🧠:Okay, so I need to find the area of a right triangle where the sum of the legs is given as 'l', and the altitude from the right angle is 'h'. Hmm, let's start by recalling some basic properties of right triangles. First, in a right triangle, the area can be expressed in two ways: one using the legs as the base and height, and another using the hypotenuse and the altitude from the right angle. Let me note that down. If the legs are 'a' and 'b', then the area is (a*b)/2. Also, the hypotenuse would be sqrt(a² + b²), right? And the altitude 'h' relates to the area as well because the area can also be written as (hypotenuse * h)/2. So, both expressions should be equal. Given that the sum of the legs a + b = l, and the altitude is h. The goal is to find the area, which is (a*b)/2. But I need to express this area in terms of 'l' and 'h' instead of 'a' and 'b'. Let me try to set up some equations. From the area expressions:1) (a*b)/2 = (hypotenuse * h)/2. So, a*b = hypotenuse * h.But hypotenuse is sqrt(a² + b²). So, substituting that in:a*b = sqrt(a² + b²) * h.Also, we know that a + b = l. Maybe I can express a² + b² in terms of (a + b)². Let's recall that (a + b)² = a² + 2ab + b². Therefore, a² + b² = (a + b)² - 2ab = l² - 2ab.Substituting back into the hypotenuse expression:sqrt(l² - 2ab) * h = ab.So, sqrt(l² - 2ab) = (ab)/h.If I square both sides to eliminate the square root:l² - 2ab = (a²b²)/h².Hmm, this seems a bit complicated. Let me rearrange the equation:Multiply both sides by h²:h²(l² - 2ab) = a²b².But I need to solve for ab, since the area is (ab)/2. Let me denote x = ab. Then the equation becomes:h²(l² - 2x) = x².Expanding that:h²l² - 2h²x = x².Bring all terms to one side:x² + 2h²x - h²l² = 0.This is a quadratic equation in terms of x. Let's write it as:x² + 2h²x - h²l² = 0.Using the quadratic formula, x = [-2h² ± sqrt((2h²)^2 + 4*1*h²l²)]/2.Simplify the discriminant:(2h²)^2 + 4*1*h²l² = 4h^4 + 4h²l² = 4h²(h² + l²).So, sqrt(4h²(h² + l²)) = 2h*sqrt(h² + l²).Therefore, x = [-2h² ± 2h*sqrt(h² + l²)]/2.Divide numerator and denominator by 2:x = [-h² ± h*sqrt(h² + l²)].Since x = ab, which is a positive quantity, we discard the negative solution. So,x = [-h² + h*sqrt(h² + l²)].Wait, but that gives x as a positive value? Let's check:sqrt(h² + l²) is greater than h, so h*sqrt(h² + l²) is greater than h². Therefore, -h² + h*sqrt(h² + l²) is positive. So, that's good.Therefore, ab = x = h(-h + sqrt(h² + l²)).But we need the area, which is ab/2. So,Area = (1/2)*h(-h + sqrt(h² + l²)).Hmm, that seems a bit messy. Let me see if there's another way to approach this problem.Alternatively, maybe using some substitution for a and b. Let me let a + b = l and ab = 2*Area. So, if I can express ab in terms of h and l, that would work.Wait, from the earlier equation: a*b = sqrt(a² + b²) * h. And we have a² + b² = l² - 2ab. So:ab = sqrt(l² - 2ab) * h.So, square both sides:(ab)^2 = h²(l² - 2ab).Which is the same equation as before. So, same quadratic equation. So, solving for ab gives the same result.Alternatively, maybe we can parameterize the legs a and b. Let me set a = l/2 + t and b = l/2 - t. Then, a + b = l, and t is some parameter. Then, the product ab = (l/2 + t)(l/2 - t) = (l²)/4 - t².Then, the hypotenuse is sqrt(a² + b²). Let me compute a² + b²:a² + b² = (l/2 + t)² + (l/2 - t)² = 2*(l²/4 + t²) = l²/2 + 2t².Then, the altitude h is given by the area over half the hypotenuse. Wait, altitude h = (a*b)/ (hypotenuse/2) = 2ab / hypotenuse.Wait, area is (a*b)/2, and also (hypotenuse * h)/2. Therefore, (a*b)/2 = (hypotenuse * h)/2 => h = (a*b)/hypotenuse.So, h = ab / sqrt(a² + b²). Substituting ab = (l²)/4 - t² and sqrt(a² + b²) = sqrt(l²/2 + 2t²).So, h = [(l²/4 - t²)] / sqrt(l²/2 + 2t²).This seems more complicated. Maybe not helpful.Alternatively, let's think about the relation between h, the legs, and the hypotenuse. In a right triangle, the altitude h relates to the legs as 1/h² = 1/a² + 1/b². Is that correct? Let me verify.Yes, in a right triangle, the altitude h from the right angle satisfies 1/h² = 1/a² + 1/b². Let me recall how that is derived.Since area is (a*b)/2, and also (c*h)/2, where c is hypotenuse. So, h = (a*b)/c. Then, 1/h² = c²/(a²b²). But c² = a² + b². Therefore, 1/h² = (a² + b²)/(a²b²) = 1/b² + 1/a². So, yes, 1/h² = 1/a² + 1/b².So, we have another equation: 1/h² = 1/a² + 1/b². And we also have a + b = l.So, now we have two equations:1) a + b = l2) 1/h² = 1/a² + 1/b²We need to find the area, which is (a*b)/2. Let me denote S = a*b. Then, the area is S/2. Our goal is to find S.From equation 1, a + b = l. From equation 2, 1/h² = (a² + b²)/(a²b²) = ( (a + b)^2 - 2ab ) / (S²) = (l² - 2S)/S².Therefore, 1/h² = (l² - 2S)/S².Multiplying both sides by S²:S²/h² = l² - 2S.Multiply both sides by h²:S² = h²(l² - 2S).Bring all terms to one side:S² + 2h²S - h²l² = 0.Which is the same quadratic equation as before, where S = ab. So, solving for S gives the same result as before. Therefore, the quadratic equation leads us to S = [ -2h² ± sqrt(4h^4 + 4h²l²) ] / 2. Which simplifies to S = [ -2h² ± 2h*sqrt(h² + l²) ] / 2 = -h² ± h*sqrt(h² + l²). Since S must be positive, we take the positive root: S = -h² + h*sqrt(h² + l²). Therefore, the area is S/2 = [ -h² + h*sqrt(h² + l²) ] / 2.Hmm, this expression seems a bit complicated. Maybe we can simplify it further. Let's factor out h from the numerator:S = h( -h + sqrt(h² + l²) ). Then, Area = (h/2)( -h + sqrt(h² + l²) ).Alternatively, rationalize the numerator. Let me see:Multiply numerator and denominator by (sqrt(h² + l²) + h):[ (-h + sqrt(h² + l²) ) * (sqrt(h² + l²) + h) ] / (sqrt(h² + l²) + h) )But that would be:[ (sqrt(h² + l²))² - h² ] / (sqrt(h² + l²) + h) )= [ (h² + l² - h² ) ] / (sqrt(h² + l²) + h)= l² / (sqrt(h² + l²) + h)Therefore, Area = (h/2) * [ l² / (sqrt(h² + l²) + h) ].So, Area = (h l²) / [ 2 (sqrt(h² + l²) + h ) ].Hmm, this might be a simpler expression. Let me check if this is equivalent to the previous one.Yes, because:Original expression: (h/2)( -h + sqrt(h² + l²) )Multiply numerator and denominator by (sqrt(h² + l²) + h):(h/2) [ (-h + sqrt(h² + l²))(sqrt(h² + l²) + h) ] / (sqrt(h² + l²) + h)The numerator becomes (sqrt(h² + l²))² - h² = h² + l² - h² = l².So, (h/2)(l²) / (sqrt(h² + l²) + h) ) = (h l²)/(2 (sqrt(h² + l²) + h )).So, that's the same as before. Therefore, Area = (h l²)/(2 (sqrt(h² + l²) + h )).Alternatively, we can rationalize the denominator in another way. Let me see:Multiply numerator and denominator by (sqrt(h² + l²) - h):[ h l² (sqrt(h² + l²) - h) ] / [ 2 ( (sqrt(h² + l²) + h)(sqrt(h² + l²) - h) ) ]Denominator becomes (h² + l²) - h² = l². So,[ h l² (sqrt(h² + l²) - h) ] / (2 l² ) ) = [ h (sqrt(h² + l²) - h) ] / 2.Which brings us back to the original expression. So, both forms are equivalent.Therefore, the area can be written in two equivalent forms:1) Area = [ h (sqrt(h² + l² ) - h ) ] / 22) Area = (h l² ) / [ 2 ( sqrt(h² + l² ) + h ) ]Either form is acceptable, but perhaps the first one is simpler. Alternatively, maybe we can write it in terms of l and h without square roots in the denominator, but I think both forms are correct.Is there a way to express this in terms of l and h more elegantly? Let me think.Alternatively, perhaps express the area as (l² - (a - b)^2 ) / 4, since (a + b)^2 = l² and (a - b)^2 = a² - 2ab + b². Therefore, ab = [l² - (a - b)^2 ] / 4. But I don't know if that helps here.Alternatively, since we have two expressions for the area: one in terms of ab and another in terms of h and hypotenuse. But I think the earlier quadratic approach is the most straightforward, even though the solution looks a bit complicated.Alternatively, let's check if the answer can be expressed as h*l/2. But no, because h is the altitude to the hypotenuse, not to a leg. So, that would not be correct. For example, if h is the altitude to the hypotenuse, then the area is (hypotenuse * h)/2, but we need to relate hypotenuse to l. But since hypotenuse is sqrt(a² + b²) and a + b = l, perhaps we can find another relation.Wait, hypotenuse squared is a² + b² = (a + b)^2 - 2ab = l² - 2ab. So, hypotenuse = sqrt(l² - 2ab). But from the area, hypotenuse = (2 * area)/h. Therefore, sqrt(l² - 2ab) = (2 * area)/h. But area is (ab)/2. Therefore, sqrt(l² - 2ab) = (2*(ab)/2)/h = ab/h. So, sqrt(l² - 2ab) = ab/h. Which is the same equation as before. So, we end up with the same quadratic.Therefore, perhaps there's no simpler way, and we have to accept that the area is expressed in terms of h and l with a square root. Let me verify with specific numbers to see if the formula makes sense.Suppose, for example, that h = l. Let's pick h = l = 2. Then, what would the area be?Using the formula Area = [ h (sqrt(h² + l² ) - h ) ] / 2.Plugging h = 2, l = 2:Area = [2 (sqrt(4 + 4) - 2)] / 2 = [2 (sqrt(8) - 2)] / 2 = (2*(2√2 - 2))/2 = (4√2 - 4)/2 = 2√2 - 2 ≈ 2.828 - 2 = 0.828.Let me check if this makes sense. If h = l = 2, then the altitude is 2 and the sum of legs is 2. Let's suppose the triangle has legs a and b, with a + b = 2 and altitude h = 2. Let's try to find a and b.From the area expressions: (a*b)/2 = (c*2)/2 => ab = c, where c is the hypotenuse. But c = sqrt(a² + b²). So, ab = sqrt(a² + b²). Also, a + b = 2.Let me square ab = sqrt(a² + b²):(ab)^2 = a² + b².But a² + b² = (a + b)^2 - 2ab = 4 - 2ab.Therefore, (ab)^2 = 4 - 2ab.Let x = ab:x² + 2x - 4 = 0.Solving: x = [-2 ± sqrt(4 + 16)]/2 = [-2 ± sqrt(20)]/2 = [-2 ± 2*sqrt(5)]/2 = -1 ± sqrt(5). Since ab is positive, x = -1 + sqrt(5) ≈ 1.236. Therefore, area is x/2 ≈ 0.618, but according to our formula, we got approximately 0.828. Wait, this is a discrepancy. So, either my example is invalid or I made a mistake.Wait, hold on. If h = l = 2, but in reality, can such a triangle exist? Let's check.Suppose a + b = 2 and altitude h = 2. Then, the altitude in a right triangle is given by h = (ab)/c, where c is the hypotenuse. So, h = (ab)/c = 2.But c = sqrt(a² + b²). So, ab = 2c. Also, a + b = 2.From a + b = 2, let's denote b = 2 - a. Then, ab = a(2 - a) = 2a - a². Then, c = sqrt(a² + (2 - a)^2) = sqrt(2a² -4a +4).Therefore, ab = 2c => 2a - a² = 2*sqrt(2a² -4a +4).Let me square both sides:(2a - a²)^2 = 4*(2a² -4a +4)Expand left side:4a² -4a³ +a^4 = 8a² -16a +16Bring all terms to left side:a^4 -4a³ +4a² -8a² +16a -16 = 0Simplify:a^4 -4a³ -4a² +16a -16 = 0Let me factor this equation. Maybe try rational roots. Possible roots are ±1, ±2, ±4, ±8, ±16.Test a = 2: 16 -32 -16 +32 -16 = 16 -32= -16 -16= -32 +32= 0. Wait, 16 -32= -16; -16 -16= -32; -32 +32= 0; 0 -16= -16. Not zero.Wait, let's compute again:a=2: 2^4 -4*2^3 -4*2^2 +16*2 -16 = 16 -32 -16 +32 -16 = (16 -32) + (-16 +32) + (-16) = (-16) + (16) + (-16) = -16 ≠0.a=1: 1 -4 -4 +16 -16= (1-4)= -3; (-4+16)=12; (-16)= -16. Total: -3 +12 -16= -7≠0.a=4: 256 -256 -64 +64 -16= (256-256)=0; (-64 +64)=0; -16≠0.a= -1: 1 +4 -4 -16 -16= -31≠0.Hmm, perhaps no rational roots. Maybe this equation has no real roots? Let's check discriminant or see if the original equation is possible.If ab = 2c, and a + b =2, with a and b positive. Let's see if such a triangle exists.Suppose a approaches 0, then b approaches 2. Then, c approaches sqrt(0 + 4) = 2. Then, ab approaches 0, which would make h = ab/c approach 0. But h is supposed to be 2. So, as a increases from 0, h increases, but is there a point where h=2?Wait, when a = b =1, then c = sqrt(2). Then, h = (1*1)/sqrt(2) ≈ 0.707. So, h is less than 2. If we need h =2, then with a + b =2, this seems impossible because the maximum possible value of h occurs when a = b =1, giving h ≈0.707, which is way less than 2. Therefore, such a triangle with a + b =2 and h=2 does not exist. So, my initial example was invalid because such a triangle cannot exist. Therefore, the formula giving Area ≈0.828 is also invalid, which suggests that when h > maximum possible h for given l, the formula gives a positive area, but in reality, no such triangle exists. Therefore, the formula is correct mathematically, but physically, h cannot exceed a certain value for given l.Therefore, in the problem statement, they must assume that h is a possible value for given l. So, assuming that such a triangle exists, the formula is correct.So, returning to the problem, the answer is Area = [ h (sqrt(h² + l² ) - h ) ] / 2. Alternatively, written as (h/2)(sqrt(h² + l²) - h). Alternatively, factoring h/2:Area = (h sqrt(h² + l² ) - h² ) / 2.But perhaps the problem expects a different form? Let me check if this can be simplified further or expressed in another way.Wait, another approach: Let’s denote t = a + b = l, and we know that in a right triangle, the altitude h = ab/c, where c is hypotenuse. Also, c = sqrt(a² + b²). We can use the relationship between t, h, and the area. Let’s express everything in terms of t and h.We have:h = ab / c => ab = h c.But c = sqrt(a² + b²) = sqrt(t² - 2ab). Therefore:ab = h sqrt(t² - 2ab).Square both sides:(ab)^2 = h² (t² - 2ab).Let S = ab, then:S² = h² t² - 2 h² S.Bring all terms to left:S² + 2 h² S - h² t² = 0.Which is the same quadratic equation as before. Solving for S gives:S = [ -2 h² ± sqrt(4 h^4 + 4 h² t² ) ] / 2 = [ -2 h² ± 2 h sqrt( h² + t² ) ] / 2 = -h² ± h sqrt( h² + t² ).Since S must be positive, take the positive root:S = -h² + h sqrt( h² + t² ).Thus, the area is S/2 = [ -h² + h sqrt( h² + t² ) ] / 2.Given that t = l, substitute back:Area = [ h sqrt( h² + l² ) - h² ] / 2 = h [ sqrt( h² + l² ) - h ] / 2.So, that's the result. I think this is the simplest form unless there's an alternative expression.Alternatively, factor h from the numerator:Area = (h/2)(sqrt(h² + l²) - h).This seems to be the most compact form.Let me check dimensions. If l and h are lengths, then the area has dimension of length squared. The expression h*(sqrt(h² + l²) - h) is h*(something with dimension length), so h*length gives length squared. Then divided by 2, so dimensions are correct.Alternatively, we can also rationalize the expression:sqrt(h² + l²) - h = [ (sqrt(h² + l²) - h) * (sqrt(h² + l²) + h) ] / (sqrt(h² + l²) + h) ) = (h² + l² - h² ) / (sqrt(h² + l²) + h ) ) = l² / (sqrt(h² + l²) + h ).Therefore, Area = (h/2) * (l² / (sqrt(h² + l²) + h )) = (h l² ) / [ 2 (sqrt(h² + l²) + h ) ].So, another form is Area = (h l² ) / [ 2 (sqrt( h² + l² ) + h ) ].Either form is acceptable. Depending on the required answer format, both are correct. But maybe the first form is preferable.Therefore, after going through the quadratic solution and verifying the steps, the area of the triangle is (h/2)(sqrt(h² + l²) - h).Final AnswerThe area of the triangle is boxed{dfrac{h}{2} left( sqrt{h^2 + l^2} - h right)}.

✅:Given a right triangle where the sum of the legs is ( l ) and the altitude from the right angle is ( h ), we need to find the area of the triangle.1. Define Variables and Basic Relationships: - Let the legs be ( a ) and ( b ). - The sum of the legs: ( a + b = l ). - The area of the triangle can be expressed as ( frac{1}{2}ab ) and also as ( frac{1}{2} times text{hypotenuse} times h ).2. Hypotenuse Calculation: - The hypotenuse ( c ) is ( sqrt{a^2 + b^2} ). - Using the relationship ( (a + b)^2 = a^2 + 2ab + b^2 ), we get ( a^2 + b^2 = l^2 - 2ab ).3. Area Relationships: - From the area expressions: ( ab = sqrt{a^2 + b^2} times h ). - Substitute ( a^2 + b^2 = l^2 - 2ab ): ( ab = sqrt{l^2 - 2ab} times h ).4. Quadratic Equation: - Square both sides: ( (ab)^2 = h^2(l^2 - 2ab) ). - Let ( S = ab ): ( S^2 + 2h^2S - h^2l^2 = 0 ).5. Solve the Quadratic Equation: - Using the quadratic formula ( S = frac{-2h^2 pm sqrt{4h^4 + 4h^2l^2}}{2} ). - Simplify: ( S = -h^2 + hsqrt{h^2 + l^2} ).6. Area Calculation: - The area is ( frac{S}{2} ): ( frac{1}{2}(-h^2 + hsqrt{h^2 + l^2}) ).7. Final Expression: - Factor out ( h ): ( frac{h}{2}(sqrt{h^2 + l^2} - h) ).Thus, the area of the triangle is (boxed{dfrac{h}{2} left( sqrt{h^2 + l^2} - h right)}).

🔑:Given:- The sum of the legs (catheti) of a right triangle is ( l ).- The height dropped from the vertex of the right angle is ( h ).We are required to find the area ( S ) of the triangle. Let's denote the legs of the right triangle ( ABC ) as ( AC = x ) and ( BC = y ).1. Using the Pythagorean Theorem: The hypotenuse ( AB ) can be expressed as: [ AB = sqrt{x^2 + y^2} ]2. Expressing the sum of the legs: We are given that: [ x + y = l ] 3. Rewriting the hypotenuse using the area relationship: Using the identity ( (a + b)^2 = a^2 + 2ab + b^2 ), we have: [ (x + y)^2 = x^2 + y^2 + 2xy implies l^2 = x^2 + y^2 + 2xy ] Therefore, [ x^2 + y^2 = l^2 - 2xy implies AB = sqrt{x^2 + y^2} = sqrt{l^2 - 2xy} ]4. Using the formula of the triangle area: The area ( S ) of the triangle can be expressed using the given height ( h ): [ S = frac{1}{2} times CD times AB ] Here, ( CD = h ) is the height, and ( AB = sqrt{l^2 - 2xy} ).5. Setting up the equation: Substitute into the area formula: [ S = frac{1}{2} h cdot sqrt{l^2 - 2xy} ]6. Relationship between ( S ) and ( h ): We know: [ S = frac{1}{2} h cdot sqrt{l^2 - 4S} ]7. Solve for ( S ): To isolate ( S ), we equate: [ frac{1}{2} h cdot sqrt{l^2 - 4S} = S ] Squaring both sides to eliminate the square root, we get: [ left(frac{1}{2} h cdot sqrt{l^2 - 4S}right)^2 = S^2 ] [ frac{1}{4} h^2 cdot (l^2 - 4S) = S^2 ] [ frac{1}{4} h^2 l^2 - h^2 S = S^2 ]8. Re-arranging and solving the quadratic equation: Multiply by 4 to clear the denominator: [ h^2 l^2 - 4h^2 S = 4S^2 ] [ 4S^2 + 4h^2 S - h^2 l^2 = 0 ] Set this up as a standard quadratic equation ( A S^2 + B S + C = 0 ): [ S^2 + h^2 S - frac{h^2 l^2}{4} = 0 ] Using the quadratic formula ( S = frac{-B pm sqrt{B^2 - 4AC}}{2A} ) where ( A = 1 ), ( B = h^2 ), and ( C = -frac{h^2 l^2}{4} ): [ S = frac{-h^2 pm sqrt{h^4 + h^2 l^2}}{2} ] [ S = frac{-h^2 pm h^2 sqrt{1 + frac{l^2}{h^2}}}{2} ] [ S = frac{-h^2 pm h sqrt{h^2 + l^2}}{2} ] Only the positive solution is meaningful here: [ S = frac{1}{2} h left(sqrt{l^2 + h^2} - hright) ]Conclusion:[boxed{frac{1}{2} h left( sqrt{l^2 + h^2} - h right)}]

🧠:Okay, let's see. I need to find integer solutions a and b to the equation 17a + 6b = 13, and then determine the smallest possible positive value for a - b. Hmm, alright. So first, since a and b are integers, this is a linear Diophantine equation. I remember that to solve such equations, we can use the Extended Euclidean Algorithm. Let me try to recall how that works.First, the equation 17a + 6b = 13. For this equation to have integer solutions, the greatest common divisor (gcd) of 17 and 6 must divide 13. Let's check the gcd of 17 and 6. The gcd of 17 and 6 is 1 because 17 is a prime number and 6 is 2×3, neither of which divide 17. Since 1 divides 13, solutions exist. Good.Now, to find particular solutions, I can use the Extended Euclidean Algorithm to express 1 as a linear combination of 17 and 6, and then scale it up to get 13. Let me try that.Starting with the Euclidean algorithm for 17 and 6:17 divided by 6 is 2 with a remainder of 5. So, 17 = 2×6 + 5.Then, 6 divided by 5 is 1 with a remainder of 1. So, 6 = 1×5 + 1.Next, 5 divided by 1 is 5 with a remainder of 0. So, gcd is 1.Now, working backwards to express 1 as a combination:From the second step, 1 = 6 - 1×5.But 5 from the first step is 17 - 2×6. Substitute that into the equation:1 = 6 - 1×(17 - 2×6) = 6 - 17 + 2×6 = 3×6 - 17.So, 1 = (-1)×17 + 3×6. Therefore, multiplying both sides by 13 to get the equation equal to 13:13 = (-13)×17 + 39×6.Therefore, one particular solution is a = -13, b = 39. Let me check: 17×(-13) + 6×39 = -221 + 234 = 13. Yep, that works.But the general solution for a linear Diophantine equation ax + by = c is given by x = x0 + (b/gcd)t and y = y0 - (a/gcd)t, where t is an integer. Since gcd(17,6) is 1, the general solution here is:a = -13 + 6tb = 39 - 17tfor integer t. Let me verify this. Plugging into 17a + 6b:17(-13 + 6t) + 6(39 - 17t) = 17×(-13) + 17×6t + 6×39 - 6×17t.The terms with t cancel out: 17×6t - 6×17t = 0. Then, -221 + 234 = 13. Correct. So the general solution is as above.Now, we need to find integer solutions a and b, so a and b must be integers. The expressions for a and b are linear in t, so as t varies over all integers, we get all possible solutions.Our goal is to find the smallest possible positive value of a - b. Let's compute a - b in terms of t.a - b = (-13 + 6t) - (39 - 17t) = -13 + 6t - 39 + 17t = (-13 - 39) + (6t + 17t) = -52 + 23t.So, a - b = -52 + 23t. We need this to be positive and as small as possible. So, we have:-52 + 23t > 0Solving for t:23t > 52t > 52/2352 divided by 23 is approximately 2.2609. Since t must be an integer, the smallest integer t satisfying this is t = 3.Wait, wait. Wait, 52 divided by 23 is 2.2609, so t must be greater than 2.2609, so the next integer is 3. So t ≥ 3. Therefore, the smallest possible positive value occurs when t=3. Let's compute a - b for t=3:-52 + 23×3 = -52 + 69 = 17.But wait, let's check if t=3 gives a positive a - b of 17, but maybe there's a smaller positive value for some higher t? Wait, no. Because as t increases, 23t increases, so a - b increases. Therefore, t=3 gives the minimal positive a - b. Hmm.Wait, but maybe there's a mistake here. Let me check for t=2. If t=2, a - b = -52 + 23×2 = -52 + 46 = -6, which is negative. t=3 gives 17, which is positive. Then, t=4 would give -52 + 23×4 = -52 + 92 = 40. So as t increases, a - b increases by 23 each time. So the minimal positive value is 17. But let's see if there's a possibility of a smaller positive value for some other t. Wait, but according to the formula, once you pass t=3, it's 17, then 40, 63, etc., so 17 is the minimal positive. However, I need to confirm if all possible solutions are covered by this parametrization.Wait, but perhaps there are other solutions where t is less than 3? But since t must be integer, and t needs to be at least 3 to make a - b positive. But let's check another way.Alternatively, perhaps there's another particular solution where a - b is smaller. Wait, the general solution is a = -13 + 6t and b = 39 -17t. So different values of t can give different a and b. Let me check a few values of t to see.t=3:a = -13 + 18 = 5b = 39 - 51 = -12a - b = 5 - (-12) = 17t=4:a = -13 + 24 = 11b = 39 - 68 = -29a - b = 11 - (-29) = 40t=5:a = -13 + 30 = 17b = 39 - 85 = -46a - b = 17 - (-46) = 63t=2:a = -13 + 12 = -1b = 39 - 34 = 5a - b = -1 - 5 = -6t=1:a = -13 + 6 = -7b = 39 - 17 = 22a - b = -7 - 22 = -29t=0:a = -13b=39a - b = -52t=-1:a = -13 -6 = -19b = 39 +17=56a - b = -19 -56= -75So, as t increases beyond 3, a - b increases by 23 each time. So for t=3, a - b is 17, which is the first positive value, and all higher t give larger values. Therefore, 17 is the smallest positive value. Therefore, the answer is 17.But wait, let me check again. Maybe I missed something here. Let's think again.Suppose there exists another solution where a - b is smaller than 17 but still positive. How can that be? Since the general solution is a = -13 +6t and b = 39 -17t, then a - b is -52 +23t. So for a - b to be positive, 23t > 52, so t > 52/23 ≈2.26. Hence, t must be at least 3. Therefore, the smallest possible positive value is when t=3, giving 17. Therefore, 17 is indeed the minimal positive value.But let's check another way. Maybe find all possible a and b that satisfy the equation and compute a - b for each to see the minimum positive value.But since the solutions are given parametrically, and a - b is linear in t, the minimal positive occurs at the smallest t where the expression becomes positive. Since it's linear and increasing with t, the first positive value is the minimal. Therefore, 17 is correct.Alternatively, perhaps the user wants a different approach? Let me try another method to confirm.Suppose we set x = a - b. We need to find the smallest positive x where there exist integers a and b such that 17a +6b =13 and x = a - b. Let's express a in terms of x and b: a = x + b. Substitute into the equation:17(x + b) + 6b = 1317x +17b +6b =1317x +23b =13So now we have 23b =13 -17xSo, 23b =13 -17x => b = (13 -17x)/23Since b must be an integer, (13 -17x) must be divisible by 23. Therefore, 13 -17x ≡0 mod23So, -17x ≡ -13 mod23Multiply both sides by -1: 17x ≡13 mod23We need to solve 17x ≡13 mod23First, find the inverse of 17 mod23. Since 17 and 23 are coprime, the inverse exists.Find an integer k such that 17k ≡1 mod23.Let's compute:17*1=17 mod23=1717*2=34 mod23=1117*3=17+11=28 mod23=517*4=5+17=22 mod23=2217*5=22+17=39 mod23=39-23=1617*6=16+17=33 mod23=1017*7=10+17=27 mod23=417*8=4+17=21 mod23=2117*9=21+17=38 mod23=1517*10=15+17=32 mod23=917*11=9+17=26 mod23=317*12=3+17=20 mod23=2017*13=20+17=37 mod23=1417*14=14+17=31 mod23=817*15=8+17=25 mod23=217*16=2+17=19 mod23=1917*17=19+17=36 mod23=1317*18=13+17=30 mod23=717*19=7+17=24 mod23=1Ah! So 17*19 ≡1 mod23. Therefore, inverse of 17 mod23 is 19.Therefore, x ≡13*19 mod23Compute 13*19: 13*20=260 -13=247247 divided by 23: 23*10=230, 247-230=17, so 247≡17 mod23.Therefore, x ≡17 mod23. So the solutions for x are x=17 +23k, where k is integer.But we need the smallest positive x. So the minimal positive solution is x=17. Then the next ones would be 17+23=40, 17+46=63, etc., and for negative k: 17-23= -6, which is negative. Therefore, the smallest positive x is 17. Therefore, this confirms the previous result.Hence, the answer is 17.

✅:To solve the equation (17a + 6b = 13) for integers (a) and (b), we first use the Extended Euclidean Algorithm to find a particular solution. The steps are as follows:1. Apply the Euclidean Algorithm: - (17 = 2 cdot 6 + 5) - (6 = 1 cdot 5 + 1) - Back-substitute to express (1) as a combination: (1 = 6 - 1 cdot 5 = 6 - 1 cdot (17 - 2 cdot 6) = 3 cdot 6 - 17).2. Scale to match the equation: Multiply by 13: (13 = (-13) cdot 17 + 39 cdot 6). Thus, a particular solution is (a_0 = -13), (b_0 = 39).3. General solution: The solutions are given by: (a = -13 + 6t) (b = 39 - 17t) where (t) is an integer.4. Express (a - b): Substitute the general solutions: (a - b = (-13 + 6t) - (39 - 17t) = -52 + 23t).5. Find the smallest positive (a - b): Solve (-52 + 23t > 0): (23t > 52 implies t > frac{52}{23} approx 2.26). The smallest integer (t) is 3. Substituting (t = 3): (a - b = -52 + 23 cdot 3 = 17).Verification: For (t = 3), (a = 5) and (b = -12). Check: (17 cdot 5 + 6 cdot (-12) = 85 - 72 = 13). Thus, (a - b = 5 - (-12) = 17).Conclusion: The smallest possible positive value for (a - b) is (boxed{17}).

🔑:We are given the Diophantine equation:[ 17a + 6b = 13 ]Our goal is to find the smallest possible positive value for ( a - b ). Here is the step-by-step derivation:1. Rewriting the equation: Notice that we can rewrite the original equation in a more useful form: [ 17(a - b) + 23b = 13 ] This transformation is done by adding and subtracting ( 17b ) simultaneously: [ 17a + 6b = 17(a - b) + 17b + 6b = 17(a - b) + 23b ]2. Working modulo ( 23 ): To simplify the equation, we consider it modulo ( 23 ): [ 17(a - b) + 23b equiv 13 pmod{23} ] Since ( 23b equiv 0 pmod{23} ), this reduces to: [ 17(a - b) equiv 13 pmod{23} ]3. Finding the inverse modulo ( 23 ): To isolate ( a - b ), we need the multiplicative inverse of ( 17 ) modulo ( 23 ), which is a number ( x ) such that: [ 17x equiv 1 pmod{23} ] Using the Extended Euclidean Algorithm, we find the inverse of ( 17 ) modulo ( 23 ) is ( -4 ), because: [ 17 cdot (-4) equiv -68 equiv 1 pmod{23} ]4. Solving for ( a - b ): With the multiplicative inverse ( -4 ), we multiply both sides of the congruence: [ 17(a - b) equiv 13 pmod{23} ] [ (a - b) equiv 13 cdot (-4) pmod{23} ] [ (a - b) equiv -52 pmod{23} ]5. Simplifying the congruence: Simplify ( -52 mod 23 ): [ -52 div 23 = -2 quad text{remainder} , -6 ] So, we have: [ (a - b) equiv 17 pmod{23} ]Thus, the smallest possible positive value for ( a - b ) is ( 17 ).6. Verification: To verify, we need to find integers ( a ) and ( b ) such that: [ a - b = 17 ] [ a = b + 17 ] Substitute into the initial equation: [ 17(b + 17) + 6b = 13 ] [ 17b + 289 + 6b = 13 ] [ 23b = 13 - 289 ] [ 23b = -276 ] [ b = -12 ] Therefore, ( a = b + 17 = -12 + 17 = 5 ).Conclusion:[ boxed{17} ]

🧠:Okay, let's try to solve this problem step by step. The question is asking for the number of values among S₁, S₂, ..., S₂₀₁₇ that are divisible by 2 but not by 4. First, let's make sure I understand the definitions correctly. The sequence aₙ is given as the sum from 1 to n, which is the nth triangular number. So, aₙ = 1 + 2 + ... + n = n(n + 1)/2. Then, Sₘ is the sum of the first m triangular numbers. So, Sₘ = a₁ + a₂ + ... + aₘ. We need to find how many of these S₁ to S₂₀₁₇ are divisible by 2 but not by 4. In other words, numbers congruent to 2 mod 4.Alright, let's start by finding a general formula for Sₘ. Since each aₙ is n(n + 1)/2, then Sₘ is the sum from n=1 to m of [n(n + 1)/2]. Let's write that out:Sₘ = Σₙ=1^m [n(n + 1)/2] = (1/2) Σₙ=1^m (n² + n) = (1/2)(Σn² + Σn)We know that Σn from 1 to m is m(m + 1)/2, and Σn² from 1 to m is m(m + 1)(2m + 1)/6. Plugging these into the formula:Sₘ = (1/2)[m(m + 1)(2m + 1)/6 + m(m + 1)/2] Let me compute that step by step. First, expand the terms inside the brackets:First term: [m(m + 1)(2m + 1)/6]Second term: [m(m + 1)/2]To add these two terms, we need a common denominator. The common denominator for 6 and 2 is 6. So, convert the second term to sixths:Second term becomes [3m(m + 1)/6]So now, adding the two terms:[m(m + 1)(2m + 1) + 3m(m + 1)] / 6Factor out m(m + 1):[m(m + 1)(2m + 1 + 3)] / 6 = [m(m + 1)(2m + 4)] / 6Simplify 2m + 4 to 2(m + 2):[m(m + 1) * 2(m + 2)] / 6 = [2m(m + 1)(m + 2)] / 6Divide numerator and denominator by 2:[m(m + 1)(m + 2)] / 3Therefore, Sₘ = m(m + 1)(m + 2)/6. Wait, no. Wait, let me check again. Let's see:Original expression after factoring was [m(m + 1)(2m + 4)] / 6. Then factoring out 2 from 2m + 4 gives 2(m + 2). So:[m(m + 1)*2(m + 2)] / 6 = [2m(m + 1)(m + 2)] / 6. Then 2 divided by 6 is 1/3. So Sₘ = m(m + 1)(m + 2)/6.Yes, that's correct. So, the sum of the first m triangular numbers is the tetrahedral number, which is given by m(m + 1)(m + 2)/6. So, Sₘ = m(m + 1)(m + 2)/6. That formula seems right. Let me verify with small m:For m = 1, S₁ = a₁ = 1. Using the formula: 1*2*3/6 = 6/6 = 1. Correct.For m = 2, S₂ = a₁ + a₂ = 1 + 3 = 4. Formula: 2*3*4/6 = 24/6 = 4. Correct.For m = 3, S₃ = 1 + 3 + 6 = 10. Formula: 3*4*5/6 = 60/6 = 10. Correct. Okay, so the formula is valid.So Sₘ = m(m + 1)(m + 2)/6. Now, the problem is to find the number of Sₘ (from m=1 to m=2017) that are divisible by 2 but not by 4. That is, numbers congruent to 2 modulo 4.So, first, let's analyze the divisibility of Sₘ by 2 and 4. Let's find when Sₘ is even (divisible by 2) but not divisible by 4.First, let's note that Sₘ = m(m + 1)(m + 2)/6. Since m, m + 1, m + 2 are three consecutive integers, one of them must be divisible by 3, so the division by 6 here is valid (since in the product of three consecutive integers, there's at least one multiple of 2 and one multiple of 3, hence divisible by 6).But we need to analyze the power of 2 in Sₘ. Let's consider the factors of 2 in the numerator and denominator. The denominator is 6 = 2 * 3, so we have a factor of 2 in the denominator. The numerator is m(m + 1)(m + 2). Since m, m + 1, m + 2 are three consecutive integers, exactly one of them is divisible by 2, and exactly one is divisible by 4 (or two of them may be even if one is a multiple of 4, but since they are consecutive, only one is even, except when one is a multiple of 4 and the others are odd. Wait, actually, in three consecutive numbers, there must be one even number, but depending on m, that even number could be divisible by 2 or 4. Let's think.Wait, m, m + 1, m + 2 are consecutive. So exactly one of them is even. So that even number is either congruent to 0 mod 4 or 2 mod 4. So, depending on m mod 4, the even term among the three can be 0 mod 4 or 2 mod 4. Let's see:If m is even:Case 1: m ≡ 0 mod 4. Then m is divisible by 4. So m is even, m + 1 is odd, m + 2 is even? Wait, no. If m is even, then m is divisible by 2, but m + 2 is m + 2. If m ≡ 0 mod 4, then m + 2 ≡ 2 mod 4. So, in this case, the product m(m + 1)(m + 2) would have m divisible by 4 and m + 2 divisible by 2. So the total number of factors of 2 would be 2 (from m) + 1 (from m + 2) = 3. Then, divided by 2 (from denominator 6), gives 3 - 1 = 2 factors of 2. So Sₘ would have 2² in its prime factorization, so divisible by 4.Case 2: m ≡ 2 mod 4. Then m is even, m ≡ 2 mod 4. Then m + 2 ≡ 4 mod 4 ≡ 0 mod 4. So m is divisible by 2, m + 2 is divisible by 4. So factors of 2: m contributes 1, m + 2 contributes 2, total 3. Divided by 2, gives 3 - 1 = 2 factors of 2. So again, Sₘ is divisible by 4.If m is odd:Then m + 1 is even. If m is odd, m ≡ 1 mod 4 or 3 mod 4.Subcase 1: m ≡ 1 mod 4. Then m + 1 ≡ 2 mod 4. So m + 1 is divisible by 2 but not by 4. So the product m(m + 1)(m + 2) would have m + 1 contributing 1 factor of 2, and m and m + 2 are both odd. So total factors of 2: 1. Divided by 2 (from denominator 6), gives 1 - 1 = 0 factors of 2. Wait, but Sₘ would then be odd? Wait, but m, m + 1, m + 2: if m is odd, then m + 1 is even. So product has exactly one factor of 2. Then divide by 6: denominator is 2 * 3. So numerator has one 2 and some 3. So Sₘ would be (product) / 6 = (something with one 2) divided by 2 * 3. So factors of 2: (1 - 1) = 0. So Sₘ is odd? Wait, let's check with m = 1: S₁ = 1, which is odd. m = 3: S₃ = 10. Wait, 10 is even. Wait, something's wrong here.Wait, m=3 is odd. Let's compute S₃ using the formula: 3*4*5/6 = 60/6 = 10. So 10 is even. But according to the previous reasoning, if m is odd (3 ≡ 3 mod 4), m + 1 = 4, which is divisible by 4. Wait, m + 1 when m=3 is 4, which is divisible by 4, not 2 mod 4. So perhaps my earlier analysis was wrong.Wait, if m is odd:If m ≡ 1 mod 4: m = 4k + 1. Then m + 1 = 4k + 2 ≡ 2 mod 4. So even, divisible by 2 but not by 4. Then the product m(m + 1)(m + 2) has m + 1 divisible by 2. So total factors of 2: 1.But m + 2 = 4k + 3, which is odd.So numerator has 1 factor of 2, denominator 6 has 1 factor of 2. So Sₘ would have (1 - 1) = 0 factors of 2, meaning it's odd? But when m=1, S₁=1, which is odd. When m=3, let's see:Wait, m=3 is 3 ≡ 3 mod 4. Then m + 1 = 4, which is 0 mod 4. So m + 1 is divisible by 4, contributing two factors of 2. Then, the product m(m + 1)(m + 2) would have m + 1 contributing 2 factors of 2, and m and m + 2 are odd. So total factors of 2: 2. Then divided by 2 (from denominator 6), gives 2 - 1 = 1 factor of 2. Therefore, Sₘ is even but not divisible by 4. Which matches S₃=10, which is 2 mod 4.Wait, so maybe I need to correct my analysis. Let's redo it.Case 1: m is even.If m is even, then m or m + 2 is divisible by 4. Wait, in three consecutive numbers, if m is even, then either m or m + 2 is divisible by 4, depending on m mod 4.If m ≡ 0 mod 4: Then m is divisible by 4, m + 1 is odd, m + 2 is divisible by 2. So the product has 4 * 2 = 8, which is 2^3. Divided by 6 (which is 2 * 3), so 2^3 / 2 = 2^2, so Sₘ is divisible by 4.If m ≡ 2 mod 4: Then m is divisible by 2, m + 1 is odd, m + 2 is divisible by 4. So product has 2 * 4 = 8, again 2^3. Divided by 2, gives 2^2, so Sₘ is divisible by 4.Case 2: m is odd.Then m + 1 is even. Depending on m mod 4:If m ≡ 1 mod 4: m + 1 ≡ 2 mod 4. So m + 1 is divisible by 2 but not 4. Then product has one factor of 2. Divided by 2 (from denominator), Sₘ has 1 - 1 = 0 factors of 2, so Sₘ is odd. For example, m=1: S₁=1 (odd). m=5: S₅=35 (odd). Wait, S₅: 5*6*7/6 = 5*7 = 35. Correct.If m ≡ 3 mod 4: Then m + 1 ≡ 0 mod 4. So m + 1 is divisible by 4, contributing two factors of 2. So product has two factors of 2. Divided by 2 (denominator), gives 2 - 1 = 1 factor of 2. So Sₘ is even but not divisible by 4. For example, m=3: S₃=10 (10 ≡ 2 mod 4). m=7: S₇= 7*8*9/6 = 504/6 = 84. 84 divided by 4 is 21, no remainder. Wait, 84 is divisible by 4. Wait, this contradicts our earlier conclusion. Wait, hold on. Let me compute S₇.Wait, S₇ is 7*8*9 / 6. Let's compute that: 7*8*9 = 504. 504 divided by 6 is 84. 84 divided by 4 is 21.0, so 84 is divisible by 4. But according to the previous reasoning, when m ≡ 3 mod 4, m=7≡3 mod 4? Wait, 7 mod 4 is 3, yes. Then m + 1 = 8, which is divisible by 8 (which is 2^3). Wait, so in this case, m + 1 is 8, which is 2^3. So product m(m + 1)(m + 2) = 7*8*9. The factors of 2 here come from 8, which is 2^3. So total factors of 2 in numerator: 3. Then divide by 6 (which is 2 * 3), so 3 - 1 = 2 factors of 2. Therefore, Sₘ = 84 is divisible by 4. But that contradicts our previous conclusion.Wait, so my analysis must be wrong. Let me check again.If m is odd and m ≡ 3 mod 4, then m + 1 = 4k + 4 = 4(k + 1), which is divisible by 4. So in the numerator, m + 1 is divisible by 4, contributing two factors of 2. However, in the product m(m + 1)(m + 2), m is odd, m + 1 is divisible by 4, m + 2 is odd. So total factors of 2 in the numerator: 2. Then denominator is 6, which has one factor of 2. Therefore, Sₘ would have 2 - 1 = 1 factor of 2, meaning divisible by 2 but not by 4. But in the case of m=7, S₇=84, which is divisible by 4. So my reasoning is flawed here.Wait, why is there a discrepancy? Let's check the factors of 2 in the numerator and denominator.For m=7:Numerator: 7*8*9. 8 is 2^3. So total factors of 2: 3.Denominator: 6=2*3. So factors of 2 in denominator: 1.Therefore, factors of 2 in Sₘ: 3 - 1 = 2. So Sₘ is divisible by 2²=4. Therefore, S₇=84 is divisible by 4. But according to previous analysis, if m ≡ 3 mod 4, m + 1 is divisible by 4, so factors of 2 in numerator is 2, but here it's 3. Wait, why?Wait, 8 is 2^3, not 2^2. So if m + 1 is divisible by 8, then factors of 2 would be more. Wait, m=7, m + 1=8, which is 2^3, so numerator has 3 factors of 2. Then denominator has 1 factor of 2. So Sₘ has 2 factors of 2. So Sₘ is divisible by 4. Therefore, our previous analysis was wrong because m + 1 can sometimes be divisible by higher powers of 2. Hmm. So my initial mistake was assuming that m + 1 being divisible by 4 only contributes two factors of 2, but in reality, m + 1 could be divisible by 8, 16, etc., contributing more factors. Therefore, my approach to categorize based solely on m mod 4 is insufficient because higher powers of 2 in m + 1 can affect the total number of factors.Therefore, maybe I need a different approach. Let's instead analyze the exponent of 2 in Sₘ.Given that Sₘ = m(m + 1)(m + 2)/6. Let's write this as Sₘ = [m(m + 1)(m + 2)] / (2 * 3). We need to compute the exponent of 2 in the numerator and subtract the exponent of 2 in the denominator (which is 1) to get the exponent of 2 in Sₘ.So, let v₂(n) denote the exponent of 2 in the prime factorization of n.Then, v₂(Sₘ) = v₂(m(m + 1)(m + 2)) - v₂(2 * 3) = v₂(m) + v₂(m + 1) + v₂(m + 2) - 1.But since m, m + 1, m + 2 are three consecutive integers, exactly one of them is even. Therefore, only one of v₂(m), v₂(m + 1), v₂(m + 2) is at least 1, and the others are 0. So, suppose the even term is m + k where k ∈ {0, 1, 2}. Let me denote the even term as E.Then, v₂(Sₘ) = v₂(E) - 1.Therefore, Sₘ is divisible by 2 but not by 4 if and only if v₂(E) - 1 = 1, which implies v₂(E) = 2. So, the even term E must be divisible by 4 but not by 8. Because v₂(E) = 2 means E = 4 * odd number. Therefore, the even term E must be ≡ 4 mod 8.So, to rephrase, Sₘ is divisible by 2 but not by 4 if and only if the even number among m, m + 1, m + 2 is divisible by 4 but not by 8. Therefore, the even term must be ≡ 4 mod 8.Therefore, depending on whether m is even or odd:If m is even, then E = m or E = m + 2. Wait, but in three consecutive numbers, if m is even, then E is m or m + 2. Wait, no: if m is even, then m and m + 2 are even, but they can't both be even unless m is even and m + 2 is even, but that's only possible if m is even. Wait, no: consecutive numbers: m, m + 1, m + 2. If m is even, then m + 1 is odd, m + 2 is even. Wait, no. Wait, m even: m, m + 1 (odd), m + 2 (even). Wait, no: m is even, m + 1 is odd, m + 2 is even only if m is even? Wait, m is even: m is even, m + 1 is odd, m + 2 is even? No, m is even, m + 2 is even only if m is even. Wait, no. If m is even, then m + 2 is even. But m, m + 1, m + 2: if m is even, then m and m + 2 are two even numbers separated by an odd. So actually, m and m + 2 are both even. Wait, but that's two even numbers in three consecutive integers. Wait, but three consecutive numbers can't have two even numbers. Wait, no: for example, 2, 3, 4: two even numbers (2 and 4). Wait, yes. So, if m is even, then m and m + 2 are both even. Wait, but m and m + 2 are two apart, so if m is even, then m + 2 is even. So in three consecutive numbers m, m + 1, m + 2, if m is even, then m and m + 2 are even. Therefore, two even numbers. But wait, how is that possible? Let me check with m=2: 2, 3, 4: two evens. Similarly, m=4: 4, 5, 6: two evens. So yes, in three consecutive numbers, if the first one is even, then the third one is also even. Therefore, when m is even, two of the three terms are even. Therefore, in that case, the product m(m + 1)(m + 2) has two even numbers, which contribute multiple factors of 2.But earlier, we thought only one of them is even. Wait, this is a key mistake in the initial analysis. So, actually, when m is even, m and m + 2 are even, so two even numbers. Therefore, the total number of factors of 2 in the product is v₂(m) + v₂(m + 2). When m is odd, then only m + 1 is even, so v₂(m + 1).Therefore, correcting the previous analysis:If m is even:Then m and m + 2 are even. Let’s denote m = 2k. Then m + 2 = 2k + 2 = 2(k + 1). So, depending on k, m and m + 2 can have different factors of 2.If m is divisible by 4 (i.e., k is even), then m is divisible by 4, and m + 2 is divisible by 2 but not 4 (since m = 4k, m + 2 = 4k + 2 = 2(2k + 1)). Therefore, v₂(m) = 2, v₂(m + 2) = 1. So total v₂(numerator) = 2 + 1 = 3. Therefore, v₂(Sₘ) = 3 - 1 = 2. So Sₘ divisible by 4.If m is divisible by 2 but not by 4 (i.e., k is odd), then m = 2(2k + 1), m + 2 = 2(2k + 1 + 1) = 2(2k + 2) = 4(k + 1). So, v₂(m) = 1, v₂(m + 2) = 2. Total v₂(numerator) = 1 + 2 = 3. So again, v₂(Sₘ) = 3 - 1 = 2. So Sₘ divisible by 4.Therefore, regardless of whether m is divisible by 4 or 2 but not 4, when m is even, Sₘ is divisible by 4.If m is odd:Then only m + 1 is even. Let m = 2k + 1. Then m + 1 = 2k + 2 = 2(k + 1). So, the exponent v₂(m + 1) = v₂(k + 1) + 1.Therefore, v₂(numerator) = v₂(m + 1) = v₂(k + 1) + 1. Then v₂(Sₘ) = (v₂(k + 1) + 1) - 1 = v₂(k + 1).Therefore, Sₘ is divisible by 2 but not by 4 if and only if v₂(k + 1) = 1. That is, k + 1 is divisible by 2 but not by 4. Which means k + 1 ≡ 2 mod 4. Therefore, k ≡ 1 mod 4. Therefore, k = 4t + 1. Then, m = 2k + 1 = 2(4t + 1) + 1 = 8t + 3. Therefore, m ≡ 3 mod 8.Wait, let's verify this:If m is odd, m = 2k + 1. Then m + 1 = 2(k + 1). So, v₂(m + 1) = v₂(2(k + 1)) = 1 + v₂(k + 1). Then, v₂(Sₘ) = 1 + v₂(k + 1) - 1 = v₂(k + 1). So, Sₘ divisible by 2 but not by 4 implies v₂(k + 1) = 1. Therefore, k + 1 ≡ 2 mod 4, so k ≡ 1 mod 4. Then, m = 2k + 1 = 2(4t + 1) + 1 = 8t + 3. So m ≡ 3 mod 8.Alternatively, m ≡ 3 mod 8.Wait, let's check with m=3: k = (3 - 1)/2 = 1. k + 1 = 2, which is 2 mod 4. So v₂(k + 1) = 1. So v₂(Sₘ)=1, so Sₘ divisible by 2 but not by 4. Which matches S₃=10, which is 2 mod 4.Another example: m=11. m=11 is odd. k=(11 -1)/2=5. k + 1=6. v₂(6)=1. So v₂(S₁₁)=1. So S₁₁=11*12*13/6= 11*2*13= 286. 286 divided by 2 is 143, so 286 is 2 mod 4 (since 286=4*71 + 2). Correct.But wait, when m=7: m=7 is 7 mod 8. Let's compute S₇=7*8*9/6=7*4*3=84. 84 divided by 4 is 21, so 84 is 0 mod 4. But according to the above, m=7≡7 mod 8, which is 7 mod 8. But according to our previous conclusion, m needs to be ≡3 mod 8. So why is S₇ divisible by 4?Wait, m=7 is odd. Let's check:k=(7 -1)/2=3. k +1=4. v₂(4)=2. So v₂(S₇)=v₂(k +1)=2. Therefore, S₇ is divisible by 2^2=4. But according to our previous reasoning, when m is odd, Sₘ is divisible by 2 but not by 4 only if v₂(k +1)=1, which corresponds to k +1=2 mod4, i.e., k=1 mod4. In this case, k=3, so k +1=4, which is 0 mod4. Therefore, v₂(k +1)=2, hence Sₘ divisible by 4. So, our conclusion was that Sₘ is divisible by 2 but not by 4 only when v₂(k +1)=1, i.e., k +1≡2 mod4, leading to m≡3 mod8. Therefore, m=3,11,19,... are the ones where Sₘ is divisible by 2 but not by 4. While m=7,15,23,... would have Sₘ divisible by 4.Therefore, for m odd, Sₘ is divisible by 2 but not by 4 iff m ≡3 mod8.Similarly, when m is even, Sₘ is divisible by 4. When m is odd and m ≡3 mod8, Sₘ is 2 mod4. When m is odd and m≡7 mod8, Sₘ is 0 mod4.Therefore, the values of Sₘ that are divisible by 2 but not by4 occur when m ≡3 mod8.So now, the problem reduces to counting the number of integers m between 1 and 2017 inclusive such that m ≡3 mod8.Therefore, the answer is the number of integers m in [1,2017] where m ≡3 mod8.To compute this, we can find the smallest and largest numbers in this range congruent to 3 mod8.The first such number is 3, then 11, 19,..., up to the largest ≤2017.Let’s find the largest m ≡3 mod8 such that m ≤2017.Let’s compute 2017 divided by8. 8*252=2016, so 2016 is divisible by8. Therefore, 2017=8*252 +1. Therefore, the largest number ≤2017 congruent to3 mod8 is 2017 - ( (2017 -3) mod8 )Wait, alternatively, since 2017=8*252 +1. So the previous number is 8*252 +1 -1 =8*252=2016. So the previous number congruent to3 mod8 would be 8*251 +3=2008 +3=2011. Let's check 2011 +8=2019>2017. So yes, 2011 is the largest number ≤2017 congruent to3 mod8.Now, the sequence is 3,11,19,...,2011.This is an arithmetic progression with first term a=3, common difference d=8, last term l=2011.The number of terms n is given by n=((l -a)/d)+1= ((2011 -3)/8)+1= (2008/8)+1=251 +1=252.Therefore, there are 252 numbers between1 and2017 congruent to3 mod8.But wait, let's verify with another method.Compute the number of terms: m=8k +3 ≤2017.So, 8k +3 ≤2017 →8k ≤2014 →k ≤2014/8=251.75. Since k must be integer, k≤251. Therefore, k=0 to251, inclusive. That's252 terms.Yes, that's correct. Therefore, the number of such m is252.Hence, the number of Sₘ divisible by2 but not by4 is252.But let me double-check with a small example. For m up to, say, 11.Numbers ≡3 mod8 up to11:3,11. So two numbers. Let's compute S₃=10, which is2 mod4. S₁₁=286, which is286/2=143, so286=2*143, which is2 mod4 (since143 is odd). Correct. So two values. Indeed, according to formula, n= ((11 -3)/8)+1=(8/8)+1=1 +1=2. Correct. So the formula works.Similarly, up tom=19: numbers are3,11,19. Three numbers. n= ((19 -3)/8)+1=16/8 +1=2 +1=3. Correct. So the formula is valid.Therefore, the final answer should be252.Final Answerboxed{252}

✅:To solve the problem, we need to find the number of values among ( S_1, S_2, ldots, S_{2017} ) that are divisible by 2 but not by 4. First, we derive the formula for ( S_m ). The sequence ( a_n ) is the ( n )-th triangular number, given by ( a_n = frac{n(n+1)}{2} ). The sum ( S_m ) of the first ( m ) triangular numbers is:[S_m = sum_{k=1}^m a_k = sum_{k=1}^m frac{k(k+1)}{2} = frac{1}{2} sum_{k=1}^m (k^2 + k)]Using the formulas for the sum of squares and the sum of the first ( m ) natural numbers, we get:[S_m = frac{1}{2} left( frac{m(m+1)(2m+1)}{6} + frac{m(m+1)}{2} right) = frac{1}{2} left( frac{m(m+1)(2m+1 + 3)}{6} right) = frac{m(m+1)(m+2)}{6}]Next, we analyze the divisibility of ( S_m ) by powers of 2. We need ( S_m ) to be divisible by 2 but not by 4, which means ( S_m equiv 2 pmod{4} ). Considering the product ( m(m+1)(m+2) ), which are three consecutive integers, one of them must be even. We need the even term to be divisible by 4 but not by 8. This happens when the even term is ( 4 pmod{8} ).For ( m ) to be such that the even term is ( 4 pmod{8} ), we find that ( m ) must be congruent to 3 modulo 8. This is because if ( m equiv 3 pmod{8} ), then ( m+1 equiv 4 pmod{8} ), which is divisible by 4 but not by 8.We then count the numbers ( m ) in the range from 1 to 2017 that are congruent to 3 modulo 8. These numbers form an arithmetic sequence: 3, 11, 19, ..., 2011. The number of terms in this sequence is calculated by:[text{Number of terms} = leftlfloor frac{2017 - 3}{8} rightrfloor + 1 = leftlfloor frac{2014}{8} rightrfloor + 1 = 251 + 1 = 252]Thus, the number of values among ( S_1, S_2, ldots, S_{2017} ) that are divisible by 2 but not by 4 is:[boxed{252}]

🔑:1. Summation of series for ( a_n ): The given series for ( a_n ) is the sum of the first ( n ) natural numbers. Therefore: [ a_n = 1 + 2 + cdots + n ] The formula for the sum of the first ( n ) natural numbers is: [ a_n = frac{n(n+1)}{2} ]2. Summation of series for ( S_m ): Let ( S_m ) be the sum of the first ( m ) terms of the sequence ( a_n ). Therefore: [ S_m = a_1 + a_2 + cdots + a_m ] Substitute ( a_n ) with its formula: [ S_m = sum_{n=1}^{m} frac{n(n+1)}{2} ]3. Simplify ( S_m ) using sum of squares and sum of first ( n ) natural numbers: Recognize that: [ sum_{n=1}^{m} n^2 = frac{m(m+1)(2m+1)}{6} ] and [ sum_{n=1}^{m} n = frac{m(m+1)}{2} ] Therefore, [ S_m = frac{1}{2} sum_{n=1}^{m} n(n+1) ] This can be further expanded: [ S_m = frac{1}{2} left( sum_{n=1}^{m} n^2 + sum_{n=1}^{m} n right) ] Substitute the given summations: [ S_m = frac{1}{2} left[ frac{m(m+1)(2m+1)}{6} + frac{m(m+1)}{2} right] ] Combine and factor out common terms: [ S_m = frac{1}{2} left[ frac{m(m+1)(2m+1) + 3m(m+1)}{6} right] ] Simplifying the numerator: [ S_m = frac{1}{2} left[ frac{m(m+1)(2m+1) + 3m(m+1)}{6} right] = frac{1}{2} left[ frac{m(m+1)(2m+4)}{6} right] ] [ = frac{1}{2} left[ frac{m(m+1)(m+2)}{3} right] = frac{m(m+1)(m+2)}{6} ]4. Finding the condition for ( S_m ): Notice that ( S_m ) should be divisible by 2 but not divisible by 4. This translates to: [ S_m = frac{m(m+1)(m+2)}{6} ] For ( S_m ) to be divisible by 2 but not 4, it implies ( S_m mod 4 = 2 ): [ frac{m(m+1)(m+2)}{6} equiv 2 pmod{4} ] Considering the periodic nature of ( S_m mod 8 ): [ S_m = frac{m(m+1)(m+2)}{6} ] We analyze modulo ( 8 ) calculations and see that this occurs when ( m mod 8 = 3 ).5. Calculating number of qualifying terms up to 2017: Since ( 2017 = 8 times 252 + 1 ), we have each complete cycle of 8 numbers contributing one qualifying term: [ text{Number of qualifying terms} = 252 ] Conclusion:[boxed{252}]

🧠:Okay, so I need to find the value of ( k ) for which the quadratic polynomial ( x^2 + 2(k-9)x + (k^2 + 3k + 4) ) is a perfect square. Hmm, let's think. A perfect square trinomial usually looks like ( (x + a)^2 ), which expands to ( x^2 + 2ax + a^2 ). Comparing that to the given polynomial, maybe I can set up equations based on the coefficients?Let me write down the given polynomial again: ( x^2 + 2(k-9)x + (k^2 + 3k + 4) ). If this is a perfect square, it should be equal to ( (x + m)^2 ) for some constant ( m ). Expanding ( (x + m)^2 ) gives ( x^2 + 2mx + m^2 ). So, equating the coefficients of the corresponding terms from both expressions:1. The coefficient of ( x ) in the given polynomial is ( 2(k - 9) ), and in the perfect square, it's ( 2m ). Therefore, ( 2m = 2(k - 9) ), which simplifies to ( m = k - 9 ).2. The constant term in the given polynomial is ( k^2 + 3k + 4 ), and in the perfect square, it's ( m^2 ). So, ( m^2 = k^2 + 3k + 4 ).Since we already have ( m = k - 9 ), substitute that into the second equation:( (k - 9)^2 = k^2 + 3k + 4 )Let me expand the left side:( k^2 - 18k + 81 = k^2 + 3k + 4 )Now, subtract ( k^2 ) from both sides:( -18k + 81 = 3k + 4 )Combine like terms. Let's bring all terms to one side:( -18k - 3k + 81 - 4 = 0 )Which simplifies to:( -21k + 77 = 0 )Then, solving for ( k ):( -21k = -77 )Divide both sides by -21:( k = frac{77}{21} )Simplify the fraction. Both 77 and 21 are divisible by 7:( 77 ÷ 7 = 11 )( 21 ÷ 7 = 3 )So, ( k = frac{11}{3} )Wait, let me check if this works. Substitute ( k = 11/3 ) back into the original polynomial and see if it's a perfect square.First, compute ( 2(k - 9) ):( 2(11/3 - 9) = 2(11/3 - 27/3) = 2(-16/3) = -32/3 )Then, the polynomial becomes:( x^2 - (32/3)x + (k^2 + 3k + 4) )Calculate ( k^2 + 3k + 4 ) with ( k = 11/3 ):( (11/3)^2 + 3*(11/3) + 4 )First term: ( 121/9 )Second term: ( 33/3 = 11 )Third term: 4, which is 36/9So, converting all to ninths:121/9 + 99/9 + 36/9 = (121 + 99 + 36)/9 = 256/9So, the constant term is 256/9.Now, if the polynomial is a perfect square, it should be equal to ( (x + m)^2 ), where m is half the coefficient of x, divided by 2. Wait, the coefficient of x is -32/3, so m would be half of that, which is -16/3. Then, squaring m should give the constant term.Compute ( (-16/3)^2 = 256/9 ), which matches the constant term. So that checks out. Therefore, ( k = 11/3 ) is indeed the correct value.But just to make sure I didn't make any arithmetic mistakes, let me go through the steps again.Starting with the equation ( (k - 9)^2 = k^2 + 3k + 4 ):Left side: ( k^2 - 18k + 81 )Right side: ( k^2 + 3k + 4 )Subtracting right side from left side:( (k^2 - 18k + 81) - (k^2 + 3k + 4) = -21k + 77 = 0 )Yes, that's correct. Then solving for k:-21k + 77 = 0 → k = 77/21 = 11/3. Correct.Another way to approach this might be by using the discriminant condition. For a quadratic to be a perfect square, it must have a double root, so the discriminant should be zero.The discriminant D of ( ax^2 + bx + c ) is ( b^2 - 4ac ). Let's compute that.Given the polynomial ( x^2 + 2(k - 9)x + (k^2 + 3k + 4) ), so a = 1, b = 2(k - 9), c = k^2 + 3k + 4.Discriminant D = [2(k - 9)]^2 - 4 * 1 * (k^2 + 3k + 4)Compute this:= 4(k - 9)^2 - 4(k^2 + 3k + 4)Factor out 4:= 4[ (k - 9)^2 - (k^2 + 3k + 4) ]Expand (k - 9)^2:= 4[ k^2 - 18k + 81 - k^2 - 3k - 4 ]Simplify inside the brackets:= 4[ -21k + 77 ]Set discriminant D = 0:4(-21k + 77) = 0 → -21k + 77 = 0 → k = 77/21 = 11/3.Same result. So this confirms the answer is correct. Therefore, the value of k is 11/3.Final AnswerThe value of ( k ) is boxed{dfrac{11}{3}}.

✅:To determine the value of ( k ) for which the polynomial ( x^2 + 2(k-9)x + (k^2 + 3k + 4) ) can be represented as a perfect square, we start by noting that a perfect square trinomial has the form ( (x + m)^2 ). This expands to ( x^2 + 2mx + m^2 ). By comparing coefficients, we equate:1. The coefficient of ( x ): ( 2(k - 9) = 2m ), which simplifies to ( m = k - 9 ).2. The constant term: ( k^2 + 3k + 4 = m^2 ).Substituting ( m = k - 9 ) into the constant term equation gives:[(k - 9)^2 = k^2 + 3k + 4]Expanding and simplifying:[k^2 - 18k + 81 = k^2 + 3k + 4]Subtracting ( k^2 + 3k + 4 ) from both sides:[-21k + 77 = 0]Solving for ( k ):[k = frac{77}{21} = frac{11}{3}]To verify, we check the discriminant of the quadratic polynomial. For the polynomial to be a perfect square, the discriminant must be zero:[[2(k - 9)]^2 - 4(k^2 + 3k + 4) = 0]Simplifying:[4(k - 9)^2 - 4(k^2 + 3k + 4) = 0][(k - 9)^2 - (k^2 + 3k + 4) = 0][-21k + 77 = 0]This confirms ( k = frac{11}{3} ).Thus, the value of ( k ) is boxed{dfrac{11}{3}}.

🔑:To determine the value of ( k ) for which the polynomial ( x^{2} + 2(k-9)x + left(k^{2} + 3k + 4right) ) can be expressed as a perfect square, we start by ensuring that its discriminant is zero. A quadratic polynomial ( ax^2 + bx + c ) can be represented as a perfect square if its discriminant is zero. The discriminant ( Delta ) of a quadratic polynomial ( ax^2 + bx + c ) is given by:[Delta = b^2 - 4ac]For the given polynomial ( x^{2} + 2(k-9)x + left(k^{2} + 3k + 4right) ):- ( a = 1 )- ( b = 2(k - 9) )- ( c = k^{2} + 3k + 4 )Let's calculate the discriminant ( Delta ):[Delta = [2(k-9)]^2 - 4 cdot 1 cdot (k^{2} + 3k + 4)]Simplify the terms:[Delta = 4(k-9)^2 - 4(k^{2} + 3k + 4)]Expand ( (k-9)^2 ):[(k-9)^2 = k^2 - 18k + 81]Therefore,[4(k-9)^2 = 4(k^2 - 18k + 81) = 4k^2 - 72k + 324]So, the discriminant becomes:[Delta = 4k^2 - 72k + 324 - 4(k^2 + 3k + 4)]Distribute and combine like terms:[Delta = 4k^2 - 72k + 324 - 4k^2 - 12k - 16]Simplify further:[Delta = -72k - 12k + 324 - 16 = -84k + 308]For the polynomial to be a perfect square, the discriminant must be zero:[-84k + 308 = 0]Solve for ( k ):[-84k = -308][k = frac{308}{84} = frac{77}{21} = frac{11}{3}]Thus, the value of ( k ) is ( frac{11}{3} ).[boxed{frac{11}{3}}]

🧠:Okay, so I need to prove that the locus of the intersection points of the perpendiculars dropped from the ends of a segment of constant length a, which slides its ends along the sides of an angle, is an arc of a circle. Hmm, let me start by visualizing the problem. There's an angle, let's say with vertex O, and two sides forming that angle. A segment of length a has its endpoints moving along these two sides. At each position of the segment, we drop perpendiculars from each end onto the sides of the angle, and then look at where these two perpendiculars intersect. The claim is that all these intersection points lie on a circular arc.First, I should probably draw a diagram to get a better sense. Let me imagine an angle, maybe right angle for simplicity, but maybe it's a general angle. The segment is sliding such that one end is on one side of the angle, and the other end is on the other side. As it slides, the segment moves, changing its position but keeping its length a constant. At each position, if I drop a perpendicular from each end to the opposite side, these two perpendiculars should intersect at some point, and all such points should form a circle.Wait, actually, when they say "perpendiculars dropped from the ends of the segment onto the sides of the angle," does that mean each end is dropping a perpendicular onto the side it's not on? For example, if the segment has endpoints on the two sides of the angle, then from each endpoint, we drop a perpendicular to the other side? That seems right. So if the angle is formed by two lines, say OA and OB, with vertex O, then the segment has one end on OA and the other on OB. From the end on OA, we drop a perpendicular to OB, and from the end on OB, we drop a perpendicular to OA. The intersection of these two perpendiculars is the point we are interested in. The locus of these intersections as the segment slides is supposed to be a circular arc.Alright, so let me formalize this. Let me set up coordinates. Maybe place the vertex O at the origin. Let the angle be θ, so the two sides can be represented as the positive x-axis and a line making an angle θ with the x-axis. Let me denote the two sides as OA (x-axis) and OB (line y = tanθ x). The segment of length a has endpoints P on OA and Q on OB. Let me parameterize the positions of P and Q. Let’s say point P is at (p, 0) on OA, and point Q is at (q cosθ, q sinθ) on OB, where p and q are parameters such that the distance between P and Q is a.So, the distance between P and Q is sqrt[(q cosθ - p)^2 + (q sinθ - 0)^2] = sqrt[(q cosθ - p)^2 + (q sinθ)^2] = a. Squaring both sides: (q cosθ - p)^2 + (q^2 sin²θ) = a². Let's expand that: q² cos²θ - 2 p q cosθ + p² + q² sin²θ = a². Then, combining terms: q² (cos²θ + sin²θ) - 2 p q cosθ + p² = a². Since cos²θ + sin²θ = 1, this simplifies to q² - 2 p q cosθ + p² = a². So, we have a relationship between p and q: p² - 2 p q cosθ + q² = a².Now, the next step is to find the equations of the perpendiculars dropped from P and Q onto the opposite sides. Let me start with the perpendicular from P to OB. Since OB has the equation y = tanθ x, which is y = (sinθ/cosθ) x. The slope of OB is tanθ, so the slope of the perpendicular is -cotθ. Therefore, the equation of the perpendicular from P(p, 0) to OB is y - 0 = -cotθ (x - p). Similarly, the perpendicular from Q(q cosθ, q sinθ) to OA (the x-axis) is a vertical line if OA is the x-axis. Wait, dropping a perpendicular from Q to OA (x-axis) is just a vertical line from Q down to the x-axis, right? Because OA is horizontal. So the perpendicular from Q to OA would be the vertical line x = q cosθ, intersecting OA at (q cosθ, 0). But actually, the perpendicular from Q to OA is the vertical line because OA is the x-axis. Wait, no. The perpendicular to OA (x-axis) is vertical, so yes, from Q(q cosθ, q sinθ), the perpendicular to OA is x = q cosθ. But that just drops straight down to (q cosθ, 0). Similarly, the perpendicular from P(p, 0) to OB is the line with slope -cotθ passing through (p, 0). Let me write that equation again: y = -cotθ (x - p).So, the intersection point of these two perpendiculars is where x = q cosθ (from the perpendicular from Q) and y = -cotθ (x - p). Plugging x = q cosθ into the second equation: y = -cotθ (q cosθ - p). Let's compute that: y = -cotθ (q cosθ - p) = - (cosθ / sinθ)(q cosθ - p) = (-q cos²θ / sinθ + p cosθ / sinθ). Therefore, the coordinates of the intersection point are (q cosθ, (-q cos²θ + p cosθ)/ sinθ ). Let me write that as (X, Y), where X = q cosθ and Y = (p cosθ - q cos²θ)/ sinθ.So, X = q cosθ, which gives q = X / cosθ. Plugging this into Y: Y = (p cosθ - (X / cosθ) cos²θ)/ sinθ = (p cosθ - X cosθ)/ sinθ = cosθ (p - X)/ sinθ. But X = q cosθ, and from the original relation between p and q: p² - 2 p q cosθ + q² = a². Let me substitute q = X / cosθ into that equation. Then we have p² - 2 p (X / cosθ) cosθ + (X / cosθ)^2 = a². Simplifying: p² - 2 p X + X² / cos²θ = a². That is, (p - X)^2 + X² (1 / cos²θ - 1) = a². Because (p - X)^2 = p² - 2 p X + X², so the equation becomes (p - X)^2 + X² (1 / cos²θ - 1) = a². Since 1 / cos²θ - 1 = tan²θ, this simplifies to (p - X)^2 + X² tan²θ = a².But from earlier, we also have Y = cosθ (p - X)/ sinθ. Let's solve for p - X. From Y = [cosθ (p - X)] / sinθ, multiply both sides by sinθ: Y sinθ = cosθ (p - X). Then, p - X = (Y sinθ) / cosθ = Y tanθ. Therefore, p - X = Y tanθ. Substitute this into the equation (p - X)^2 + X² tan²θ = a²: (Y tanθ)^2 + X² tan²θ = a². Factor out tan²θ: tan²θ (Y² + X²) = a². Therefore, X² + Y² = a² / tan²θ. Which is the equation of a circle centered at the origin with radius a / tanθ. Wait, but is that correct?Wait, X and Y here are coordinates of the intersection point. So the equation X² + Y² = (a / tanθ)^2 implies that all such points lie on a circle of radius a / tanθ centered at the origin. But the origin is the vertex of the angle. However, is this the case? Let me verify with a specific example. Suppose θ is 90 degrees, so tanθ is 1, and the radius would be a. Let's see. If the angle is a right angle, then the sides are the x and y axes. The segment slides with endpoints on the x and y axes, maintaining length a. The perpendiculars from each end to the opposite axis would just be the vertical and horizontal lines from those points. Their intersection would be the same as the point itself. Wait, no. Wait, if the segment is between (p, 0) and (0, q) such that p² + q² = a². Then, dropping a perpendicular from (p,0) to the y-axis is the line x = p, and dropping a perpendicular from (0, q) to the x-axis is the line y = q. Their intersection is (p, q). But the locus of (p, q) with p² + q² = a² is indeed a circle of radius a. But in this case, θ is 90 degrees, so tanθ = 1, so a / tanθ = a. So the equation X² + Y² = a², which matches. So in this case, it's a quarter-circle in the first quadrant.But in the general case, when θ is not 90 degrees, the equation X² + Y² = (a / tanθ)^2 is a circle of radius a / tanθ. However, since the original angle is θ, the movement of the segment is confined within that angle, so the locus would be an arc of that circle, bounded by the angle's sides. Therefore, the locus is indeed an arc of a circle. Therefore, the proof is complete.Wait, let me check again. If θ is arbitrary, say θ = 60 degrees, then tanθ = √3, so the radius is a / √3. Let me see if this makes sense. If the angle is 60 degrees, and the segment slides with length a, the intersection points of the perpendiculars lie on a circle of radius a / √3. Hmm. Let me take a specific position. Suppose the segment is aligned along the bisector of the angle. Wait, but in this case, maybe when p = q? Let me see. If p = q, then from the equation p² - 2 p q cosθ + q² = a², if p = q, then p² - 2 p² cosθ + p² = a² ⇒ 2 p² (1 - cosθ) = a² ⇒ p = a / sqrt(2(1 - cosθ)). Then, the coordinates X = q cosθ = p cosθ, Y = (p cosθ - q cos²θ)/ sinθ. Since p = q, Y = (p cosθ - p cos²θ)/ sinθ = p cosθ (1 - cosθ)/ sinθ. Substitute p = a / sqrt(2(1 - cosθ)): Y = [a / sqrt(2(1 - cosθ))] * cosθ (1 - cosθ)/ sinθ. Simplify numerator: a cosθ (1 - cosθ) / sqrt(2(1 - cosθ)) = a cosθ sqrt(1 - cosθ) / sqrt(2). Then Y = [a cosθ sqrt(1 - cosθ) / sqrt(2)] / sinθ. Since sinθ = sqrt(1 - cos²θ) = sqrt((1 - cosθ)(1 + cosθ)), so sqrt(1 - cosθ) cancels out partially: Y = [a cosθ / sqrt(2)] / sqrt(1 + cosθ). Multiply numerator and denominator by sqrt(1 + cosθ): Y = [a cosθ sqrt(1 + cosθ) / sqrt(2)] / (1 + cosθ) ) = [a cosθ / sqrt(2(1 + cosθ))]. Hmm, this seems complicated. Alternatively, compute X and Y for this case: X = p cosθ = [a / sqrt(2(1 - cosθ))] cosθ, Y = [p cosθ (1 - cosθ)] / sinθ. Let me compute X² + Y²:X² = [a² / (2(1 - cosθ))] cos²θ,Y² = [a² cos²θ (1 - cosθ)^2 / (2(1 - cosθ) sin²θ)] = [a² cos²θ (1 - cosθ) / (2 sin²θ)].So X² + Y² = [a² cos²θ / (2(1 - cosθ))] + [a² cos²θ (1 - cosθ) / (2 sin²θ)].Factor out a² cos²θ / 2:= (a² cos²θ / 2)[ 1 / (1 - cosθ) + (1 - cosθ)/ sin²θ ]Note that sin²θ = 1 - cos²θ = (1 - cosθ)(1 + cosθ). So substitute:= (a² cos²θ / 2)[ 1 / (1 - cosθ) + (1 - cosθ)/[(1 - cosθ)(1 + cosθ)] ]Simplify the second term: (1 - cosθ)/[(1 - cosθ)(1 + cosθ)] = 1 / (1 + cosθ)Thus,= (a² cos²θ / 2)[ 1 / (1 - cosθ) + 1 / (1 + cosθ) ]Combine the fractions:= (a² cos²θ / 2)[ (1 + cosθ + 1 - cosθ) / ( (1 - cosθ)(1 + cosθ) ) ]= (a² cos²θ / 2)[ 2 / (1 - cos²θ) ]= (a² cos²θ / 2)( 2 / sin²θ )= (a² cos²θ) / sin²θBut according to our previous result, X² + Y² should be equal to (a / tanθ)^2 = (a cosθ / sinθ)^2 = a² cos²θ / sin²θ. So this checks out. Therefore, even in the case where p = q, the point (X, Y) lies on the circle X² + Y² = (a / tanθ)^2. Therefore, the previous derivation is correct.Therefore, the locus is indeed a circle of radius a / tanθ centered at the vertex of the angle. However, since the original segment slides with its endpoints on the sides of the angle, the intersection points can't go beyond the angle's sides. Hence, the locus is not the full circle but an arc of it. The arc's extent would depend on the angle θ. For example, in the case of a right angle, it's a quarter-circle; for an acute angle, it's a smaller arc, and for an obtuse angle, a larger arc, but still a portion of the circle.Therefore, the proof is concluded by showing that the coordinates (X, Y) of the intersection point satisfy the equation of a circle, hence the locus is a circular arc.Final AnswerThe locus is an arc of a circle, specifically boxed{an arc of a circle}.

✅:To prove that the locus of the intersection points of the perpendiculars dropped from the ends of a segment of constant length ( a ) onto the sides of an angle is an arc of a circle, we start by setting up a coordinate system with the vertex of the angle at the origin. The sides of the angle are represented as the x-axis and a line making an angle ( theta ) with the x-axis.1. Parameterize the Endpoints: - Let ( P ) be on the x-axis at ( (p, 0) ). - Let ( Q ) be on the line ( y = tantheta x ) at ( (q costheta, q sintheta) ). - The distance between ( P ) and ( Q ) is ( a ), leading to the equation: [ p^2 - 2pq costheta + q^2 = a^2 ]2. Equations of Perpendiculars: - The perpendicular from ( P ) to the line ( y = tantheta x ) has the equation ( y = -cottheta (x - p) ). - The perpendicular from ( Q ) to the x-axis is the vertical line ( x = q costheta ).3. Intersection of Perpendiculars: - Solving the equations of the perpendiculars, the intersection point ( (X, Y) ) is found to be: [ X = q costheta ] [ Y = frac{costheta (p - q costheta)}{sintheta} ]4. Relate ( p ) and ( q ): - Substitute ( q = frac{X}{costheta} ) into the distance equation: [ (p - X)^2 + X^2 tan^2theta = a^2 ] - Using ( Y = frac{costheta (p - X)}{sintheta} ), we find ( p - X = Y tantheta ).5. Derive the Circle Equation: - Substitute ( p - X = Y tantheta ) into the equation: [ (Y tantheta)^2 + X^2 tan^2theta = a^2 ] - Factor out ( tan^2theta ): [ tan^2theta (X^2 + Y^2) = a^2 ] - Simplify to get: [ X^2 + Y^2 = left( frac{a}{tantheta} right)^2 ]This equation represents a circle centered at the origin with radius ( frac{a}{tantheta} ). Since the segment slides within the angle, the locus is an arc of this circle.[boxed{an arc of a circle}]

🔑:1. Understand the Problem: - We are given a segment of length a that slides with its ends along the sides of an angle. - We need to prove that the geometric locus of the points where perpendiculars at the ends of the segment intersect is an arc of a circle.2. Setup the Diagram: - Let the angle be angle AOB with vertex O. - Let the ends of the segment be P and Q, sliding along OA and OB respectively. - Draw perpendiculars from P to OB and from Q to OA, intersecting each other at point R.3. Establish Key Relationships: - By construction, P lies on OA and Q lies on OB. - Let d_1 be the distance from P to O, and d_2 be the distance from Q to O. - Note that PQ = a.4. Use the Right Triangle Properties: - Consider the right triangles formed by the perpendiculars, triangle PRO and triangle QRO. - Because RO is perpendicular to both OA and OB, triangles PRO and QRO are right triangles.5. Sum of Squares of Distances in Right Triangles: - Using the Pythagorean Theorem in triangle PRO and triangle QRO: [ d_1^2 + h^2 = a^2 ] [ d_2^2 + h^2 = a^2 ] where h is the vertical distance from R to line OQ.6. Equal Perpendicular Distances: - Given RO = RO, the lengths derived from these two distances should be equal.7. Geometric Locus as Circle: - Since the relationship must always hold true, the point R must always be a fixed distance, a, from O. - Therefore, the geometric location of R describes an arc of a circle with radius a and center O.8. Conclusion: - The geometric locus of the points of intersection of the perpendiculars is an arc of a circle with radius a centered at the vertex of the angle.[boxed{text{Arc of a circle}}]

🧠:Okay, so I need to solve this problem: Given a real number α, if the set {sin α, cos 2α} is equal to the set {cos α, sin 2α}, then find all angles α that satisfy this condition. Alright, let me start by understanding what this means. First, the problem is saying that these two sets are equal. Since sets don't have order or duplicate elements, this means that each element in the first set must be in the second set, and vice versa. So, either sin α equals cos α and cos 2α equals sin 2α, or sin α equals sin 2α and cos 2α equals cos α. But since sets are unordered, we have to consider both possibilities. Let me write that down:Case 1: sin α = cos α and cos 2α = sin 2αCase 2: sin α = sin 2α and cos 2α = cos αBut wait, there might be another case where sin α equals cos 2α and cos α equals sin 2α. Wait, no. Wait, the set {a, b} equals {c, d} if either (a = c and b = d) or (a = d and b = c). So, actually, there are two possibilities for the equality of the sets:Either:1. sin α = cos α and cos 2α = sin 2αOR2. sin α = sin 2α and cos 2α = cos αSo, those are the two cases I need to consider. Let me tackle each case separately.Starting with Case 1: sin α = cos α and cos 2α = sin 2α.First, solve sin α = cos α. Let's recall that sin α = cos α when tan α = 1, because dividing both sides by cos α (assuming cos α ≠ 0) gives tan α = 1. The solutions for this are α = π/4 + kπ, where k is any integer.Now, we also need to check if these solutions satisfy the second equation of Case 1: cos 2α = sin 2α. Let's substitute α = π/4 + kπ into this equation.First, compute 2α: 2α = π/2 + 2kπ. Then, cos(π/2 + 2kπ) = 0, and sin(π/2 + 2kπ) = 1. But 0 ≠ 1, so this equation does not hold. Therefore, there are no solutions in Case 1. Hmm, that's interesting. So Case 1 gives us no solutions.Moving on to Case 2: sin α = sin 2α and cos 2α = cos α.First, solve sin α = sin 2α. Let's recall that sin A = sin B implies that A = B + 2πn or A = π - B + 2πn for some integer n.So, applying this to sin α = sin 2α:Either:1. α = 2α + 2πn ⇒ -α = 2πn ⇒ α = -2πnOR2. α = π - 2α + 2πn ⇒ 3α = π + 2πn ⇒ α = π/3 + 2πn/3Similarly, solving cos 2α = cos α. The general solution for cos A = cos B is A = B + 2πm or A = -B + 2πm for some integer m.So:Either:1. 2α = α + 2πm ⇒ α = 2πmOR2. 2α = -α + 2πm ⇒ 3α = 2πm ⇒ α = 2πm/3Therefore, the solutions for sin α = sin 2α are α = -2πn or α = π/3 + 2πn/3, and the solutions for cos 2α = cos α are α = 2πm or α = 2πm/3.But since in Case 2, both equations need to be satisfied simultaneously, we need to find α that satisfies both sin α = sin 2α and cos 2α = cos α.So, let me consider the intersection of the solutions from both equations.First, from sin α = sin 2α, solutions are α = -2πn and α = π/3 + 2πn/3.From cos 2α = cos α, solutions are α = 2πm and α = 2πm/3.We need to find α such that they are in both solution sets.So, let's check the first set of solutions from sin α = sin 2α:1. α = -2πn: Is this in the solution set of cos 2α = cos α?Check if α = -2πn satisfies cos 2α = cos α.Compute cos 2α: cos(-4πn) = cos(4πn) = 1.Compute cos α: cos(-2πn) = cos(2πn) = 1.So, 1 = 1, which is true. So α = -2πn is indeed a solution for cos 2α = cos α. Therefore, α = -2πn is part of the solution. But since n is any integer, α = 2πk (where k = -n) is equivalent to α = 2πk. So this solution corresponds to α = 2πk, which is the same as the first solution from the cosine equation.Similarly, check the second set of solutions from sin α = sin 2α: α = π/3 + 2πn/3.We need to check if these α satisfy cos 2α = cos α.So, let's substitute α = π/3 + 2πn/3 into cos 2α and see if it equals cos α.Compute 2α: 2(π/3 + 2πn/3) = 2π/3 + 4πn/3.Compute cos 2α: cos(2π/3 + 4πn/3) = cos(2π/3 + 4πn/3). Let's note that 4πn/3 can be written as 2π*(2n)/3. Since cosine has a period of 2π, cos(2π/3 + 4πn/3) = cos(2π/3 + 4πn/3 - 2πk) where k is chosen such that the angle is within [0, 2π). But perhaps it's better to compute for specific n.Alternatively, let's compute cos(2π/3 + 4πn/3) and compare it to cos(π/3 + 2πn/3).First, cos(2α) = cos(2π/3 + 4πn/3). Let me see if this equals cos(α) = cos(π/3 + 2πn/3).Let me take n as an integer. Let's check for different n modulo 3, since the angle repeats every 3 steps.Let's take n = 0: Then α = π/3. Then cos 2α = cos(2π/3) = -1/2, and cos α = cos(π/3) = 1/2. So -1/2 ≠ 1/2. Not equal.n = 1: α = π/3 + 2π/3 = π. Then cos 2α = cos(2π) = 1, cos α = cos π = -1. 1 ≠ -1.n = 2: α = π/3 + 4π/3 = 5π/3. Then cos 2α = cos(10π/3) = cos(10π/3 - 2π*1) = cos(4π/3) = -1/2. Cos α = cos(5π/3) = 1/2. Again, -1/2 ≠ 1/2.n = 3: α = π/3 + 6π/3 = π/3 + 2π = 7π/3. Cos 2α = cos(14π/3) = cos(14π/3 - 4π) = cos(2π/3) = -1/2, and cos α = cos(7π/3) = cos(π/3) = 1/2. Still not equal.Similarly, n = -1: α = π/3 - 2π/3 = -π/3. Cos 2α = cos(-2π/3) = cos(2π/3) = -1/2, cos α = cos(-π/3) = cos(π/3) = 1/2. Again, not equal.So it seems that for α = π/3 + 2πn/3, cos 2α never equals cos α. Therefore, these solutions do not satisfy the second equation in Case 2. Therefore, the only solutions in Case 2 are α = 2πk where k is an integer.But wait, in the first part of Case 2 solutions, sin α = sin 2α gives α = -2πn, which is equivalent to α = 2πk (since n is an integer, k = -n). So these solutions correspond to α = 2πk. Then, checking cos 2α = cos α, we already saw that α = 2πk gives cos 2α = cos 4πk = 1, and cos α = cos 2πk = 1. So both equations are satisfied. Therefore, the solutions here are α = 2πk.But we also need to check if there are any other solutions that might overlap. Wait, but we considered all possibilities for Case 2. So in Case 2, the only solutions are α = 2πk. So then, the set of all solutions is α = 2πk for integer k.But wait, let me verify with an example. Take α = 0. Then the first set is {sin 0, cos 0} = {0, 1}, and the second set is {cos 0, sin 0} = {1, 0}, so they are equal. Similarly, α = 2π: same as α = 0. So that works. What about α = π/2? Let's check if it's a solution. First set: {sin π/2, cos π} = {1, -1}, second set: {cos π/2, sin π} = {0, 0}. These are not equal. So π/2 is not a solution. So seems like only multiples of 2π work.But wait, maybe there are other solutions. Let me check α = π. Then first set: {sin π, cos 2π} = {0, 1}, second set: {cos π, sin 2π} = {-1, 0}. These sets are {0,1} and {-1,0}, which are not equal. So α = π is not a solution. Wait, but according to our previous conclusion, α = 2πk. So α = 2π, 4π, etc., but not π. So, α = 0, 2π, 4π, etc., which indeed works. So, only the even multiples of π.Wait, but wait, what about α = π/2. Let's see. If α = π/2, first set: {1, cos π} = {1, -1}; second set: {0, sin π} = {0, 0}. Not equal. So no. What about α = π/4? First set: {sin π/4, cos π/2} = {√2/2, 0}; second set: {cos π/4, sin π/2} = {√2/2, 1}. These sets are {√2/2, 0} vs {√2/2, 1}. Not equal. So that's not a solution.Wait, so according to our analysis, the only solutions are α = 2πk. But let me check another angle, say α = -π. Then first set: {sin(-π), cos(-2π)} = {0, 1}; second set: {cos(-π), sin(-2π)} = {-1, 0}. These are {0,1} and {-1, 0}. Not equal. So α = -π is not a solution. But according to our solution, α = 2πk. So negative multiples of 2π would also work, like α = -2π. Let's check α = -2π. First set: {sin(-2π), cos(-4π)} = {0,1}; second set: {cos(-2π), sin(-4π)} = {1,0}. So {0,1} and {1,0} are equal. So that works. So the solution is all integer multiples of 2π.But wait, let me check another angle. Suppose α = π/3 + 2πn/3. Wait, even though these didn't satisfy the second equation, but maybe there's an overlap. For example, if n = 3m, then α = π/3 + 2πm. But then, that's π/3 + 2πm. Let's take m = 0: α = π/3. Check first set: {sin π/3, cos 2π/3} = {√3/2, -1/2}; second set: {cos π/3, sin 2π/3} = {1/2, √3/2}. These are the same sets, since both are {√3/2, -1/2} and {1/2, √3/2} but wait, {√3/2, -1/2} vs {1/2, √3/2}. These are not equal. Because -1/2 is not equal to 1/2. So the sets are different. So even if α = π/3 + 2πm, it doesn't work. So yeah, only multiples of 2π.Wait a minute, but when we considered Case 2, we found α = 2πk. But is that the only solution? Let me check another angle. Suppose α = 0. Then, as before, the sets are {0,1} and {1,0}, which are equal. What about α = π/2. Wait, already checked, not a solution.Wait, but let's check another angle. Let's suppose α = 2π/3. Then:First set: {sin(2π/3), cos(4π/3)} = {√3/2, -1/2}Second set: {cos(2π/3), sin(4π/3)} = {-1/2, -√3/2}These sets are {√3/2, -1/2} and {-1/2, -√3/2}. Are these equal? Since sets are unordered, they are equal if they have the same elements. √3/2 ≈ 0.866, -1/2 ≈ -0.5, and -√3/2 ≈ -0.866. So the first set has elements 0.866 and -0.5, the second set has -0.5 and -0.866. These are different. Therefore, not equal. So α = 2π/3 is not a solution.Therefore, so far, only α = 2πk seems to work. But maybe there are other solutions. Let me think again.Wait, the problem says "the set of all angles α". So in radians, angles are modulo 2π. So maybe the general solution is α = 2πk, where k is integer.But maybe there are other solutions within 0 to 2π besides 0. Let's check.Suppose α = 0: works.α = π/2: doesn't work.α = π: doesn't work.α = 3π/2: doesn't work.α = 2π: same as 0.So in 0 to 2π, only α = 0 works. So the general solution is α = 2πk.But wait, let me check α = π/6. First set: {sin π/6, cos π/3} = {1/2, 1/2}. So the set is {1/2}, because both elements are the same. Second set: {cos π/6, sin π/3} = {√3/2, √3/2}. So the set is {√3/2}. These are not equal.Wait, so even if there was a case where both elements in the set are equal, but in that case, the other set must also have the same element twice. So for example, if sin α = cos 2α and cos α = sin 2α, but in that case, the sets would each have two equal elements. But that's another case. Wait, in the problem statement, the sets are {sin α, cos 2α} and {cos α, sin 2α}. So they can have duplicate elements if sin α = cos 2α or cos α = sin 2α, but as sets, duplicates are ignored. Wait, no. Wait, in sets, duplicates are not considered. So even if sin α = cos 2α, the set {sin α, cos 2α} would just be a singleton set. Similarly for the other set. So if the two sets are equal, then either both are singleton sets with the same element, or both have two distinct elements which are the same.But in our initial analysis, we considered the cases where each element in one set matches the other. So maybe we need to consider another case where sin α = cos 2α and cos α = sin 2α. Wait, but in the set notation, the order doesn't matter, so the equality of the sets can occur in two ways: either each element matches one in the other set, regardless of order. So that includes the possibilities:1. sin α = cos α and cos 2α = sin 2α2. sin α = sin 2α and cos α = cos 2αOR3. sin α = cos 2α and cos α = sin 2αWait, I think I missed this third case. Because the sets {a, b} and {c, d} are equal if a and b are the same as c and d in any order. So in addition to the two cases I considered initially (matching first elements and second elements), there's also the possibility that a matches d and b matches c.So, actually, the two cases are:Either:1. sin α = cos α and cos 2α = sin 2α (Case 1)OR2. sin α = sin 2α and cos 2α = cos α (Case 2)OR3. sin α = cos 2α and cos α = sin 2α (Case 3)Wait, but in set theory, since order doesn't matter, the three cases are:Either both elements match in some order, so:Either:- sin α and cos 2α are the same as cos α and sin 2α, which can happen in two ways:1. sin α = cos α and cos 2α = sin 2α (Case 1)2. sin α = sin 2α and cos 2α = cos α (Case 2)But actually, there's another way: if sin α = cos 2α and cos α = sin 2α, then the sets would be equal. So this is a third case. Therefore, in total, there are three cases to consider. However, in my initial analysis, I considered only two. That might be the mistake here.Therefore, I need to include this third case:Case 3: sin α = cos 2α and cos α = sin 2α.So now, I need to check this case as well.Let me proceed to analyze Case 3.Case 3: sin α = cos 2α and cos α = sin 2α.So we have two equations:1. sin α = cos 2α2. cos α = sin 2αLet me try to solve these equations.First, equation 1: sin α = cos 2α.We can use the identity cos 2α = 1 - 2 sin² α. So:sin α = 1 - 2 sin² αBring all terms to one side:2 sin² α + sin α - 1 = 0This is a quadratic in sin α. Let me solve it.Let x = sin α. Then:2x² + x - 1 = 0Solutions:x = [-1 ± √(1 + 8)] / 4 = [-1 ± 3]/4So x = (-1 + 3)/4 = 2/4 = 1/2 or x = (-1 - 3)/4 = -4/4 = -1.Therefore, sin α = 1/2 or sin α = -1.Similarly, equation 2: cos α = sin 2α.Using the identity sin 2α = 2 sin α cos α:cos α = 2 sin α cos αBring all terms to one side:cos α - 2 sin α cos α = 0Factor:cos α (1 - 2 sin α) = 0Therefore, either cos α = 0 or 1 - 2 sin α = 0 ⇒ sin α = 1/2.So, from equation 2, we have two possibilities:1. cos α = 02. sin α = 1/2Now, let's combine the solutions from equations 1 and 2.From equation 1: sin α = 1/2 or sin α = -1.From equation 2: cos α = 0 or sin α = 1/2.Therefore, the solutions for the system are the intersection of these.Case 3a: sin α = 1/2.Then from equation 2, this satisfies sin α = 1/2. So we need to check if sin α = 1/2 and cos α = sin 2α.Wait, but in equation 2, if sin α = 1/2, then cos α can be either positive or negative sqrt(1 - sin² α) = sqrt(3)/2 or -sqrt(3)/2. Then sin 2α = 2 sin α cos α = 2*(1/2)*cos α = cos α. Therefore, equation 2 becomes cos α = cos α, which is always true. Therefore, for sin α = 1/2, equation 2 is satisfied. Therefore, all solutions to sin α = 1/2 satisfy both equations 1 and 2.Similarly, sin α = -1 from equation 1. Let's check if this satisfies equation 2.If sin α = -1, then α = 3π/2 + 2πk. Then cos α = 0. So equation 2 becomes cos α = sin 2α. Compute sin 2α: sin(3π + 4πk) = sin(3π) = 0. Therefore, cos α = 0 and sin 2α = 0, so 0 = 0. Therefore, sin α = -1 also satisfies both equations.Therefore, solutions for Case 3 are:1. sin α = 1/2 ⇒ α = π/6 + 2πk or 5π/6 + 2πk, where k is an integer.2. sin α = -1 ⇒ α = 3π/2 + 2πk, where k is an integer.So these are the solutions for Case 3.But we need to check if these solutions satisfy both equations in Case 3. Let's verify.First, take α = π/6. Then:Equation 1: sin π/6 = 1/2, cos 2*(π/6) = cos π/3 = 1/2. So 1/2 = 1/2, holds.Equation 2: cos π/6 = √3/2, sin 2*(π/6) = sin π/3 = √3/2. So √3/2 = √3/2, holds.Similarly, α = 5π/6:Equation 1: sin 5π/6 = 1/2, cos 2*(5π/6) = cos(5π/3) = 1/2. So 1/2 = 1/2, holds.Equation 2: cos 5π/6 = -√3/2, sin 2*(5π/6) = sin(5π/3) = -√3/2. So -√3/2 = -√3/2, holds.Now, take α = 3π/2:Equation 1: sin 3π/2 = -1, cos 2*(3π/2) = cos 3π = -1. So -1 = -1, holds.Equation 2: cos 3π/2 = 0, sin 2*(3π/2) = sin 3π = 0. So 0 = 0, holds.Therefore, all these solutions are valid for Case 3.Therefore, combining all cases:From Case 2, we had α = 2πk.From Case 3, we have α = π/6 + 2πk, 5π/6 + 2πk, and 3π/2 + 2πk.But wait, we need to check if these solutions from Case 3 actually make the original sets equal.Original problem: {sin α, cos 2α} = {cos α, sin 2α}For Case 3, where sin α = cos 2α and cos α = sin 2α, so the sets become {sin α, sin α} and {cos α, cos α}. But since sets don't have duplicates, they are just {sin α} and {cos α}. For these sets to be equal, sin α must equal cos α. So unless sin α = cos α, the sets would not be equal. Wait, hold on. If in Case 3, sin α = cos 2α and cos α = sin 2α, but if sin α ≠ cos α, then the sets {sin α, cos 2α} = {sin α, sin α} = {sin α} and {cos α, sin 2α} = {cos α, cos α} = {cos α}. Therefore, for these sets to be equal, we must have sin α = cos α.But in Case 3, we have sin α = cos 2α and cos α = sin 2α. But if sin α = cos α, then this would mean that we have an additional condition. So actually, in Case 3, the equations sin α = cos 2α and cos α = sin 2α may lead to sin α = cos α, but we need to check if that's the case.Wait, let's check for the solutions we found in Case 3:For α = π/6 + 2πk: sin α = 1/2, cos α = √3/2. So sin α ≠ cos α. Therefore, the sets would be {1/2, 1/2} = {1/2} and {√3/2, √3/2} = {√3/2}. These are not equal. Wait, but according to our earlier verification, they do satisfy the equations, but do they satisfy the original set equality?Wait, this is a critical mistake here. In Case 3, solving sin α = cos 2α and cos α = sin 2α ensures that {sin α, cos 2α} = {sin α, sin α} = {sin α} and {cos α, sin 2α} = {cos α, cos α} = {cos α}. Therefore, for these sets to be equal, we need sin α = cos α. But in our solutions for Case 3, sin α = 1/2 and cos α = √3/2 or -√3/2 (for α = 5π/6), which are not equal. Similarly, for α = 3π/2, sin α = -1 and cos α = 0. So sin α ≠ cos α. Therefore, the sets {sin α} and {cos α} would not be equal unless sin α = cos α. Therefore, in addition to solving the equations in Case 3, we also need to impose that sin α = cos α. So, this adds another condition.Therefore, in Case 3, not only do we need sin α = cos 2α and cos α = sin 2α, but also sin α = cos α. Therefore, these equations together must hold:1. sin α = cos 2α2. cos α = sin 2α3. sin α = cos αSo this is a system of three equations. Let's solve this.From equation 3: sin α = cos α ⇒ tan α = 1 ⇒ α = π/4 + πk.So substitute α = π/4 + πk into equations 1 and 2.First, compute sin α and cos α:sin(π/4 + πk) = sin π/4 cos πk + cos π/4 sin πk = (√2/2)(cos πk) + (√2/2)(0) = (√2/2)(-1)^kSimilarly, cos α = cos(π/4 + πk) = cos π/4 cos πk - sin π/4 sin πk = (√2/2)(-1)^k - (√2/2)(0) = (√2/2)(-1)^kTherefore, sin α = cos α = (√2/2)(-1)^k.Now, check equation 1: sin α = cos 2α.Compute cos 2α:cos 2α = cos(2*(π/4 + πk)) = cos(π/2 + 2πk) = 0.But sin α = (√2/2)(-1)^k ≠ 0 unless k is such that (-1)^k = 0, which is impossible. Therefore, equation 1 is not satisfied. Therefore, there are no solutions in Case 3 where sin α = cos α. Therefore, even though we found solutions for equations 1 and 2 in Case 3, they don't satisfy the original set equality unless sin α = cos α, which in turn makes equations 1 and 2 impossible. Therefore, there are no solutions in Case 3.Wait, this is a contradiction. Earlier, we found solutions for equations 1 and 2 in Case 3, but when we check if the sets are equal, they aren't because sin α ≠ cos α. Therefore, those solutions are invalid. Therefore, there are no solutions in Case 3. Therefore, my earlier conclusion was incorrect because I didn't consider that even if equations 1 and 2 hold, the set equality requires sin α = cos α. Therefore, in reality, Case 3 doesn't provide any valid solutions.Therefore, going back, the only valid solutions are from Case 2: α = 2πk.But wait, in Case 2, when we considered α = 2πk, the sets {sin α, cos 2α} and {cos α, sin 2α} become {0, 1} and {1, 0}, which are equal. Therefore, α = 2πk are valid solutions.But what about other angles where sin α = sin 2α and cos α = cos 2α, which would lead to {sin α, cos 2α} = {sin α, cos α} and {cos α, sin 2α} = {cos α, sin α}, which would be equal sets. Wait, but in this case, if sin α = sin 2α and cos α = cos 2α, then the sets {sin α, cos 2α} = {sin α, cos α} and {cos α, sin 2α} = {cos α, sin α}, which are the same sets. Therefore, in this case, the sets are equal. But in our earlier analysis of Case 2, we found that the solutions are α = 2πk.But in addition, perhaps other angles where both sin α = sin 2α and cos α = cos 2α hold. Let me check.From Case 2, solving sin α = sin 2α and cos α = cos 2α.We had solutions α = 2πk.But are there other solutions?Solve sin α = sin 2α and cos α = cos 2α.From sin α = sin 2α, solutions are α = 2πk or α = π/3 + 2πk/3.From cos α = cos 2α, solutions are α = 2πk or α = 2πk/3.The intersection of these solutions is α = 2πk.Therefore, only multiples of 2π satisfy both equations. Therefore, indeed, only α = 2πk.Therefore, combining all cases, the only solutions are α = 2πk, where k is any integer.But wait, let me check another angle that might satisfy both equations. For example, α = 0: sin 0 = 0, sin 0 = 0; cos 0 = 1, cos 0 = 1. So yes, works.What about α = 2π/3. From earlier, sin α = √3/2, sin 2α = sin 4π/3 = -√3/2. Not equal. Similarly, cos α = -1/2, cos 2α = cos 4π/3 = -1/2. Wait, cos α = cos 2α here. So for α = 2π/3, cos α = cos 2α = -1/2. But sin α = √3/2, sin 2α = -√3/2. So sin α ≠ sin 2α. Therefore, not a solution.Therefore, the only solutions are α = 2πk.Therefore, the answer should be all real numbers α such that α is an integer multiple of 2π, i.e., α = 2πk for some integer k.But wait, let's confirm this with another example. Take α = 4π. Then:{sin 4π, cos 8π} = {0, 1}{cos 4π, sin 8π} = {1, 0}Yes, these sets are equal.Another example: α = -2π. Same as α = 0. Works.But let me check α = π/2 again. Doesn't work. So indeed, only multiples of 2π.Therefore, the solution set is α = 2πk, k ∈ ℤ.But let me verify once more by considering the original equations.Original problem: {sin α, cos 2α} = {cos α, sin 2α}For this to hold, each element in the first set must be in the second set. Therefore, sin α must be either cos α or sin 2α, and cos 2α must be the other one.So either:1. sin α = cos α and cos 2α = sin 2αOR2. sin α = sin 2α and cos 2α = cos αWe saw that Case 1 leads to no solution. Case 2 leads to α = 2πk.Alternatively, perhaps I made a mistake in not considering that the mapping could be cross-wise. Let me think.Suppose that sin α = sin 2α and cos 2α = cos α. From earlier, this gives α = 2πk.Alternatively, suppose that sin α = cos 2α and cos α = sin 2α. As discussed, this leads to solutions only when sin α = cos α, which then leads to inconsistency in equations. Hence, no solutions.Therefore, the only valid solutions are α = 2πk.But let me verify with α = π/3, which was a solution to equations in Case 3 but not to set equality. As we saw, it doesn't satisfy the set equality. Therefore, indeed, only α = 2πk works.Therefore, the final answer is α = 2πk, where k is any integer. In boxed form, that would be α = 2πk, but typically, such answers are written with a specific interval, but since the problem says "the set of all angles α", and angles in real numbers can be any real number, the solution is all real numbers congruent to 0 modulo 2π. So, written as:α = 2πk, k ∈ ℤOr in boxed form:boxed{alpha = 2kpi} quad text{for some integer } kBut since the problem might expect the answer in terms of specific solutions within [0, 2π) and then generalizing, but since they didn't specify an interval, the answer is all integer multiples of 2π.Alternatively, using the general solution notation:boxed{alpha = 2kpi} quad text{where } k in mathbb{Z}Yes, that's standard.Final AnswerThe set of all angles ( alpha ) is (boxed{2kpi}) for some integer ( k ).

✅:Given ( alpha in mathbf{R} ), if the set ( {sin alpha, cos 2alpha} = {cos alpha, sin 2alpha} ), we need to find all angles ( alpha ) that satisfy this condition. Analysis1. Case 1: ( sin alpha = cos alpha ) and ( cos 2alpha = sin 2alpha ) - Solving ( sin alpha = cos alpha ) gives ( alpha = frac{pi}{4} + kpi ). - Substituting ( alpha = frac{pi}{4} + kpi ) into ( cos 2alpha = sin 2alpha ) results in a contradiction (0 ≠ 1). Thus, no solutions here.2. Case 2: ( sin alpha = sin 2alpha ) and ( cos 2alpha = cos alpha ) - Solving ( sin alpha = sin 2alpha ): - ( alpha = 2kpi ) or ( alpha = frac{pi}{3} + frac{2kpi}{3} ). - Solving ( cos 2alpha = cos alpha ): - ( alpha = 2kpi ) or ( alpha = frac{2kpi}{3} ). - Intersection of solutions: ( alpha = 2kpi ).3. Case 3: ( sin alpha = cos 2alpha ) and ( cos alpha = sin 2alpha ) - Solving ( sin alpha = cos 2alpha ) and ( cos alpha = sin 2alpha ) leads to solutions ( alpha = frac{pi}{6} + 2kpi ), ( frac{5pi}{6} + 2kpi ), or ( frac{3pi}{2} + 2kpi ). - However, these solutions do not satisfy ( sin alpha = cos alpha ), hence invalid for set equality. ConclusionThe only valid solutions are from Case 2, where ( alpha = 2kpi ) for some integer ( k ).[boxed{2kpi} quad text{for some integer } k]

🔑:Let us analyze the given condition: [{sin alpha, cos 2 alpha}={cos alpha, sin 2 alpha}]1. Consider Possible Cases: Since the two sets must be equal, we consider the permutations: 1. sin alpha = cos alpha and cos 2alpha = sin 2alpha 2. sin alpha = sin 2alpha and cos 2alpha = cos alpha 2. Analyze Case 1: [ sin alpha = cos alpha ] This means: [ tan alpha = 1 implies alpha = frac{pi}{4} + kpi, ; k in mathbb{Z} ] Now, consider the second part: [ cos 2alpha = sin 2alpha ] This means: [ cos 2alpha = sin 2alpha ] Utilizing the double angle identities, we get: [ cos 2alpha = sin 2alpha implies cos 2alpha = cosleft(frac{pi}{2} - 2alpharight) ] Since cos theta = cos beta implies: [ 2alpha = 2npi pm left(frac{pi}{2} - 2alpharight) implies 4alpha = 2npi + frac{pi}{2} ; text{or} ; 4alpha = 2npi - left(frac{pi}{2} - 2alpha right) ] Solving for alpha: [ 4alpha = 2npi + frac{pi}{2} implies alpha = frac{npi}{2} + frac{pi}{8} ]3. Analyze Case 2: [ sin alpha = sin 2alpha quad text{and} quad cos 2alpha = cos alpha ] Starting with sin alpha = sin 2alpha, we have: [ sin alpha = 2 sin alpha cos alpha implies 2 sin alpha cos alpha - sin alpha = 0 implies sin alpha (2 cos alpha - 1) = 0 ] Therefore, sin alpha = 0 or cos alpha = frac{1}{2} If sin alpha = 0: [ alpha = kpi ] If cos alpha = frac{1}{2}: [ alpha = pm frac{pi}{3} + 2kpi ] Now, cos 2alpha = cos alpha: [ 2alpha = 2npi pm alpha implies alpha = 2kpi ]Combining all the possible values derived: The values must be periodic with alpha=2kpi, since they must satisfy both cases 1 and 2 altogether, and thus:[alpha = 2kpi, ; k in mathbb{Z}] Conclusion:[boxed{{alpha mid alpha = 2kpi, k in mathbf{Z}}}]