🧠:Alright, so I have this problem here: Given a set T of right triangles, each of which has an altitude to the hypotenuse that is numerically equal to 1. I need to prove that from each such triangle, one can cut out a circle whose area is greater than 0.5. Hmm, okay. Let me try to think through this step by step.First, let me recall some properties of right triangles. In a right triangle, the altitude to the hypotenuse relates the two segments into which the hypotenuse is divided. Also, I remember that the length of this altitude can be expressed in terms of the legs of the triangle. Wait, the problem states that the altitude to the hypotenuse is equal to 1. So, for every triangle in set T, the altitude is 1. We need to show that in each such triangle, there exists a circle with area greater than 0.5. Since the area of a circle is πr², a circle with area greater than 0.5 would have a radius r such that πr² > 0.5, so r > sqrt(0.5/π). Calculating that, sqrt(0.5/π) is approximately sqrt(0.159) ≈ 0.398. So, the radius needs to be just over 0.4. Therefore, if we can inscribe or fit a circle with radius greater than 0.4 in the triangle, then its area would be greater than 0.5.But the problem says "cut out a circle", which might mean that we can remove a circle from the triangle, not necessarily inscribed or anything. But probably the largest possible circle that can fit inside the triangle. Wait, the largest circle that can fit inside a right triangle is the incircle. The radius of the incircle of a right triangle is given by r = (a + b - c)/2, where a and b are the legs and c is the hypotenuse. Alternatively, there's a formula for the inradius of a right triangle: r = (a + b - c)/2. Alternatively, I also recall that the inradius of any triangle is equal to the area divided by the semiperimeter. For a right triangle, area is (a*b)/2, semiperimeter is (a + b + c)/2. So, r = (a*b)/(a + b + c). Let me check both formulas.Let me take a simple right triangle, say with legs 3 and 4, hypotenuse 5. Then, inradius using (a + b - c)/2 would be (3 + 4 -5)/2 = 2/2 = 1. Alternatively, area is 6, semiperimeter is (3 + 4 + 5)/2 = 6, so r = 6/6 = 1. Okay, so both formulas give the same result. So, the inradius is indeed (a + b - c)/2 or area over semiperimeter.Therefore, if we can express the inradius in terms of the given altitude to the hypotenuse, which is 1, then perhaps we can find a relationship. So, in our problem, each triangle has an altitude to the hypotenuse equal to 1. Let's denote the legs as a and b, hypotenuse c, altitude h = 1.First, let's recall that in a right triangle, the altitude to the hypotenuse is related to the legs by the formula h = (a*b)/c. Since h = 1, this gives us a*b = c. So, the product of the legs is equal to the hypotenuse. Interesting. So, c = a*b.But also, in a right triangle, by the Pythagorean theorem, a² + b² = c². So substituting c = a*b into this, we get a² + b² = (a*b)². Let me write that down:a² + b² = (a*b)^2.So, this is a key equation here. Let's see if we can manipulate this.Alternatively, perhaps we can express the inradius in terms of a and b. Since the inradius r = (a + b - c)/2. But c = sqrt(a² + b²), right? Wait, but in our case, c is also equal to a*b, from the altitude formula. Wait, so in our problem, c is equal to a*b. But in a right triangle, c must also equal sqrt(a² + b²). Therefore, we have:sqrt(a² + b²) = a*b.So, squaring both sides, we get:a² + b² = a²*b².So, that's the equation that relates a and b for these triangles. Therefore, any such triangle must satisfy a² + b² = a²*b².Now, the inradius is r = (a + b - c)/2. Since c = a*b, so:r = (a + b - a*b)/2.Therefore, the inradius is (a + b - a*b)/2. So, we need to show that this inradius is greater than sqrt(0.5/π), which is approximately 0.398. But wait, the problem states that the area of the circle must be greater than 0.5. So, π*r² > 0.5 => r > sqrt(0.5/π) ≈ 0.398. Therefore, if we can show that the inradius r > 0.398, then the area of the incircle is greater than 0.5. Alternatively, perhaps the inradius is even larger, so maybe we can show that the inradius is at least 0.5, which would give area π*(0.5)^2 ≈ 0.785, which is greater than 0.5. But maybe the inradius is not necessarily that large. Alternatively, maybe there is a larger circle that can be cut out, not just the incircle. Wait, but the incircle is the largest circle that fits inside the triangle. So, if the incircle has area greater than 0.5, then that's the circle we need. If not, maybe there is another circle? Wait, but the incircle is tangent to all three sides, so it's the largest possible circle that can fit inside. So, if the incircle's area is greater than 0.5, then we're done. If not, then we have to see if another circle can be cut out which is larger. But I think the problem is referring to the incircle. Maybe the key is to show that the inradius is greater than sqrt(0.5/π), hence the area is greater than 0.5.Alternatively, maybe not. Let's check. Let's see if the inradius is always greater than some value. Let me consider the expression for r: r = (a + b - a*b)/2. We need to find the minimum possible value of r given the constraint a² + b² = a²*b². So, we need to minimize (a + b - a*b)/2 subject to a² + b² = a²*b². If we can show that the minimum of r is greater than sqrt(0.5/π), then it would hold for all such triangles. Alternatively, if the minimum is less, then perhaps there exists a triangle where the incircle is too small, but maybe there's another circle we can cut out. Hmm.Wait, maybe the incircle is the only circle we can "cut out" in the sense of fitting entirely within the triangle. So, perhaps the problem is indeed about the inradius. Let me proceed under that assumption.So, the problem reduces to showing that for any right triangle with altitude to hypotenuse equal to 1, the inradius is greater than sqrt(0.5/π) ≈ 0.398. Alternatively, maybe even better, maybe we can show that the inradius is at least 0.5, which would make the area π*(0.5)^2 = π/4 ≈ 0.785 > 0.5. Then we would be done.So, let's see. Let's try to find the minimal possible inradius given the constraints.Given that a² + b² = a²*b² and r = (a + b - a*b)/2.First, let's try to express r in terms of a single variable. Let me set x = a and y = b. Then, we have x² + y² = x²*y² and r = (x + y - x*y)/2.But this seems a bit complicated. Maybe we can find a substitution. Let's consider that x² + y² = x²*y². Let me divide both sides by x²*y²:1/(y²) + 1/(x²) = 1.Let me set u = 1/x and v = 1/y. Then, the equation becomes u² + v² = 1. So, (u, v) lies on the unit circle. Interesting.Also, the inradius r = (x + y - x*y)/2. Let's express this in terms of u and v.Since x = 1/u and y = 1/v, so:r = (1/u + 1/v - (1/u)(1/v))/2 = ( (v + u)/uv - 1/(uv) ) / 2 = ( (u + v - 1)/uv ) / 2 = (u + v - 1)/(2uv).But since u² + v² = 1, we can parameterize u and v as u = cosθ, v = sinθ for some θ in (0, π/2). Therefore:r = (cosθ + sinθ - 1)/(2 cosθ sinθ).So, now, we need to find the minimum value of r(θ) = (cosθ + sinθ - 1)/(2 cosθ sinθ) for θ ∈ (0, π/2).Hmm, this seems manageable. Let's compute this function.First, let me denote t = θ. So,r(t) = (cos t + sin t - 1)/(2 cos t sin t).We can write this as:r(t) = [ (cos t + sin t) - 1 ] / (2 cos t sin t ).Perhaps we can simplify the numerator. Let's compute cos t + sin t. We know that cos t + sin t = sqrt(2) sin(t + π/4). So, the maximum value is sqrt(2), and the minimum value is 1 when t = 0 or π/2, but since t is in (0, π/2), cos t + sin t ranges from 1 to sqrt(2). However, here we have cos t + sin t - 1, so the numerator ranges from 0 to sqrt(2) - 1.So, the numerator is non-negative, which makes sense because inradius should be positive.Therefore, r(t) is non-negative. Our goal is to find the minimal value of r(t).To find the minimum of r(t), we can take the derivative and set it equal to zero.Let me compute dr/dt.First, let me denote:Let N = cos t + sin t - 1D = 2 cos t sin tThen, r(t) = N/D.So, dr/dt = (N’ D - N D’) / D².Compute N’:N’ = -sin t + cos tD = 2 cos t sin t = sin(2t)D’ = 2(cos^2 t - sin^2 t) = 2 cos(2t)Therefore,dr/dt = [ (-sin t + cos t)(2 cos t sin t) - (cos t + sin t - 1)(2 cos 2t) ] / (4 cos² t sin² t )Simplify numerator:First term: (-sin t + cos t)(2 cos t sin t) = 2 cos t sin t (-sin t + cos t)Second term: - (cos t + sin t - 1)(2 cos 2t )So, the entire numerator is:2 cos t sin t (-sin t + cos t) - 2 (cos t + sin t - 1) cos 2tFactor out 2:2 [ cos t sin t (-sin t + cos t) - (cos t + sin t - 1) cos 2t ]Let me compute each part:First part: cos t sin t (-sin t + cos t) = cos t sin t (cos t - sin t)Second part: - (cos t + sin t - 1) cos 2tSo, the numerator becomes:2 [ cos t sin t (cos t - sin t) - (cos t + sin t - 1) cos 2t ]This looks complicated. Maybe it's better to substitute some trigonometric identities here.First, note that cos 2t = cos² t - sin² t = 1 - 2 sin² t = 2 cos² t - 1.Alternatively, let's try to express cos t sin t (cos t - sin t):Let me write cos t sin t = (sin 2t)/2Then, cos t sin t (cos t - sin t) = (sin 2t)/2 (cos t - sin t)But I don't see an immediate identity here.Alternatively, let's expand the terms:cos t sin t (cos t - sin t) = cos² t sin t - cos t sin² tHmm, not sure.Alternatively, let's make a substitution x = t. Let's denote s = sin t, c = cos t. Then, the numerator becomes:2 [ c * s (c - s) - (c + s - 1)(c² - s²) ]Since cos 2t = c² - s².So:First term: c * s (c - s) = c² s - c s²Second term: - (c + s - 1)(c² - s²) = - (c + s -1)(c - s)(c + s )So, the entire expression inside the brackets is:c² s - c s² - (c + s -1)(c - s)(c + s )Let me expand the second term:(c + s -1)(c - s)(c + s ) = (c + s -1)(c² - s² )= (c + s)(c² - s²) - (c² - s² )= (c^3 + c^2 s - c s² - s^3 ) - c² + s²Wait, no:Wait, (c + s -1)(c² - s² ) = (c + s)(c² - s² ) - 1*(c² - s² )= c(c² - s² ) + s(c² - s² ) - c² + s²= c³ - c s² + s c² - s³ - c² + s²Combine like terms:c³ - c s² + s c² - s³ - c² + s²= c³ - c² + (-c s² + s c² ) + (- s³ + s² )= c³ - c² + c² s ( -s + 1 ) + s² (-s + 1 )Hmm, this seems messy. Maybe there's another approach. Alternatively, maybe try specific angles to see where the minimum might occur.Alternatively, note that the function r(t) is symmetric in a way if we swap a and b, which corresponds to θ and π/2 - θ. Therefore, perhaps the minimum occurs at θ = π/4. Let's check.At θ = π/4, cos θ = sin θ = √2/2 ≈ 0.707.Compute r(π/4):Numerator: cos(π/4) + sin(π/4) - 1 = √2/2 + √2/2 - 1 = √2 - 1 ≈ 0.4142Denominator: 2 cos(π/4) sin(π/4) = 2*(√2/2)*(√2/2) = 2*(0.5) = 1Thus, r(π/4) = (√2 -1)/1 ≈ 0.4142. Which is approximately 0.4142, which is just over 0.398. So, at θ=π/4, the inradius is approximately 0.4142, which is greater than sqrt(0.5/π) ≈ 0.398. So, that's promising. But is this the minimal value?Wait, let's check θ approaching 0. As θ approaches 0, cosθ approaches 1, sinθ approaches 0. Then, r(t) = (cosθ + sinθ -1)/(2 cosθ sinθ ). As θ approaches 0, numerator approaches (1 + 0 -1 ) = 0, denominator approaches 0. So, we have 0/0. Let's apply L’Hospital’s Rule.Compute limit as θ→0 of [cosθ + sinθ -1] / [2 cosθ sinθ ]Differentiate numerator: -sinθ + cosθDifferentiate denominator: 2*(-sinθ sinθ + cosθ cosθ ) = 2(cos²θ - sin²θ )At θ=0, numerator derivative: -0 +1=1Denominator derivative: 2(1 -0)=2Thus, the limit is 1/2=0.5. So, as θ approaches 0, r(t) approaches 0.5. Similarly, as θ approaches π/2, same thing. So, near θ=0 or θ=π/2, the inradius approaches 0.5. But at θ=π/4, it is approximately 0.4142. Therefore, the minimal value of r(t) occurs at θ=π/4, and it is equal to √2 -1 ≈0.4142. Therefore, the inradius is always between approximately 0.4142 and 0.5 in these triangles.Therefore, since √2 -1 ≈0.4142 is greater than sqrt(0.5/π)≈0.398, then the inradius is always greater than approximately 0.4142, which is greater than 0.398. Therefore, the area of the incircle is π*(√2 -1)^2 ≈ π*(0.1716) ≈0.539, which is greater than 0.5. Therefore, in every such triangle, the incircle has area greater than 0.5. Hence, the conclusion follows.Wait, wait, hold on. Let me compute (√2 -1)^2. (√2 -1)^2 = 2 - 2√2 +1 = 3 - 2√2 ≈3 - 2.828 ≈0.172. Then, π*0.172 ≈0.540, yes, which is greater than 0.5. Therefore, even the minimal inradius gives an area of approximately 0.54, which is greater than 0.5. Therefore, in every triangle in set T, the inradius is at least √2 -1, which leads to an incircle area greater than 0.5. Hence, the proof is complete.But let me verify this calculation once more. If r_min = √2 -1 ≈0.4142, then area is π*(0.4142)^2 ≈3.1416*0.1716≈0.539. Yes, that's correct. So, 0.539 >0.5. Therefore, even the smallest possible incircle in such triangles has area ~0.54, which is above 0.5. Therefore, the statement is proven.Therefore, the answer is that in each triangle, the inradius is at least √2 -1, which leads to an area greater than 0.5. Hence, the incircle suffices.But let me check if there could be a mistake here. For example, when I found the minimal inradius at θ=π/4, is that correct? Let's see.We saw that as θ approaches 0 or π/2, the inradius approaches 0.5. At θ=π/4, it is approximately 0.4142. So, actually, the inradius has a minimum at θ=π/4. So, the minimal possible inradius is √2 -1≈0.4142. Therefore, in any triangle of set T, the inradius is at least 0.4142, which gives an area over 0.5. Therefore, the incircle's area is always greater than 0.5. Therefore, the proof is done.Alternatively, maybe I need to formalize this more. Let me structure the proof step by step.1. For any right triangle with altitude to hypotenuse h=1, we have h = (a*b)/c =1, so c =a*b.2. By Pythagoras, a² + b² =c² = (a*b)^2.3. The inradius r = (a + b -c)/2. Substituting c =a*b, we get r=(a + b -a*b)/2.4. We need to show that r > sqrt(0.5/π) ≈0.398.5. By substituting variables u=1/a, v=1/b, we derive that u² +v²=1, leading to parameterization u=cosθ, v=sinθ.6. Expressing r in terms of θ gives r(θ)=(cosθ + sinθ -1)/(2 cosθ sinθ).7. Analyzing this function, we find that the minimum occurs at θ=π/4, giving r=√2 -1≈0.4142.8. Calculating the area with this minimal radius gives π*(√2 -1)^2≈0.539>0.5.Therefore, in every such triangle, the incircle has area exceeding 0.5, hence the required circle exists.I think that covers it. So, the key steps were realizing the relationship between the altitude and the legs/hypotenuse, expressing the inradius in terms of the legs, converting the problem into a trigonometric optimization problem, finding the minimum inradius, and verifying that even the minimal inradius gives a sufficiently large area. Therefore, the conclusion holds.

✅:To prove that from each right triangle ( T ) with an altitude to the hypotenuse numerically equal to 1, one can cut out a circle with area greater than 0.5, follow these steps:1. Key Relationships: - Let the legs be ( a ) and ( b ), hypotenuse ( c ), and altitude ( h = 1 ). - In a right triangle, the altitude to the hypotenuse is ( h = frac{ab}{c} ). Given ( h = 1 ), this implies ( ab = c ). - By the Pythagorean theorem: ( a^2 + b^2 = c^2 ). Substituting ( c = ab ), we get ( a^2 + b^2 = (ab)^2 ).2. Inradius Formula: - The inradius ( r ) of a right triangle is ( r = frac{a + b - c}{2} ). - Substituting ( c = ab ), the inradius becomes ( r = frac{a + b - ab}{2} ).3. Optimization via Parameterization: - Introduce ( u = frac{1}{a} ) and ( v = frac{1}{b} ). The constraint ( a^2 + b^2 = (ab)^2 ) transforms to ( u^2 + v^2 = 1 ), representing the unit circle. - Parameterize ( u = costheta ), ( v = sintheta ), leading to ( a = frac{1}{costheta} ), ( b = frac{1}{sintheta} ).4. Expressing Inradius in Terms of ( theta ): - Substitute ( a ) and ( b ) into the inradius formula: [ r(theta) = frac{frac{1}{costheta} + frac{1}{sintheta} - frac{1}{costheta sintheta}}{2} = frac{costheta + sintheta - 1}{2costheta sintheta}. ]5. Finding the Minimum Inradius: - Analyze ( r(theta) ) over ( theta in (0, frac{pi}{2}) ). At ( theta = frac{pi}{4} ): [ costheta = sintheta = frac{sqrt{2}}{2}, quad rleft(frac{pi}{4}right) = frac{sqrt{2} - 1}{1} approx 0.4142. ] - As ( theta ) approaches 0 or ( frac{pi}{2} ), ( r(theta) ) approaches 0.5. The minimum occurs at ( theta = frac{pi}{4} ).6. Area Calculation: - The minimal inradius ( r_{text{min}} = sqrt{2} - 1 approx 0.4142 ). - The corresponding area is ( pi r_{text{min}}^2 approx pi(0.4142)^2 approx 0.539 ), which exceeds 0.5.Conclusion: Every right triangle in set ( T ) has an incircle with radius greater than ( sqrt{0.5/pi} approx 0.398 ), ensuring its area exceeds 0.5. Thus, such a circle can always be cut out.Final Answerboxed{text{Such a circle with area greater than } 0.5 text{ always exists.}}

🔑:1. Let T be the given set of right triangles where the altitude dropped to the hypotenuse is precisely 1. We need to show that for any such triangle, it is possible to inscribe a circle whose area is greater than 0.5.2. Let us consider a right triangle ABC with legs AD=|CD|=1, angle hat{A}=alpha, and |ON| = r where ON is the radius of the inscribed circle.3. First, we need to express various sides of triangle ABC in terms of alpha: - Using trigonometric identities, we have: [ |AD| = cot alpha, quad |DB| = tan alpha, quad |AC| = frac{1}{sin alpha}, quad |BC| = frac{1}{cos alpha} ]4. Therefore, the hypotenuse length |AB| is given by: [ |AB| = tan alpha + cot alpha = frac{1}{sin alpha cos alpha} ]5. The perimeter of the triangle can be found by summing up the lengths of the sides: [ |AB| = |AP| + |PB| = |AN| + |BM| = left( |AC| - r right) + left( |BC| - r right) = frac{1}{sin alpha} + frac{1}{cos alpha} - 2r ]6. Equate the two expressions for |AB|: [ frac{1}{sin alpha cos alpha} = frac{1}{sin alpha} + frac{1}{cos alpha} - 2r ]7. To solve for r, rearrange the equation: [ 2r = frac{1}{sin alpha} + frac{1}{cos alpha} - frac{1}{sin alpha cos alpha} ]8. Simplify the equation: [ r = frac{1}{2} left( frac{1}{sin alpha} + frac{1}{cos alpha} - frac{1}{sin alpha cos alpha} right) = frac{1}{2} left( frac{sin alpha - (1 - cos alpha)}{sin alpha cos alpha} right) ]9. Further simplification: [ r = frac{1}{2} left( frac{2sin frac{alpha}{2} cos frac{alpha}{2} - 2sin^2 frac{alpha}{2}}{2sin frac{alpha}{2} cos frac{alpha}{2} left(cos^2 frac{alpha}{2} - sin^2 frac{alpha}{2} right)} right) ]10. Finally, we simplify to: [ r = frac{1}{1 + sin alpha + cos alpha} = frac{1}{1 + sqrt{2} cos left(alpha - frac{pi}{4}right)} ]11. Note that the radius and hence the area of the inscribed circle will be minimized when alpha = frac{pi}{4}, because at this value cosleft(alpha - frac{pi}{4}right) achieves its maximum value.12. Evaluate r and the area S of the inscribed circle: [ cos left(alpha - frac{pi}{4}right) = 1 Rightarrow r = frac{1}{1 + sqrt{2}} = sqrt{2} - 1 ] [ S = pi (sqrt{2} - 1)^2 ]13. Calculate the area: [ S = pi (2 - 2sqrt{2} + 1) = pi (3 - 2sqrt{2}) ]14. We find that this area is indeed greater than (0.5): Conclusion:[boxed{S > 0.5}]

🧠:Okay, so I need to show that these three propositions are equivalent. That means I have to prove that each one implies the others. Let me start by recalling what each statement means in the context of group theory.First, proposition 1: x * H = y * H. This is saying that the left cosets of H in G by x and y are equal. Proposition 2: (x * H) ∩ (y * H) ≠ ∅. So the intersection of these two cosets is not empty, meaning they share at least one element. Proposition 3: x⁻¹ * y is an element of H. That seems like a condition on the relationship between x and y relative to the subgroup H.Since we need to show equivalence between all three, I can approach this by showing a cyclic chain of implications: 1 implies 2, 2 implies 3, and 3 implies 1. Alternatively, maybe it's easier to show 1 ⇨ 2, 1 ⇨ 3, then 2 ⇨ 3 and 3 ⇨1? Wait, but since equivalence requires each implies the others, perhaps it's better to do 1 ⇨ 2, 2 ⇨ 3, 3 ⇨1. Let me see.Starting with 1 ⇒ 2: If x * H = y * H, then their intersection is certainly not empty because they are the same set. Since H is a subgroup, it contains the identity element, so x * e = x is in x * H, and y is in y * H. But if x * H equals y * H, then x is in y * H and y is in x * H, so their intersection is the entire set, which is non-empty. So 1 ⇒ 2 is straightforward.Next, 2 ⇒ 3: If (x * H) ∩ (y * H) ≠ ∅, then there exists some element g in G such that g is in both x * H and y * H. So, there exist h1, h2 in H such that g = x * h1 and g = y * h2. Therefore, x * h1 = y * h2. Let's solve for x⁻¹ * y. If I multiply both sides on the left by x⁻¹, we get h1 = x⁻¹ * y * h2. Then, multiplying both sides on the right by h2⁻¹ (which exists because H is a subgroup, so h2 has an inverse in H), we get h1 * h2⁻¹ = x⁻¹ * y. Since H is a subgroup, h1 * h2⁻¹ is in H, so x⁻¹ * y is in H. Therefore, 2 ⇒ 3 holds.Now, 3 ⇒ 1: If x⁻¹ * y is in H, then we need to show that x * H = y * H. Let's start by showing x * H ⊆ y * H. Take any element x * h in x * H. Since x⁻¹ * y is in H, let's denote h0 = x⁻¹ * y. Then, y = x * h0. So, x * h = x * h = y * h0⁻¹ * h. Since h0 is in H, h0⁻¹ is in H, and H is closed under multiplication, h0⁻¹ * h is in H. Let's call that h'. Therefore, x * h = y * h', which is in y * H. Thus, x * H ⊆ y * H.Similarly, to show y * H ⊆ x * H, note that since x⁻¹ * y is in H, then y⁻¹ * x = (x⁻¹ * y)⁻¹ is also in H. So, using the same logic, any y * h can be written as x * (y⁻¹ * x)⁻¹ * h, which is in x * H. Therefore, y * H ⊆ x * H, so x * H = y * H. Hence, 3 ⇒ 1.Therefore, since we have 1 ⇒ 2, 2 ⇒ 3, and 3 ⇒ 1, all three propositions are equivalent.Wait, but let me double-check each step to make sure I didn't skip anything.For 1 ⇒ 2: If two cosets are equal, their intersection is obviously non-empty, which is correct. For 2 ⇒ 3: The existence of a common element leads to x⁻¹ * y being in H. The steps seem right: equate the two expressions for g, manipulate to solve for x⁻¹ * y, and since H is a subgroup, the product of elements and inverses remain in H. So that's good.For 3 ⇒ 1: Starting with x⁻¹ * y ∈ H, then y = x * h0 for some h0 in H. Therefore, any element of y * H is x * h0 * h, which is in x * H, hence y * H ⊆ x * H. Similarly, since y⁻¹ * x = h0⁻¹ is in H, then x = y * h0⁻¹, so x * H is y * h0⁻¹ * H = y * H, so x * H ⊆ y * H. Therefore, equality holds. That seems correct.Is there any case where these implications might not hold? Let's consider examples.Take G as the symmetric group S3, H as a subgroup, say {e, (12)}, and x and y as elements of S3. Suppose x = e and y = (12). Then x⁻¹ * y = y = (12) which is in H. Then x * H = H, y * H = H, so 1 holds. Their intersection is H, which is non-empty. If x and y are such that x⁻¹ * y is not in H, say x = e, y = (13). Then x⁻¹ * y = (13), which is not in H. Then x * H = {e, (12)}, y * H = {(13), (132)}. Their intersection is empty, which aligns with 2 being false. If 2 holds, then 3 must hold, which in the case where they do intersect, like x = (12), y = e, then (12)*H = H, and e*H = H, so their intersection is H. Then x⁻¹ * y = (12)⁻¹ * e = (12) * e = (12) ∈ H. So that checks out.Another example: Let G be integers mod 6 under addition, H = {0, 2, 4}. Let x = 1, y = 3. Then x + H = {1, 3, 5}, y + H = {3, 5, 1}. So x + H = y + H. Their intersection is the entire set, non-empty. x⁻¹ + y = (-1) + 3 = 2, which is in H. So all three conditions hold. If x =1, y=2. Then x + H = {1,3,5}, y + H = {2,4,0}. Intersection is empty. x⁻¹ + y = (-1) + 2 =1, which is not in H. So 3 fails, which aligns with 1 and 2 failing.Thus, the implications hold in these examples.Therefore, the logic seems sound. Each condition implies the others through the chain of implications, so the three propositions are equivalent.

✅:To demonstrate the equivalence of the three propositions concerning a finite group ( G ), a subgroup ( H subseteq G ), and elements ( x, y in G ), we establish a cyclic chain of implications:1. ( x * H = y * H ) 2. ( (x * H) cap (y * H) neq emptyset ) 3. ( x^{-1} * y in H ) Proof:1. ( 1 implies 2 ): If ( x * H = y * H ), the cosets are identical. Since ( H ) contains the identity ( e ), ( x = x * e in x * H ) and ( y = y * e in y * H ). Thus, their intersection is non-empty (as they are the same set).2. ( 2 implies 3 ): Suppose ( g in (x * H) cap (y * H) ). Then ( g = x * h_1 = y * h_2 ) for some ( h_1, h_2 in H ). Rearranging: [ x^{-1} * y = h_1 * h_2^{-1}. ] Since ( H ) is a subgroup, ( h_1 * h_2^{-1} in H ), so ( x^{-1} * y in H ).3. ( 3 implies 1 ): If ( x^{-1} * y = h_0 in H ), then ( y = x * h_0 ). For any ( h in H ): [ y * h = x * h_0 * h in x * H implies y * H subseteq x * H. ] Similarly, ( y^{-1} * x = h_0^{-1} in H ), so ( x = y * h_0^{-1} ), and for any ( h in H ): [ x * h = y * h_0^{-1} * h in y * H implies x * H subseteq y * H. ] Hence, ( x * H = y * H ). Conclusion: The cyclic implications ( 1 implies 2 implies 3 implies 1 ) establish the equivalence of all three propositions.Final Answer boxed{x * H = y * H text{ if and only if } (x * H) cap (y * H) neq emptyset text{ if and only if } x^{-1} * y in H}

🔑:1. Montrer que x * H = y * H implies (x * H) cap (y * H) neq emptyset:Si x * H = y * H, alors par définition de sous-groupe, cela signifie que chaque élément de x * H appartient également à y * H et vice-versa.En prenant l'élément neutre e_G in H, nous avons :[ x * e_G in x * H ]Ainsi, comme x * H = y * H, on obtient :[ x in y * H ]Il en résulte que :[ x in (x * H) cap (y * H) ]Cela prouve que :[ (x * H) cap (y * H) neq emptyset ]2. Montrer que (x * H) cap (y * H) neq emptyset implies x^{-1} * y in H:Supposons qu'il existe un élément z in G tel que :[ z in (x * H) cap (y * H) ]Alors par définition, z appartient à la fois à x * H et y * H, ce qui signifie :[ exists h_1, h_2 in H text{ tels que } z = x * h_1 = y * h_2 ]On a donc :[ x * h_1 = y * h_2 ]On peut isoler x en multipliant à gauche par x^{-1} :[ h_1 = x^{-1} * y * h_2 ]Étant donné que H est un sous-groupe, il est stable par l'opération de groupe, alors :[ x^{-1} * y = h_1 * h_2^{-1} ]Puisque h_1, h_2 in H et que H est un sous-groupe (donc fermé par l'opération), il existe :[ h_1 * h_2^{-1} in H ]Ce qui montre :[ x^{-1} * y in H ]3. Montrer que x^{-1} * y in H implies x * H = y * H:Si x^{-1} * y in H, posons x^{-1} * y = h avec h in H. Par définition, nous avons :[ y = x * h ]Maintenant, nous prouvons que y * H subseteq x * H:Prenons un élément arbitraire de y * H, soit y * h' in y * H avec h' in H. Alors :[ y * h' = (x * h) * h' ]Comme H est un sous-groupe, il est fermé sous l'opération de groupe, ainsi h * h' in H. Donc :[ y * h' in x * H ]De cette manière, nous avons montré :[ y * H subseteq x * H ]De la même façon, pour montrer x * H subseteq y * H :Soit un élément arbitraire de x * H, x * h'' in x * H pour h'' in H. Sachant que y = x * h, nous avons aussi :[ x = y * h^{-1} ]Donc :[ x * h'' = (y * h^{-1}) * h'' ]Étant donné que h^{-1} * h'' in H (car H est un sous-groupe), :[ x * h'' in y * H ]Ainsi :[ x * H subseteq y * H ]La double inclusion montre que :[ x * H = y * H ] Conclusion:[ boxed{x * H = y * H iff (x * H) cap (y * H) neq emptyset iff x^{-1} * y in H} ]

🧠:Okay, let's see. The problem is to write the equation of a plane that passes through point A and is perpendicular to the vector BC. The points given are A(-3, 1, 0), B(6, 3, 3), and C(9, 4, -2). Hmm, so first, I need to recall how to find the equation of a plane when given a point and a normal vector. Right, the general equation of a plane is ax + by + cz + d = 0, where (a, b, c) is the normal vector to the plane. Since the plane is perpendicular to vector BC, that vector must be the normal vector of the plane. So my first step should be to find vector BC. Vector BC is calculated by subtracting the coordinates of point B from point C. Let me do that. So, point C is (9, 4, -2) and point B is (6, 3, 3). Therefore, vector BC is (9 - 6, 4 - 3, -2 - 3) which simplifies to (3, 1, -5). So the normal vector to the plane is (3, 1, -5). Now, the plane passes through point A(-3, 1, 0). To write the equation, I can use the point-normal form of the plane equation. The formula is: a(x - x₀) + b(y - y₀) + c(z - z₀) = 0, where (a, b, c) is the normal vector and (x₀, y₀, z₀) is the point the plane passes through. Plugging in the values, we get: 3(x - (-3)) + 1(y - 1) + (-5)(z - 0) = 0. Let me simplify this step by step. First, simplify the terms inside the parentheses:- For the x-component: (x - (-3)) becomes (x + 3)- For the y-component: (y - 1) stays the same- For the z-component: (z - 0) simplifies to zSo the equation becomes: 3(x + 3) + 1(y - 1) - 5z = 0. Now, distribute the coefficients:- 3 times (x + 3) is 3x + 9- 1 times (y - 1) is y - 1- -5 times z is -5zPutting it all together: 3x + 9 + y - 1 - 5z = 0. Combine like terms. The constants 9 and -1 add up to 8. So:3x + y - 5z + 8 = 0. Wait, is that right? Let me check the arithmetic again. 3(x + 3) is 3x + 9. Then 1*(y - 1) is y - 1. Then -5z. So adding those: 3x + 9 + y - 1 -5z. Yes, 9 -1 is 8. So 3x + y -5z +8 = 0. Alternatively, we can write it as 3x + y -5z = -8. But usually, the equation is written with the constant term on the right side. But both forms are correct as long as they are equivalent. Let me verify this equation with the given point A. Plugging A(-3, 1, 0) into the equation: 3*(-3) + 1 -5*0 = -9 +1 +0 = -8. Yes, that's equal to the right side, which is -8. So the equation satisfies the point A. Now, just to ensure that the normal vector is indeed BC. We calculated BC as (3,1,-5), which we used as the normal vector. That's correct because the plane is supposed to be perpendicular to BC, so the normal vector is parallel to BC. Alternatively, another way to check is to take another point on the plane and see if it satisfies the equation, but we don't have another point here. However, since we correctly used the point-normal form and verified that point A satisfies the equation, it should be correct. Wait, but just to make sure, perhaps I should check if vector BC is indeed the normal vector. Let me recall that the normal vector is perpendicular to every vector lying on the plane. So, if I take any vector from point A to another arbitrary point on the plane, their dot product with BC should be zero. But since we don't have another point, maybe another approach is not necessary here. Alternatively, maybe I can check if the normal vector is correct. Vector BC is (3,1,-5), which is correct because from B(6,3,3) to C(9,4,-2), the change in x is +3, change in y is +1, and change in z is -5. So yes, that's correct. Therefore, the equation 3x + y -5z +8 =0 is the correct equation of the plane. Alternatively, sometimes the equation is written as 3x + y -5z = -8. Both forms are equivalent. Depending on the desired format, either is acceptable. The problem doesn't specify, so either form should be okay, but perhaps it's preferable to write it in the standard form ax + by + cz + d =0, so 3x + y -5z +8 =0. Alternatively, multiplying both sides by -1 would give -3x -y +5z -8 =0, but that's not necessary unless they want the leading coefficient to be positive. But since the normal vector is (3,1,-5), which already has the first component positive, the equation as we have is standard. Therefore, the final answer should be 3x + y -5z +8 =0. Let me just write that in a box.

✅:To find the equation of the plane passing through point ( A(-3, 1, 0) ) and perpendicular to the vector ( overrightarrow{BC} ):1. Determine vector ( overrightarrow{BC} ): [ overrightarrow{BC} = C - B = (9 - 6, 4 - 3, -2 - 3) = (3, 1, -5) ] This vector serves as the normal vector to the plane.2. Use the point-normal form of the plane equation: [ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 ] Substituting ( (a, b, c) = (3, 1, -5) ) and point ( A(-3, 1, 0) ): [ 3(x + 3) + 1(y - 1) - 5(z - 0) = 0 ]3. Expand and simplify: [ 3x + 9 + y - 1 - 5z = 0 implies 3x + y - 5z + 8 = 0 ]4. Verify by substituting point ( A ): [ 3(-3) + 1 - 5(0) = -9 + 1 = -8 quad (text{matches the constant term}) ]Final Answer:[boxed{3x + y - 5z + 8 = 0}]

🔑:1. Begin by determining the vector overrightarrow{BC}: [ overrightarrow{BC} = { C_x - B_x, C_y - B_y, C_z - B_z } ] Given the coordinates of points B(6, 3, 3) and C(9, 4, -2), we compute: [ overrightarrow{BC} = { 9 - 6, 4 - 3, -2 - 3 } = { 3, 1, -5 } ]2. Since the vector overrightarrow{BC} is perpendicular to the plane we seek, it serves as the normal vector to the plane.3. The general equation of a plane that passes through a point A(x_0, y_0, z_0) and has a normal vector vec{N} = {A, B, C} is given by: [ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 ] Here, A(-3, 1, 0) is the point, and vec{N} = {3, 1, -5} from the previous step.4. Substituting the coordinates of point A and the components of vec{N} into the plane equation: [ 3(x - (-3)) + 1(y - 1) - 5(z - 0) = 0 ] Simplify inside the parentheses: [ 3(x + 3) + 1(y - 1) - 5(z) = 0 ] Expand and combine like terms: [ 3x + 9 + y - 1 - 5z = 0 ] [ 3x + y - 5z + 8 = 0 ] Conclusion:The equation of the plane that passes through point A and is perpendicular to vector overrightarrow{BC} is:[boxed{3x + y - 5z + 8 = 0}]

🧠:Okay, so I need to solve these two parts about the function f_n(x). Let me start with part (1). The problem states that f_{n+1}(x) equals y times f_n(x) minus f_{n-1}(x), where y is defined as x + 1/x. Hmm, this seems like a recurrence relation. Maybe I can derive it by manipulating the given expression for f_n(x).First, let me recall the definition of f_n(x). It's (x^{n+1} - x^{-n-1}) divided by (x - x^{-1}). So, f_{n}(x) = (x^{n+1} - x^{-n-1}) / (x - x^{-1}). Similarly, f_{n+1}(x) would be (x^{n+2} - x^{-n-2}) / (x - x^{-1}), and f_{n-1}(x) is (x^{n} - x^{-n}) / (x - x^{-1}).The claim is that f_{n+1}(x) = y f_n(x) - f_{n-1}(x). Let's substitute y with x + 1/x. Then, y f_n(x) would be (x + 1/x) * [(x^{n+1} - x^{-n-1}) / (x - x^{-1})]. So, multiplying that out, let's compute the numerator:(x + 1/x)(x^{n+1} - x^{-n-1}) = x*(x^{n+1} - x^{-n-1}) + (1/x)*(x^{n+1} - x^{-n-1})Calculating each term:First term: x*x^{n+1} = x^{n+2}, and x*(-x^{-n-1}) = -x^{-n}Second term: (1/x)*x^{n+1} = x^{n}, and (1/x)*(-x^{-n-1}) = -x^{-n-2}So altogether, the numerator becomes x^{n+2} - x^{-n} + x^{n} - x^{-n-2}Therefore, y f_n(x) = [x^{n+2} - x^{-n} + x^{n} - x^{-n-2}] / (x - x^{-1})Now, subtract f_{n-1}(x) from this. f_{n-1}(x) is (x^n - x^{-n}) / (x - x^{-1}). So:y f_n(x) - f_{n-1}(x) = [x^{n+2} - x^{-n} + x^{n} - x^{-n-2} - x^n + x^{-n}] / (x - x^{-1})Simplify numerator:x^{n+2} - x^{-n} + x^{n} - x^{-n-2} - x^n + x^{-n} = x^{n+2} - x^{-n-2}Therefore, y f_n(x) - f_{n-1}(x) = (x^{n+2} - x^{-n-2}) / (x - x^{-1}) = f_{n+1}(x), which is exactly what we needed to prove. So part (1) is done. That went smoothly.Now moving on to part (2), which requires using mathematical induction to prove that f_n(x) has a certain expression in terms of y, with coefficients involving combinations. The expression alternates signs and the coefficients are binomial coefficients. For even n, it ends with (-1)^{n/2}, and for odd n, it ends with a term involving C_{(n+1)/2}^{(n-1)/2} times y.First, let me parse the given expression. For even n:f_n(x) = y^n - C(n-1,1) y^{n-2} + ... + (-1)^i C(n-2i, i) y^{n-2i} + ... + (-1)^{n/2}Wait, but the combination notation here is a bit unclear. The problem writes C_{n-1}^{1} y^{n-2} etc. In standard notation, C(n, k) is "n choose k". Here, it's written as C_{n-1}^{1}, which would be equivalent to C(n-1, 1), right? Similarly, later terms have C_{n-2i}^{n}... Wait, no, in the problem statement, it's written as C_{n-2i}^{n}. Wait, that seems inverted. Wait, in the problem statement, it's:For even n:y^{n} - C_{n-1}^{1} y^{n-2} + ... + (-1)^i C_{n-2i}^{n} y^{n-2i} + ... + (-1)^{n/2}Wait, C_{n-2i}^{n} would be equivalent to C(n, n-2i) which is the same as C(n, 2i). But that notation is a bit confusing. Let me check the original problem again.The user wrote:For even n:y^{n} - C_{n-1}^{1} y^{n-2} + ... + (-1)^i C_{n-2i}^{n} y^{n-2i} + ... + (-1)^{n/2}Wait, maybe that's a typo? Because C_{n - 2i}^{n} would be "n choose n - 2i", which is the same as C(n, 2i). But in the case of the first term, when i=1, it's C_{n - 2*1}^{n} = C(n - 2, n), but n - 2 is less than n, so that would be zero? Wait, no. Wait, perhaps the problem has a typo, or maybe the notation is different.Wait, hold on. Let me check the problem statement again.The user wrote:For even n:y^{n} - C_{n-1}^{1} y^{n-2} + ... + (-1)^i C_{n - 2i}^{n} y^{n - 2i} + ... + (-1)^{n/2}But for i=1, term is - C_{n -1 }^{1} y^{n-2}. Then for general i, it's (-1)^i C_{n - 2i}^{n} y^{n - 2i}Wait, but when n is even, and i goes up to n/2, the last term is when i = n/2, so n - 2i = n - n = 0, so C_0^n, which is 0 choose n, which is zero unless n=0. But n is a natural number, so this seems problematic.Wait, maybe the notation is flipped? Maybe it's C_{n}^{n - 2i}? Because C_{n}^{n - 2i} is equal to C(n, n - 2i) = C(n, 2i). Alternatively, if the upper index is the top, and the lower is the bottom, then C_{n - 2i}^{n} would be C(n, n - 2i). Hmm, maybe the notation here is C^{k}_{m} = C(m, k). So C_{n - 2i}^{n} would mean C(n, n - 2i). But that's equivalent to C(n, 2i). So perhaps in standard terms, the coefficients are C(n, 2i). Let's check for small n.Wait, maybe testing for small n can help. Let's take n=2 (even). Then according to the formula:f_2(x) = y^2 - C_{1}^{1} y^{0} = y^2 - 1*1 = y^2 -1. Let's compute f_2(x) directly.Given f_n(x) = (x^{n+1} - x^{-n-1})/(x - x^{-1})For n=2: f_2(x) = (x^3 - x^{-3})/(x - x^{-1})Multiply numerator and denominator by x to make it easier: (x^4 - x^{-2})/(x^2 -1). Wait, but perhaps a better approach is to note that x^3 - x^{-3} = (x - x^{-1})(x^2 + 1 + x^{-2}). Let me check:(x - x^{-1})(x^2 + 1 + x^{-2}) = x^3 + x + x^{-1} - x - x^{-1} - x^{-3} = x^3 - x^{-3}, yes. So f_2(x) = (x^3 - x^{-3})/(x - x^{-1}) = x^2 + 1 + x^{-2} = (x + x^{-1})^2 - 1 = y^2 -1. Which matches the formula for n=2.Similarly, for n=3 (odd), according to the formula:f_3(x) = y^3 - C_{2}^{1} y^{1} + (-1)^{(3-1)/2} C_{(3+1)/2}^{(3-1)/2} y. Wait, the formula for odd n is:y^n - C_{n-1}^{1} y^{n-2} + ... + (-1)^{(n-1)/2} C_{(n+1)/2}^{(n-1)/2} ySo for n=3:y^3 - C_{2}^{1} y^1 + (-1)^{1} C_{2}^{1} y = y^3 - 2y + (-1)*2y = y^3 - 4y? Wait, that doesn't seem right. Wait, let me compute f_3(x) directly.f_3(x) = (x^4 - x^{-4})/(x - x^{-1}). Let's compute this.Multiply numerator and denominator by x: (x^5 - x^{-3})/(x^2 -1). Hmm, maybe better to factor numerator.x^4 - x^{-4} = (x^2)^2 - (x^{-2})^2 = (x^2 - x^{-2})(x^2 + x^{-2}) = (x - x^{-1})(x + x^{-1})(x^2 + x^{-2})Therefore, f_3(x) = (x^4 - x^{-4})/(x - x^{-1}) = (x + x^{-1})(x^2 + x^{-2}) = y*(y^2 - 2) = y^3 - 2y.But according to the given formula for n=3 (odd):It should be y^3 - C_{2}^{1} y^{1} + ... + (-1)^{(3-1)/2} C_{(3+1)/2}^{(3-1)/2} y.So (n-1)/2 = 1, so the last term is (-1)^1 C_{2}^{1} y = -2y. So the expression is y^3 - 2y -2y? Wait, no. Wait, let me check again.Wait, the formula for odd n is:y^n - C_{n-1}^1 y^{n-2} + ... + (-1)^i C_{n - i}^{n - i} y^{n - 2i} + ... + (-1)^{(n-1)/2} C_{(n+1)/2}^{(n-1)/2} yWait, for n=3, the terms are:y^3 - C_{2}^1 y^{1} + (-1)^{(3-1)/2} C_{(3+1)/2}^{(3-1)/2} yBut (n-1)/2 = 1, so the last term is (-1)^1 C_{2}^{1} y = -2y. Therefore, the expression is y^3 - 2y -2y = y^3 -4y. But from direct computation, it's y^3 - 2y. So there's a discrepancy here. That suggests the formula in the problem might have a typo, or my understanding is incorrect.Wait, maybe I misapplied the formula. Let's check the problem statement again.For odd n:f_n(x) = y^n - C_{n-1}^1 y^{n-2} + ... + (-1)^i C_{n - i}^{n - i} y^{n - 2i} + ... + (-1)^{(n-1)/2} C_{(n+1)/2}^{(n-1)/2} yWait, the coefficients here are C_{n - i}^{n - i}. That would be C(n - i, n - i) which is 1. But that can't be, because for n=3 and i=1: C_{3 -1}^{3 -1} = C(2,2) =1. Then the term would be (-1)^1 *1 * y^{3 - 2*1} = -1 *1* y = -y. So the expression becomes y^3 - C_{2}^1 y^{1} + ... + (-1)^{1} C_{2}^{1} y. Wait, but n=3, so the last term is i= (n-1)/2 =1. So terms are for i=1: (-1)^1 C_{3 -1}^{3 -1} y^{3 - 2*1} = -1*C(2,2)*y^1 = -1*1*y = -y. Then the expression is y^3 - C_{2}^1 y^{1} + (-1)^1 C_{2}^{2} y^{3 -2*1} = y^3 - 2y - y = y^3 -3y. But when we computed directly, f_3(x) is y^3 -2y. So something's wrong.Wait, this suggests that either the formula in the problem is incorrect, or my calculation is wrong. Let me check f_3(x) again.Original definition: f_n(x) = (x^{n+1} - x^{-n-1}) / (x - x^{-1})For n=3: (x^4 - x^{-4}) / (x - x^{-1})Multiply numerator and denominator by x: (x^5 - x^{-3}) / (x^2 -1). But perhaps another approach:Note that x^4 - x^{-4} = (x^2)^2 - (x^{-2})^2 = (x^2 - x^{-2})(x^2 + x^{-2})Then, x^2 - x^{-2} = (x - x^{-1})(x + x^{-1})Therefore, x^4 -x^{-4} = (x -x^{-1})(x +x^{-1})(x^2 +x^{-2})Divide by (x -x^{-1}): f_3(x) = (x + x^{-1})(x^2 + x^{-2}) = y*(y^2 - 2) = y^3 -2y. So this is correct. But according to the given formula, for n=3, the expression is y^3 - C_{2}^1 y^{1} + (-1)^1 C_{2}^{1} y = y^3 -2y -2y = y^3 -4y, which is incorrect. Therefore, there must be a problem with the formula as stated.Alternatively, perhaps the coefficients are different. Wait, in the problem statement, for odd n, the general term is (-1)^i C_{n -i}^{n -i} y^{n -2i}. Wait, C_{n -i}^{n -i} is 1, as it's choosing all elements. So that can't be right. That suggests each term after the first has coefficient -something, but with coefficients being 1? That doesn't align with the actual calculation.Wait, maybe the problem statement has a typo, and instead of C_{n -i}^{n -i}, it's supposed to be C_{n -i}^{i} or something else. Let me think.Alternatively, looking back at the even case, for i=1, the coefficient is C_{n -1}^{1}, which would be C(n -1, 1). For even n, the next term is C_{n -2i}^{n}... Hmm, this is confusing.Wait, let me check part (1) result again. If we can establish a recurrence relation, maybe the expression in part (2) is a solution to that recurrence, so we can use induction based on the recurrence.Given that f_{n+1} = y f_n - f_{n-1}, with base cases maybe. Let's see. For induction, we need base cases and then assume it's true for n and n-1, then prove for n+1.First, let's check for n=1. For n=1, f_1(x) = (x^2 - x^{-2}) / (x -x^{-1}) = (x^2 -x^{-2})/(x -x^{-1}) = x + x^{-1} = y. According to the formula, for n=1 (odd), it should be y^1 - ... + (-1)^{(1-1)/2} C_{(1+1)/2}^{(1-1)/2} y. Wait, n=1 is odd, (1-1)/2=0, so the last term is (-1)^0 C_{1}^{0} y = 1*1*y = y. So f_1(x) = y, which matches.For n=2 (even), as before, f_2(x)=y^2 -1. According to the formula, y^2 - C_{1}^1 y^{0} = y^2 -1, which matches.For n=3, according to the recurrence relation: f_3 = y f_2 - f_1 = y*(y^2 -1) - y = y^3 - y - y = y^3 -2y, which matches the direct computation. But according to the given formula, with n=3 (odd):y^3 - C_{2}^1 y^{1} + (-1)^1 C_{2}^{1} y = y^3 -2y -2y = y^3 -4y, which is wrong. Therefore, there's a problem with the formula as presented in the problem.This suggests that the coefficients in the problem's statement might be incorrectly typeset. Alternatively, perhaps the combination terms are written with a different convention.Wait, let's look again at the problem statement for part (2):For even n:y^{n} - C_{n-1}^{1} y^{n-2} + ... + (-1)^i C_{n-2i}^{n} y^{n-2i} + ... + (-1)^{n/2}For odd n:y^{n} - C_{n-1}^{1} y^{n-2} + ... + (-1)^i C_{n-i}^{n-i} y^{n-2i} + ... + (-1)^{(n-1)/2} C_{(n+1)/2}^{(n-1)/2} yWait, in the even case, the coefficients are C_{n - 2i}^{n}, but in the odd case, it's C_{n - i}^{n -i}.Alternatively, maybe the coefficients are supposed to be C(n - i, i). For example, in the even case, when n=2, term is -C_{1}^{1} which is -1, so it's y^2 -1, which matches. For n=4 (even), according to the formula:y^4 - C_{3}^1 y^{2} + C_{4 - 2*2}^{n} y^{0}*(-1)^2. Wait, n=4, i=1: -C_{3}^1 y^{2}; i=2: +C_{4 - 4}^{4} y^{0} = C(0,4)=0, but that term is supposed to be +1. Wait, this is confusing.Alternatively, maybe the combination terms are C(n - i, i). For n=2, i=1: C(2 -1,1)=C(1,1)=1, which gives -1, matching. For n=4, i=1: C(4 -1,1)=C(3,1)=3; i=2: C(4 -2,2)=C(2,2)=1, so terms would be y^4 -3y^2 +1, which is (y^2 -1)^2 - y^2, but not sure. Wait, let's compute f_4(x).f_4(x) = (x^5 -x^{-5})/(x -x^{-1})Multiply numerator and denominator by x: (x^6 -x^{-4})/(x^2 -1). Alternatively, factor numerator:x^5 -x^{-5} = (x -x^{-1})(x^4 +x^2 +1 +x^{-2} +x^{-4})Therefore, f_4(x) = x^4 +x^2 +1 +x^{-2} +x^{-4} = (x^4 +x^{-4}) + (x^2 +x^{-2}) +1 = (y^4 -4y^2 +2) + (y^2 -2) +1 = y^4 -3y^2 +1. So indeed, if the formula for even n=4 is y^4 -3y^2 +1, which would correspond to coefficients C_{3}^1=3 and the constant term 1. So in terms of the problem's notation, for even n=4:y^4 - C_{3}^1 y^{2} + C_{0}^4 y^{0}*(-1)^2. Wait, but C_{0}^4 is zero. Hmm, unless the combination terms are different.Alternatively, if the coefficients are C(n -i, i). For n=4, i=1: C(4 -1,1)=C(3,1)=3; i=2: C(4 -2,2)=C(2,2)=1. So the terms would be y^4 -3y^2 +1, which matches f_4(x)= y^4 -3y^2 +1. Therefore, perhaps the combination notation in the problem is C_{n -i}^{i} instead of C_{n -2i}^{n} for even n.Looking back at the problem statement for even n:"...(-1)^i C_{n-2i}^{n} y^{n-2i}..."If instead, this is supposed to be C(n -i, i), then for n even, the coefficient for term i is C(n -i, i). For n=4, i=1: C(3,1)=3; i=2: C(2,2)=1. So that works. Similarly, for n=2, i=1: C(1,1)=1, which gives the -1 coefficient. So maybe the problem's combination notation is C_{n - i}^{i} instead of C_{n -2i}^{n}. That would make sense.Similarly, for odd n=3, the coefficients would be C(n -i, i). For n=3, i=1: C(2,1)=2; i=2: C(1,2)=0, which is invalid. Wait, but for odd n, the last term is when i=(n-1)/2. For n=3, (n-1)/2=1, so only i=1. So the expression would be y^3 - C(2,1) y^{1} = y^3 -2y, which matches. So for odd n=5, it would be y^5 - C(4,1)y^3 + C(3,2)y^1 - ... up to i=(5-1)/2=2. Then terms would be y^5 -4y^3 +3y.Wait, let me check f_5(x). f_5(x)=(x^6 -x^{-6})/(x -x^{-1})Factor numerator: x^6 -x^{-6} = (x^3)^2 - (x^{-3})^2 = (x^3 -x^{-3})(x^3 +x^{-3}) = (x -x^{-1})(x^2 +1 +x^{-2})(x^3 +x^{-3})Divide by (x -x^{-1}): f_5(x) = (x^2 +1 +x^{-2})(x^3 +x^{-3}) = (y^2 -1)(y^3 -3y) = y^5 -3y^3 -y^3 +3y = y^5 -4y^3 +3y. Which matches y^5 - C(4,1)y^3 + C(3,2)y = y^5 -4y^3 +3y. Therefore, the coefficients are indeed C(n -i, i) for each term i.Therefore, the problem's combination notation is likely a misprint or different convention. The correct coefficients are C(n -i, i). So in the problem statement, for even n, the coefficient C_{n -2i}^{n} should be C(n -i, i), and for odd n, C_{n -i}^{n -i} should be C(n -i, i). Therefore, assuming this correction, the formula in the problem is:For even n:f_n(x) = y^n - C(n-1,1) y^{n-2} + C(n-2,2) y^{n-4} - ... + (-1)^{n/2}For odd n:f_n(x) = y^n - C(n-1,1) y^{n-2} + C(n-2,2) y^{n-4} - ... + (-1)^{(n-1)/2} C((n+1)/2, (n-1)/2) yBut with the combinations written as C_{n -i}^{i} or similar. Given this, the problem's combination notation is non-standard and likely mistyped, but the correct coefficients are C(n -i, i). Therefore, proceeding under this assumption.Thus, to use induction, let's consider the base cases n=1 and n=2.Base case n=1 (odd): f_1(x) = y. According to the formula, it should be y^1 - ... + (-1)^{0} C(1,0) y. Wait, but according to the formula for odd n=1, the last term is (-1)^{(1-1)/2} C_{(1+1)/2}^{(1-1)/2} y = (-1)^0 C(1,0) y = 1*1*y = y. So f_1(x)=y, which matches.For n=2 (even): f_2(x)=y^2 -1. According to the formula, it's y^2 - C(1,1) y^0 = y^2 -1, which matches.Now assume that for some k ≥2, the formula holds for f_k(x) and f_{k-1}(x). Then, using the recurrence f_{k+1}(x) = y f_k(x) - f_{k-1}(x). If we can show that substituting the formulas for f_k and f_{k-1} into the right-hand side gives the formula for f_{k+1}, then the induction step holds.Let's consider two cases: when k+1 is even or odd.Case 1: k+1 is even. Then k is odd, and k-1 is even.Assume f_k(x) follows the odd formula, and f_{k-1}(x) follows the even formula.Then f_{k+1}(x) = y * [y^k - C(k-1,1) y^{k-2} + ... + (-1)^{(k-1)/2} C((k+1)/2, (k-1)/2) y] - [y^{k-1} - C(k-2,1) y^{k-3} + ... + (-1)^{(k-1)/2 -1/2} ...]. Wait, this is getting messy. Maybe a better approach is to track how the terms combine.Alternatively, note that the expression for f_n(x) is similar to the expansion of Chebyshev polynomials or some related orthogonal polynomials, which satisfy similar recurrences. However, perhaps it's better to proceed step by step.Suppose for induction that f_n(x) for even n is given by y^n - C(n-1,1)y^{n-2} + C(n-2,2)y^{n-4} - ... + (-1)^{n/2}, and for odd n by y^n - C(n-1,1)y^{n-2} + ... + (-1)^{(n-1)/2} C((n+1)/2, (n-1)/2) y.Assume for n=k (odd), f_k(x) = y^k - C(k-1,1)y^{k-2} + ... + (-1)^{(k-1)/2} C((k+1)/2, (k-1)/2)y.And for n=k-1 (even), f_{k-1}(x) = y^{k-1} - C(k-2,1)y^{k-3} + ... + (-1)^{(k-1)/2 - 1/2} ... Wait, since k-1 is even, the last term is (-1)^{(k-1)/2}. For example, if k=3 (odd), then k-1=2 (even), last term is (-1)^1 = -1.Then f_{k+1}(x) = y*f_k(x) - f_{k-1}(x).Multiplying y into f_k(x):y^{k+1} - C(k-1,1)y^{k-1} + ... + (-1)^{(k-1)/2} C((k+1)/2, (k-1)/2) y^2Subtracting f_{k-1}(x):- [y^{k-1} - C(k-2,1)y^{k-3} + ... + (-1)^{(k-1)/2}]Thus, combining terms:y^{k+1} - [C(k-1,1) +1] y^{k-1} + [C(k-2,1) + C(...)] y^{k-3} - ... + (-1)^{(k-1)/2} C(...) y^2 - (-1)^{(k-1)/2}Wait, this seems complicated. Let's compute the coefficients term by term.First term: y^{k+1}Second term: -C(k-1,1)y^{k-1} - (-y^{k-1}) = [-C(k-1,1) +1] y^{k-1}Third term: + ... Hmm, need to see the pattern.Alternatively, recall that the coefficients in the expression correspond to the number of ways to select certain terms, related to combinatorial interpretations. Alternatively, using the properties of binomial coefficients.Alternatively, note that the coefficients C(n -i, i) satisfy the recurrence C(n -i, i) + C(n -i -1, i -1) = C(n -i +1, i). Wait, that's Pascal's identity.But in our case, the coefficient of y^{k+1 -2i} in f_{k+1}(x) would be obtained by y*f_k(x) contributing (-1)^i C(k -i, i) y^{k+1 -2i} and -f_{k-1}(x) contributing (-1)^{i} C(k -1 -i, i) y^{k+1 -2i -2}? Wait, maybe not.Alternatively, when multiplying y*f_k(x), each term in f_k(x) is of the form (-1)^i C(k -i, i) y^{k - 2i}, so multiplying by y gives (-1)^i C(k -i, i) y^{k+1 -2i}.Similarly, subtracting f_{k-1}(x), which has terms (-1)^j C((k-1) -j, j) y^{(k-1)-2j}.Thus, the term y^{m} in f_{k+1}(x) would come from two sources:1. y * term y^{m -1} in f_k(x): coefficient (-1)^{i} C(k -i, i) where m -1 = k -2i ⇒ i = (k - (m -1))/22. - term y^{m} in f_{k-1}(x): coefficient (-1)^{j+1} C((k-1)-j, j) where m = (k -1) -2j ⇒ j = ((k -1) -m)/2Therefore, combining these contributions:For each m, the coefficient is:(-1)^{(k - m +1)/2} C(k - (k - m +1)/2, (k - m +1)/2) + (-1)^{((k -1 - m)/2)+1} C((k-1) - ((k -1 - m)/2), ((k -1 - m)/2))This seems too convoluted. Maybe instead, since we know the expression for f_{k+1}(x) should follow the same pattern as f_k(x) and f_{k-1}(x), with coefficients C((k+1) -i, i), we can use Pascal's identity.Note that C(n -i, i) + C(n -i -1, i -1) = C(n -i +1, i). So if we can express the coefficients in f_{k+1}(x) as the sum of coefficients from y*f_k(x) and -f_{k-1}(x), then Pascal's identity would apply.For example, the coefficient of y^{k+1 -2i} in f_{k+1}(x) is (-1)^i [C(k -i, i) + C(k -1 - (i -1), i -1))] = (-1)^i [C(k -i, i) + C(k -i, i -1)] = (-1)^i C(k -i +1, i). Which is exactly the coefficient in the expression for f_{k+1}(x).Therefore, by induction, if f_k(x) has coefficients (-1)^i C(k -i, i) and f_{k-1}(x) has coefficients (-1)^i C((k-1)-i, i), then their combination via the recurrence y*f_k - f_{k-1} will give coefficients (-1)^i [C(k -i, i) + C(k -i -1, i -1)] = (-1)^i C(k -i +1, i), which matches the coefficients for f_{k+1}(x).Thus, the induction holds, and the formula is valid.Therefore, the problem's combination notation likely has a typo, but with the corrected interpretation, the induction works. Therefore, part (2) can be proven by mathematical induction using the recurrence from part (1) and verifying the base cases.Final Answer(1) boxed{f_{n+1}(x) = y f_n(x) - f_{n-1}(x)}.(2) boxed{f_{n}(x) = sum_{i=0}^{lfloor n/2 rfloor} (-1)^i binom{n - i}{i} y^{n - 2i}}.

✅:To solve the given problems, we start by proving the recurrence relation for ( f_{n+1}(x) ) and then use mathematical induction to prove the explicit formula for ( f_n(x) ). Part (1)Proof of the recurrence relation:Given ( f_n(x) = frac{x^{n+1} - x^{-n-1}}{x - x^{-1}} ), we need to show that ( f_{n+1}(x) = y f_n(x) - f_{n-1}(x) ) where ( y = x + frac{1}{x} ).1. Compute ( y f_n(x) ): [ y f_n(x) = left( x + frac{1}{x} right) frac{x^{n+1} - x^{-n-1}}{x - x^{-1}} ]2. Expand the numerator: [ left( x + frac{1}{x} right)(x^{n+1} - x^{-n-1}) = x^{n+2} - x^{-n} + x^{n} - x^{-n-2} ]3. Subtract ( f_{n-1}(x) ): [ y f_n(x) - f_{n-1}(x) = frac{x^{n+2} - x^{-n} + x^{n} - x^{-n-2}}{x - x^{-1}} - frac{x^n - x^{-n}}{x - x^{-1}} ]4. Simplify the numerator: [ x^{n+2} - x^{-n-2} ]5. Conclude: [ y f_n(x) - f_{n-1}(x) = frac{x^{n+2} - x^{-n-2}}{x - x^{-1}} = f_{n+1}(x) ]Thus, the recurrence relation is proven:[boxed{f_{n+1}(x) = y f_n(x) - f_{n-1}(x)}] Part (2)Proof by mathematical induction:We use induction to prove the explicit formula for ( f_n(x) ).Base Cases:- For ( n = 1 ) (odd): [ f_1(x) = frac{x^2 - x^{-2}}{x - x^{-1}} = x + x^{-1} = y ] The formula gives ( y ), which matches.- For ( n = 2 ) (even): [ f_2(x) = frac{x^3 - x^{-3}}{x - x^{-1}} = x^2 + 1 + x^{-2} = (x + x^{-1})^2 - 1 = y^2 - 1 ] The formula gives ( y^2 - 1 ), which matches.Inductive Step:Assume the formula holds for ( f_k(x) ) and ( f_{k-1}(x) ). We need to show it holds for ( f_{k+1}(x) ).Using the recurrence relation:[f_{k+1}(x) = y f_k(x) - f_{k-1}(x)]- If ( k ) is even, ( f_k(x) ) and ( f_{k-1}(x) ) follow the even and odd formulas respectively.- If ( k ) is odd, ( f_k(x) ) and ( f_{k-1}(x) ) follow the odd and even formulas respectively.By expanding the terms and using the binomial coefficient identity ( binom{n - i}{i} + binom{n - i - 1}{i - 1} = binom{n - i + 1}{i} ), we verify that the coefficients combine correctly to form the terms for ( f_{k+1}(x) ).Conclusion:By induction, the formula holds for all ( n ):[boxed{f_{n}(x) = sum_{i=0}^{lfloor n/2 rfloor} (-1)^i binom{n - i}{i} y^{n - 2i}}]

🔑:# (1) To prove: ( f_{n+1}(x) = y f_n(x) - f_{n-1}(x) ) for ( n > 1 )Given:[f_n(x) = frac{x^{n+1} - x^{-n-1}}{x - x^{-1}}]where ( y = x + frac{1}{x} ).We start by considering:[y f_n(x) - f_{n-1}(x)]Breaking it down, we have:begin{align*}y f_n(x) &= left(x + frac{1}{x}right) frac{x^{n+1} - x^{-n-1}}{x - x^{-1}} = frac{left(x + frac{1}{x}right)(x^{n+1} - x^{-n-1})}{x - x^{-1}}, f_{n-1}(x) &= frac{x^n - x^{-n}}{x - x^{-1}}.end{align*}Now,[y f_n(x) - f_{n-1}(x) = frac{left(x + frac{1}{x}right)(x^{n+1} - x^{-n-1}) - (x^n - x^{-n})}{x - x^{-1}}]Expanding the numerator:[= frac{x^{n+2} + x^n - x^{-n-2} - x^{-n} + x^n - x^{-n}}{x - x^{-1}}]Simplifying the numerator:[= frac{x^{n+2} + 2x^n - x^{-n-2} - 2x^{-n}}{x - x^{-1}}]Notice that ( x^n - x^{-n} = 0 ) for the additional terms, thus:[= frac{x^{n+2} - x^{-n-2}}{x - x^{-1}} = f_{n+1}(x)]Therefore, we have shown that:[f_{n+1}(x) = y f_n(x) - f_{n-1}(x).]# (2) Using Mathematical Induction:Base Case:For ( n = 1 ):[f_1(x) = x + frac{1}{x} = y]For ( n = 2 ):[f_2(x) = x^2 + frac{1}{x^2} = left(x + frac{1}{x}right)^2 - 1 = y^2 - 1]Both cases ( n=1 ) and ( n=2 ) are consistent with our formula.Inductive Step:Assume that for any ( n leq m ), the formula holds. That is:[f_n(x) = y^n - C_{n-1}^1 y^{n-2} + cdots + (-1)^{i} C_{n-i}^{i} y^{n-2i} + cdots ]We need to show that the formula holds for ( n = m+1 ):Case I: ( m+1 ) is oddGiven:[f_m(x) = y^m - C_{m-1}^1 y^{m-2} + cdots + (-1)^{i} C_{m-i}^{i} y^{m-2i}][f_{m-1}(x) = y^{m-1} - C_{m-2}^1 y^{m-3} + cdots + (-1)^{i-1} C_{m-i}^{i-1} y^{m+1-2i}]Multiply ( f_m(x) ) by ( y ) and subtract ( f_{m-1}(x) ):[y f_m(x) - f_{m-1}(x) = y^{m+1} - cdots + (-1)^i left(C_{m-i}^{i} + C_{m-i}^{i-1}right) y^{m+1-2i} + cdots ]Using Pascal’s equation:[C_{m-i}^{i} + C_{m-i}^{i-1} = C_{m-i+1}^{i}]Thus we get:[y f_m(x) - f_{m-1}(x) = y^{m+1} - C_{m+1}^{1} y^{m-1} + cdots + (-1)^i C_{m-i}^{i} y^{m+1-2i} + cdots + (-1)^{m/2} C_{m/2+1}^{m/2} y]Hence it is proven for ( n = m+1 ).Case II: ( m+1 ) is evenThe steps are analogous to the above proof with careful attention to the index shifts due to even numbers. Omitting repetitive intermediate steps for brevity, using the same principle of combining binomial coefficients:[y f_m(x) - f_{m-1}(x) = y^{m+1} - C_{m}^{1} y^{m-1} + cdots + (-1)^{(m+1)/2} ]Thus, by induction, the formula holds for all ( n ). Conclusion:[boxed{f_{n+1}(x) = y f_n(x) - f_{n-1}(x)}]

🧠:Okay, let's try to figure out on which floor Dima lives. The problem says he's in a nine-story building. When he takes the elevator down from his floor to the first floor, it takes 1 minute. But when he goes up, he can't reach his floor's button, so he presses a lower floor button, then walks the rest. The whole trip up takes 1 minute 10 seconds. The elevator's speed is the same going up and down, and Dima walks up twice as slowly as the elevator. Hmm, so we need to find his floor.First, let's parse all the information. The building has 9 floors. Let's denote Dima's floor as floor N. So, N is between 2 and 9 because he can't reach the button himself when going up, so he must live above the first floor.When he goes down from floor N to 1, the elevator takes 1 minute. So, the time taken to go down is 60 seconds. The elevator's speed is constant. Let's denote the elevator's speed as v floors per second. Then, the number of floors between N and 1 is (N - 1). So, the time taken to go down is (N - 1)/v = 60 seconds. Therefore, we can write that equation as:(N - 1)/v = 60 --> equation 1Now, when he goes up, he takes the elevator to some lower floor, let's say floor K, which he can reach, and then walks up from K to N. The total time for the trip up is 1 minute 10 seconds, which is 70 seconds. The elevator takes the same time per floor as when going down, right? So, the elevator going up from 1 to K takes K - 1 floors. So, the time in the elevator going up is (K - 1)/v. Then, Dima walks from K to N, which is (N - K) floors. His walking speed is twice as slow as the elevator's speed. Since the elevator's speed is v floors per second, Dima's walking speed is v/2 floors per second. Therefore, the time walking is (N - K)/(v/2) = 2(N - K)/v.So, total time going up is elevator time + walking time:(K - 1)/v + 2(N - K)/v = 70 secondsWe can factor out 1/v:[ (K - 1) + 2(N - K) ] / v = 70But from equation 1, we know that (N - 1)/v = 60, so 1/v = 60/(N - 1). Substitute that into the equation:[ (K - 1) + 2(N - K) ] * (60/(N - 1)) ) = 70Let me compute the numerator first:(K - 1) + 2(N - K) = K -1 + 2N - 2K = -K -1 + 2N = 2N - K -1Therefore:(2N - K -1) * 60 / (N - 1) = 70Multiply both sides by (N - 1):60(2N - K -1) = 70(N -1)Divide both sides by 10:6(2N - K -1) = 7(N -1)Expand both sides:12N -6K -6 = 7N -7Bring all terms to left:12N -6K -6 -7N +7 = 0Simplify:5N -6K +1 =0So, 5N -6K = -1 --> equation 2So, now we have equation 1: (N -1)/v =60, and equation 2:5N -6K = -1.But we have another variable K here, which is the floor he takes the elevator up to. We need another equation or some constraints on K.Wait, K must be a floor that Dima can reach, i.e., he can press the button for K. So, K must be less than N, since he can't reach his own floor's button. Also, K must be an integer between 1 and N-1. Since he starts from floor 1 when going up, K is the floor he takes the elevator to, then walks the rest. So, K must be an integer such that 1 ≤ K < N.But how do we find K? Since K has to be an integer, and N is also an integer between 2 and 9, perhaps we can find possible integer solutions for equation 2:5N -6K = -1.Let me rearrange equation 2 to express K in terms of N:5N +1 =6KThus:K=(5N +1)/6Since K must be an integer, (5N +1) must be divisible by 6.So, 5N +1 ≡0 mod6Which implies 5N ≡ -1 mod6But 5 ≡5 mod6, and -1 ≡5 mod6, so:5N ≡5 mod6Divide both sides by 5 (since 5 and 6 are coprime, the inverse of 5 mod6 is 5, because 5*5=25≡1 mod6)Thus, N ≡5*5 mod6 ≡25 mod6 ≡1 mod6Therefore, N ≡1 mod6But N is between 2 and 9, inclusive. So, possible N values are 7, since 7≡1 mod6 (because 7-6=1). Also, 1 is congruent 1 mod6, but N must be at least 2. So, only N=7 satisfies N≡1 mod6 in the range 2-9.Wait, wait. Let me check:If N ≡1 mod6, then possible N values in 2-9 would be 7, since 1 mod6 is 1, then adding 6 gives 7. So, N=7.Therefore, N=7. Then, K=(5*7 +1)/6=(35 +1)/6=36/6=6. So, K=6.So, according to this, Dima lives on floor 7. He takes the elevator to floor 6, then walks up one floor to 7. Let's check if this satisfies the time constraints.First, the downward trip: from 7 to 1, 6 floors. Time is (7-1)/v=6/v=60 seconds. So, v=6/60=0.1 floors per second.Then, the upward trip: elevator from 1 to 6, which is 5 floors. Time is (6-1)/v=5/0.1=50 seconds. Then, walking from 6 to 7: 1 floor. Walking speed is v/2=0.05 floors per second. Time=1/0.05=20 seconds. Total time:50+20=70 seconds, which matches the given 1 minute 10 seconds. So, this works.Wait, but let me confirm if there are other possible solutions. Suppose N=7 is the only solution. Let's check if other N could also satisfy the equation.Suppose N=7, K=6.If N=1, but he can't be on first floor. N=2: check if N=2 can satisfy 5N -6K = -1. Then 10 -6K =-1 → 6K=11→ K≈1.83, not integer. N=3: 15 -6K =-1 →6K=16→K≈2.66, not integer. N=4:20-6K=-1→6K=21→K=3.5, not integer. N=5:25-6K=-1→6K=26→K≈4.33, not integer. N=6:30 -6K=-1→6K=31→K≈5.16, no. N=7:35 -6K=-1→6K=36→K=6. N=8:40 -6K=-1→6K=41→≈6.83. N=9:45-6K=-1→6K=46→≈7.66. So, only N=7 gives integer K=6. Therefore, N=7 is the only solution.Therefore, Dima lives on the 7th floor.But wait, let me just cross-verify. Let me check with N=7, K=6. Elevator down: 6 floors in 60 seconds, so 0.1 floors per second. Up: elevator to 6th floor, which is 5 floors, taking 50 seconds. Then walking 1 floor at 0.05 floors per second: 20 seconds. Total 70 seconds. Correct. So, this works. So, answer is 7th floor.I think that's solid. Let me check once again for any possible mistakes.The key steps were setting up the equations for time down and time up. Then realizing that K must be an integer, leading to the modular equation. Solving that gives N=7. Checked all possible N from 2-9, only N=7 satisfies. So, it's correct.

✅:Dima lives on the 7th floor. Here's the breakdown:1. Downward Trip: - Time = 60 seconds. - Floors descended: (N - 1). - Elevator speed: (v = frac{N - 1}{60}) floors/second.2. Upward Trip: - Total time = 70 seconds. - Elevator ascends to floor (K) (where (K < N)), taking (frac{K - 1}{v}) seconds. - Dima walks from (K) to (N) at half the elevator speed ((v/2)), taking (frac{2(N - K)}{v}) seconds. - Total time equation: [ frac{K - 1 + 2(N - K)}{v} = 70 ] - Substituting (v = frac{N - 1}{60}): [ frac{2N - K - 1}{N - 1} cdot 60 = 70 quad Rightarrow quad 5N - 6K = -1. ]3. Solving for (N) and (K): - (5N + 1 = 6K) implies (5N + 1) must be divisible by 6. - Modular arithmetic: (5N equiv 5 pmod{6} Rightarrow N equiv 1 pmod{6}). - Valid (N) (2–9): (N = 7). - (K = frac{5(7) + 1}{6} = 6).4. Verification: - Downward: (7 to 1) (6 floors) in (6 / 0.1 = 60) seconds. - Upward: Elevator to 6th floor (5 floors) in (5 / 0.1 = 50) seconds, then walking 1 floor at (0.05) floors/second (20 seconds). Total: (50 + 20 = 70) seconds.Answer: Dima lives on the 7th floor.

🔑:1. Assume Variables: - Let ( n ) be the total number of floors in the building. Given ( n = 9 ). - Let ( k ) be the floor where Dima can reach the button on the lift. - Let ( F ) be the floor where Dima lives. We need to determine ( F ).2. Setup Known Facts: - The lift takes ( 1 ) minute to descend from the 9th floor to the 1st floor. Thus, it descends ( n-1 = 8 ) floors in ( 60 ) seconds. - The total time for Dima to ascend from the 1st floor to his floor ( F ) is ( 1 ) minute ( 10 ) seconds or ( 70 ) seconds.3. Relative Speed Analysis: - The lift ascends at a certain speed, let this speed be ( s ) floors per second. - Dima ascends at half that speed: ( frac{s}{2} ) floors per second.4. Portion of the Path: - When descending, the lift takes ( frac{1}{6} ) of the total time for Dima for the portion he walks. Thus, ( frac{1}{6} times 70 = frac{70}{6} = 11.67 ) seconds approximately. - Equivalently, ( F - k ) is the number of floors Dima walks, taking ( 2 times 11.67 = 23.33 ) seconds.5. Equating Lift and Walking Times: - Since the lift's downward journey takes ( 1 ) minute (( 60 ) seconds), ( frac{70 - 60}{2} ) equates to some whole part of the journey ( F - k ).6. Calculate Floors Walked: - Dima reaches ( frac{10 seconds}{s floors per second} = 10 frac{floors}{s} times frac{s}{1} = 10 floors ) - Therefore, the extra floors Dima walks: - ( F - k = 10 floors ) - Let ( k = text{closest floor down from 9 allowed}. Therefore, ( k = 5 ).7. Determine Floors: - The floors Dima equals ( F = k + 5 = 5 + 2 = 7 ). Conclusion:The floor where Dima lives:(boxed{7})

🧠:Okay, so I have this problem where there are an equal number of black and white checkers arranged around a circle. We need to consider all triples of consecutive checkers and count how many triples have a majority of white (A) and a majority of black (B). The goal is to prove that A is at most three times B, meaning A ≤ 3B. First, let me make sure I understand the problem correctly. We have a circle with, say, n black and n white checkers. That makes a total of 2n checkers arranged around the circle. For each consecutive triple (so overlapping triples, since it's a circle), we check if the majority are white or black. Then we count A as the number of white-majority triples and B as the number of black-majority triples. The claim is that A can't be more than three times B. Hmm. So, since there are 2n checkers, there are also 2n triples of consecutive checkers (each starting at a different position). Each triple is either majority white, majority black, or a tie. But since the total number of checkers is even (2n), each triple has three checkers, which is odd, so there can't be a tie. Therefore, every triple is either in A or B, so A + B = 2n. But we need to show that A ≤ 3B. Let's see. If A + B = 2n, then substituting B = 2n - A into the inequality gives A ≤ 3(2n - A), which simplifies to A ≤ 6n - 3A, then 4A ≤ 6n, so A ≤ (6n)/4 = (3n)/2. Wait, but this seems like a different inequality. Wait, maybe this approach is not correct. Wait, maybe this substitution is valid, but perhaps we need a different approach because the relation A ≤ 3B must hold regardless of the arrangement. Wait, but maybe there is a problem here. If A + B = 2n, then A ≤ 3B is equivalent to A ≤ 3(2n - A) → 4A ≤ 6n → A ≤ (3/2)n. So if we can show that A cannot exceed 3n/2, then that would prove the inequality. But does that hold? Let's test with small cases. Suppose n = 1, so two checkers: one black and one white. Wait, but then arranged in a circle. But each triple would require three checkers, but n=1 gives 2 checkers. Wait, so n must be at least 2? Wait, the problem says "a certain number of black and the same number of white checkers". So n can be any positive integer. But when n=1, we have 1 black and 1 white, so two checkers. Then, how do we form triples? Each triple is three consecutive checkers, but with two checkers, this would wrap around. Wait, in a circle of two checkers, each triple would consist of the two checkers plus one more, which would overlap. Wait, actually, in a circle of 2 checkers, each triple is the two checkers plus the first one again. So each triple is the same two checkers repeated. But that seems odd. Maybe the problem requires n ≥ 2? Or maybe the problem assumes that there are enough checkers to form distinct triples. Hmm. Maybe I should consider n ≥ 2. Let's try n=2: two black and two white checkers arranged in a circle. So total checkers: 4. Each triple is three consecutive checkers. So let's list all triples:If the circle is B, W, B, W. Then the triples are:1. B, W, B (two blacks, one white: majority black)2. W, B, W (two whites, one black: majority white)3. B, W, B (same as first)4. W, B, W (same as second)So in this case, A = 2, B = 2. Then A = 2 ≤ 3B = 6. Which holds. If arranged as B, B, W, W. Then the triples are:1. B, B, W (majority black)2. B, W, W (majority white)3. W, W, B (majority white)4. W, B, B (majority black)So here, A = 2, B = 2. Again, same. If arranged as B, W, W, B:Triples:1. B, W, W (majority white)2. W, W, B (majority white)3. W, B, B (majority black)4. B, B, W (majority black)So A = 2, B = 2. Hmm. So in these cases, A equals B, so A = B ≤ 3B is trivially true. But maybe there are other arrangements where A is larger. Let's try arranging three whites and one black. Wait, but n must be equal. So n=2, two blacks and two whites. Let me try another arrangement. Let's arrange B, W, W, W. Wait, no, that would be three whites and one black, which isn't equal. So back to n=2. Let's try arranging B, B, B, W. Wait, that's three blacks and one white, which is unequal. So not allowed. So maybe with n=2, it's impossible to get A higher than B. Wait, but in the previous examples, A and B are equal. So maybe in larger n, A can be higher? Let me try n=3: three blacks and three whites. Let's arrange them alternately: B, W, B, W, B, W. Then each triple would be B, W, B (two B, one W), W, B, W (two W, one B), etc. So each triple alternates between majority B and majority W. So there are 6 triples. So three of each. So A=3, B=3. So A=3 ≤ 3B=9. True. Another arrangement: group all blacks together and all whites together: BBBWWW. Let's check the triples:1. BBB: majority B2. BBW: majority B3. BWW: majority W4. WWW: majority W5. WWB: majority W6. WBB: majority BSo the triples: BBB (B), BBW (B), BWW (W), WWW (W), WWB (W), WBB (B). So counts: B: 3, W: 3. Again A=3, B=3. Hmm.Wait, maybe if we arrange them differently. Let's try BBWBWW. Let's check the triples:1. BBW: majority B2. BWB: majority B? B, W, B: two Bs, one W. So majority B.3. WBW: majority W? W, B, W: two Ws, one B. Majority W.4. BWW: majority W5. WWB: majority W? W, W, B: two Ws. Majority W.6. WBB: majority BSo counts: A (majority W): triples 3,4,5: three. B (majority B): triples 1,2,6: three. Again equal. Hmm.Wait, maybe it's always equal? But the problem says to prove A ≤ 3B, which would be trivial if A = B. But maybe there's a case where A is larger. Let me try a different arrangement. Let's have more white triples. For n=3. Let's arrange as WWWBBB. The triples:1. WWW: majority W2. WWB: majority W3. WBB: majority B4. BBB: majority B5. BBW: majority B6. BWW: majority WSo counts: A = 3 (triples 1,2,6), B = 3 (triples 3,4,5). Still equal. Hmm. Maybe even if we interleave them. How about WBWBWB. Then triples:1. WBW: majority W (two Ws)2. BWB: majority B (two Bs)3. WBW: majority W4. BWB: majority B5. WBW: majority W6. BWB: majority BSo A=3, B=3. Still equal. Wait, maybe when n is even versus odd? Let me check n=4. Let's try arranging 4 B and 4 W. Maybe arrange them as WWWWBBBB. Let's check triples:1. WWW: W2. WWW: W3. WWB: W4. WBB: B5. BBB: B6. BBB: B7. BBW: B8. BWW: WSo counts: A = triples 1,2,3,8: 4; B = triples 4,5,6,7: 4. Equal again. Wait, maybe if we arrange them with more alternating. Let's try WBWBWBWB. Then each triple is WBW (W), BWB (B), WBW (W), BWB (B), etc. So A=4, B=4. Equal. Wait, how about a different arrangement. Let's try to cluster some whites and spread blacks. For example, WWBWWBBBB. Wait, n=4? Wait, 4 whites and 4 blacks. Let me count: WWBWWBBB. Wait, that's 5 W and 3 B. Not equal. Hmm. Let's do WWBWBBWB. Let's check:Positions: W, W, B, W, B, B, W, B. Let's list triples:1. W, W, B: W2. W, B, W: W3. B, W, B: B4. W, B, B: B5. B, B, W: B6. B, W, B: B7. W, B, B: B8. B, B, W: BWait, that can't be. Wait, let me list them properly. Wait, the circle is W, W, B, W, B, B, W, B.Triple 1: W, W, B (W)Triple 2: W, B, W (W)Triple 3: B, W, B (B)Triple 4: W, B, B (B)Triple 5: B, B, W (B)Triple 6: B, W, B (B)Triple 7: W, B, B (B)Triple 8: B, B, W (B)Wait, so A = 2 (triples 1 and 2), B = 6. Then A = 2 ≤ 3B = 18, which is true, but here A is much less than B. Hmm. But maybe the other way. Let's arrange more whites. Wait, but we have equal numbers. Let me try an arrangement where whites are arranged such that they form more triples. Hmm. For example, if I have alternating W, W, B, B, W, W, B, B. Let's check triples:Positions: W, W, B, B, W, W, B, B.Triple 1: W, W, B (W)Triple 2: W, B, B (B)Triple 3: B, B, W (B)Triple 4: B, W, W (W)Triple 5: W, W, B (W)Triple 6: W, B, B (B)Triple 7: B, B, W (B)Triple 8: B, W, W (W)So A = triples 1,4,5,8: 4; B = triples 2,3,6,7: 4. Again equal. Hmm. Wait, maybe there's a problem with my approach. In all these small cases, A and B are equal. But the problem states that A ≤ 3B. So maybe in some arrangements, A can be higher than B, but not more than three times. But in my small cases, it's equal. Maybe in larger n? Let's try n=4, but with a different arrangement. Let's try WWBWBBWB. Wait, check if it's equal. Let's list:Positions: W, W, B, W, B, B, W, B.Triple 1: W, W, B (W)Triple 2: W, B, W (W)Triple 3: B, W, B (B)Triple 4: W, B, B (B)Triple 5: B, B, W (B)Triple 6: B, W, B (B)Triple 7: W, B, B (B)Triple 8: B, B, W (B)Wait, same as before. A=2, B=6. So here, A is much less than B. So the maximum A seems to be when arranged alternately. But in that case, A and B are equal. So maybe A can never exceed B? But the problem says to prove A ≤ 3B. Which is a weaker statement. So if A is always equal to B, then certainly A ≤ 3B. But maybe in some cases, A can be higher. Wait, perhaps in the problem statement, the checkers are placed around the circle, but not necessarily alternately. Maybe if we arrange the checkers in such a way that there are clusters of white triples. Let's try n=3 again, with 3 B and 3 W. Let's arrange them as WWWBBB. Then triples:1. WWW: W2. WWB: W3. WBB: B4. BBB: B5. BBW: B6. BWW: WSo A=3, B=3. Equal. Hmm. Wait, maybe if we arrange them as WWBWBB. Let's check:Positions: W, W, B, W, B, B.Triples:1. W, W, B: W2. W, B, W: W3. B, W, B: B4. W, B, B: B5. B, B, W: B6. B, W, W: WSo A = triples 1,2,6: 3; B = triples 3,4,5:3. Equal again. Wait, maybe I need to find a configuration where A is greater than B. Let's try n=4, 4W and 4B. Let's arrange as WWWBBWWBB. Wait, that's 5 W and 5 B? No, 4W and 4B. Let me arrange as WWWWBBBB. Wait, that's 4W followed by 4B. Then triples:1. WWW: W2. WWW: W3. WWB: W4. WBB: B5. BBB: B6. BBB: B7. BBW: B8. BWW: WSo A = 4 (1,2,3,8); B = 4 (4,5,6,7). Equal. Alternatively, arrange as WWBWBBWB. Let me check:Positions: W, W, B, W, B, B, W, B.Triples:1. W, W, B: W2. W, B, W: W3. B, W, B: B4. W, B, B: B5. B, B, W: B6. B, W, B: B7. W, B, B: B8. B, B, W: BA = 2, B = 6. So here, A is smaller. Wait, maybe the maximum A occurs when the white checkers are as spread out as possible. Wait, but when spread out alternately, A and B are equal. If clustered, A might be less. Hmm. Wait, perhaps the key is that each white checker is part of multiple triples. Let's think about each checker's contribution. Each checker is part of three triples: the one starting at its position, the one before, and the one after. Wait, no. Each checker is part of three consecutive triples. For example, in a circle, checker i is part of triple (i-1, i, i+1), triple (i, i+1, i+2), and triple (i-2, i-1, i). Wait, no, in a circle, each triple is three consecutive checkers. Each checker is part of exactly three triples: the triple starting at its previous position, the triple starting at its position, and the triple starting at the next position. Wait, no. Let's clarify. For a circle of 2n checkers, each checker is in three different triples. For example, in a circle of 6 checkers labeled 1 to 6, checker 1 is in triples (6,1,2), (1,2,3), and (2,3,4). Wait, no. Wait, each triple is three consecutive checkers. So for checker 1, the triples that include it are (2n, 1, 2), (1, 2, 3), and (2n-1, 2n, 1). So yes, each checker is part of exactly three triples. Now, since there are n white checkers and n black checkers, each contributes to three triples. Let’s denote that. The total number of white checkers in all triples is 3n, and same for black. But each triple has three checkers, so total checkers across all triples is 3*2n = 6n. Which matches 3n white + 3n black = 6n. Now, each triple that's majority white (A) has at least two white checkers, and each majority black (B) has at least two black checkers. Let’s compute the total number of white checkers across all triples. For each A triple, it has either 2 or 3 white checkers. Similarly, each B triple has 0 or 1 white checkers. Let’s denote the number of triples with 3 white as A3 and those with 2 white as A2. Similarly, B1 for triples with 1 white and B0 for triples with 0 white. Then A = A3 + A2, B = B0 + B1. The total white checkers in all triples is 3A3 + 2A2 + 1B1 + 0B0 = 3A3 + 2A2 + B1. Similarly, total black checkers is 3B0 + 2B1 + 1A2 + 0A3. But since total white checkers across all triples is 3n, we have:3A3 + 2A2 + B1 = 3n.Similarly, total black checkers:3B0 + 2B1 + A2 = 3n.But since A = A3 + A2 and B = B0 + B1, and A + B = 2n.We need to relate A and B. Let me see. From the white checkers equation:3A3 + 2A2 + B1 = 3n.From the black checkers equation:3B0 + 2B1 + A2 = 3n.But B0 + B1 = B, and A3 + A2 = A. Let me try to express everything in terms of A and B.Let’s note that B0 = B - B1. Substitute into the black equation:3(B - B1) + 2B1 + A2 = 3n → 3B - 3B1 + 2B1 + A2 = 3n → 3B - B1 + A2 = 3n.Similarly, from the white equation:3A3 + 2A2 + B1 = 3n. But A3 = A - A2. Substitute:3(A - A2) + 2A2 + B1 = 3n → 3A - 3A2 + 2A2 + B1 = 3n → 3A - A2 + B1 = 3n.Now we have two equations:1. 3B - B1 + A2 = 3n2. 3A - A2 + B1 = 3nLet’s add these two equations:(3B - B1 + A2) + (3A - A2 + B1) = 6n → 3B + 3A = 6n → A + B = 2n. Which we already knew. So that doesn't help. Alternatively, subtract one equation from the other. Let's subtract equation 1 from equation 2:(3A - A2 + B1) - (3B - B1 + A2) = 0 → 3A - A2 + B1 - 3B + B1 - A2 = 0 → 3A - 3B - 2A2 + 2B1 = 0.Hmm. Not sure if helpful. Let’s try to express variables in terms of A and B. Let’s denote A2 as x, then A3 = A - x. Similarly, B1 = y, so B0 = B - y.From the white equation:3(A - x) + 2x + y = 3n → 3A - 3x + 2x + y = 3n → 3A - x + y = 3n.From the black equation:3(B - y) + 2y + x = 3n → 3B - 3y + 2y + x = 3n → 3B - y + x = 3n.So we have:1. 3A - x + y = 3n2. 3B + x - y = 3nLet’s add equations 1 and 2:3A - x + y + 3B + x - y = 6n → 3A + 3B = 6n → A + B = 2n. Again redundant.Subtract equation 2 from equation 1:(3A - x + y) - (3B + x - y) = 0 → 3A - x + y - 3B - x + y = 0 → 3A - 3B - 2x + 2y = 0 → 3(A - B) = 2(x - y).So 3(A - B) = 2(x - y). We need to relate this to our goal of proving A ≤ 3B. Let’s see. If we can find a relation between x and y that allows us to bound A in terms of B. Note that x and y are non-negative integers. x is the number of A triples with exactly 2 white checkers, and y is the number of B triples with exactly 1 white checker. From the equation 3(A - B) = 2(x - y). So, A - B = (2/3)(x - y). If x ≥ y, then A - B is non-negative, so A ≥ B. If x < y, then A < B. But we need to prove A ≤ 3B. Let’s manipulate the equation. A - B = (2/3)(x - y) → A = B + (2/3)(x - y).To prove A ≤ 3B, substitute:B + (2/3)(x - y) ≤ 3B → (2/3)(x - y) ≤ 2B → x - y ≤ 3B.But since x and y are counts, x ≤ A (since x is the number of A triples with 2 whites, which can't exceed A). Similarly, y ≥ 0. So x - y ≤ x ≤ A. But A = B + (2/3)(x - y). This seems circular. Maybe another approach. Alternatively, from the white equation: 3A - x + y = 3n. And from the black equation: 3B + x - y = 3n. So both equal 3n. Therefore:3A - x + y = 3B + x - y → 3A - 3B = 2x - 2y → 3(A - B) = 2(x - y). Same as before.Alternatively, let's express x and y in terms of A and B. Let’s solve the system:From 3A - x + y = 3n and 3B + x - y = 3n.Let’s call these equations (1) and (2). Adding them gives A + B = 2n, which is known.Let’s solve for x and y. From equation (1): y = 3n - 3A + x. From equation (2): x = 3n - 3B + y. Substitute y from equation (1) into equation (2):x = 3n - 3B + (3n - 3A + x) → x = 6n - 3B - 3A + x → 0 = 6n - 3B - 3A → 6n = 3A + 3B → 2n = A + B. Again, redundant.Not helpful. Maybe think about the total number of white checkers. Let’s recall that 3A3 + 2A2 + B1 = 3n. Also, A3 + A2 = A. Similarly, B1 ≤ B. So 3A3 + 2A2 = 3A - A2. So 3A - A2 + B1 = 3n. Therefore, B1 = 3n - 3A + A2. But B1 is non-negative, so 3n - 3A + A2 ≥ 0. Also, B1 ≤ B = 2n - A. Therefore:3n - 3A + A2 ≤ 2n - A → 3n - 3A + A2 ≤ 2n - A → n - 2A + A2 ≤ 0 → A2 ≤ 2A - n.But A2 is non-negative, so 2A - n ≥ 0 → A ≥ n/2. But A can vary. Not sure. Alternatively, considering that each white checker is in three triples, and each black checker is in three triples. Let’s think about the contribution of each checker. For a white checker, how many triples it contributes to that are majority white. Similarly for black. Alternatively, let's use double counting. Consider each white checker. It can be in triples that are majority white or majority black. Let’s denote for each white checker, the number of majority white triples it is part of. Similarly for black checkers. But this might get complicated. Let’s consider that each majority white triple has at least two white checkers. So the total number of "white contributions" to A triples is at least 2A. Similarly, each white checker is in three triples, so the total white contributions across all triples is 3n. Therefore:2A + 0*B ≤ 3n → 2A ≤ 3n → A ≤ 3n/2. Similarly, for black checkers: 2B ≤ 3n → B ≤ 3n/2. But since A + B = 2n, if A ≤ 3n/2, then B ≥ 2n - 3n/2 = n/2. Therefore, A ≤ 3n/2 = 3*(n/2) ≤ 3B since B ≥ n/2. Wait, this might be the key. Wait, if A ≤ 3n/2 and B ≥ n/2, then 3B ≥ 3*(n/2) = 3n/2 ≥ A. Therefore, A ≤ 3B. Is this correct? Let me check. Since A + B = 2n, B = 2n - A. So substituting into the inequality A ≤ 3B gives A ≤ 3(2n - A) → A ≤ 6n - 3A → 4A ≤ 6n → A ≤ (3/2)n. So if we can show that A ≤ 3n/2, then since B = 2n - A ≥ 2n - 3n/2 = n/2, therefore 3B ≥ 3*(n/2) = 3n/2 ≥ A. Therefore, A ≤ 3B. But how do we show that A ≤ 3n/2? Earlier, we considered that each majority white triple has at least two white checkers, so total white contributions to A triples is at least 2A. But each white checker is in three triples, so total white contributions is 3n. Therefore, 2A ≤ 3n → A ≤ 3n/2. Yes! That's it. Similarly, 2B ≤ 3n → B ≤ 3n/2. But since A + B = 2n, we can relate A and B. So since 2A ≤ 3n → A ≤ 3n/2. And since B = 2n - A, then substituting:A ≤ 3n/2 → 2n - B ≤ 3n/2 → -B ≤ -n/2 → B ≥ n/2. Therefore, 3B ≥ 3*(n/2) = 3n/2 ≥ A. Hence, A ≤ 3B. This seems to work. So the key idea is that each majority white triple requires at least two white checkers, and since each white checker can only be in three triples, the total number of such contributions (2A) cannot exceed the total available white contributions (3n). Therefore, A ≤ 3n/2. Similarly, B ≤ 3n/2. Then using the relation A + B = 2n, we substitute B = 2n - A into the inequality A ≤ 3B. Let me verify this with an example. Take n=3. So 3n/2 = 4.5. So A ≤ 4.5, meaning A ≤4. But in our earlier examples with n=3, A was 3. So it's within the bound. If somehow we can get A=4, then B=2, so 3B=6 ≥4. But in reality, in the case where A=4, B=2, does that hold? Let's try to construct such a case. Wait, with n=3, 3 W and 3 B. Let's try arranging checkers such that there are 4 majority white triples. Let me attempt: Arrange as WWWBWBB. Wait, but that's 7 checkers. n=3.5, which is invalid. n must be integer. Let me try n=4. 4W and 4B. Let's arrange as WWWWBBBB. Then as before, A=4, B=4. So A=4, B=4. So 3B=12 ≥4. Alternatively, arrange to maximize A. Suppose we arrange as WWWBBWWB. Let's count:Positions: W, W, W, B, B, W, W, B.Triples:1. W, W, W: W2. W, W, B: W3. W, B, B: B4. B, B, W: B5. B, W, W: W6. W, W, B: W7. W, B, B: B8. B, B, W: BSo A = triples 1,2,5,6: 4; B = triples 3,4,7,8:4. Still equal. Hmm. It seems challenging to get A > B. But according to the earlier logic, even if A=3n/2, which for n=4 is 6, but A cannot exceed 2n=8. Wait, for n=4, 3n/2=6. So A cannot exceed 6. But A + B=8, so if A=6, then B=2. Let's see if this is possible. Arrange 4W and 4B such that there are 6 majority white triples. Let's try:Arrange as WWWWBWBWBB. Wait, check count. Wait, 4W and 4B. Let's try a different approach. Let's place W's such that every other triple is W. Maybe WBWBWBWW. Wait:Positions: W, B, W, B, W, B, W, W.Triples:1. W, B, W: W (2W, 1B)2. B, W, B: B3. W, B, W: W4. B, W, B: B5. W, B, W: W6. B, W, W: W (2W, 1B)7. W, W, B: W8. W, B, W: WSo A = triples 1,3,5,6,7,8:6; B = triples 2,4:2. So A=6, B=2. Here, A=6 and B=2. Then 6 ≤ 3*2=6. So equality holds. Ah! So in this case, A=3B. Therefore, the bound is tight. Let me check this arrangement. Positions: W, B, W, B, W, B, W, W. Let's list all triples:1. W, B, W → W, B, W: two Ws → majority W.2. B, W, B → B, W, B: two Bs → majority B.3. W, B, W → W, B, W: two Ws → W.4. B, W, B → B, W, B: two Bs → B.5. W, B, W → W, B, W: two Ws → W.6. B, W, W → B, W, W: two Ws → W.7. W, W, B → W, W, B: two Ws → W.8. W, B, W → W, B, W: two Ws → W.So indeed, A=6 and B=2. So here, A=3B. Thus, the inequality A ≤3B is tight. So the key idea is that since each majority white triple requires at least two white checkers, the total number of white checkers across all triples is at least 2A. But since each white checker is counted three times (once in each triple it's part of), the total white checkers across all triples is exactly 3n. Hence, 2A ≤3n → A ≤3n/2. Similarly, 2B ≤3n → B ≤3n/2. Then, since A + B =2n, substituting B=2n -A into A ≤3B gives A ≤3(2n -A) → A ≤6n -3A →4A ≤6n →A ≤(3/2)n, which is the same as before. Thus, the inequality A ≤3B holds. Therefore, the proof follows from these inequalities. Final Answerboxed{A leqslant 3B}

✅:Given an equal number of black and white checkers arranged around a circle, we need to prove that the number of triples of consecutive checkers where the majority are white (denoted ( A )) is at most three times the number of triples where the majority are black (denoted ( B )), i.e., ( A leq 3B ).1. Total Triples and Majority Check: - There are ( 2n ) checkers ( ( n ) black and ( n ) white) arranged in a circle. - Each triple of consecutive checkers must be majority white or majority black since there are no ties (as each triple has three checkers, an odd number). - Therefore, ( A + B = 2n ).2. Key Inequalities: - Each majority white triple requires at least two white checkers, and each majority black triple requires at least two black checkers. - The total number of white checkers across all triples is ( 3n ) (each white checker is part of three triples). - Similarly, the total number of black checkers across all triples is ( 3n ).3. Bounding ( A ) and ( B ): - The total number of white checkers in all majority white triples is at least ( 2A ). - Since the total number of white checkers across all triples is ( 3n ), we have ( 2A leq 3n ), leading to ( A leq frac{3n}{2} ). - Similarly, for black checkers, ( 2B leq 3n ), leading to ( B leq frac{3n}{2} ).4. Relating ( A ) and ( B ): - Using ( A + B = 2n ), substitute ( B = 2n - A ) into the inequality ( A leq 3B ): [ A leq 3(2n - A) ] Simplifying this: [ A leq 6n - 3A implies 4A leq 6n implies A leq frac{3n}{2} ] - Since ( A leq frac{3n}{2} ) and ( B = 2n - A ), we get ( B geq frac{n}{2} ). Therefore, ( 3B geq frac{3n}{2} geq A ).Thus, we conclude that ( A leq 3B ).[boxed{A leqslant 3B}]

🔑:1. Let N be the number of black checkers and, as given, there are also N white checkers on the circle.2. Denote A as the number of triplets predominantly consisting of white checkers and B as that of predominantly black checkers. Additionally, note that each triplet consists of three consecutive checkers.3. Given a triplet: - A triplet predominantly black can have from 2 to 3 black checkers. - A triplet predominantly white can have from 1 to 3 white checkers. - But since we're considering each checkers being part of three triplets, we count each checker up to three times,4. Therefore, if a black checker is part of a triplet that is predominantly black, it is counted at most 3 times in these formations.5. Similarly, each white checker in predominantly white triplets is counted at most 3 times.6. Summarizing the total number of black and white checkers taking part in these triplets, we establish: [ N leq frac{3B + A}{3} ]7. This implies: [ 3N leq 3B + A ]8. As N is half of the total number of checkers: [ 2N = A + B implies A + B leq frac{2}{3}(3B + A) ]9. Simplifying further: [ 2N leq frac{6B + 2A}{3} ]10. Therefore it follows that: [ 3(A + B) leq 6B + 2A quad text{or} quad 3A + 3B leq 6B + 2A ]11. Rearrange the terms to isolate A: [ 3A + 3B - 2A - 6B leq 0 ]12. This simplifies to: [ A leq 3B ]13. Conclusion: [ boxed{A leq 3B} ]