's a typo. Alternatively, maybe the coefficient is (sqrt(x) + 2/x²)^n, then sum is (1 + 2)^n=3^n=1024. But that's not possible. Alternatively, maybe it's (sqrt(x) - 1/x²)^n, sum is (1 -1)^n=0. Not 1024.Wait, perhaps the problem is mistyped, and the binomial is (sqrt(x) + 2/x^2)^n, but the sum is 1024. Then (1 + 2)^n =3^n=1024. But 3^n=1024. Since 3^6=729, 3^7=2187. So that's not possible. Alternatively, if the binomial is (x - 2/x^2)^n, then setting x=1 gives (1 - 2)^n=(-1)^n. Again, same problem.Alternatively, if the binomial is (sqrt(x) + 2/x^3)^n, then x=1 gives (1 + 2)^n=3^n=1024. Still not possible.Alternatively, maybe the user intended (sqrt(x) - 2/x^3)^n. Then, sum of coefficients is (1 - 2)^n=(-1)^n. Still not 1024.Alternatively, maybe the original problem had different exponents. For example, (sqrt(x) - 2/x)^n. Then sum is (1 - 2)^n=(-1)^n.Alternatively, maybe (x - 2/x)^n. Then sum is (1 - 2)^n=(-1)^n.Alternatively, maybe (x^{1/2} - 2x^{-1})^n. Sum is (1 - 2)^n=(-1)^n.Alternatively, perhaps the problem is not from an English source, and there's a mistranslation. For example, maybe "sum of the binomial coefficients" actually refers to the sum of the terms in the expansion, not the coefficients. But even then, when x=1, the sum is (-1)^n. Alternatively, if x is set to another value to make the sum 1024. But how?Alternatively, perhaps the problem is asking for the sum of the coefficients of the terms with integer exponents. Let's explore that.In the expansion of (√x - 2/x²)^n, each term is C(n,k)*(√x)^{n - k}*(-2/x²)^k = C(n,k)*(-2)^k*x^{(n - k)/2 - 2k} = C(n,k)*(-2)^k*x^{(n -5k)/2}.So the exponent of x is (n -5k)/2. For a term to have an integer exponent, (n -5k)/2 must be integer. So n -5k must be even. So n must be congruent to 5k mod 2. But 5k is congruent to k mod 2, since 5 is odd. So n -k must be even, so n and k have the same parity.But the problem is asking for the sum of the coefficients of all terms. Wait, maybe the problem is mistyped, and instead of "sum of the binomial coefficients", it should be "sum of the coefficients of the terms with integer exponents". Then, that sum is 1024. So, in that case, we need to sum over all k where (n -5k)/2 is integer, the coefficients C(n,k)*(-2)^k. Then, this sum is 1024. That could be possible, but it's more complicated.But how would I approach that? Let's think. If n and k have the same parity, then n -5k is even. So, for each k such that k ≡ n mod 2, we include the term C(n,k)*(-2)^k. The sum over those terms is 1024. But this seems complicated. Let's try n=10. Then, sum over k even (since n=10 is even). So sum_{k even} C(10,k)*(-2)^k. Let's compute that. The sum over even k: C(10,0)*(-2)^0 + C(10,2)*(-2)^2 + ... + C(10,10)*(-2)^10. This is similar to [ (1 + (-2))^10 + (1 - (-2))^10 ] / 2 = [ (-1)^10 + 3^10 ] / 2 = [1 + 59049]/2 = 59050/2 = 29525. Which is way larger than 1024. So n=10 is too big.Alternatively, if n=5. Then sum over k odd and even? Wait, n=5 is odd. So sum over k odd. C(5,1)*(-2)^1 + C(5,3)*(-2)^3 + C(5,5)*(-2)^5 = 5*(-2) + 10*(-8) + 1*(-32) = -10 -80 -32 = -122. Not 1024.Alternatively, maybe n=0, sum is 1. Not 1024. Hmm.Alternatively, n=2. Sum over k even: k=0,2. C(2,0)*1 + C(2,2)*(-2)^2 = 1 + 1*4=5. Not 1024. Hmm.This approach is not leading me anywhere. Maybe the problem is indeed incorrect or there's a typo. Alternatively, perhaps I need to reinterpret the problem.Wait, going back, maybe the problem is saying "the sum of the binomial coefficients of all terms in its expansion is 1024". Now, binomial coefficients usually refer to C(n,k), so maybe the problem is asking for the sum of C(n,k) for k=0 to n, which is 2^n. Then, 2^n=1024, which is 2^10, so n=10. But in the expansion of (√x -2/x²)^n, the coefficients of the terms are C(n,k)*(-2)^k. So the sum of C(n,k) is 2^n, and the sum of coefficients is (-1)^n. So if the problem says the sum of the binomial coefficients (C(n,k)) is 1024, then n=10. If it's the sum of the coefficients (including the constants), then (-1)^n=1024, which is impossible. Therefore, maybe the problem is using "binomial coefficients" to refer to C(n,k), not the entire coefficient. Then, answer is n=10. Then, moving to part (2), find the constant term. Let's check if n=10 makes sense for part (2).So assuming n=10.(2) Find the constant term in the expansion.In the expansion of (√x -2/x²)^10, the general term is C(10,k)*(√x)^{10 -k}*(-2/x²)^k = C(10,k)*(-2)^k*x^{(10 -k)/2 -2k} = C(10,k)*(-2)^k*x^{(10 -5k)/2}.We need the exponent of x to be 0 for the constant term. Therefore:(10 -5k)/2 = 0 => 10 -5k = 0 => 5k =10 => k=2.Therefore, the constant term is C(10,2)*(-2)^2 = 45*4=180. So that would be 180.Then part (3), find the number of rational terms in the expansion.A rational term is one where the exponent of x is an integer. So (10 -5k)/2 must be integer. Which implies that 10 -5k is even. So 10 is even, 5k must be even. Since 5 is odd, k must be even. Therefore, k must be even numbers from 0 to 10. So k=0,2,4,6,8,10. That's 6 terms. Therefore, the number of rational terms is 6.But this all assumes that n=10, which we got from part (1) assuming that "sum of binomial coefficients" is 2^n=1024. However, the problem states "the sum of the binomial coefficients of all terms in its expansion is 1024". If binomial coefficients here refer to the C(n,k), then yes, sum is 2^n=1024, so n=10. But if binomial coefficients refer to the coefficients including the constants, then it's impossible. Therefore, the problem must mean the sum of the binomial coefficients C(n,k), which is 2^n=1024, so n=10. Then parts (2) and (3) follow.Therefore, the answers would be:1) n=102) constant term=1803) 6 rational terms.But to confirm, let's re-examine the problem statement again.Original problem: "the sum of the binomial coefficients of all terms in its expansion is 1024."If "binomial coefficients" refer to the combinatorial coefficients C(n,k), then the sum is indeed 2^n=1024, so n=10. If they refer to the coefficients in the expansion (including constants), then the sum is (-1)^n=1024 which is impossible. Therefore, the problem must be using the first definition. Therefore, the answer is n=10.Therefore, I think the answer is n=10, constant term 180, and 6 rational terms.Final Answer(1) boxed{10}(2) boxed{180}(3) boxed{6}</think>Given the binomial expansion ((sqrt{x} - frac{2}{x^2})^n), we need to solve three parts:1. Find the value of (n): - The sum of the binomial coefficients in the expansion is given as 1024. The sum of the binomial coefficients (C(n, k)) for a binomial expansion ((a + b)^n) is (2^n). Setting this equal to 1024, we get (2^n = 1024). Solving for (n), we find (n = 10).2. Find the constant term in the expansion: - The general term in the expansion of ((sqrt{x} - frac{2}{x^2})^{10}) is given by: [ C(10, k) cdot (sqrt{x})^{10 - k} cdot left(-frac{2}{x^2}right)^k = C(10, k) cdot (-2)^k cdot x^{frac{10 - 5k}{2}} ] - To find the constant term, the exponent of (x) must be zero: [ frac{10 - 5k}{2} = 0 implies 10 - 5k = 0 implies k = 2 ] - Substituting (k = 2) into the term, we get: [ C(10, 2) cdot (-2)^2 = 45 cdot 4 = 180 ]3. Find the number of rational terms in the expansion: - A term is rational if the exponent of (x) is an integer. The exponent is (frac{10 - 5k}{2}). For this to be an integer, (10 - 5k) must be even. Since 10 is even, (5k) must also be even, implying (k) must be even. The even values of (k) range from 0 to 10, which are 0, 2, 4, 6, 8, 10. This gives 6 rational terms. Final Answers1. (boxed{10})2. (boxed{180})3. (boxed{6})]

🔑:(1) The sum of the binomial coefficients of all terms is 1024, which means 2^n=1024. Solving this, we get n=10; (2) The general term of the expansion of ( sqrt {x}- frac {2}{x^{2}})^{n} is T_{r+1} = C_{10}^{r} (sqrt {x})^{10-r}(frac{-2}{x^{2}})^{r} = C_{10}^{r} (-2)^{r}x^{frac{10-5r}{2}}, Let frac{10-5r}{2}=0, then r=2, Thus, the constant term is: C_{10}^{2}(-2)^{2}=180; (3) Rational terms occur when frac{10-5r}{2} is an integer, which happens for r=0, 2, 4, 6, 8, 10. Therefore, there are boxed{6} rational terms.

🧠:[<think>Alright, let's try to tackle this problem. So, the question is about finding all two-digit numbers that satisfy a certain condition. Let me parse the problem again to make sure I understand it correctly.We have a two-digit number. When we add this number to the product of its tens and units digits, the result is exactly equal to the square of the sum of its tens and units digits. Hmm, okay. Let's break that down step by step.First, let's represent the two-digit number in terms of its digits. Let’s say the tens digit is 'a' and the units digit is 'b'. Then, the actual number can be written as 10a + b. For example, if the number is 34, then a = 3 and b = 4, so the number is 10*3 + 4 = 34. Got that part.Now, according to the problem, when we add this number (10a + b) to the product of its tens and units digits (which is a*b), the result should equal the square of the sum of its digits. The sum of the digits is a + b, so the square of that sum is (a + b)^2. Therefore, the equation we need to solve is:10a + b + a*b = (a + b)^2Let me write that out:10a + b + ab = (a + b)^2Let me make sure I got that right. The problem states, "when added to the product of its tens and units digits, is exactly equal to the square of the sum of its tens and units digits." So, number + (product of digits) = (sum of digits squared). Yep, that matches the equation I wrote.Now, let's simplify this equation. Let's expand the right-hand side (RHS):(a + b)^2 = a^2 + 2ab + b^2So substituting back into the equation:10a + b + ab = a^2 + 2ab + b^2Let me rearrange the equation to bring all terms to one side:10a + b + ab - a^2 - 2ab - b^2 = 0Simplify like terms:10a + b - a^2 - ab - b^2 = 0Let me write this as:- a^2 - b^2 - ab + 10a + b = 0Alternatively, multiplying both sides by -1 to make the quadratic terms positive:a^2 + b^2 + ab - 10a - b = 0Hmm, okay. Let's see if we can rearrange or factor this equation. Alternatively, maybe express it in terms of a or b and find possible integer solutions since a and b are digits. Since a is the tens digit, it must be between 1 and 9, inclusive, and b is the units digit, so between 0 and 9.Alternatively, maybe we can bring all terms to one side and see if we can factor or find a pattern. Let's consider:Original equation: 10a + b + ab = (a + b)^2Let me rearrange it as:(a + b)^2 - ab - 10a - b = 0Alternatively, perhaps expanding (a + b)^2:a^2 + 2ab + b^2 - ab - 10a - b = 0Which simplifies to:a^2 + ab + b^2 - 10a - b = 0Hmm, same as before. So the equation is a quadratic in both a and b. Maybe we can write this as a quadratic equation in one variable by expressing one variable in terms of the other.Alternatively, since a and b are digits with limited ranges (a from 1-9, b from 0-9), perhaps we can substitute possible values and check for solutions. But that might take longer. Maybe first try to manipulate the equation algebraically.Let me see:Starting again from 10a + b + ab = (a + b)^2Let me subtract 10a + b + ab from both sides to get:0 = (a + b)^2 - 10a - b - abWhich is the same as before. Let me factor where possible.Looking at (a + b)^2 - ab. That's a^2 + 2ab + b^2 - ab = a^2 + ab + b^2. So:a^2 + ab + b^2 - 10a - b = 0Hmm. Maybe group terms:a^2 - 10a + ab + b^2 - b = 0Take the terms with a:a^2 -10a + abAnd the terms with b:b^2 - bSo:a(a -10 + b) + b(b -1) = 0Hmm, not sure if that helps. Let's see. Maybe factor differently. Let's consider:a^2 + ab -10a + b^2 - b = 0Perhaps group a^2 -10a and b^2 - b, then the ab term:a(a -10) + b(b -1) + ab =0Wait, combining:a(a -10 + b) + b(b -1) =0Same as before. Maybe not particularly helpful. Let's consider rearranging the original equation:10a + b + ab = (a + b)^2Let me bring all terms to the left-hand side:10a + b + ab - (a + b)^2 = 0Expand (a + b)^2:10a + b + ab - a^2 - 2ab - b^2 = 0Simplify:10a + b - a^2 - ab - b^2 =0Which is the same as before. Let's try to rearrange terms:-a^2 +10a -ab + b - b^2 =0Multiply through by -1:a^2 -10a +ab -b + b^2 =0Hmm. Maybe group terms:a^2 -10a + b^2 -b + ab =0Alternatively, (a^2 + b^2) + (ab -10a -b) =0Still not obvious. Maybe try completing the square? Let's see.Looking at a^2 -10a, we can complete the square for a:a^2 -10a = (a -5)^2 -25Similarly, for b^2 -b:b^2 -b = (b -0.5)^2 -0.25So substitute these into the equation:(a -5)^2 -25 + (b -0.5)^2 -0.25 + ab =0So:(a -5)^2 + (b -0.5)^2 + ab =25.25Hmm, this introduces squares of (a -5) and (b -0.5), but also the cross term ab. Maybe not helpful, since a and b are integers, but the squares are fractions. Not sure. Maybe another approach.Alternatively, let's treat this as a quadratic equation in one variable. Let's suppose we fix a and solve for b, or vice versa.Let me consider treating the equation as a quadratic in b:Original equation: a^2 + ab + b^2 -10a -b =0Rewrite as:b^2 + (a -1)b + (a^2 -10a) =0So quadratic in b: coefficients are:Coefficient of b^2: 1Coefficient of b: (a -1)Constant term: (a^2 -10a)Therefore, solving for b:b = [ - (a -1) ± sqrt( (a -1)^2 -4 *1* (a^2 -10a) ) ] / 2Simplify the discriminant:D = (a -1)^2 -4(a^2 -10a)Expand (a -1)^2: a^2 -2a +1So D = a^2 -2a +1 -4a^2 +40aCombine like terms:a^2 -4a^2 = -3a^2-2a +40a = 38aAnd +1So D = -3a^2 +38a +1Therefore, the discriminant must be a perfect square for b to be an integer. So:D = -3a^2 +38a +1 must be a non-negative perfect square.But since a is an integer from 1 to 9, let's compute D for each a from 1 to 9 and see if it's a perfect square. If yes, then we can find integer solutions for b. Otherwise, no.Let me compute D for each a:a=1:D = -3(1)^2 +38(1) +1 = -3 +38 +1 = 36. 36 is a perfect square (6^2). Okay.a=2:D = -3(4) +76 +1 = -12 +76 +1 = 65. 65 is not a perfect square.a=3:D = -3(9) +114 +1 = -27 +114 +1 = 88. Not a perfect square.a=4:D = -3(16) +152 +1 = -48 +152 +1 = 105. Not a perfect square.a=5:D = -3(25) +190 +1 = -75 +190 +1 = 116. Not a perfect square.a=6:D = -3(36) +228 +1 = -108 +228 +1 = 121. 121 is a perfect square (11^2). Good.a=7:D = -3(49) +266 +1 = -147 +266 +1 = 120. Not a perfect square.a=8:D = -3(64) +304 +1 = -192 +304 +1 = 113. Not a perfect square.a=9:D = -3(81) +342 +1 = -243 +342 +1 = 100. 100 is a perfect square (10^2). Okay.So for a=1,6,9, the discriminant D is a perfect square. Let's check these cases.Case 1: a=1, D=36Compute b:b = [ - (a -1) ± sqrt(D) ] /2 = [ -(1 -1) ±6 ]/2 = [0 ±6]/2So b = (0 +6)/2 = 3 or (0 -6)/2 = -3But b is a digit, so it must be between 0 and 9. Therefore, b=3 is valid, b=-3 is invalid.Therefore, for a=1, b=3. The number is 13. Let's check if this satisfies the original condition.Original condition: 13 + (1*3) = (1 +3)^213 +3 = 16, and (4)^2=16. Yes, that works.Case 2: a=6, D=121Compute b:b = [ -(6 -1) ±11 ]/2 = [ -5 ±11 ]/2So two possibilities:b = (-5 +11)/2 =6/2=3b = (-5 -11)/2 = -16/2=-8Again, b must be between 0 and 9, so b=3 is valid, b=-8 is invalid.Therefore, for a=6, b=3. The number is 63. Let's check the original condition.63 + (6*3) =63 +18=81. Sum of digits:6+3=9, 9^2=81. So 81=81. Correct.Case3: a=9, D=100Compute b:b = [ -(9 -1) ±10 ]/2 = [ -8 ±10 ]/2So two possibilities:b = (-8 +10)/2 =2/2=1b = (-8 -10)/2 = -18/2=-9Again, b must be between 0 and 9, so b=1 is valid, b=-9 invalid.Therefore, for a=9, b=1. The number is 91. Let's check.91 + (9*1)=91 +9=100. Sum of digits:9+1=10, 10^2=100. 100=100. Correct.So the possible numbers are 13, 63, 91.Wait, but let me check if there are other possible values of a. Wait, we checked a=1 to a=9, only a=1,6,9 gave perfect squares for D. So these are the only possibilities.But let me just confirm by testing all possible a and b. Since a from 1-9, b from 0-9. Let's check all possible combinations where the equation holds.Alternatively, since the equation is 10a + b + ab = (a + b)^2, we can compute both sides for each two-digit number and check. But that would be tedious, but maybe we can check if there are other numbers that we might have missed.Wait, according to our algebra, only a=1,6,9 give valid b's, leading to numbers 13,63,91. Let me verify with another approach.Suppose we take the original equation:10a + b + ab = (a + b)^2Let me rearrange this as:10a + b = (a + b)^2 - abExpand the right-hand side:(a + b)^2 -ab = a^2 + 2ab + b^2 -ab =a^2 +ab +b^2Therefore:10a + b = a^2 +ab +b^2Which is the same as before.Alternatively, perhaps we can write the equation as:10a + b = a^2 +ab +b^2Then, 10a = a^2 +ab +b^2 -b10a = a^2 +ab +b(b -1)Not sure. Alternatively, factor a out of some terms:10a = a(a + b) + b(b -1)Hmm. So 10a = a(a + b) + b(b -1)Hmm, maybe we can write this as:10a -a(a + b) = b(b -1)Factor left side:a[10 - (a + b)] = b(b -1)So:a(10 -a -b) = b(b -1)Interesting. So now we have:a(10 - a - b) = b(b -1)Since a and b are digits (a from 1-9, b from 0-9), perhaps we can analyze possible values here.Note that the right-hand side, b(b -1), is non-negative only when b >=1 (since if b=0, it's 0*(-1)=0, which is zero. For b=1, 1*0=0, for b=2, 2*1=2, etc. So RHS can be 0 or positive.Left-hand side is a(10 -a -b). Let's see. Since a is positive (1-9), and 10 -a -b. Let's note that 10 -a -b must be non-negative for the left-hand side to be non-negative (since a is positive). Therefore:10 -a -b >=0 => a + b <=10So possible values of a and b must satisfy a + b <=10.Moreover, the equation is a(10 -a -b) = b(b -1)Since a and b are integers, perhaps we can iterate over possible values of a and b where a + b <=10 and find solutions.Let's try this approach.Since a ranges from 1 to 9, and b from 0 to 9, but with a + b <=10.Let me consider possible values of a from 1 to 9:For each a, b can range from 0 to (10 -a).For each (a, b), check if a(10 -a -b) = b(b -1)Let's start with a=1:a=1, b from 0 to 9 (since a +b <=10, but a=1, so b can be up to 9, but let's check:Equation: 1*(10 -1 -b) = b(b -1) => 1*(9 -b) = b^2 -bSo 9 -b = b^2 -b => 9 = b^2 => b=3 or b=-3. Since b is non-negative digit, b=3. Which gives us number 13, as before.a=2:Equation: 2*(10 -2 -b)=b(b -1) => 2*(8 -b)=b^2 -bLeft: 16 -2bRight: b^2 -bSet equal: 16 -2b = b^2 -b => b^2 +b -16 =0Solutions: b = [-1 ± sqrt(1 +64)]/2 = [-1 ± sqrt(65)]/2. Not integers. So no solution.a=3:3*(10 -3 -b)=b(b -1) => 3*(7 -b)=b^2 -b21 -3b =b^2 -b => b^2 +2b -21=0Solutions: b = [-2 ± sqrt(4 +84)]/2 = [-2 ± sqrt(88)]/2 = [-2 ± 2*sqrt(22)]/2 = [-1 ± sqrt(22)]. Not integers. No solution.a=4:4*(10 -4 -b)=b(b -1) =>4*(6 -b)=b^2 -b24 -4b = b^2 -b => b^2 +3b -24=0Solutions: b = [-3 ± sqrt(9 +96)]/2 = [-3 ± sqrt(105)]/2. Not integers.a=5:5*(10 -5 -b)=b(b -1) =>5*(5 -b)=b^2 -b25 -5b = b^2 -b =>b^2 +4b -25=0Solutions: b = [-4 ± sqrt(16 +100)]/2 = [-4 ± sqrt(116)]/2. sqrt(116) is irrational. No solution.a=6:6*(10 -6 -b)=b(b -1) =>6*(4 -b)=b^2 -b24 -6b = b^2 -b =>b^2 +5b -24=0Solutions: b = [-5 ± sqrt(25 +96)]/2 = [-5 ± sqrt(121)]/2 = [-5 ±11]/2Thus, b=(6)/2=3 or b=(-16)/2=-8. Only b=3 is valid. So number is 63. As before.a=7:7*(10 -7 -b)=b(b -1) =>7*(3 -b)=b^2 -b21 -7b =b^2 -b =>b^2 +6b -21=0Solutions: b=[-6 ± sqrt(36 +84)]/2 = [-6 ± sqrt(120)]/2 = [-6 ± 2*sqrt(30)]/2= [-3 ± sqrt(30)]. Not integers.a=8:8*(10 -8 -b)=b(b -1) =>8*(2 -b)=b^2 -b16 -8b =b^2 -b =>b^2 +7b -16=0Solutions: b=[-7 ± sqrt(49 +64)]/2 = [-7 ± sqrt(113)]/2. Not integers.a=9:9*(10 -9 -b)=b(b -1) =>9*(1 -b)=b^2 -b9 -9b =b^2 -b =>b^2 +8b -9=0Solutions: b=[-8 ± sqrt(64 +36)]/2 = [-8 ± sqrt(100)]/2 = [-8 ±10]/2Thus, b=(2)/2=1 or b=(-18)/2=-9. Only b=1 is valid. Number is 91. As before.So this method also gives us the numbers 13,63,91. Therefore, confirming the previous result.Therefore, the two-digit numbers satisfying the given condition are 13, 63, and 91.Final AnswerThe two-digit numbers are boxed{13}, boxed{63}, and boxed{91}.</think>Given a two-digit number, we need to find all such numbers that satisfy the condition: the number plus the product of its tens and units digits equals the square of the sum of its digits. Let the tens digit be ( a ) and the units digit be ( b ). The number can be represented as ( 10a + b ). The given condition translates to the equation:[10a + b + ab = (a + b)^2]Expanding and simplifying the equation:[10a + b + ab = a^2 + 2ab + b^2][10a + b + ab - a^2 - 2ab - b^2 = 0][-a^2 - ab - b^2 + 10a + b = 0][a^2 + ab + b^2 - 10a - b = 0]We solve this quadratic equation by considering ( a ) and ( b ) as digits ( ( a ) from 1 to 9, ( b ) from 0 to 9). We check the discriminant of the quadratic equation in ( b ):[b^2 + (a - 1)b + (a^2 - 10a) = 0]The discriminant ( D ) must be a perfect square:[D = (a - 1)^2 - 4(a^2 - 10a)][D = -3a^2 + 38a + 1]Checking values of ( a ) from 1 to 9, we find that ( D ) is a perfect square for ( a = 1, 6, 9 ):- For ( a = 1 ), ( D = 36 ), leading to ( b = 3 ) (number 13).- For ( a = 6 ), ( D = 121 ), leading to ( b = 3 ) (number 63).- For ( a = 9 ), ( D = 100 ), leading to ( b = 1 ) (number 91).Verification confirms these numbers satisfy the original condition. Thus, the two-digit numbers are (boxed{13}), (boxed{63}), and (boxed{91}).,<think>Okay, let's try to figure out this problem. So, we have a two-digit number. Let me denote the tens digit as 'a' and the units digit as 'b'. Then, the two-digit number itself can be represented as 10a + b. The problem states that when this number is added to the product of its tens and units digits, the result is exactly equal to the square of the sum of its tens and units digits. Let me write that out in equation form to make it clearer.So, the number is 10a + b. The product of its tens and units digits is a * b. Adding those together gives (10a + b) + (a * b). This should equal the square of the sum of the digits. The sum of the digits is (a + b), so the square of that is (a + b)^2. Therefore, the equation we need is:10a + b + a*b = (a + b)^2Hmm, okay. Let's simplify both sides and see if we can find a relationship between a and b. Let's expand the right-hand side first:(a + b)^2 = a^2 + 2ab + b^2So the equation becomes:10a + b + ab = a^2 + 2ab + b^2Let's move all terms to one side to set the equation to zero. Subtract (10a + b + ab) from both sides:0 = a^2 + 2ab + b^2 - 10a - b - abSimplify the terms:a^2 + (2ab - ab) + b^2 - 10a - b = 0Which simplifies to:a^2 + ab + b^2 - 10a - b = 0Hmm, so we have:a^2 + ab + b^2 - 10a - b = 0This is a quadratic in terms of a and b. Since a and b are digits, they can only take integer values from 0 to 9. However, since it's a two-digit number, a can't be 0. So a is from 1 to 9, and b is from 0 to 9.Given that we have two variables here, maybe we can express one variable in terms of the other and then check possible values.Alternatively, rearrange the equation to make it more manageable. Let's try grouping terms.Looking at the equation:a^2 + ab + b^2 - 10a - b = 0Let me see if I can factor some parts. Hmm, not obvious. Maybe rearrange terms:a^2 - 10a + ab + b^2 - b = 0Take a common factor of a from the first two terms:a(a - 10) + b(a) + b^2 - b = 0Wait, that might not help much. Alternatively, group terms with a and terms with b:a^2 -10a + ab + b^2 - b = 0Take a^2 -10a + ab = a(a -10 + b)And then b^2 - b = b(b -1)So we have:a(a -10 + b) + b(b -1) = 0Hmm, not sure if that helps. Let me try a different approach. Let's consider that since a and b are digits, their possible values are limited, so perhaps we can iterate through possible a and b values. Though maybe there's a smarter way.Alternatively, let's rewrite the original equation:10a + b + ab = (a + b)^2Let's bring all terms to one side:(a + b)^2 - ab -10a - b = 0Expanding (a + b)^2 as before:a^2 + 2ab + b^2 -ab -10a -b = 0Which simplifies to:a^2 + ab + b^2 -10a -b = 0Same equation as before. Maybe we can express this as a quadratic in a or b. Let's try treating it as a quadratic in a.So, arranging terms:a^2 + (b -10)a + (b^2 - b) = 0This is a quadratic equation in terms of a. For each b, we can solve for a and see if we get integer solutions between 1 and 9.Alternatively, perhaps for each a from 1 to 9, we can solve for b from 0 to 9.But since solving a quadratic might be a bit tedious, but maybe manageable.Alternatively, let's see if we can factor this equation or find some relationship.Looking at the equation again:10a + b + ab = (a + b)^2Alternatively, let's move all terms to the right-hand side:(a + b)^2 - ab -10a -b = 0Wait, another thought. Let's write (a + b)^2 - ab = a^2 + 2ab + b^2 - ab = a^2 + ab + b^2So, the equation is:a^2 + ab + b^2 = 10a + b + abWait, the left side is a^2 + ab + b^2, and the right side is 10a + b + ab. Let's subtract (ab) from both sides:a^2 + b^2 = 10a + bSo, simplifying the equation, we actually get:a^2 + b^2 = 10a + bThat's a simpler equation. Wait, how?Original equation:10a + b + ab = (a + b)^2Which expands to:10a + b + ab = a^2 + 2ab + b^2Subtract 10a + b + ab from both sides:0 = a^2 + 2ab + b^2 -10a -b -abWhich simplifies to:0 = a^2 + ab + b^2 -10a -bBut if we rearrange the original equation:10a + b + ab = a^2 + 2ab + b^2Then subtract ab from both sides:10a + b = a^2 + ab + b^2Wait, but that contradicts the previous step. Wait, no. Wait, original equation:10a + b + ab = (a + b)^2 = a^2 + 2ab + b^2So, moving all terms to the right:0 = a^2 + 2ab + b^2 -10a -b -abWhich is:0 = a^2 + ab + b^2 -10a -bBut if we subtract 10a + b from both sides:Wait, maybe I made a mistake here. Let me check again.Original equation:10a + b + ab = (a + b)^2Which is:10a + b + ab = a^2 + 2ab + b^2Subtract 10a + b + ab from both sides:0 = a^2 + 2ab + b^2 -10a -b -abWhich simplifies to:0 = a^2 + ab + b^2 -10a -bSo, the equation is a^2 + ab + b^2 -10a -b = 0But in the previous step, when I tried to subtract ab, that might have been an error.Alternatively, maybe I can rearrange the original equation as follows:From 10a + b + ab = (a + b)^2Subtract 10a + b from both sides:ab = (a + b)^2 -10a -bWhich is:ab = a^2 + 2ab + b^2 -10a -bThen subtract ab from both sides:0 = a^2 + ab + b^2 -10a -bSo, same equation again.Alternatively, let's consider the equation:a^2 + ab + b^2 -10a -b = 0Maybe we can complete the square for a or b.Let's try with a. Let's write the equation as:a^2 + (b -10)a + (b^2 - b) = 0This is a quadratic in a. The solutions for a would be:a = [-(b -10) ± sqrt((b -10)^2 - 4*1*(b^2 - b))]/2Simplify the discriminant:D = (b -10)^2 -4(b^2 - b)= b^2 -20b +100 -4b^2 +4b= -3b^2 -16b +100Hmm, since a must be a positive integer between 1 and 9, the discriminant must be a perfect square and positive. So D must be non-negative and a perfect square.So, let's compute D for each possible b (0-9) and see if it's a perfect square.Let's compute D = -3b^2 -16b +100 for b from 0 to 9:For b=0: D = -0 -0 +100 = 100. sqrt(100)=10. So possible.For b=1: D = -3(1) -16(1) +100 = -3 -16 +100=81. sqrt(81)=9.For b=2: D = -3(4) -32 +100= -12 -32 +100=56. 56 is not a perfect square.For b=3: D=-3(9)-48+100= -27 -48 +100=25. sqrt(25)=5.For b=4: D=-3(16)-64 +100= -48 -64 +100= -12. Negative, discard.For b=5: D=-3(25)-80 +100= -75 -80 +100= -55. Negative.For b=6: D=-3(36)-96 +100= -108 -96 +100= -104. Negative.For b=7: D=-3(49)-112 +100= -147 -112 +100= -159. Negative.For b=8: D=-3(64)-128 +100= -192 -128 +100= -220. Negative.For b=9: D=-3(81)-144 +100= -243 -144 +100= -287. Negative.So discriminant D is a perfect square only for b=0,1,3.So let's check those.For b=0:D=100. So a = [-(0 -10) ±10]/2 = [10 ±10]/2.Thus:a=(10+10)/2=20/2=10, which is invalid (a must be <=9)a=(10-10)/2=0/2=0, which is invalid (a must be >=1)So no solution for b=0.For b=1:D=81. So a = [-(1 -10) ±9]/2 = [9 ±9]/2.Thus:a=(9+9)/2=18/2=9a=(9-9)/2=0/2=0, invalid.So a=9. Let's check if this satisfies the original equation.Number is 10a + b = 90 +1=91Product of digits: 9*1=9Sum of digits:9+1=10, square is 100Check:91 +9=100. 91+9=100. Yes, 100=100. So 91 is a solution.Next, b=3:D=25. So a = [-(3 -10) ±5]/2 = [7 ±5]/2.Thus:a=(7+5)/2=12/2=6a=(7-5)/2=2/2=1So two possible a's:6 and1.Check a=6, b=3:Number is 10*6 +3=63Product of digits:6*3=18Sum of digits:6+3=9, square is81Check:63 +18=81. 63+18=81. Yes, 81=81. So 63 is a solution.Check a=1, b=3:Number is10*1 +3=13Product of digits:1*3=3Sum of digits:1+3=4, square is16Check:13 +3=16. 13+3=16. Yes, 16=16. So 13 is a solution.So the solutions are a=9,b=1 (91), a=6,b=3 (63), and a=1,b=3 (13). Wait, but a=1,b=3 gives 13. Let me check again if that's correct.Yes, 13: digits 1 and 3. 13 + (1*3) =13 +3=16. Sum of digits:1+3=4. 4 squared is16. So yes, that's correct.Now, are there any other possible b's? We saw that for b=2, D=56, which isn't a perfect square. For b=4 onwards, D becomes negative. So only b=0,1,3 give possible D, and among them, only b=1,3 yield valid a's.Wait, but when we considered b=0, the a's were 10 and 0, which are invalid, so no solution there.So the numbers are 13,63,91.Wait, but let's check if there are any other possibilities we might have missed.Wait, perhaps in the process of solving the quadratic, but since we considered all possible b's (0-9) and found only b=1,3 yield valid a's, and we found the numbers 13,63,91. Let me verify each:13:1*3 =313+3=16(1+3)^2=16. Correct.63:6*3=1863+18=81(6+3)^2=81. Correct.91:9*1=991+9=100(9+1)^2=100. Correct.So these three numbers satisfy the condition.Wait, but let's check another number, say 33. Just to see.33: 3*3=9. 33+9=42. Sum of digits is6, square is36. 42≠36. So no.Or 43:43 +12=55. (4+3)^2=49. 55≠49.So no. So seems like 13,63,91 are the only ones.Alternatively, perhaps 86:86 +48=134. Sum of digits 14, square 196. 134≠196. Nope.Another way: since we already did the discriminant approach and found only those three, maybe that's all.Hence, the answer should be 13,63,91.Final AnswerThe two-digit numbers are boxed{13}, boxed{63}, and boxed{91}.</think>Given a two-digit number, we need to find all such numbers that when added to the product of its tens and units digits, the result is equal to the square of the sum of its digits. Let the tens digit be (a) and the units digit be (b). The number can be represented as (10a + b). The problem translates to the equation:[10a + b + ab = (a + b)^2]Expanding and simplifying the equation:[10a + b + ab = a^2 + 2ab + b^2][a^2 + ab + b^2 - 10a - b = 0]We solve this quadratic equation by considering the discriminant for each possible value of (b) (0 to 9) and checking if it results in a valid integer solution for (a) (1 to 9). The discriminant (D) for the quadratic equation in (a) is:[D = (b - 10)^2 - 4(b^2 - b)]Calculating (D) for each (b) from 0 to 9, we find that (D) is a perfect square only for (b = 1) and (b = 3). Solving for (a) in these cases:- For (b = 1), (a = 9)- For (b = 3), (a = 6) and (a = 1)Checking these solutions:- (a = 9), (b = 1) gives the number 91, which satisfies (91 + 9 times 1 = 100 = (9 + 1)^2)- (a = 6), (b = 3) gives the number 63, which satisfies (63 + 6 times 3 = 81 = (6 + 3)^2)- (a = 1), (b = 3) gives the number 13, which satisfies (13 + 1 times 3 = 16 = (1 + 3)^2)Thus, the two-digit numbers are (boxed{13}), (boxed{63}), and (boxed{91}).]

🔑:From the problem, we can derive the equation10 x+y+x y=(x+y)^{2} text {. }Rearranging the above equation into a quadratic equation with x as the main variable, we getx^{2}+(y-10) x+left(y^{2}-yright)=0 text {. }If the required x exists, thenbegin{array}{l}Delta=(y-10)^{2}-4left(y^{2}-yright) geqslant 0 Rightarrow 3 y^{2}+16 y-100 leqslant 0 .end{array}When y geqslant 4, the above inequality does not hold.Therefore, we only need to discuss the cases y=0,1,2,3, and we find that the numbers satisfying the condition are 13,63,91.

🧠:[<think>Okay, so the problem is: given n real numbers (where n is at least 3) whose squares sum to 1, we need to find the maximum possible value of the minimum distance between any two numbers. In other words, arrange the numbers on the real line such that the smallest gap between any pair is as large as possible, but their squares still add up to 1. Hmm, interesting.Let me try to rephrase. We have numbers a1, a2, ..., an. The minimum of |ai - aj| for all i < j is called m. We need to maximize this m given that the sum of their squares is 1. So, we want the numbers to be as spread out as possible, but not so spread out that their squares can't sum to 1. Got it.First thought: To maximize the minimal distance between any two points, they should be equally spaced. Like, if you have points on a line, the way to maximize the smallest gap is to space them evenly. But here, there's a constraint on the sum of squares. So maybe arranging them symmetrically around 0?Wait, if they are equally spaced, but symmetrically placed around 0, then maybe we can have pairs like -k, -k + m, ..., 0, ..., k - m, k? But for n numbers, if n is even, maybe we can split them into pairs, and if odd, there's a middle one at 0. Hmm, but the minimal distance would be m, and we need to maximize m such that the sum of squares is 1.Alternatively, perhaps the maximum minimal distance occurs when the numbers are arranged in an arithmetic progression? Let me consider that.Suppose the numbers are arranged in an arithmetic progression with common difference d. Then the minimal distance between any two consecutive numbers is d. So the minimal distance m would be d. If we can model the numbers as being spaced at intervals of d, then the sum of their squares would be 1. But since they need to be symmetrically placed around 0 for the sum of squares to be minimized? Wait, no, actually, to make the sum of squares as small as possible given the spacing, but here we have a fixed sum of squares. Wait, no, our constraint is that the sum of squares is 1. So given that, we need to arrange the numbers such that their spacing is maximized.Wait, maybe if the numbers are equally spaced around 0, symmetric. Let's try an example with small n, like n=3.For n=3, suppose we have three numbers: -a, 0, a. Then the minimal distance is a. Sum of squares: 2a² =1, so a = 1/√2. Then minimal distance is 1/√2 ≈0.707. But wait, but maybe we can do better by spacing them differently. Suppose instead they are equally spaced at intervals of m. Let me see. Let's say the three numbers are x, x + m, x + 2m. But since they need to be symmetrically placed? Wait, no, because if they are not symmetric, the sum of squares might be larger. Maybe the symmetric arrangement is optimal.Wait, for three numbers, if we set them as -m/2, 0, m/2. Then the minimal distance is m/2. Sum of squares: (m²/4) + 0 + (m²/4) = m²/2. So m²/2 =1 → m = √2 ≈1.414. But then the minimal distance between them is m/2 ≈0.707, same as before. So actually, same result. Hmm.But maybe there's a better arrangement. Suppose we place them at points -a, b, a, where a > b >0. Then minimal distance is min(a - b, b + a). Let's assume that a - b is the minimal distance, so we need to maximize a - b, such that 2a² + b² =1. Then set m = a - b. To maximize m, we can set up equations. Let m = a - b, so b = a - m. Then substitute into the sum: 2a² + (a - m)² =1. But then we have two variables a and m. Hmm, but how do we relate them? Maybe take derivative.Alternatively, perhaps the maximum minimal distance is achieved when all the distances are equal? That is, the points are equally spaced. But in 1D, equally spaced points would have the same minimal distance between consecutive pairs. But how does that interact with the sum of squares?Wait, maybe if we arrange the numbers in a symmetric arithmetic progression. Let's say for n numbers, if n is even, they are at positions ±(m/2), ±(3m/2), ..., ±((n-1)m/2). If n is odd, they are at 0, ±m, ±2m, ..., ±((n-1)/2)m). Then sum of squares would be 2*( (m/2)^2 + (3m/2)^2 + ... + ((n-1)m/2)^2 ) for even n, or 2*(m^2 + (2m)^2 + ... + ((n-1)/2 m)^2 ) + 0 for odd n.Wait, let's compute that for even n. Let n = 2k. Then the numbers are ±(m/2), ±(3m/2), ..., ±((2k -1)m)/2. The sum of squares is 2* [ (m/2)^2 + (3m/2)^2 + ... + ((2k -1)m/2)^2 ] = 2*(m²/4)(1 + 9 + ... + (2k -1)^2 ). The sum inside is the sum of squares of the first k odd numbers. The formula for the sum of squares of first k odd numbers is k(2k -1)(2k +1)/3. Wait, is that correct? Let's check for k=1: 1²=1. For k=2: 1 +9=10. Formula: 2(4 -1)(4 +1)/3 = 2*3*5/3=10. Correct. For k=3: 1 +9 +25=35. Formula: 3(6 -1)(6 +1)/3= 3*5*7/3=35. Yes, so the formula is k(2k -1)(2k +1)/3. So the sum of squares becomes 2*(m²/4)*(k(2k -1)(2k +1)/3) = (m²/2)*(k(2k -1)(2k +1)/3). Then set this equal to 1 and solve for m.Similarly, for odd n. Let n=2k+1. The numbers are 0, ±m, ±2m, ..., ±km. Sum of squares: 0 + 2*(m² + (2m)^2 + ... + (km)^2 ) = 2*m²*(1 + 4 + ... + k² ). The sum inside is the sum of squares up to k, which is k(k +1)(2k +1)/6. Therefore, total sum is 2*m²*(k(k +1)(2k +1)/6 ) = (m²/3)*k(k +1)(2k +1). Set equal to 1 and solve for m.So then, in both cases, we can write m in terms of n. Let me write it for general n.But before that, is this the optimal arrangement? That is, arranging the numbers in an arithmetic progression symmetric around 0? Maybe. But maybe there's a different arrangement where the minimal distance is larger. Wait, but how?Suppose instead of spacing them equally, we cluster some numbers closer and spread others apart. But then the minimal distance would be determined by the closest pair. So to maximize the minimal distance, all pairs should be at least m apart. Therefore, the optimal arrangement should have all consecutive pairs exactly m apart. So equally spaced. But in 1D, equally spaced points would form an arithmetic progression. But if we want them to be symmetric, then yes, they would be as I described above.Therefore, perhaps the maximum minimal distance is achieved when the numbers are equally spaced in a symmetric arithmetic progression. Then, the sum of squares is calculated as above, and we can solve for m.Therefore, for even n=2k, the sum is (m²/2)*(k(2k -1)(2k +1)/3)=1. So m²= 2*3 / [k(2k -1)(2k +1)] = 6 / [k(2k -1)(2k +1)]. Therefore, m= sqrt(6 / [k(2k -1)(2k +1)]). Similarly, for odd n=2k+1, sum is (m²/3)*k(k +1)(2k +1)=1. So m²= 3 / [k(k +1)(2k +1)], hence m= sqrt(3 / [k(k +1)(2k +1)]).But let's check with n=3. For n=3, k=1 (since n=2k+1=3). Then m= sqrt(3 / [1*2*3])= sqrt(3/6)=sqrt(1/2)=1/√2≈0.707. Which matches the earlier example. So that seems right.Similarly, for n=4 (k=2), m= sqrt(6 / [2*(4 -1)*(4 +1)/3])? Wait, wait, let me substitute correctly. For n=2k=4, k=2. So m= sqrt(6 / [2*(2*2 -1)(2*2 +1)])=sqrt(6 / [2*3*5])=sqrt(6/30)=sqrt(1/5)=1/√5≈0.447.Wait, but let's check with actual points. For n=4, the numbers would be ±m/2, ±3m/2. Then sum of squares is 2*( (m/2)^2 + (3m/2)^2 )= 2*( m²/4 +9m²/4 )=2*(10m²/4)=5m². Set equal to 1, so m=1/√5≈0.447. So the minimal distance is m=1/√5≈0.447. But wait, in this case, the minimal distance is m. Wait, but between the closest points, which would be between -3m/2 and -m/2, which is a distance of m. Similarly between -m/2 and m/2 is m, and between m/2 and 3m/2 is m. So yes, all consecutive pairs are m apart. So m is indeed the minimal distance. So that seems correct.But now, is this the maximum possible m? Or could there be a different arrangement where the minimal distance is larger? For example, if we place points not in arithmetic progression but in some other way. Let's see.Suppose for n=3, instead of placing them at -1/√2, 0, 1/√2, maybe place them at -a, b, a. Then the minimal distance is min(a - b, b + a). Wait, if we set a - b = b + a, then that would require -b = b, so b=0. So that gives us the symmetric case. If instead, we have a - b > b + a, then minimal distance would be b + a. But since a and b are positive, b + a is larger than a - b (if b>0). Therefore, the minimal distance would be a - b. To maximize that, we need to maximize a - b, but under 2a² + b² =1. So let's set up the optimization.Let m = a - b. Then b = a - m. Substitute into the equation: 2a² + (a - m)² =1. Expand: 2a² + a² - 2am + m² =1 → 3a² -2am +m² =1. We need to maximize m. So, we can treat this as an equation in a for given m. To have real solutions, the discriminant must be non-negative. The equation is quadratic in a: 3a² -2ma + (m² -1)=0. Discriminant D=4m² -12(m² -1)=4m² -12m² +12= -8m² +12. For real solutions, D≥0 → -8m² +12 ≥0 →8m² ≤12 →m² ≤12/8=3/2→m≤√(3/2)≈1.225. But since in our symmetric case, m=1/√2≈0.707. So this suggests that m can be up to 1.225, but that can't be, because in that case, the numbers would be too spread out. Wait, but we have the constraint that 2a² + b²=1, so even if m is large, a and b would have to be scaled accordingly.Wait, perhaps there's a mistake here. Let's re-examine. If we set m = a - b, and then express in terms of a, then 3a² -2am +m²=1. To find real a for given m, the discriminant must be non-negative, which gives us m ≤ sqrt(3/2). But this m is not the minimal distance, but rather a variable we defined as a - b. However, the minimal distance could also be between -a and b, which is a + b. Wait, in our previous setup, the points are -a, b, a. So the distances are:From -a to b: b + aFrom b to a: a - bFrom -a to a: 2aSo the minimal distance is the minimum of (a + b, a - b, 2a). But since a and b are positive, and if we want to maximize the minimal distance, we need all these distances to be at least m. But if we set m = a - b, then a + b must also be ≥ m. Since a + b ≥ a - b (as b ≥0), so as long as m = a - b, then the minimal distance is m, and a + b is ≥ m. Also, 2a must be ≥ m. Since a = m + b, 2(m + b) ≥ m → m + 2b ≥0, which is always true.But the key is, by setting m = a - b, and then solving for a in terms of m, we can see how large m can be. But this might not correspond to the maximum minimal distance, since there could be other configurations where the minimal distance is between other pairs.Wait, maybe I need to formalize this. Let's consider three points: left, middle, right. Let them be -a, b, a, with -a < b < a. Then the distances are:Left to middle: b + aMiddle to right: a - bLeft to right: 2aSo the minimal distance is the smallest of these three. To maximize the minimal distance, we need all three distances to be at least m, and we want the maximum m such that this is possible. So we need:b + a ≥ ma - b ≥ m2a ≥ mBut the minimal distance is the minimum of these three. To have the minimal distance as large as possible, we need all three distances to be equal. That is:b + a = a - b = 2aBut solving these equations:From b + a = a - b: 2b =0 → b=0From a - b =2a: -b =a → b= -a. But since b is between -a and a, and positive, this is impossible unless a=0, which can't be. Therefore, it's impossible for all three distances to be equal. Therefore, the next best thing is to set two of them equal and greater than the third. But in this case, the minimal distance will be the third one. Wait, but we need to maximize the minimal distance.Alternatively, maybe set the two smallest distances to be equal. Let's suppose that the minimal distance is m, and the two smallest distances are equal.In this case, the three distances are:Left-middle: b + aMiddle-right: a - bLeft-right: 2aAssuming that b is closer to the right, so a - b is smaller than b + a. Wait, if b is closer to the right, then middle is at b, so a - b is the distance from middle to right, which would be smaller than left to middle (b + a) if b is positive. For example, if a=1 and b=0.5, then left-middle is 1.5, middle-right is 0.5, and left-right is 2. So the minimal distance is 0.5. If we want to maximize the minimal distance, we need to increase m=0.5 as much as possible.But if we set a - b = m and b + a = m, then as before, that implies b=0. But then the distances become left-middle = a, middle-right =a, left-right=2a. So minimal distance is a. So to have minimal distance m, set a = m. Then sum of squares is 2m² +0=2m²=1 → m=1/√2≈0.707. Which is the symmetric case.Alternatively, if we allow b to be negative? Wait, but then the middle point could be to the left of the origin. But the problem doesn't specify any order on the numbers, so maybe arranging them in a different order. But regardless, the distances would be similar.Wait, perhaps the symmetric case is indeed optimal. Because if you try to shift the middle point to the right, you decrease the middle-right distance but increase the left-middle distance. However, the minimal distance becomes smaller, so to maximize the minimal distance, it's better to have the middle point at the center, making the distances equal on both sides. So for three points, the optimal arrangement is symmetric with the middle point at 0.Similarly, for higher n, arranging the points symmetrically in an arithmetic progression might be optimal. Then, the minimal distance m is determined by the spacing between consecutive points, and the sum of squares can be calculated accordingly.So, generalizing this, for n points, the maximum minimal distance m is achieved when the points are symmetrically placed around 0 in an arithmetic progression with common difference m. Then, the positions are either ±m/2, ±3m/2, ..., ±(n-1)m/2 for even n, or 0, ±m, ±2m, ..., ±k m for odd n=2k+1.Calculating the sum of squares for even n=2k:Sum = 2 * [ (m/2)^2 + (3m/2)^2 + ... + ((2k -1)m/2)^2 ] = 2*(m²/4)[1² + 3² + ... + (2k -1)^2 ]The sum of squares of the first k odd numbers is k(2k -1)(2k +1)/3. Therefore, Sum = 2*(m²/4)*(k(2k -1)(2k +1)/3) = (m²/2)*(k(2k -1)(2k +1)/3)Set this equal to 1:(m²/2)*(k(2k -1)(2k +1)/3) =1 ⇒ m² = 6 / [k(2k -1)(2k +1)]Therefore, m = sqrt(6 / [k(2k -1)(2k +1)] )Similarly, for odd n=2k+1:Sum = 0 + 2*[m² + (2m)^2 + ... + (k m)^2 ] = 2m²[1² +2² + ... +k² ] = 2m²*(k(k +1)(2k +1)/6) = (m²/3)*k(k +1)(2k +1)Set equal to 1:(m²/3)*k(k +1)(2k +1)=1 ⇒ m²=3/[k(k +1)(2k +1)] ⇒ m= sqrt(3/[k(k +1)(2k +1)] )Therefore, combining both cases, the maximum minimal distance m is:For even n=2k: m = sqrt(6 / [k(2k -1)(2k +1)] )For odd n=2k+1: m = sqrt(3 / [k(k +1)(2k +1)] )Now, let's check with n=3, which is odd. For n=3, k=1:m= sqrt(3/[1*2*3])=sqrt(1/2)=1/√2≈0.707. Correct.For n=4, even, k=2:m= sqrt(6/[2*(4 -1)*(4 +1)])=sqrt(6/[2*3*5])=sqrt(6/30)=sqrt(1/5)=1/√5≈0.447.Another example, n=5 (odd, k=2):m= sqrt(3/[2*3*5])=sqrt(3/30)=sqrt(1/10)=1/√10≈0.316.Similarly, n=6 (even, k=3):m= sqrt(6/[3*5*7])=sqrt(6/105)=sqrt(2/35)=√(2)/√35≈0.246.Okay, seems consistent.But is this actually the maximum possible? Let's see.Suppose for n=3, instead of placing them symmetrically at -1/√2, 0, 1/√2, what if we place them at three equally spaced points with minimal distance m, but not symmetric. For example, points at 0, m, 2m. Then sum of squares is 0 + m² +4m²=5m²=1⇒m=1/√5≈0.447. But the minimal distance is m≈0.447, which is less than the symmetric case of 1/√2≈0.707. Therefore, the symmetric arrangement gives a larger minimal distance. Therefore, symmetric is better.Therefore, the conclusion is that for maximum minimal distance, the points should be symmetrically arranged in an arithmetic progression. Therefore, the formula derived above should give the maximum m.Therefore, the answer is:For even n=2k: m= sqrt(6 / [k(2k -1)(2k +1)] )For odd n=2k+1: m= sqrt(3 / [k(k +1)(2k +1)] )But we can write this in terms of n.For even n=2k:k = n/2So m= sqrt(6 / [(n/2)(n -1)(n +1)/2] )= sqrt(6 / [ (n(n² -1))/4 ]) = sqrt(24 / [n(n² -1)])= 2*sqrt(6 / [n(n² -1)])Wait, let's see:Original expression:m= sqrt(6 / [k(2k -1)(2k +1)] )Since k = n/2,m= sqrt(6 / [ (n/2)(n -1)(n +1) ]) = sqrt(6 * 2 / [n(n -1)(n +1)]) = sqrt(12 / [n(n² -1)])= 2*sqrt(3 / [n(n² -1)])Wait, but 6*2=12. sqrt(12 / [n(n² -1)]) = sqrt(4*3 / [n(n² -1)])= 2*sqrt(3 / [n(n² -1)]). Yes.Similarly, for odd n=2k+1:k=(n-1)/2So m= sqrt(3 / [k(k +1)(2k +1)] )= sqrt(3 / [ ((n-1)/2)( (n+1)/2 )(n) ]) = sqrt(3 / [ (n(n² -1))/4 ]) = sqrt(12 / [n(n² -1)])= same as even case? Wait, wait:Wait, let's compute it step by step.For odd n=2k+1:k=(n-1)/2Therefore, k(k +1)(2k +1) = ( (n-1)/2 )( (n+1)/2 )(n )Multiply them:[(n-1)(n+1)n]/4 = [n(n² -1)]/4Therefore, m= sqrt(3 / [n(n² -1)/4 ]) = sqrt(12 / [n(n² -1)])= same as even case.Therefore, regardless of whether n is even or odd, the maximum minimal distance m is sqrt(12 / [n(n² -1)])= 2*sqrt(3 / [n(n² -1)]).Wait, but let's check for n=3:sqrt(12 / [3*(9 -1)])= sqrt(12 /24)= sqrt(0.5)=1/√2. Correct.For n=4:sqrt(12 / [4*(16 -1)])=sqrt(12 /60)=sqrt(0.2)=sqrt(1/5)=1/√5. Correct.Therefore, the general formula for any n≥3 is m= sqrt(12 / [n(n² -1)]). Which simplifies to 2*sqrt(3)/sqrt(n(n² -1)), but can also be written as 2*sqrt(3)/(sqrt(n(n² -1))).But sqrt(12)=2*sqrt(3), so yes.Alternatively, rationalizing, 2*sqrt(3)/sqrt(n(n² -1)).But perhaps we can write it as sqrt(12/(n(n² -1))). However, the problem asks for the maximum value of m, so this would be the answer.But let me check again for n=3:sqrt(12 / (3*(9 -1)))=sqrt(12/24)=sqrt(0.5)=1/√2≈0.707. Correct.n=4: sqrt(12/(4*15))=sqrt(12/60)=sqrt(0.2)=√(1/5)=1/√5≈0.447. Correct.n=5: sqrt(12/(5*24))=sqrt(12/120)=sqrt(0.1)=√(1/10)=1/√10≈0.316. Correct.Yes, so the general formula is m= sqrt(12/(n(n² -1))) = 2*sqrt(3/(n(n² -1))).Therefore, the maximum value of m is sqrt(12/(n(n² -1))).But let's write it in the simplest radical form. sqrt(12) is 2*sqrt(3), so:m= 2*sqrt(3)/sqrt(n(n² -1))= 2*sqrt(3)/sqrt(n³ -n).Alternatively, rationalizing the denominator, but maybe it's simplest as sqrt(12/(n(n² -1))).Alternatively, factor out:sqrt(12/(n(n² -1)))=sqrt( (4*3)/(n(n² -1)) )=2*sqrt(3/(n(n² -1))).But both forms are acceptable. However, the problem might expect the answer in terms of a single square root. Let me compute for general n:m= sqrt(12 / [n(n² -1)])= sqrt( (4*3) / [n(n² -1)] )= sqrt(4 / [n(n² -1)] ) * sqrt(3)= 2/sqrt(n(n² -1)) * sqrt(3)= 2*sqrt(3)/sqrt(n(n² -1)).Alternatively, writing as:m= 2√3 / √(n(n² -1)).But depending on the problem's preference. However, in the boxed answer, we can present it as:boxed{sqrt{dfrac{12}{n(n^{2}-1)}}}Which is equivalent and perhaps simpler.Yes, that's the answer. Therefore, after going through the reasoning, the maximum value of m is the square root of 12 divided by n(n squared minus 1). This works for both even and odd n, as we've verified with examples.Final AnswerThe maximum value of ( m ) is boxed{sqrt{dfrac{12}{n(n^{2}-1)}}}.</think>Given ( n geq 3 ) real numbers ( a_1, a_2, ldots, a_n ) such that their squares sum to 1, we need to find the maximum value of the minimum distance ( m ) between any two numbers. To maximize the minimum distance ( m ), the numbers should be arranged symmetrically around 0 in an arithmetic progression. For even ( n ), the numbers are placed at positions ( pm frac{m}{2}, pm frac{3m}{2}, ldots, pm frac{(n-1)m}{2} ). For odd ( n ), the numbers are placed at ( 0, pm m, pm 2m, ldots, pm frac{(n-1)m}{2} ).The sum of squares for even ( n = 2k ) is calculated as:[2 left( left( frac{m}{2} right)^2 + left( frac{3m}{2} right)^2 + cdots + left( frac{(2k-1)m}{2} right)^2 right) = frac{m^2}{2} cdot frac{k(2k-1)(2k+1)}{3}]Setting this equal to 1, we solve for ( m ):[m = sqrt{frac{6}{k(2k-1)(2k+1)}}]For odd ( n = 2k+1 ), the sum of squares is:[2 left( m^2 + (2m)^2 + cdots + (km)^2 right) = frac{m^2}{3} cdot k(k+1)(2k+1)]Setting this equal to 1, we solve for ( m ):[m = sqrt{frac{3}{k(k+1)(2k+1)}}]Both cases can be generalized to:[m = sqrt{frac{12}{n(n^2 - 1)}}]Thus, the maximum value of ( m ) is:[boxed{sqrt{dfrac{12}{n(n^{2}-1)}}}],<think>Okay, so I need to find the maximum value of the minimum distance between any two numbers in a set of real numbers a₁, a₂, ..., aₙ, given that their squares add up to 1. Let me try to unpack that step by step.First, the problem states that for n numbers (where n is at least 3), we define m as the smallest absolute difference between any two distinct numbers in the set. Our goal is to maximize this m, under the constraint that the sum of their squares is 1. So we're looking for the arrangement of numbers where the closest any two numbers are to each other is as large as possible, while still keeping the sum of squares equal to 1.Let me think of this as an optimization problem. We want to maximize m, which is the minimum pairwise distance, subject to the sum of squares constraint. Since m is the minimum distance, all other pairwise distances must be at least m, but we want the configuration where this minimal distance is as large as possible.I recall that in such optimization problems, especially with constraints, using symmetry might be helpful. If we arrange the numbers symmetrically, perhaps equally spaced around some interval, that might maximize the minimal distance. But since these are real numbers, not necessarily on a circle or something, maybe arranging them on a line with equal spacing?Wait, but how does the sum of squares come into play here? If we have numbers spaced equally, the sum of their squares would depend on their positions. So maybe we need to place them in such a way that they are as spread out as possible, but scaled so that the sum of squares is 1.Let me consider an example. Suppose n=3. If I have three numbers, to maximize the minimal distance between them, maybe they should be equally spaced. But how does scaling affect this? Let's say we have three points: -d, 0, +d. The sum of squares would be 2d². But if we set 2d²=1, then d=√(1/2). The minimal distance here is d, which is between -d and 0, and 0 and +d. But wait, in this case, the minimal distance is d, but the distance between -d and +d is 2d. So the minimal distance is d. But is this the maximum possible minimal distance?Alternatively, maybe arranging them as three equally spaced points over an interval. Let's say they are placed at -a, 0, +a. But that's the same as the previous case. Alternatively, maybe spacing them as a, b, c such that b - a = m and c - b = m. So three equally spaced points with spacing m. Then the positions would be x, x + m, x + 2m. But since we can place them symmetrically, maybe center them around 0. So for three points, symmetric around 0. Then the points would be -m, 0, +m. But then the distances are m (from -m to 0) and 2m (from -m to +m). The minimal distance is m. Then the sum of squares would be (-m)^2 + 0^2 + m^2 = 2m². So 2m²=1 implies m=√(1/2). So minimal distance m=√(1/2). But is this the maximum possible?Wait, but maybe if we don't center them around 0, but spread them out more? For example, if we have three points equally spaced with spacing m, but not symmetric around 0. Let's say they are at a, a + m, a + 2m. Then the sum of squares is a² + (a + m)^2 + (a + 2m)^2. To minimize the sum of squares, we might need to adjust a. Wait, but our goal is to maximize m under the constraint that the sum of squares is 1. So perhaps the minimal sum of squares for a given m would correspond to the optimal a. Hmm, but I need to maximize m such that there exists a configuration with sum of squares 1. So maybe the minimal sum of squares for a given m is achieved when the numbers are as closely packed as possible? Wait, perhaps the opposite.Wait, if I fix m, then arrange the numbers such that they are spaced at least m apart, then the sum of squares depends on their positions. To minimize the sum of squares for a given m, you would want to cluster them as closely as possible, but since they have to be at least m apart, the minimal sum would be when they are spaced exactly m apart. But in our problem, we need the sum of squares to be 1. So to maximize m, perhaps we need to arrange the numbers in a way that the sum of squares is exactly 1, and m is as large as possible. So maybe the optimal configuration is when the numbers are equally spaced with the maximum possible m, such that their sum of squares equals 1.Alternatively, maybe arranging them symmetrically around 0 to minimize the sum of squares. For example, if n is even, we can have pairs of points symmetrically placed around 0. For odd n, we can have symmetric points and 0 in the middle. Maybe this symmetry helps in reducing the sum of squares, allowing m to be larger.Let me test this with n=3. If we have points at -a, 0, +a, then the sum of squares is 2a². To make this equal to 1, a=√(1/2). Then the minimal distance is a, as before. However, if we arrange them differently, maybe with two points close to each other and the third far away, but the sum of squares constraint limits how far the third can be. For example, suppose we have two points at -b and +b, and the third point at c. The minimal distance would be the smaller of |2b| and |c - b| (assuming c > b). Wait, but if we set c as far as possible, but the sum of squares would be 2b² + c² = 1. To maximize the minimal distance, which is min(2b, c - b). So we need to maximize min(2b, c - b) under 2b² + c² = 1.Let me set m = min(2b, c - b). To maximize m, set 2b = c - b = m. Then c = 3b. Then substitute into the sum of squares: 2b² + (3b)^2 = 2b² + 9b² = 11b² = 1 => b = 1/√11, c = 3/√11. Then the minimal distance m = 2b = 2/√11 ≈ 0.603. Compare this with the symmetric case where m = √(1/2) ≈ 0.707. Wait, so the symmetric case gives a larger m. So that's better. Therefore, arranging the points symmetrically around 0 gives a better m. So maybe symmetry is the key here.Similarly, for n=4, if we arrange them as two pairs: -a, -a + m, a - m, a. Wait, but maybe equally spaced points. For four points, equally spaced on a line. Let's say -1.5m, -0.5m, 0.5m, 1.5m. Then the minimal distance is m between consecutive points. The sum of squares would be 2*( (1.5m)^2 + (0.5m)^2 ) = 2*(2.25m² + 0.25m²) = 2*2.5m² = 5m². Setting that equal to 1 gives m = 1/√5 ≈ 0.447. Alternatively, arranging them as two pairs at -a and +a. For four points, two at -a and two at +a. Then the minimal distance is 2a. The sum of squares is 4a². So 4a²=1 => a=0.5. Then minimal distance is 2a=1. But wait, that's larger than the previous case. Wait, but in this case, the minimal distance is the distance between the two points at -a and +a, which is 2a. However, the distance between the two points at -a is 0, but since we need distinct points, maybe the two points at -a are actually the same? Wait, but the problem states "n given real numbers a₁, a₂, ..., aₙ". If they are allowed to be equal, then the minimal distance could be zero, but in the problem statement, the minimum is taken over 1 ≤ i < j ≤ n. So if two points are equal, their distance is zero, which would make m=0. But since we want to maximize m, we need all points to be distinct. Therefore, in the case of four points, we cannot have two points at the same position. Therefore, arranging them as two pairs is invalid because that would result in two pairs with zero distance. So for four points, we need to have four distinct points. Therefore, perhaps equally spaced points as -1.5m, -0.5m, 0.5m, 1.5m. Then the sum of squares is 5m²=1, so m=1/√5≈0.447. Alternatively, another arrangement?Alternatively, arrange them symmetrically but with different spacing. For four points, maybe -a, -b, +b, +a, with a > b. The minimal distance would be the smaller of (b - (-a))? Wait, no. Wait, the points are ordered as -a, -b, +b, +a. Then the distances are between -a and -b: a - b, between -b and +b: 2b, between +b and +a: a - b. So the minimal distance is the minimum of (a - b) and 2b. To maximize this minimum, set a - b = 2b => a = 3b. Then the sum of squares is (-3b)^2 + (-b)^2 + b^2 + (3b)^2 = 9b² + b² + b² + 9b² = 20b² =1 => b=1/√20=√5/10≈0.2236. Then minimal distance m=2b=2/√20=1/√5≈0.447, same as before. So whether we arrange them as equally spaced or in another symmetric way, the minimal distance is the same.But in the previous case, when I thought about arranging two pairs, that's invalid because duplicates aren't allowed. So the maximum minimal distance for n=4 is 1/√5. However, wait, if we instead arrange the four points at -c, -c + m, c - m, c. Then the minimal distance is m. Let's calculate the sum of squares. Let the points be -c, -c + m, c - m, c. Then the sum of squares is c² + (c - m)^2 + (c - m)^2 + c². Wait, no: the four points are -c, -c + m, c - m, c. So their squares are (-c)^2, (-c + m)^2, (c - m)^2, c². So sum is 2c² + 2(c - m)^2. Let's set this equal to 1:2c² + 2(c - m)^2 =1. Let's also note that the distance between -c + m and c - m is (c - m) - (-c + m) = 2c - 2m. To have all distances at least m, this distance must be at least m. So 2c - 2m ≥ m => 2c ≥ 3m => c ≥ (3/2)m.But we need to set up the equations. Let me try to maximize m. Let's assume that all consecutive distances are exactly m. So between -c and -c + m, then between -c + m and c - m, which is a distance of (c - m) - (-c + m) = 2c - 2m. Wait, but if we want all consecutive distances to be at least m, then the distance between -c + m and c - m must be at least m. So 2c - 2m ≥ m => 2c ≥ 3m => c ≥ (3/2)m. But perhaps if we set this distance equal to m as well, then 2c - 2m = m => 2c = 3m => c = (3/2)m. Then substituting into the sum of squares:2c² + 2(c - m)^2 = 2*( (9/4)m² ) + 2*( (3/2 m - m)^2 ) = 2*(9/4 m²) + 2*( (1/2 m)^2 ) = (9/2)m² + 2*(1/4 m²) = (9/2 + 1/2)m² = 10/2 m² = 5m². So 5m²=1 => m=1/√5≈0.447, same as before. So whether you arrange them with equal spacing or in another symmetric way, the maximum minimal distance is the same. So symmetry seems to be giving the optimal configuration here.Therefore, perhaps for any n, arranging the numbers symmetrically around 0 with equal spacing between consecutive points gives the maximum minimal distance m. Then the sum of squares can be calculated based on that arrangement, allowing us to solve for m.Let me formalize this. Suppose we have n numbers arranged symmetrically around 0 with equal spacing m. For even n, we can have pairs at ±(m/2), ±(3m/2), ..., ±((n/2 - 1/2)m). For odd n, we have 0 in the middle and pairs at ±m, ±2m, ..., ±((n-1)/2 m). Wait, but let's check.Wait, if n is even, say n=4, then positions would be -1.5m, -0.5m, 0.5m, 1.5m (as before). For n=6, they would be -2.5m, -1.5m, -0.5m, 0.5m, 1.5m, 2.5m. The spacing between consecutive points is m. The sum of squares would be 2*( (2.5m)^2 + (1.5m)^2 + (0.5m)^2 ) for n=6. For n=4, it was 2*( (1.5m)^2 + (0.5m)^2 ) = 5m².Similarly, for n=3 (odd), the positions would be -m, 0, +m. The sum of squares is 2m². For n=5, positions would be -2m, -m, 0, +m, +2m. The sum of squares is 2*( (2m)^2 + m² ) = 2*(4m² + m²)=10m².Wait, so in general, for n numbers arranged symmetrically with spacing m between consecutive points, the sum of squares can be calculated as follows:If n is odd: there are (n-1)/2 pairs of points at distances m, 2m, ..., ((n-1)/2)m from the center, plus the center point at 0. So sum of squares is 2*(m² + (2m)^2 + ... + ((n-1)/2 m)^2 ) + 0².If n is even: there are n/2 pairs of points at distances (m/2), (3m/2), ..., ((n-1)/2 + 1/2)m from the center. The sum of squares is 2*( (m/2)^2 + (3m/2)^2 + ... + (( (2k - 1)/2 )m)^2 ), where k = n/2.So for even n, the positions are ±( (2i - 1)/2 )m for i = 1 to n/2. Therefore, the sum of squares is 2 * Σ_{i=1}^{n/2} [ ((2i - 1)/2 * m )^2 ].Similarly, for odd n, the positions are 0 and ±im for i=1 to (n-1)/2. The sum of squares is 2 * Σ_{i=1}^{(n-1)/2} (im)^2.Therefore, for both cases, the sum of squares can be written as 2 * m² * Σ_{k=1}^m k², where the upper limit of the sum depends on n. Wait, no. Let me formalize it.For odd n = 2k + 1:Sum of squares = 2 * (m² + (2m)^2 + ... + (km)^2 ) = 2m²(1² + 2² + ... + k² )For even n = 2k:Sum of squares = 2 * [ ( (1/2 m)^2 + (3/2 m)^2 + ... + ( (2k - 1)/2 m )^2 ) ] = 2m² * [ (1/2)^2 + (3/2)^2 + ... + ( (2k - 1)/2 )^2 ]Which simplifies to 2m² * Σ_{i=1}^k ( (2i - 1)/2 )² = 2m² * (1/4) Σ_{i=1}^k (2i - 1)^2 = (m²/2) Σ_{i=1}^k (4i² - 4i + 1 ) = (m²/2)(4 Σi² -4 Σi + Σ1 )Calculating this sum might be a bit tedious, but perhaps there's a formula for the sum of squares of odd numbers.Alternatively, recall that Σ_{i=1}^k (2i - 1)^2 = (4k³ - k)/3. Let me verify this for small k:For k=1: 1²=1, (4*1 -1)/3=3/3=1. Correct.For k=2: 1² +3²=1+9=10, (4*8 -2)/3= (32 -2)/3=30/3=10. Correct.For k=3:1+9+25=35, (4*27 -3)/3=(108 -3)/3=105/3=35. Correct. So yes, the formula Σ_{i=1}^k (2i - 1)^2 = (4k³ - k)/3.Therefore, for even n=2k:Sum of squares = 2m² * [ (1/2)^2 + (3/2)^2 + ... + ( (2k -1)/2 )^2 ] = 2m² * (1/4) Σ_{i=1}^k (2i -1)^2 = (2m² /4) * (4k³ -k)/3 = (m²/2) * (4k³ -k)/3 = m² (4k³ -k)/6.But since n=2k, then k =n/2, so substituting:Sum of squares = m² (4(n/2)^3 -n/2)/6 = m² (4(n³/8) -n/2)/6 = m² (n³/2 -n/2)/6 = m² ( (n³ -n)/2 ) /6 = m² (n³ -n)/12.Alternatively, perhaps better to keep it in terms of k for now.Similarly, for odd n=2k +1:Sum of squares = 2m² Σ_{i=1}^k i² = 2m² * [k(k +1)(2k +1)/6]. Using the formula for the sum of squares.Therefore, depending on whether n is even or odd, we can express the sum of squares in terms of m and n, then set it equal to 1 to solve for m.Thus, for maximum m, the configuration is symmetric with equally spaced points, and the sum of squares gives us the equation to solve for m.So for even n=2k:Sum = m² (4k³ -k)/6 =1, so m= sqrt(6/(4k³ -k)).But since n=2k, we can write m in terms of n:k =n/2, so m= sqrt(6/(4*(n/2)^3 -n/2 ))=sqrt(6/( (n³)/2 -n/2 ))=sqrt(6/( (n³ -n)/2 ))=sqrt(12/(n³ -n ))=sqrt(12/(n(n² -1)))=sqrt(12/(n(n-1)(n+1))).Similarly, for odd n=2k +1:Sum = 2m² [k(k +1)(2k +1)/6] =1 => m² = 6/(2k(k +1)(2k +1)) => m= sqrt(6/(2k(k +1)(2k +1))).But n=2k +1, so k=(n-1)/2. Substitute:m= sqrt(6/(2*( (n-1)/2 )*( (n-1)/2 +1 )*(n) ))=sqrt(6/( (n-1)/1 * (n+1)/2 *n )).Wait, let's compute step by step:Denominator: 2k(k +1)(2k +1) with k=(n-1)/2.So 2k = n -1, (k +1)= (n -1)/2 +1 = (n +1)/2, and (2k +1)=n.Therefore denominator becomes (n -1)* (n +1)/2 *n. So:m= sqrt(6/( (n -1)*(n +1)/2 *n ))=sqrt(12/(n(n² -1))).Which is the same expression as for even n!Wait, that's interesting. For both even and odd n, after substitution, we get m= sqrt(12/(n(n² -1))).Therefore, regardless of whether n is even or odd, the maximum minimal distance m is sqrt(12/(n(n² -1))).Wait, let's check with n=3:n=3: m= sqrt(12/(3*(9 -1)))=sqrt(12/(3*8))=sqrt(12/24)=sqrt(0.5)=1/√2≈0.707. Which matches our earlier example.For n=4: sqrt(12/(4*(16 -1)))=sqrt(12/(4*15))=sqrt(12/60)=sqrt(0.2)=1/√5≈0.447. Which also matches the previous result.Similarly, n=5: sqrt(12/(5*(25 -1)))=sqrt(12/(5*24))=sqrt(12/120)=sqrt(0.1)=1/√10≈0.316.This seems consistent.Therefore, the general formula for the maximum minimal distance m is sqrt(12/(n(n² -1))). However, let's simplify sqrt(12) as 2*sqrt(3), so:m= 2*sqrt(3/(n(n² -1))) = 2*sqrt{3}/sqrt{n(n² -1)}.But the problem asks for the maximum value of m. So the answer should be sqrt(12/(n(n² -1))) which simplifies to 2*sqrt(3/(n(n² -1))).Alternatively, rationalizing:sqrt(12) = 2*sqrt(3), so:sqrt(12/(n(n² -1))) = 2*sqrt(3/(n(n² -1))).But perhaps they want it written as sqrt(12/(n(n² -1))).Alternatively, factor numerator and denominator:12 = 4*3, so sqrt(4*3/(n(n² -1))) = 2*sqrt(3/(n(n² -1))).Either way is acceptable, but let's check the original problem statement for the format. The problem says "Find the maximum value of the above m...", so likely they want it in simplest radical form. Therefore, writing it as 2 times the square root of 3 divided by the square root of n(n² -1). But let's see:sqrt(12/(n(n² -1))) = sqrt(12)/sqrt(n(n² -1)) = (2*sqrt(3))/sqrt(n(n² -1)) = 2*sqrt(3)/sqrt(n(n² -1)).Alternatively, if we rationalize further:sqrt(12/(n(n² -1))) = sqrt{(12)/(n(n-1)(n+1))}.But perhaps it's acceptable as is.Alternatively, let's confirm for n=3:sqrt(12/(3*(9 -1)))=sqrt(12/24)=sqrt(0.5)=1/√2, which is correct.Similarly, for n=4:sqrt(12/(4*15))=sqrt(12/60)=sqrt(1/5)=1/√5.So the formula works.Therefore, the maximum value of m is sqrt(12/(n(n² -1))). Which can also be written as 2*sqrt{3}/(sqrt{n(n² -1)}). Either form is correct.Therefore, the answer is sqrt(12/(n(n² -1))) or 2*sqrt{3/(n(n² -1))}.But let's check units. The problem states that a₁, ..., aₙ are real numbers, so the answer is dimensionless. So both forms are acceptable.But perhaps the problem expects the answer in the form with the square root in the numerator. Let me check.Original problem statement: Find the maximum value of m. Since it's a math competition problem, they often rationalize denominators. However, in the example given (example 26), it's possible that the answer is left in a form like sqrt(12/(n(n² -1))) or simplified.Alternatively, note that 12/(n(n² -1)) can be written as 12/(n(n-1)(n+1)). Therefore, sqrt(12/(n(n-1)(n+1))) = sqrt(12)/sqrt(n(n-1)(n+1)) = (2*sqrt(3))/sqrt(n(n-1)(n+1)).Alternatively, factor 12 as 4*3:sqrt(4*3/(n(n-1)(n+1))) = 2*sqrt(3/(n(n-1)(n+1))).Either way, both forms are equivalent.But I think the answer is typically written as sqrt{12} over sqrt{n(n² -1)}, so sqrt(12)/sqrt{n(n² -1)}. But since sqrt(12) is 2*sqrt(3), it's better to write 2*sqrt{3}/sqrt{n(n² -1)}. However, in the example with n=3, that would be 2*sqrt{3}/sqrt{3*8}=2*sqrt{3}/(sqrt{24})=2*sqrt{3}/(2*sqrt{6})=sqrt{3}/sqrt{6}=1/sqrt{2}, which is correct.Alternatively, simplifying 2*sqrt{3}/sqrt{n(n² -1)} as 2*sqrt{3/(n(n² -1))} is also acceptable.Given that the problem is from a Chinese competition (since it's example 26 from a problem set), and Chinese competition problems often prefer simplified radical forms with rational denominators. But in this case, the denominator has n(n² -1), which is already factored, so it might be acceptable to leave it as sqrt(12/(n(n² -1))).Alternatively, to rationalize it further:sqrt(12/(n(n² -1))) = sqrt(12)/sqrt(n(n² -1)) = (2*sqrt{3})/sqrt{n(n² -1)}.But unless there's a further simplification, this is as simple as it gets.Therefore, the answer should be sqrt{12} divided by sqrt{n(n² -1)}, which is equal to 2 times sqrt{3} divided by sqrt{n(n² -1)}. However, since sqrt{n(n² -1)} can be written as sqrt{n(n-1)(n+1)}, but that's not necessarily simpler.Alternatively, if we write sqrt{n(n² -1)} = sqrt{n(n-1)(n+1)}, which is the same thing.Therefore, the answer is 2*sqrt{3}/sqrt{n(n-1)(n+1)}.Alternatively, combining terms:2*sqrt{3}/(sqrt{n(n² -1)}) = 2*sqrt{3}/(sqrt{n} * sqrt{n² -1}) = 2*sqrt{3}/(sqrt{n} * sqrt{(n -1)(n +1)}) ) = 2*sqrt{3}/(sqrt{n(n -1)(n +1)}).But in any case, these are all equivalent forms.Therefore, the maximum value of m is 2*sqrt{3}/sqrt{n(n² -1)}.But let's check another example. For n=5:2*sqrt{3}/sqrt{5*(25 -1)}=2*sqrt{3}/sqrt{5*24}=2*sqrt{3}/sqrt{120}=2*sqrt{3}/(2*sqrt{30})=sqrt{3}/sqrt{30}=sqrt{3}/(sqrt{30})=sqrt{3}/(sqrt{3*10})=1/sqrt{10}=√10/10≈0.316, which matches the earlier calculation. So that's correct.Therefore, the answer is 2*sqrt{3}/sqrt{n(n² -1)}. Therefore, written with LaTeX, it would be boxed{dfrac{2sqrt{3}}{sqrt{n(n^{2}-1)}}}.Alternatively, as boxed{sqrt{dfrac{12}{n(n^{2}-1)}}}.Since both forms are equivalent, but the problem might prefer the simplest radical form. In the example for n=3, the answer is 1/√2, which is equivalent to sqrt{1/2}, but in the formula here, sqrt{12/(3*8)}=sqrt{1/2}=1/√2. So in that case, the formula gives the answer in the form sqrt{12/(n(n² -1))}. However, when simplified, for n=3, it's written as 1/√2. Similarly, for n=4, sqrt{12/(4*15)}=sqrt{1/5}=1/√5.Therefore, perhaps the answer is better written as sqrt{12/(n(n² -1))} because it generalizes nicely. Let me confirm.For n=3: sqrt{12/(3*8)}=sqrt{12/24}=sqrt{0.5}=1/√2.For n=4: sqrt{12/(4*15)}=sqrt{12/60}=sqrt{0.2}=sqrt{1/5}=1/√5.For n=5: sqrt{12/(5*24)}=sqrt{12/120}=sqrt{0.1}=sqrt{1/10}=1/√10.Therefore, it's equivalent to 1/sqrt{n(n² -1)/12} but written as sqrt{12/(n(n² -1))}.Alternatively, if we rationalize the denominator for n=3:sqrt{12/(3*8)}=sqrt{12/24}=sqrt{1/2}=√(1/2)=√2/2.But in the general case, it's more straightforward to write sqrt{12/(n(n² -1))}.Alternatively, maybe the problem expects the answer in terms of a simplified fraction inside the square root. However, since 12 and n(n² -1) might not have common factors, depending on n.But perhaps the answer is simply sqrt{dfrac{12}{n(n^2 -1)}}.Alternatively, if we factor numerator and denominator:12 = 4 * 3,n(n² -1)=n(n -1)(n +1).But there's no further simplification unless n has specific factors. So, I think the answer is boxed{sqrt{dfrac{12}{n(n^{2}-1)}}}.But let's check the initial problem statement. It says "Find the maximum value of the above m under the condition that a₁² + a₂² + ... + aₙ² = 1."Therefore, the answer is the square root of 12 divided by the square root of n(n² -1). So boxed as boxed{sqrt{dfrac{12}{n(n^2 - 1)}}}.Yes, this seems correct.Final AnswerThe maximum value of ( m ) is boxed{sqrt{dfrac{12}{n(n^2 - 1)}}}.</think>To find the maximum value of the minimum distance ( m ) between any two numbers in a set of real numbers ( a_1, a_2, ldots, a_n ) under the constraint that ( a_1^2 + a_2^2 + cdots + a_n^2 = 1 ), we consider arranging the numbers symmetrically around 0 with equal spacing. For both even and odd ( n ), the optimal configuration is when the numbers are equally spaced. The sum of squares for such a configuration is calculated and set to 1, leading to the solution for ( m ).For even ( n ):- The positions are symmetric pairs around 0 with spacing ( m ).- The sum of squares is calculated and simplified to find ( m ).For odd ( n ):- The positions include 0 and symmetric pairs around 0 with spacing ( m ).- The sum of squares is similarly calculated and simplified.The key steps involve:1. Arranging the numbers symmetrically around 0 with equal spacing.2. Calculating the sum of squares for this configuration.3. Setting the sum of squares equal to 1 and solving for ( m ).The resulting maximum value of ( m ) is given by:[boxed{sqrt{dfrac{12}{n(n^2 - 1)}}}]]

🔑:Let's assume a_{1} leqslant a_{2} leqslant cdots leqslant a_{n}. We estimate sum_{1 leqslant i<j leqslant n}left(a_{i}-a_{j}right)^{2} from two aspects.On one hand,begin{array}{c}sum_{1 leqslant i=j leqslant i}left(a_{i}-a_{j}right)^{2}=(n-1) sum_{i=1}^{n} a_{i}^{2}-2 sum_{1{ }_{1} i<j leqslant n} a_{i} a_{j} =n sum_{i=1}^{n} a_{i}^{2}-left(sum_{i=1}^{n} a_{i}right)^{2} leqslant n .end{array}On the other hand, since a_{2}-a_{1} geqslant m, a_{3}-a_{2} geqslant m, cdots, a_{n}-a_{n-1} geqslant m, for 1 leqslant i<j leqslant n, we have a_{j}-a_{i} geqslant(j-i) m.Thus,begin{array}{l}sum_{1 leqslant i<j leqslant n}left(a_{i}-a_{j}right)^{2} geqslant m^{2} sum_{1 leqslant i<j leqslant n}(i-j)^{2} =m^{2} sum_{k=1}^{n-1}(n-k) k^{2}=frac{m^{2}}{12} n^{2}left(n^{2}-1right) .end{array}From the above two aspects, we knowTherefore,begin{array}{l}n geqslant frac{m^{2}}{12} n^{2}left(n^{2}-1right) m leqslant sqrt{frac{12}{nleft(n^{2}-1right)}} .end{array}Furthermore, when sum_{i=1}^{n} a_{i}=0, and a_{1}, a_{2}, cdots, a_{n} form an arithmetic sequence, the above inequality holds with equality. Hence, the maximum value of m is sqrt{frac{12}{nleft(n^{2}-1right)}}.

🧠:[<think>Okay, so I need to find the equation of the hyperbola given some conditions. Let me start by recalling what I know about hyperbolas. The standard equation for a hyperbola centered at the origin opening horizontally is (frac{x^2}{a^2} - frac{y^2}{b^2} = 1), where (a) and (b) are positive constants. The foci of this hyperbola are located at ((pm c, 0)), where (c^2 = a^2 + b^2).The problem states that the left focus is (F), so the coordinates of (F) should be ((-c, 0)). Then, there's a point (A) on the asymptote of the hyperbola. The asymptotes of a hyperbola (frac{x^2}{a^2} - frac{y^2}{b^2} = 1) are the lines (y = pm frac{b}{a}x). So point (A) lies on one of these two lines.It also mentions that triangle (OAF) is an equilateral triangle with side length 2, where (O) is the origin. Let me visualize this. The origin (O), the left focus (F(-c, 0)), and point (A) on an asymptote form an equilateral triangle. All sides must be length 2, so the distances (OA), (AF), and (OF) should each be 2.First, let's note that the distance from the origin (O) to the focus (F(-c, 0)) is (OF = sqrt{(-c - 0)^2 + (0 - 0)^2} = c). Since the triangle is equilateral with side length 2, this means (OF = 2), so (c = 2). Therefore, (c = 2), and since (c^2 = a^2 + b^2), we have (a^2 + b^2 = 4).Next, point (A) is on one of the asymptotes. Let's consider both possibilities: either (A) is on (y = frac{b}{a}x) or (y = -frac{b}{a}x). Let me assume it's on the positive asymptote first. The coordinates of (A) can be written as ((t, frac{b}{a}t)) for some parameter (t). Alternatively, if it's on the negative asymptote, it would be ((t, -frac{b}{a}t)). Since the problem doesn't specify which asymptote, I might need to check both possibilities, but maybe the equilateral triangle condition will restrict it to one.Since (OA = 2), the distance from the origin to point (A) must be 2. Let's calculate that. If (A) is ((t, frac{b}{a}t)), then (OA = sqrt{t^2 + left(frac{b}{a}tright)^2} = sqrt{t^2 left(1 + frac{b^2}{a^2}right)} = |t| sqrt{1 + frac{b^2}{a^2}}). Since this equals 2, we have (|t| sqrt{1 + frac{b^2}{a^2}} = 2). Let's square both sides to eliminate the square root: (t^2 left(1 + frac{b^2}{a^2}right) = 4).Similarly, the distance from (A) to (F(-c, 0)) must also be 2. Let's compute (AF). The coordinates of (A) are ((t, frac{b}{a}t)) and (F) is ((-c, 0)). The distance (AF) is (sqrt{(t + c)^2 + left(frac{b}{a}t - 0right)^2}). This should also equal 2. So:[sqrt{(t + c)^2 + left(frac{b}{a}tright)^2} = 2]Squaring both sides:[(t + c)^2 + left(frac{b}{a}tright)^2 = 4]Now, we already know that (c = 2), so substituting that in:[(t + 2)^2 + left(frac{b}{a}tright)^2 = 4]Earlier, we had an equation from (OA = 2):[t^2 left(1 + frac{b^2}{a^2}right) = 4]Let me denote (k = frac{b}{a}). Then, (k = frac{b}{a}), so (b = ka). Then, since (c^2 = a^2 + b^2 = a^2 + k^2a^2 = a^2(1 + k^2)). But since (c = 2), then (a^2(1 + k^2) = 4). So (a^2 = frac{4}{1 + k^2}). Hmm, maybe this substitution will help.But let's see. Let me try to express both equations in terms of (t) and (k) (or (t) and (a), but maybe (k) is better).From the first equation (OA = 2):[t^2 (1 + k^2) = 4]From the second equation (AF = 2):[(t + 2)^2 + (kt)^2 = 4]Let me expand the second equation:[t^2 + 4t + 4 + k^2t^2 = 4]Combine like terms:[(1 + k^2)t^2 + 4t + 4 - 4 = 0][(1 + k^2)t^2 + 4t = 0]Factor out (t):[t[(1 + k^2)t + 4] = 0]So either (t = 0) or ((1 + k^2)t + 4 = 0). If (t = 0), then point (A) is at the origin, but that can't form a triangle with (O) and (F). So the other possibility is:[(1 + k^2)t + 4 = 0 implies t = -frac{4}{1 + k^2}]Now, from the first equation (t^2(1 + k^2) = 4), substitute (t = -4/(1 + k^2)):[left(-frac{4}{1 + k^2}right)^2 (1 + k^2) = 4][frac{16}{(1 + k^2)^2} cdot (1 + k^2) = 4][frac{16}{1 + k^2} = 4][16 = 4(1 + k^2)][4 = 1 + k^2][k^2 = 3 implies k = sqrt{3} text{ or } k = -sqrt{3}]But since (k = frac{b}{a}) and both (a) and (b) are positive, (k) must be positive. Therefore, (k = sqrt{3}), so (frac{b}{a} = sqrt{3}), hence (b = asqrt{3}).Now, recall that (c = 2) and (c^2 = a^2 + b^2). Substitute (b = asqrt{3}):[4 = a^2 + (asqrt{3})^2 = a^2 + 3a^2 = 4a^2][4a^2 = 4 implies a^2 = 1 implies a = 1]Then, (b = asqrt{3} = sqrt{3}), so (b^2 = 3).Therefore, the equation of the hyperbola is:[frac{x^2}{1^2} - frac{y^2}{(sqrt{3})^2} = 1 implies frac{x^2}{1} - frac{y^2}{3} = 1]Looking at the answer choices:A) (frac{x^2}{4} - frac{y^2}{12} = 1)B) (frac{x^2}{12} - frac{y^2}{4} = 1)C) (frac{x^2}{3} - y^2 = 1)D) (x^2 - frac{y^2}{3} = 1)Wait, my result is (frac{x^2}{1} - frac{y^2}{3} = 1), which is option D. But let me double-check to make sure I didn't make a mistake.Wait, so I concluded that (a^2 = 1), so (a = 1), and (b^2 = 3). Therefore, plugging into the standard form, (frac{x^2}{1} - frac{y^2}{3} = 1), which is exactly option D. So the answer should be D.But let me check if there's a possibility that point (A) is on the other asymptote, (y = -frac{b}{a}x). Let's see. If I had assumed point (A) is on the negative asymptote, then coordinates would be ((t, -frac{b}{a}t)). Let's see if that affects the result.Repeating the same steps with (A(t, -kt)), where (k = frac{b}{a}):Distance OA is still (sqrt{t^2 + (-kt)^2} = |t|sqrt{1 + k^2}), same as before. So the OA equation would be the same. The distance AF would be (sqrt{(t + c)^2 + (-kt - 0)^2} = sqrt{(t + c)^2 + (kt)^2}), which is the same as before. So even if point (A) is on the negative asymptote, the equations would be the same because squaring eliminates the sign. Therefore, regardless of which asymptote (A) is on, we end up with the same equations. Therefore, the result is still the same.Therefore, the hyperbola's equation is (x^2 - frac{y^2}{3} = 1), which is option D.Wait, but let me check the answer choices again. Option D is written as (x^2 - frac{y^2}{3} = 1), which matches. So the answer must be D.But let me just verify once more because sometimes there might be a calculation mistake.We found that (c = 2), (a = 1), (b = sqrt{3}). Then, (c^2 = a^2 + b^2 = 1 + 3 = 4), which checks out. The asymptotes should have slopes (pm frac{b}{a} = pm sqrt{3}), so the asymptotes are (y = pm sqrt{3}x). Then, point (A) is on one of these asymptotes, and we found (t = -frac{4}{1 + k^2}), where (k = sqrt{3}), so (1 + k^2 = 4), so (t = -1). Therefore, coordinates of (A) would be ((-1, sqrt{3}(-1)) = (-1, -sqrt{3})) if on the negative asymptote, or ((-1, sqrt{3})) if on the positive asymptote. Wait, but wait:Wait, when we found (t = -frac{4}{1 + k^2}), and (1 + k^2 = 4), so (t = -1). So if point (A) is on the positive asymptote, (y = sqrt{3}x), then coordinates are ((-1, sqrt{3}*(-1)) = (-1, -sqrt{3})), but that's actually on the negative asymptote. Wait, maybe I made a mistake here.Wait, if (k = sqrt{3}), the asymptotes are (y = sqrt{3}x) and (y = -sqrt{3}x). If point (A) is on (y = sqrt{3}x), then coordinates are ((t, sqrt{3}t)). But we found (t = -1), so that would be ((-1, -sqrt{3})), which is actually on (y = sqrt{3}x) only if (t = -1), but substituting (t = -1), we get (y = sqrt{3}*(-1) = -sqrt{3}), which is actually on the asymptote (y = -sqrt{3}x). Wait, so there's a confusion here.Wait, if the asymptote is (y = sqrt{3}x), then when (t = -1), the y-coordinate is (sqrt{3}*(-1) = -sqrt{3}). But that's the same as the asymptote (y = -sqrt{3}x) with (t = 1). Wait, no. Let's clarify.If we parametrize the positive asymptote as (y = sqrt{3}x), then any point on it can be written as ((t, sqrt{3}t)). Similarly, the negative asymptote can be written as ((t, -sqrt{3}t)). If (t = -1), then on the positive asymptote, the point is ((-1, -sqrt{3})), but that's actually the same as the point ((1, -sqrt{3})) on the negative asymptote with (t = 1). Wait, no. Each asymptote is a line extending infinitely in both directions. So the point ((-1, -sqrt{3})) lies on the negative asymptote (y = -sqrt{3}x), since (-sqrt{3}*(-1) = sqrt{3}), which isn't correct. Wait, actually, substituting (x = -1) into (y = -sqrt{3}x), we get (y = -sqrt{3}*(-1) = sqrt{3}). So ((-1, -sqrt{3})) is not on (y = -sqrt{3}x). Wait, confusion here.Wait, perhaps better to check: For the positive asymptote (y = sqrt{3}x), when (x = -1), (y = sqrt{3}*(-1) = -sqrt{3}). So the point ((-1, -sqrt{3})) is on the positive asymptote. Similarly, the point ((-1, sqrt{3})) would be on the negative asymptote (y = -sqrt{3}x), since substituting (x = -1) into (y = -sqrt{3}x) gives (y = -sqrt{3}*(-1) = sqrt{3}). So, points on the asymptotes can have negative (x) or (y) values depending on the parameter.But regardless, in our calculation, we found that (t = -1), so the coordinates of (A) are ((-1, pm sqrt{3}(-1)) = (-1, mp sqrt{3})). So if we take the positive asymptote, it's ((-1, -sqrt{3})), and the negative asymptote would have ((-1, sqrt{3})).Now, let's check if triangle (OAF) with (O(0,0)), (F(-2,0)), and (A(-1, -sqrt{3})) is equilateral with side length 2.Compute (OA): distance from (0,0) to (-1, -√3):√[(-1)^2 + (-√3)^2] = √[1 + 3] = √4 = 2. Correct.Distance (OF) is from (0,0) to (-2,0): 2. Correct.Distance (AF): from (-2,0) to (-1, -√3):√[(-1 + 2)^2 + (-√3 - 0)^2] = √[(1)^2 + ( -√3)^2] = √[1 + 3] = √4 = 2. Correct.Similarly, if point (A) is (-1, √3), check the distances:OA: √[(-1)^2 + (√3)^2] = √[1 + 3] = 2.AF: distance from (-2,0) to (-1, √3):√[(-1 + 2)^2 + (√3 - 0)^2] = √[1 + 3] = 2.So both points give equilateral triangles. However, in both cases, the hyperbola is the same. Therefore, regardless of which asymptote point (A) is on, the hyperbola's equation is (x^2 - y^2/3 = 1), which is option D.Therefore, the answer should be D.Final Answerboxed{D}</think>Given the hyperbola (frac{x^2}{a^2} - frac{y^2}{b^2} = 1) with its left focus (F) and point (A) on the asymptote of the hyperbola, we need to determine the equation of the hyperbola given that (triangle OAF) is an equilateral triangle with side length 2.1. Identify the coordinates of the left focus (F): - The foci of the hyperbola are ((pm c, 0)) where (c^2 = a^2 + b^2). - Since (F) is the left focus, its coordinates are ((-c, 0)). - The distance (OF) is 2, so (c = 2). Therefore, (c^2 = 4).2. Relate (a) and (b) using (c^2 = a^2 + b^2): - (a^2 + b^2 = 4).3. Determine the coordinates of point (A): - Point (A) lies on the asymptote (y = pm frac{b}{a}x). - Assume (A) is on the asymptote (y = frac{b}{a}x). The coordinates of (A) can be written as ((t, frac{b}{a}t)). - The distance (OA = 2) gives (|t| sqrt{1 + left(frac{b}{a}right)^2} = 2).4. Solve for (t): - Let (k = frac{b}{a}). Then, (|t| sqrt{1 + k^2} = 2). - The distance (AF = 2) gives (sqrt{(t + 2)^2 + (kt)^2} = 2). - Solving these equations, we find (t = -frac{4}{1 + k^2}) and substituting back, we get (k^2 = 3), so (k = sqrt{3}).5. Determine (a) and (b): - Since (frac{b}{a} = sqrt{3}), we have (b = asqrt{3}). - Using (a^2 + b^2 = 4), substitute (b = asqrt{3}) to get (a^2 + 3a^2 = 4), leading to (a^2 = 1) and (b^2 = 3).6. Write the equation of the hyperbola: - The equation is (frac{x^2}{1} - frac{y^2}{3} = 1), which corresponds to option D.(boxed{D}),<think>Okay, so I have this hyperbola equation given as x²/a² - y²/b² = 1, where a and b are positive. The left focus is F, and there's a point A on one of the asymptotes. The triangle OAF is equilateral with side length 2, and O is the origin. I need to find the equation of the hyperbola from the options provided. Let me try to break this down step by step.First, let me recall some properties of hyperbolas. For a hyperbola of the form x²/a² - y²/b² = 1, the foci are located at (±c, 0), where c² = a² + b². Since the left focus is F, its coordinates should be (-c, 0). The asymptotes of this hyperbola are the lines y = ±(b/a)x. Point A is on one of these asymptotes, so its coordinates can be represented as either (t, (b/a)t) or (t, -(b/a)t) for some real number t.Now, the triangle OAF is equilateral with side length 2. O is the origin (0,0), F is (-c, 0), and A is some point on the asymptote. Since all sides are equal, the distance between O and A, O and F, and A and F should all be 2.Let me start by finding the coordinates of point F. Since F is the left focus, it's at (-c, 0). The distance from O to F is |OF| = sqrt[(-c - 0)² + (0 - 0)²] = sqrt[c²] = c. But the problem states that the side length of the triangle is 2, so |OF| = 2. Therefore, c = 2. That means c = 2, so c² = 4 = a² + b². That's one equation relating a and b.Next, point A is on the asymptote. Let's assume it's on the positive asymptote y = (b/a)x; if not, we can check the negative one later. So coordinates of A would be (t, (b/a)t) for some t. The distance from O to A must also be 2. Let's compute that distance:|OA| = sqrt[(t - 0)² + ((b/a)t - 0)²] = sqrt[t² + (b²/a²)t²] = sqrt[t²(1 + b²/a²)] = |t|sqrt(1 + b²/a²) = 2.Since sqrt(1 + b²/a²) is positive, and |t| is non-negative, we can write |t| = 2 / sqrt(1 + b²/a²). Therefore, t = ±2 / sqrt(1 + b²/a²). Let's keep this in mind.Also, the distance from A to F must be 2. Let's compute that distance. F is at (-c, 0) = (-2, 0), and A is at (t, (b/a)t). So:|AF| = sqrt[(t - (-2))² + ((b/a)t - 0)²] = sqrt[(t + 2)² + (b²/a²)t²].This should equal 2. So we have the equation:sqrt[(t + 2)² + (b²/a²)t²] = 2.Squaring both sides:(t + 2)² + (b²/a²)t² = 4.Let me expand (t + 2)²:t² + 4t + 4 + (b²/a²)t² = 4.Combine like terms:[1 + b²/a²]t² + 4t + 4 - 4 = 0Simplifies to:[1 + b²/a²]t² + 4t = 0Factor out t:t([1 + b²/a²]t + 4) = 0So either t = 0 or [1 + b²/a²]t + 4 = 0.If t = 0, then point A would be (0, 0), which is the same as point O. But then triangle OAF would collapse to a line, which can't be an equilateral triangle. So t ≠ 0. Therefore, the other factor must be zero:[1 + b²/a²]t + 4 = 0 => t = -4 / [1 + b²/a²]But from earlier, we had |t| = 2 / sqrt(1 + b²/a²). Let's write both expressions for t. The t we found here is t = -4 / [1 + (b²/a²)]. The other expression is |t| = 2 / sqrt(1 + (b²/a²)). Let's check if these can be consistent.Let me set the absolute value of t from the second equation equal to the absolute value of t from the first equation.| -4 / [1 + (b²/a²)] | = 4 / [1 + (b²/a²)].From the first equation, |t| = 2 / sqrt(1 + (b²/a²)).So:4 / [1 + (b²/a²)] = 2 / sqrt(1 + (b²/a²))Multiply both sides by [1 + (b²/a²)] * sqrt(1 + (b²/a²)) to eliminate denominators:4 * sqrt(1 + (b²/a²)) = 2 * [1 + (b²/a²)]Divide both sides by 2:2 * sqrt(1 + (b²/a²)) = [1 + (b²/a²)]Let me let u = sqrt(1 + (b²/a²)). Then the equation becomes:2u = u²Rearranged:u² - 2u = 0u(u - 2) = 0Solutions are u = 0 or u = 2. But u = sqrt(1 + (b²/a²)) which is at least sqrt(1) = 1, so u can't be 0. Therefore, u = 2.Thus,sqrt(1 + (b²/a²)) = 2Square both sides:1 + (b²/a²) = 4 => b²/a² = 3 => b² = 3a².So now we have two equations:1. c² = a² + b² = 4 (since c = 2)2. b² = 3a²Substituting equation 2 into equation 1:a² + 3a² = 4 => 4a² = 4 => a² = 1 => a = 1 (since a > 0)Then b² = 3a² = 3*1 = 3. Therefore, the equation of the hyperbola is x²/1 - y²/3 = 1, which simplifies to x² - y²/3 = 1. Let's check the options given:A) x²/4 - y²/12 =1B) x²/12 - y²/4=1C) x²/3 - y²=1D) x² - y²/3=1So option D matches our result. Wait, but before I conclude, let me verify if point A is indeed on the asymptote and if all sides are length 2.Given a=1, b=√3, so the asymptotes are y=±√3 x. Let's check coordinates of point A.From earlier, t = -4 / [1 + (b²/a²)] = -4 / (1 + 3) = -4/4 = -1. Therefore, point A is (t, (b/a)t) = (-1, √3*(-1)) = (-1, -√3). Alternatively, if we take the negative asymptote, but since we assumed positive asymptote earlier, let's check both possibilities.But the problem states point A is on the asymptote of the hyperbola. The asymptotes are y=√3x and y=-√3x. So point A could be on either. Let's check coordinates of A.If we take t = -1, then with positive asymptote, A is (-1, -√3). Let's compute the distances:Distance OA: sqrt[(-1)^2 + (-√3)^2] = sqrt[1 + 3] = sqrt[4] = 2. Good.Distance AF: distance between (-1, -√3) and F (-2, 0). Let's compute:sqrt[(-1 - (-2))^2 + (-√3 - 0)^2] = sqrt[(1)^2 + ( -√3)^2] = sqrt[1 + 3] = sqrt[4] = 2. Perfect.Distance OF: sqrt[(-2)^2 + 0^2] = 2. Correct.So all sides are indeed 2. The triangle OAF with O(0,0), A(-1, -√3), F(-2,0) is equilateral. Let me visualize this triangle. Points O(0,0), F(-2,0), and A(-1, -√3). The distance from O to F is 2, from O to A is 2, and from A to F is 2. The coordinates check out.Wait, but the problem didn't specify which asymptote, but in our calculation, point A is on the asymptote y = -√3 x. Let's confirm that. The point (-1, -√3) lies on y = -√3 x, since substituting x = -1 gives y = -√3*(-1) = √3, which is not the case. Wait, wait a second. Wait, hold on. If the asymptote is y = (b/a)x, then with a=1, b=√3, the asymptote is y = √3 x. If point A is (-1, -√3), plugging into y = √3 x: √3*(-1) = -√3, which is correct? Wait, no. Wait, point (-1, -√3) would satisfy y = √3 x if when x = -1, y = √3*(-1) = -√3. So yes, that's exactly the case. So point A is on the asymptote y = √3 x. Wait, but when x is negative, y is negative. So even though the asymptote is y = √3 x, which for positive x goes up, for negative x, it goes down. So the point (-1, -√3) is indeed on the asymptote y = √3 x. So that's correct.Alternatively, if we considered the other asymptote y = -√3 x, then plugging x = -1, we'd get y = -√3*(-1) = √3, which would be the point (-1, √3), but that point would not satisfy the distance requirements. Let's check. If A was (-1, √3), then distance OA would still be sqrt[1 + 3] = 2. Distance AF would be sqrt[(-1 + 2)^2 + (√3 - 0)^2] = sqrt[1 + 3] = 2. So actually, that point is also possible. Wait, but then we have two possible points A: (-1, -√3) and (-1, √3). But in our calculation earlier, t was found to be -1, leading to y = (b/a)t = √3*(-1) = -√3. So perhaps the solution assumes the lower asymptote. However, the problem states point A is on the asymptote, not specifying which one. However, in our case, since we started by assuming the positive asymptote (y = (b/a)x), but t turned out negative, so the point is on the lower part of that asymptote.But in either case, both points (-1, √3) and (-1, -√3) are on the asymptotes, and both would form an equilateral triangle with O and F. Wait, but if A is (-1, √3), then the triangle OAF would have points O(0,0), F(-2,0), and A(-1, √3). Let's check the distances again. OA is 2, AF is sqrt[(-1 + 2)^2 + (√3 - 0)^2] = sqrt[1 + 3] = 2, and OF is 2. So both positions of A are valid. However, the orientation of the triangle would differ. Depending on the position of A, the triangle could be above or below the x-axis. But the problem doesn't specify, so either is acceptable. However, in our calculation, we found t = -1, leading to A being (-1, -√3), which is on the asymptote y = √3 x. Wait, but if the asymptote is y = √3 x, then for x negative, y is negative, so that's correct. Alternatively, if the asymptote was y = -√3 x, then for x negative, y would be positive. But since we derived b/a = √3, the asymptotes are y = ±√3 x, so both possibilities are covered.In any case, the key result is that a² = 1, b² = 3, leading to the hyperbola equation x² - y²/3 = 1, which is option D.But let me check the answer options again. Option D is x² - y²/3 =1. Yes, that's what we derived. Let me verify once again with the given values.Given a² =1, b²=3, c² = a² + b² = 4, so c=2. So foci at (±2,0). Left focus is (-2,0), which is point F. Point A is (-1, ±√3), which is on the asymptotes y=±√3 x. The triangle with vertices at (0,0), (-2,0), and (-1, ±√3) is indeed equilateral with all sides length 2. So everything checks out. Therefore, the correct answer should be D.But wait, let me check the options again. The options are:A) x²/4 - y²/12=1B) x²/12 - y²/4=1C) x²/3 - y²=1D) x² - y²/3=1Yes, D is x² - y²/3=1, which is equivalent to x²/1 - y²/3=1, which is what we have.Just to be thorough, let me see if any of the other options could also satisfy the conditions, in case I made a mistake.Take option A: x²/4 - y²/12=1. Here, a²=4, b²=12, so c²=4+12=16, c=4. Then the left focus would be at (-4,0). The asymptotes would be y=±(b/a)x=±(sqrt(12)/2)x=±(2*sqrt(3)/2)x=±sqrt(3)x. So asymptotes y=±sqrt(3)x. Then point A would be on asymptote. Let's attempt the same steps. Suppose triangle OAF is equilateral with sides 2. Then |OF|=c=4, but the side length is supposed to be 2. So c=2. But in option A, c=4. So this contradicts. Therefore, option A is invalid.Similarly, option B: x²/12 - y²/4=1. Here, a²=12, b²=4, so c²=12+4=16, c=4. Again, focus at (-4,0), which would give |OF|=4, not 2. So invalid.Option C: x²/3 - y²=1. Here, a²=3, b²=1, so c²=3+1=4, c=2. So left focus at (-2,0). Then asymptotes are y=±(b/a)x=±(1/sqrt(3))x. Let's check if triangle OAF can be equilateral with side length 2.Assume point A is on asymptote y=(1/sqrt(3))x. Let's go through the steps again. c=2. Then |OF|=2. Point A is (t, (1/sqrt(3))t). Distance OA=2:sqrt[t² + (1/3)t²] = sqrt[(4/3)t²] = (2/sqrt(3))|t| = 2. Therefore, |t|=sqrt(3). So t=±sqrt(3). Then point A is (sqrt(3), 1) or (-sqrt(3), -1). Now, distance AF:From (-2,0) to (sqrt(3),1):sqrt[(sqrt(3)+2)^2 + (1-0)^2] = sqrt[(3 +4sqrt(3) +4) +1] = sqrt[8 +4sqrt(3)] ≈ sqrt[8+6.928]≈sqrt[14.928]≈3.86, which is not 2. Similarly, for (-sqrt(3), -1):sqrt[(-sqrt(3)+2)^2 + (-1 -0)^2] = sqrt[(3 -4sqrt(3)+4)+1] = sqrt[8 -4sqrt(3)] ≈ sqrt[8 -6.928]≈sqrt[1.072]≈1.035, which is not 2. So neither of these gives a distance of 2. Therefore, option C is invalid.Option D, as we saw, works. Therefore, the correct answer is D.Final Answerboxed{D}</think>Given the hyperbola (frac{x^2}{a^2} - frac{y^2}{b^2} = 1) with its left focus (F) and point (A) on the asymptote of the hyperbola, where (triangle OAF) is an equilateral triangle with side length 2 (O is the origin), we need to determine the equation of the hyperbola.1. Identify the coordinates of the left focus (F): - The foci of the hyperbola are at ((pm c, 0)) where (c^2 = a^2 + b^2). - Since the side length of the equilateral triangle is 2, the distance from (O) to (F) is 2, so (c = 2).2. Relate (a) and (b) using the focus distance: - (c^2 = a^2 + b^2 = 4).3. Determine the coordinates of point (A) on the asymptote: - The asymptotes of the hyperbola are (y = pm frac{b}{a}x). - Point (A) is on one of these asymptotes, so let (A = (t, frac{b}{a}t)) or (A = (t, -frac{b}{a}t)).4. Calculate the distance from (O) to (A): - The distance (|OA| = 2) gives (sqrt{t^2 + left(frac{b}{a}tright)^2} = 2). - Simplifying, (|t| sqrt{1 + left(frac{b}{a}right)^2} = 2).5. Calculate the distance from (A) to (F): - The distance (|AF| = 2) gives (sqrt{(t + 2)^2 + left(frac{b}{a}tright)^2} = 2). - Simplifying, ((t + 2)^2 + left(frac{b}{a}tright)^2 = 4).6. Solve for (t) and relate to the previous equations: - Solving the equations derived from the distances, we find (1 + left(frac{b}{a}right)^2 = 4), leading to (frac{b^2}{a^2} = 3) or (b^2 = 3a^2).7. Substitute (b^2 = 3a^2) into (c^2 = a^2 + b^2): - (a^2 + 3a^2 = 4) gives (4a^2 = 4) or (a^2 = 1). Thus, (a = 1) and (b^2 = 3).8. Determine the equation of the hyperbola: - Substituting (a^2 = 1) and (b^2 = 3) into the standard form, we get (frac{x^2}{1} - frac{y^2}{3} = 1), which simplifies to (x^2 - frac{y^2}{3} = 1).Thus, the equation of the hyperbola is (boxed{D}).]

🔑:This problem involves applying properties of hyperbolas. Since triangle OAF is an equilateral triangle, we can find the values of a and b. Then, substitute these values into the equation of the hyperbola.Given that the hyperbola frac{{x}^{2}}{{a}^{2}}-frac{{y}^{2}}{{b}^{2}}=1 (a > 0, b > 0) has its left focus at F, and point A on the asymptote of the hyperbola. triangle OAF is an equilateral triangle with side length 2.It follows that c=2 and frac{b}{a}=sqrt{3}.Squaring both sides of frac{b}{a}=sqrt{3}, we get frac{{b}^{2}}{{a}^{2}}=3.Also, the relationship between a, b, and c in a hyperbola is given by frac{{c}^{2}-{a}^{2}}{{a}^{2}}=frac{{b}^{2}}{{a}^{2}}.Substituting frac{{b}^{2}}{{a}^{2}}=3 into frac{{c}^{2}-{a}^{2}}{{a}^{2}}=frac{{b}^{2}}{{a}^{2}}, we get frac{{c}^{2}-{a}^{2}}{{a}^{2}}=3.Solving for a, we obtain a=1.Substituting a=1 into frac{b}{a}=sqrt{3}, we obtain b=sqrt{3}.Thus, the equation of the hyperbola is {{x}^{{2}}}-frac{{{y}^{{2}}}}{{3}}={1}.Therefore, the correct answer is boxed{D}.