🧠:Alright, let me try to figure out this problem step by step. So, the question is about an ellipse given by the equation x²/9 + y²/4 = 1. The foci of this ellipse are F₁ and F₂. There's a point P on the ellipse such that the ratio of the distances from P to each focus is 2:1, meaning |PF₁| : |PF₂| = 2:1. We need to find the area of triangle PF₁F₂.First, I should recall some basic properties of an ellipse. The standard equation of an ellipse is x²/a² + y²/b² = 1, where a is the semi-major axis and b is the semi-minor axis. In this case, comparing with the given equation, a² = 9 and b² = 4, so a = 3 and b = 2. Now, the foci of an ellipse are located at (±c, 0) where c is the distance from the center to each focus, and c² = a² - b². Let me compute c:c² = a² - b² = 9 - 4 = 25? Wait, no, 9 - 4 is 5. So c = √5. Therefore, the foci F₁ and F₂ are at (-√5, 0) and (√5, 0), respectively.So, F₁ is (-√5, 0), F₂ is (√5, 0). The point P is somewhere on the ellipse, and the ratio of its distances to F₁ and F₂ is 2:1. So |PF₁| = 2|PF₂|. Our goal is to find the area of triangle PF₁F₂.Hmm, okay. Let's think about how to approach this. Since we know the coordinates of F₁ and F₂, if we can find the coordinates of point P, we can compute the area of the triangle using the coordinates. Alternatively, maybe we can use some geometric properties or vector methods. Let's explore.First, let's denote the coordinates of point P as (x, y). Since P lies on the ellipse, it must satisfy the ellipse equation: x²/9 + y²/4 = 1.Additionally, the distance conditions must hold. The distance from P to F₁ is |PF₁| = √[(x + √5)² + y²], and the distance to F₂ is |PF₂| = √[(x - √5)² + y²]. The ratio of these distances is 2:1, so:√[(x + √5)² + y²] / √[(x - √5)² + y²] = 2/1Let me square both sides to eliminate the square roots:[(x + √5)² + y²] / [(x - √5)² + y²] = 4/1Cross-multiplying:(x + √5)² + y² = 4[(x - √5)² + y²]Let me expand both sides.Left side: (x² + 2x√5 + 5) + y²Right side: 4[(x² - 2x√5 + 5) + y²] = 4x² - 8x√5 + 20 + 4y²So, setting left equal to right:x² + 2x√5 + 5 + y² = 4x² - 8x√5 + 20 + 4y²Bring all terms to the left side:x² + 2x√5 + 5 + y² - 4x² + 8x√5 - 20 - 4y² = 0Combine like terms:(1 - 4)x² + (2x√5 + 8x√5) + (5 - 20) + (1 - 4)y² = 0Calculating each term:-3x² + 10x√5 - 15 - 3y² = 0Divide both sides by -3:x² - (10/√5)x + 5 + y² = 0Wait, let's check the calculation again. Let me re-express the combination:x² - 4x² = -3x²2x√5 + 8x√5 = 10x√55 - 20 = -15y² - 4y² = -3y²So indeed, the equation becomes:-3x² + 10x√5 - 15 - 3y² = 0Dividing both sides by -3 gives:x² - (10/√5)x + 5 + y² = 0Simplify 10/√5: 10/√5 = 2√5. Because 10/√5 = (10√5)/5 = 2√5.So, the equation is:x² - 2√5 x + 5 + y² = 0But we also know that the point P lies on the ellipse, which gives x²/9 + y²/4 = 1. Let me write both equations:1) x² - 2√5 x + 5 + y² = 02) x²/9 + y²/4 = 1Perhaps we can use equation 2 to express y² in terms of x², then substitute into equation 1.From equation 2: y² = 4(1 - x²/9) = 4 - (4x²)/9Substitute into equation 1:x² - 2√5 x + 5 + [4 - (4x²)/9] = 0Simplify:x² - 2√5 x + 5 + 4 - (4x²)/9 = 0Combine like terms:x² - (4x²)/9 - 2√5 x + 9 = 0Compute x² - (4x²)/9: (9x² - 4x²)/9 = (5x²)/9So:(5x²)/9 - 2√5 x + 9 = 0Multiply both sides by 9 to eliminate denominator:5x² - 18√5 x + 81 = 0So now we have a quadratic equation in x:5x² - 18√5 x + 81 = 0Let me try to solve this quadratic for x.Quadratic formula: x = [18√5 ± √( (18√5)^2 - 4*5*81 ) ] / (2*5)Compute discriminant D:(18√5)^2 - 4*5*81First, (18√5)^2 = 324 * 5 = 16204*5*81 = 20*81 = 1620Therefore, D = 1620 - 1620 = 0Wait, discriminant is zero? That means there's only one real root, so x = [18√5]/(10) = (9√5)/5Hmm, that's interesting. So, x = (9√5)/5.But since the discriminant is zero, there's only one solution for x? That suggests that there is only one such point P on the ellipse where the ratio is 2:1. Wait, but an ellipse is symmetric, so maybe there are two points symmetric over the x-axis? Because if x is fixed, y could be positive or negative. Let me check.Given that x = (9√5)/5, then substitute back into equation 2 to find y.From equation 2: x²/9 + y²/4 = 1Compute x²:x = (9√5)/5, so x² = (81*5)/25 = 405/25 = 81/5Thus, x²/9 = (81/5)/9 = 9/5Therefore, y²/4 = 1 - 9/5 = (5 - 9)/5 = (-4)/5. Wait, that can't be. y²/4 can't be negative.Hold on, this is a problem. y²/4 = -4/5, which would imply y² is negative, which is impossible. That suggests there's a mistake in my calculations.Wait, let's check the steps again.We started with the ratio |PF₁| : |PF₂| = 2:1, leading to the equation √[(x + √5)² + y²] / √[(x - √5)² + y²] = 2, then squared both sides to get [(x + √5)^2 + y²] = 4[(x - √5)^2 + y²], which expanded to x² + 2x√5 +5 + y² = 4x² - 8x√5 +20 +4y².Then moving everything to the left: x² +2x√5 +5 + y² -4x² +8x√5 -20 -4y² =0. Combine terms: (-3x²) +10x√5 -15 -3y²=0. Divide by -3: x² - (10/√5)x +5 + y²=0. Then simplified 10/√5 to 2√5, so x² -2√5 x +5 + y²=0.Then substituted y² from the ellipse equation: y²=4 - (4x²)/9.Substituted into equation 1: x² -2√5 x +5 +4 - (4x²)/9 =0. Then combined terms:x² - (4x²)/9 = (5x²)/9So equation becomes (5x²)/9 -2√5 x +9=0. Multiply by 9: 5x² -18√5 x +81=0. Then discriminant D=(18√5)^2 -4*5*81=1620-1620=0.Hmm, so x=(18√5)/(2*5)= (9√5)/5.But substituting back into ellipse equation gives x²/9 + y²/4=1. x²= (81*5)/25=81/5. Therefore, x²/9= 81/(5*9)=9/5. Then 9/5 + y²/4=1. So y²/4=1 -9/5= -4/5. Which is impossible.This suggests that there is no real solution, but the problem states that such a point P exists. Therefore, there must be an error in my reasoning.Wait a minute, perhaps I made a mistake in the sign when moving terms. Let me check that again.Original equation after cross-multiplying:(x + √5)^2 + y² = 4[(x - √5)^2 + y²]Expanding left side: x² + 2x√5 +5 + y²Right side: 4x² -8x√5 +20 +4y²Moving everything to the left:x² +2x√5 +5 + y² -4x² +8x√5 -20 -4y²=0Wait, here when moving terms, the right side is subtracted, so the equation is left side - right side =0. So:Left side - right side = (x² +2x√5 +5 + y²) - (4x² -8x√5 +20 +4y²) =0Which becomes x² +2x√5 +5 + y² -4x² +8x√5 -20 -4y²=0Combine like terms:x² -4x²= -3x²2x√5 +8x√5=10x√55 -20= -15y² -4y²= -3y²Thus, equation: -3x² +10x√5 -15 -3y²=0Divide by -3:x² - (10/√5)x +5 + y²=0So that part is correct. Then substituting y²=4 - (4x²)/9 into this equation:x² - (10/√5)x +5 +4 - (4x²)/9=0Wait, 5 +4 is 9, so:x² - (10/√5)x +9 - (4x²)/9=0Combine x² terms:(1 - 4/9)x² = (5/9)x²Thus:(5/9)x² - (10/√5)x +9=0Multiply by 9:5x² - 90/√5 x +81=0Wait, hold on! Previously, I thought that 10/√5 is 2√5, which is correct. But here, in the term after substitution, when I substitute back into equation, the coefficient of x is -(10/√5). Wait, but after substitution, the equation is:(5/9)x² - (10/√5)x +9=0But when I multiplied by 9, I should have 5x² - (90/√5)x +81=0Wait, this is different from before. Wait, earlier I had written:After substitution:(5x²)/9 - 2√5 x + 9 = 0But that seems to be a mistake.Wait, let's clarify:Original substitution step:From equation 1, after substitution:x² -2√5 x +5 + y²=0But y²=4 - (4x²)/9So substituting:x² -2√5 x +5 +4 - (4x²)/9=0Simplify:x² - (4x²)/9 -2√5 x +9=0Which is (1 - 4/9)x² -2√5 x +9=0(5/9)x² -2√5 x +9=0Multiply by 9:5x² -18√5 x +81=0Ah, here is the confusion. The term with x is -2√5 x, so when multiplied by 9, becomes -18√5 x.Wait, but in my initial calculation, I think I had - (10/√5)x, but in reality, after substituting, the coefficient is -2√5 x.Wait, but where did the 10/√5 come from? Wait, let's retrace.Original equation after substitution was:x² -2√5 x +5 + y²=0But substituting y²=4 - (4x²)/9:x² -2√5 x +5 +4 - (4x²)/9=0Which is x² - (4x²)/9 -2√5 x +9=0So, (5/9)x² -2√5 x +9=0Multiply by 9:5x² -18√5 x +81=0Yes, this is correct. So the quadratic equation is 5x² -18√5 x +81=0But then when computing the discriminant:D = ( -18√5 )² -4*5*81 = (324*5) - 20*81 = 1620 - 1620 = 0So discriminant is zero, which leads to x=(18√5)/(10)= (9√5)/5But when substituting back into the ellipse equation, x=(9√5)/5, x²=81*5/25=81/5, so x²/9=9/5, then y²=4*(1 -9/5)=4*(-4/5)= -16/5. Wait, that's impossible.This suggests that there is a contradiction. How come?But the problem states that such a point P exists. So maybe there is a mistake in the process?Alternatively, perhaps the ratio is different. Wait, the problem says |PF1| : |PF2| = 2:1, so PF1 is twice PF2. But in my calculation, squaring the ratio leads to the equation which results in no real solution. Hmm.Alternatively, maybe I have a mistake in the direction of the ratio. Wait, if the ratio is |PF1| : |PF2| = 2:1, then PF1 = 2 PF2. But when I set up the equation, I wrote:√[(x + √5)² + y²] / √[(x - √5)² + y²] = 2/1Which is correct. Because PF1 / PF2 = 2/1.But solving this leads to an impossible y². So that suggests that maybe such a point does not exist? But the problem says "point P lies on the ellipse", so it must exist.Alternatively, maybe I made a mistake in calculating c, the distance to the foci.Wait, c² = a² - b². For an ellipse, the major axis is longer than the minor. In the given equation, x²/9 + y²/4 =1, so since 9>4, the major axis is along the x-axis, with a=3, b=2. Then c²=9 -4=5, so c=√5. So foci at (±√5, 0). That's correct.Alternatively, maybe I made a mistake in setting up the equation. Let me check again.Starting over:|PF1| = 2 |PF2|So, sqrt[(x + √5)^2 + y²] = 2 sqrt[(x - √5)^2 + y²]Square both sides:(x + √5)^2 + y² = 4[(x - √5)^2 + y²]Expand left side: x² + 2x√5 +5 + y²Right side: 4x² - 8x√5 +20 +4y²Bring all terms to left:x² +2x√5 +5 + y² -4x² +8x√5 -20 -4y²=0Simplify:-3x² +10x√5 -15 -3y²=0Divide by -3:x² - (10/√5)x +5 + y²=0As before. Then 10/√5=2√5, so x² -2√5 x +5 + y²=0Then, using ellipse equation to substitute y²=4 - (4x²)/9So:x² -2√5 x +5 +4 - (4x²)/9=0Which is:x² - (4x²)/9 -2√5 x +9=0Which is:(5/9)x² -2√5 x +9=0Multiply by 9:5x² -18√5 x +81=0Discriminant D=(18√5)^2 -4*5*81=1620 -1620=0, so x=(18√5)/10=(9√5)/5≈ 9*2.236/5≈20.124/5≈4.025But the ellipse extends from x=-3 to x=3. So x≈4.025 is outside the ellipse. Hence, that's impossible. Therefore, such a point P cannot exist on the ellipse. But the problem states that it exists. Contradiction.Wait, so maybe the problem has a typo, or perhaps the ratio is different? Alternatively, maybe the ratio is |PF2| : |PF1| =2:1 instead of |PF1|:|PF2|=2:1. Let's check.If the ratio was 1:2 instead, so |PF1| : |PF2|=1:2, then the equation would be sqrt[(x + √5)^2 + y²] / sqrt[(x - √5)^2 + y²] =1/2. Then squaring gives [(x + √5)^2 + y²]/[(x - √5)^2 + y²] =1/4. Then cross-multiplying: 4[(x + √5)^2 + y²] = (x - √5)^2 + y².Expanding left:4x² +8x√5 +20 +4y²Right:x² -2x√5 +5 + y²Bring all terms to left:4x² +8x√5 +20 +4y² -x² +2x√5 -5 -y²=0Simplify:3x² +10x√5 +15 +3y²=0Divide by 3:x² + (10/√5)x +5 + y²=0Again, 10/√5=2√5:x² +2√5 x +5 + y²=0Then substitute y²=4 -4x²/9:x² +2√5 x +5 +4 -4x²/9=0Combine terms:x² -4x²/9 +2√5 x +9=0(5x²)/9 +2√5 x +9=0Multiply by 9:5x² +18√5 x +81=0Discriminant D=(18√5)^2 -4*5*81=1620 -1620=0. So again x=(-18√5)/10=(-9√5)/5≈-4.025, which is also outside the ellipse. Hence, same problem.Therefore, regardless of the ratio direction, the solutions for x are outside the ellipse. Which suggests that there is no such point P on the ellipse with the given ratio. But the problem states that such a point exists, so this is a contradiction.Alternatively, maybe the ratio is possible but my approach is wrong.Wait, another thought: in an ellipse, the sum of distances from any point to the foci is constant and equal to 2a. In this case, 2a=6. So |PF1| + |PF2|=6. Given that |PF1| : |PF2|=2:1, let me set |PF1|=2k, |PF2|=k. Then 2k +k=6 => 3k=6 =>k=2. Hence, |PF1|=4, |PF2|=2.So the distances are 4 and 2. Therefore, maybe instead of using the ratio directly, I can use the sum and the ratio to set up equations.Let me try this approach.Given |PF1| + |PF2| =6 and |PF1|=2|PF2|. Let |PF2|=d, then |PF1|=2d. So 2d +d=3d=6 => d=2. Therefore, |PF1|=4, |PF2|=2.So now, we can write:√[(x + √5)^2 + y²] =4and√[(x - √5)^2 + y²] =2Now, squaring both equations:First equation: (x + √5)^2 + y²=16Second equation: (x - √5)^2 + y²=4Subtract the second equation from the first:[(x + √5)^2 + y²] - [(x - √5)^2 + y²] =16 -4=12Expand the left side:[x² +2x√5 +5 + y²] - [x² -2x√5 +5 + y²] =12Simplify:4x√5=12 => x√5=3 => x=3/√5= (3√5)/5≈1.3416Now, substitute x back into one of the equations to find y. Let's use the second equation:(x - √5)^2 + y²=4Compute (x - √5)^2:x=3√5/5, so x - √5= (3√5/5 -5√5/5)= (-2√5)/5Thus, (x - √5)^2=(4*5)/25=20/25=4/5Hence:4/5 + y²=4 => y²=4 -4/5=16/5 => y=±4/√5=±(4√5)/5≈±1.788Therefore, coordinates of P are (3√5/5, ±4√5/5)So, there are two possible points P, symmetric over the x-axis. The area of triangle PF₁F₂ would be the same for both.Now, let's compute the area. The coordinates of F₁ and F₂ are (-√5, 0) and (√5, 0), respectively. Point P is (3√5/5, 4√5/5) or (3√5/5, -4√5/5). The area can be calculated using the formula for the area of a triangle given coordinates of vertices.The formula is 1/2 | (x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)) |, but since two of the points are on the x-axis, perhaps there is a simpler way.Alternatively, since F₁ and F₂ are on the x-axis, the base of the triangle is the distance between F₁ and F₂, which is 2√5. The height is the y-coordinate of point P, which is 4√5/5. Therefore, the area is (1/2)*base*height=(1/2)*(2√5)*(4√5/5)= (√5)*(4√5/5)= (4*5)/5=4.Wait, that seems straightforward. Let me verify.Base is F₁F₂ distance: from -√5 to √5 on x-axis, so distance is 2√5. Height is the vertical distance from P to the x-axis, which is |y| of P. Since P has y=±4√5/5, the height is 4√5/5.Therefore, area=1/2 * base * height=1/2 *2√5 *4√5/5= (√5 *4√5)/5= (4*5)/5=4. So area is 4.Alternatively, using coordinates:Coordinates of F₁ (-√5,0), F₂ (√5,0), P (3√5/5,4√5/5)Using the formula for area:1/2 | (x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)) |Plugging in:x₁=-√5, y₁=0; x₂=√5, y₂=0; x₃=3√5/5, y₃=4√5/5Area=1/2 | (-√5)(0 -4√5/5) + √5(4√5/5 -0) +3√5/5(0 -0) |Calculate each term:First term: -√5*(-4√5/5)= (√5*4√5)/5= (4*5)/5=4Second term: √5*(4√5/5)= (4*5)/5=4Third term:0Sum:4+4=8Area=1/2 * |8|=4. Correct.Therefore, the area is 4.So the issue earlier was that I tried to set up the ratio directly by squaring, which led to an inconsistency, but by using the property of the ellipse that the sum of distances is constant, I could set up two equations and solve them, leading to a valid solution. Therefore, the key was recognizing that the sum of distances is 2a=6, and combining that with the ratio given to find the individual distances, then solving for coordinates of P.The initial approach might have gone wrong because I didn't consider the sum of distances, which is a fundamental property of an ellipse. Without incorporating that, the equations derived from the ratio alone led to an impossible solution. Hence, it was crucial to use both the ratio and the sum of distances to properly determine the coordinates of point P.In summary, by combining the ratio condition with the constant sum of distances from any point on the ellipse to the foci, we can find valid coordinates for P, and then compute the area of the triangle using basic coordinate geometry.

✅:To solve the problem, we start by noting the given ellipse equation (frac{x^2}{9} + frac{y^2}{4} = 1), which has semi-major axis (a = 3) and semi-minor axis (b = 2). The distance to each focus is (c = sqrt{a^2 - b^2} = sqrt{5}), so the foci are at (F_1 = (-sqrt{5}, 0)) and (F_2 = (sqrt{5}, 0)).Given the ratio (|PF_1| : |PF_2| = 2:1), we denote (|PF_2| = d) and (|PF_1| = 2d). Using the ellipse property that the sum of distances from any point to the foci is (2a = 6), we find:[2d + d = 6 implies d = 2 implies |PF_1| = 4, quad |PF_2| = 2]We set up the equations for these distances:[sqrt{(x + sqrt{5})^2 + y^2} = 4 quad text{and} quad sqrt{(x - sqrt{5})^2 + y^2} = 2]Squaring both equations:[(x + sqrt{5})^2 + y^2 = 16 quad text{and} quad (x - sqrt{5})^2 + y^2 = 4]Subtracting the second equation from the first:[(x + sqrt{5})^2 - (x - sqrt{5})^2 = 12 implies 4xsqrt{5} = 12 implies x = frac{3}{sqrt{5}} = frac{3sqrt{5}}{5}]Substituting (x) back into the second equation to find (y):[left(frac{3sqrt{5}}{5} - sqrt{5}right)^2 + y^2 = 4 implies left(-frac{2sqrt{5}}{5}right)^2 + y^2 = 4 implies frac{20}{25} + y^2 = 4 implies y^2 = frac{16}{5} implies y = pm frac{4sqrt{5}}{5}]Thus, point (P) has coordinates (left(frac{3sqrt{5}}{5}, pm frac{4sqrt{5}}{5}right)).The area of triangle (PF_1F_2) is calculated using the base (F_1F_2 = 2sqrt{5}) and the height from (P) to the x-axis, which is (|y| = frac{4sqrt{5}}{5}):[text{Area} = frac{1}{2} times text{base} times text{height} = frac{1}{2} times 2sqrt{5} times frac{4sqrt{5}}{5} = frac{1}{2} times frac{8 times 5}{5} = 4]The area of triangle (PF_1F_2) is (boxed{4}).

🔑:We are given the ellipse (frac{x^{2}}{9}+frac{y^{2}}{4}=1). The foci (F_{1}) and (F_{2}) of this ellipse are provided along with a point (P) on the ellipse satisfying (|P F_{1}| : |P F_{2}| = 2 : 1).1. Calculate the Major Axis Length, (2a): - The standard form of an ellipse is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1). - Here, (a = 3) and (b = 2). - The length of the major axis is (2a = 2 times 3 = 6).2. Define the Condition on Distances, (|P F_{1}| + |P F_{2}|): - By the definition of an ellipse, the sum of distances from any point on the ellipse to the two foci is constant and equal to (2a). - Hence, (|P F_{1}| + |P F_{2}| = 6).3. Determine Individual Distances (|P F_{1}|) and (|P F_{2}|): - Given the ratio (|P F_{1}| : |P F_{2}| = 2 : 1), let (|P F_{2}| = x). - Thus, (|P F_{1}| = 2x). - From (|P F_{1}| + |P F_{2}| = 6), we have: [ 2x + x = 6 quad implies quad 3x = 6 quad implies quad x = 2. ] - Therefore, (|P F_{1}| = 2x = 2 times 2 = 4) and (|P F_{2}| = x = 2).4. Calculate the Distance Between the Foci, (2c): - The distance between the foci of an ellipse is (2c), where (c = sqrt{a^2 - b^2}). - Thus, (c = sqrt{3^2 - 2^2} = sqrt{9 - 4} = sqrt{5}). - Therefore, (2c = 2 times sqrt{5} = 2sqrt{5}).5. Verification if (triangle P F_{1} F_{2}) is a Right Triangle: - Using the distances found (|P F_{1}| = 4), (|P F_{2}| = 2), and (|F_{1} F_{2}| = 2sqrt{5}), check if: [ |P F_{1}|^2 + |P F_{2}|^2 = |F_{1} F_{2}|^2 ] - Substitute the values: [ 4^2 + 2^2 = (2sqrt{5})^2. ] [ 16 + 4 = 20. ] [ 20 = 20. ] - Thus, (triangle P F_{1} F_{2}) is a right triangle.6. Calculate the Area of the Right Triangle (triangle P F_{1} F_{2}): - The area (A) of a right triangle is given by: [ A = frac{1}{2} times text{base} times text{height}. ] - Here, (A = frac{1}{2} times |P F_{1}| times |P F_{2}|): [ A = frac{1}{2} times 4 times 2 = 4. ] Conclusion:[boxed{4}]

🧠:Okay, let me try to figure out this problem. The question is: Will the quotient and remainder change if the dividend and divisor are increased by 3 times? Hmm, so if I have a division problem, say, dividend divided by divisor gives a quotient and a remainder. Then, if I multiply both the dividend and the divisor by 3, does the quotient stay the same? What about the remainder?First, I should recall the division algorithm. For any integers a (dividend) and b (divisor), there exist unique integers q (quotient) and r (remainder) such that a = bq + r, where 0 ≤ r < b. So, the original equation is a = bq + r.Now, if we multiply both a and b by 3, the new dividend is 3a and the new divisor is 3b. Let's see what happens when we divide 3a by 3b. Let's denote the new quotient as q' and the new remainder as r'. According to the division algorithm, 3a = 3b * q' + r', where 0 ≤ r' < 3b.But wait, the original equation was a = bq + r. If I substitute a into the new equation, it becomes 3*(bq + r) = 3b * q' + r'. Let's expand the left side: 3bq + 3r = 3bq' + r'.Hmm, so comparing both sides, maybe we can factor out 3b from both terms. Let's rearrange the equation: 3bq - 3bq' = r' - 3r. That simplifies to 3b(q - q') = r' - 3r. Now, this equation tells us that the left side is a multiple of 3b, and the right side is r' - 3r. But since r' is a remainder when dividing by 3b, it must be less than 3b. Similarly, r is less than b. So, 3r is less than 3b. Therefore, r' - 3r must be between -3r and r' - 0. Since r' < 3b and 3r < 3b, the difference r' - 3r can be negative or positive but the absolute value can't be too large.But the left side is 3b times (q - q'), which has to equal this difference. The only way this can happen is if both sides are zero. Because if 3b(q - q') = r' - 3r, and the right side is less than 3b in absolute value (since r' < 3b and 3r < 3b), then the left side must be a multiple of 3b, but the right side is less than 3b. The only multiple of 3b that is less than 3b in absolute value is zero. Therefore, 3b(q - q') = 0 and r' - 3r = 0. Since b is not zero (as a divisor), then q - q' = 0, which implies q = q'. And then r' = 3r.Wait, but if r' = 3r, then we need to check if this new remainder r' satisfies 0 ≤ r' < 3b. Since originally 0 ≤ r < b, multiplying by 3 gives 0 ≤ 3r < 3b. So 3r is indeed a valid remainder for the new division problem. Therefore, the quotient q remains the same, and the remainder becomes 3 times the original remainder.But let me test this with an example to make sure. Let's take a simple case. Suppose the original dividend is 10 and the divisor is 3. Then 10 divided by 3 is 3 with a remainder of 1, because 3*3 +1 =10.Now, multiply both by 3: dividend becomes 30, divisor becomes 9. Dividing 30 by 9: 9*3=27, remainder 3. So quotient is still 3, remainder is 3, which is 3 times the original remainder of 1. That works.Another example: original dividend 7, divisor 2. Quotient 3, remainder 1. Multiply both by 3: dividend 21, divisor 6. 21 divided by 6 is 3 with remainder 3. Again, quotient same, remainder tripled.But wait, what if the original remainder is zero? Let's try dividend 6, divisor 3. Quotient 2, remainder 0. Multiply both by 3: 18 divided by 9 is 2, remainder 0. So quotient same, remainder 0, which is 3*0. Still holds.Another test case where the remainder might not be a multiple of 3. Wait, no, since original remainder is r, which is an integer between 0 and b-1. Then 3r is an integer, and since r < b, 3r < 3b, so it fits in the new remainder's range. But let me check a case where maybe 3r is larger than the original divisor. Wait, no, because the new divisor is 3b, so 3r < 3b, so it's still okay.Wait, but suppose original remainder is 2, divisor is 3. Then 3r = 6. New divisor is 9, so 6 is less than 9. So remainder is 6. For example, original dividend 5, divisor 3: 5 = 3*1 + 2. Multiply by 3: 15 divided by 9 is 1, remainder 6. Correct.But what if the original remainder multiplied by 3 is equal to the new divisor? For example, original dividend 4, divisor 3. Then quotient is 1, remainder 1. Multiply by 3: dividend 12, divisor 9. 12 divided by 9 is 1 with remainder 3. But 3 is less than 9, so okay. 3 is 3*1.Wait, if original remainder is equal to divisor minus 1. Let's say original dividend is 8, divisor 5. Then 8 = 5*1 + 3. Multiply by 3: 24 divided by 15 is 1, remainder 9. Which is 3*3=9. 9 is less than 15. Correct.But wait, another case: original dividend 14, divisor 5. 14 ÷5 is 2, remainder 4. Multiply by 3: 42 ÷15. 15*2=30, 42-30=12. So remainder 12, which is 3*4=12. Correct. So in all these cases, the quotient remains the same, and the remainder is tripled.Therefore, the answer is that the quotient remains the same, and the remainder is also multiplied by 3. Therefore, both the quotient and remainder do change? Wait, no. Wait, the quotient remains the same, so it doesn't change. The remainder does change because it's multiplied by 3.Wait, the question is: Will the quotient and remainder change? So quotient stays the same, remainder changes. So the answer is: The quotient will not change, but the remainder will increase by 3 times. Therefore, the quotient remains the same, but the remainder does change.Wait, but the user is asking if both the quotient and remainder will change. So in the original problem, when both dividend and divisor are multiplied by 3, the quotient remains the same, the remainder becomes 3 times as before. So quotient does not change, remainder does change.Therefore, the answer is: The quotient remains the same, but the remainder is multiplied by 3, so they do not both change; only the remainder changes.But let me check another case where maybe the remainder after tripling could be equal to the new divisor. Wait, if original remainder is b -1, then 3*(b -1) = 3b -3. Which is less than 3b, so it's a valid remainder. But 3b -3 is still less than 3b. So even in that case, the remainder is valid.Another edge case: original remainder is 0. Then new remainder is 0, which is same as 3*0. So if original remainder was zero, then new remainder is zero. So in that case, neither quotient nor remainder changes? Wait, no, quotient stays the same in all cases. If original remainder was zero, then new remainder is zero. So in that case, both quotient and remainder remain the same. But in other cases, when remainder was non-zero, the remainder changes.Therefore, the answer is: If the original remainder is zero (i.e., the original division has no remainder), then multiplying both dividend and divisor by 3 will not change the quotient or the remainder (both stay the same). However, if the original remainder is not zero, then the quotient remains the same, but the remainder becomes three times the original remainder.Wait, but in the problem statement, it's a general question: does the quotient and remainder change when both are increased by 3 times. The answer would depend on whether the original remainder is zero or not. If the original division had a remainder, then the remainder changes. If it didn't, then neither changes.But perhaps the question is intended to be in the general case, not specifically when the remainder is zero. So in general, when you multiply both dividend and divisor by 3, the quotient stays the same, and the remainder is multiplied by 3. Therefore, the quotient does not change, but the remainder does. Hence, the answer is that the quotient remains unchanged, but the remainder is tripled.But wait, let me formalize this. Let’s denote original division: a = bq + r, 0 ≤ r < b. Multiply both a and b by 3: 3a = 3bq + 3r. Now, since 0 ≤ r < b, multiplying by 3: 0 ≤ 3r < 3b. So 3r is a valid remainder for the division of 3a by 3b, with the same quotient q. Therefore, the quotient q remains the same, and the remainder becomes 3r.Therefore, the quotient does not change, but the remainder does. So the answer is: The quotient remains the same, but the remainder is multiplied by 3, so they do not both change. Only the remainder changes.But in the case when r = 0, then remainder remains 0, so in that specific case, neither changes. But the general answer would be that the quotient doesn't change, remainder does. Therefore, to answer the question "Will the quotient and remainder change...", the answer is that the quotient will not change, but the remainder will. Therefore, they don't both change; only the remainder does.But the question is phrased as "Will the quotient and remainder change...". So if either changes, then the answer is yes. But in this case, the quotient does not change, but the remainder does. So strictly speaking, the quotient does not change, so part of them (the remainder) changes. Therefore, the answer is: The quotient will not change, but the remainder will be three times the original. Therefore, the remainder changes, but the quotient remains the same.Hence, the final answer is that the quotient stays the same, and the remainder increases threefold. Therefore, the quotient does not change, but the remainder does.

✅:When both the dividend and divisor are multiplied by 3, the quotient remains unchanged, while the remainder is tripled. Here's the reasoning:1. Original Division: For integers ( a ) (dividend) and ( b ) (divisor), there exist unique integers ( q ) (quotient) and ( r ) (remainder) such that: [ a = bq + r quad text{where } 0 leq r < b. ]2. Scaled Division: Multiply both ( a ) and ( b ) by 3: [ 3a = 3b cdot q' + r' quad text{where } 0 leq r' < 3b. ]3. Substitute ( a ): Substitute ( a = bq + r ) into the scaled equation: [ 3(bq + r) = 3b cdot q' + r' implies 3bq + 3r = 3bq' + r'. ]4. Equate Terms: For equality to hold: - ( 3bq = 3bq' implies q = q' ) (quotient remains the same). - ( r' = 3r ) (remainder triples). 5. Validity: Since ( 0 leq r < b ), multiplying by 3 gives ( 0 leq 3r < 3b ), ensuring ( r' = 3r ) is a valid remainder for the scaled divisor ( 3b ).Examples: - Original: ( 10 div 3 = 3 ) (quotient), remainder ( 1 ). Scaled: ( 30 div 9 = 3 ) (same quotient), remainder ( 3 ). - Original: ( 7 div 2 = 3 ), remainder ( 1 ). Scaled: ( 21 div 6 = 3 ), remainder ( 3 ).Conclusion: - Quotient: Unchanged. - Remainder: Tripled. Thus, only the remainder changes. Final Answer: The quotient remains the same, but the remainder becomes three times larger. boxed{text{The quotient stays unchanged; the remainder triples.}}

🔑:1. Let's denote the dividend by ( a ) and the divisor by ( b ). The division of ( a ) by ( b ) can be expressed using the division algorithm: [ a = bq + r ] where ( q ) is the quotient and ( r ) is the remainder, with ( 0 leq r < b ).2. Now, if we increase both the dividend and the divisor by 3 times, the new dividend becomes ( 3a ) and the new divisor becomes ( 3b ). We need to investigate how this affects the quotient and the remainder.3. Let's perform the division of ( 3a ) by ( 3b ) similarly: [ 3a = (3b)q' + r' ] where ( q' ) is the new quotient and ( r' ) is the new remainder, with ( 0 leq r' < 3b ).4. Notice that: [ 3a = 3(bq + r) = 3bq + 3r ]5. Comparing the expressions ( 3a = 3bq + 3r ) and ( 3a = (3b)q' + r' ), we can see that: [ q' = q quad text{(because the quotient of 3q by 3 is q)} ] [ r' = 3r ]Hence, the quotient remains the same (( q )) when both the dividend and the divisor are increased threefold. The remainder will be three times the original remainder. ConclusionThus, the quotient will not change, and the remainder will be increased threefold if the original remainder is non-zero. Specifically, if the original remainder ( r ) was zero, the new remainder ( r' ) stays zero. Therefore, the answer is:[boxed{text{The quotient does not change, the remainder changes if it was non-zero.}}]

🧠:Okay, so I need to figure out the maximum number of points that can be placed in a plane and in space such that no triangle formed by any three points has an obtuse angle. The problem also mentions that no three points can be collinear, which makes sense because otherwise, you could have infinitely many points on a line, but they wouldn't form triangles. First, let me tackle part a) which is in the plane. Let's start with small numbers and see if I can find a pattern or maybe a limit. For 3 points, it's trivial—they form a triangle. If the triangle isn't obtuse, then it's either acute or right-angled. But the question says "none of the triangles formed by these points have an obtuse angle," so right angles might be allowed? Wait, the problem says "none of the triangles formed by these points have an obtuse angle." So right angles are okay? Hmm, the problem doesn't specify excluding right angles, only obtuse. So triangles can be acute or right-angled. That's important.Wait, but actually, in some definitions, a right-angled triangle is considered a type of oblique triangle, distinct from acute and obtuse. But in any case, the problem mentions "none of the triangles... have an obtuse angle," so right angles are permitted. So triangles can be acute or right-angled, but not obtuse.So, starting with 3 points. If I place them in a triangle, as long as it's acute or right-angled, that's fine. But the key is when adding more points. Let's see.For four points. Let me think about a square. In a square, the diagonals are longer than the sides. If I connect all four points, the triangles formed by two adjacent sides and a diagonal would have a right angle. For example, in a square ABCD, triangle ABC has sides AB, BC, and AC. Since AB and BC are sides of the square and AC is the diagonal, the angle at B is 90 degrees. So that's a right-angled triangle, which is allowed. But what about triangle ACD? In a square, the diagonals are equal, so triangle ACD is also a right-angled triangle. But wait, if I consider the triangle formed by three vertices of a square, all the triangles are either right-angled or have an obtuse angle?Wait, hold on. Let's take a square. Let's look at triangle formed by three vertices. For example, if I take three consecutive vertices of a square: A, B, C. That's a right-angled triangle. If I take A, B, D, where D is the opposite corner, then triangle ABD. Let's compute the angles. AB is a side, AD is a side, and BD is the diagonal. Since AB and AD are sides of the square, they are equal, and BD is the diagonal. So triangle ABD is an isosceles right-angled triangle? Wait, no. In a square, the diagonals are equal and longer than the sides. So in triangle ABD, sides AB and AD are equal, each of length s, and BD is s√2. So the angles at A are 90 degrees, and the other two angles are 45 degrees each. So that triangle is right-angled, not obtuse. So all triangles in a square are right-angled or acute?Wait, let's check another triangle. Take three points: A, C, D in the square. AC is a diagonal, CD is a side, and AD is a side. So AC is longer than CD and AD. Let's see the angles. In triangle ACD, sides AC, CD, AD. Since AC is the diagonal, length s√2, and CD and AD are sides of length s. So the sides are s, s, s√2. The angle opposite the longest side, which is AC, would be the largest angle. Since s√2 > s, the angle opposite AC (which is angle D) is the largest. Let's calculate that angle. Using the Law of Cosines:c² = a² + b² - 2ab cosθHere, (s√2)² = s² + s² - 2*s*s*cosθ2s² = 2s² - 2s² cosθSubtract 2s² from both sides:0 = -2s² cosθTherefore, cosθ = 0, so θ = 90 degrees. So angle D is 90 degrees. So triangle ACD is right-angled. So all triangles formed by three vertices of a square are either right-angled or acute? Wait, is that true? Wait, in the square, any three points form a right-angled triangle. Let me check another combination. Let's take three points that form an isosceles triangle but not right-angled. Wait, in a square, if I take points A, B, and D. AB is a side, BD is a diagonal. Wait, triangle ABD as we saw is right-angled. Hmm. Maybe all triangles formed by three vertices of a square are right-angled? Wait, but if I take three points that are not adjacent? Wait, in a square, any three points will either form a right-angled triangle with legs as sides and hypotenuse as diagonal, or another right-angled triangle. Wait, actually, maybe all triangles in a square are right-angled. That seems possible. Let's confirm.Take points A(0,0), B(1,0), C(1,1), D(0,1). Let's pick three points: A, B, C: right-angled at B. A, B, D: right-angled at A. A, C, D: right-angled at D. B, C, D: right-angled at C. So indeed, every triangle formed by three vertices of a square is right-angled. So in a square, all triangles are right-angled, so there are no obtuse triangles. Therefore, four points in a plane can be placed such that no triangle has an obtuse angle.Wait, but is that really the case? Wait, maybe I made a mistake here. Let's check another configuration. Suppose I place four points in a plane not forming a square. For example, a regular quadrilateral, a square. But if I make a rectangle that's not a square, does that create any obtuse triangles?Take a rectangle with sides of different lengths. Let's say points A(0,0), B(2,0), C(2,1), D(0,1). Let's look at triangle A, B, C. It's right-angled at B. Triangle A, B, D: right-angled at A. Triangle A, C, D: right-angled at D. Triangle B, C, D: right-angled at C. What about triangle A, C, B? Wait, AC is the diagonal. Let's compute angles. The sides are AB=2, BC=1, AC=√(2² +1²)=√5. The sides are 2, 1, √5. The largest side is √5. Let's compute the angle opposite √5, which is angle B. Using the Law of Cosines:(√5)² = 2² + 1² - 2*2*1*cosθ5 = 4 + 1 - 4 cosθ5 = 5 - 4 cosθ0 = -4 cosθcosθ = 0θ = 90 degrees. So angle B is 90 degrees. So even in a rectangle, all triangles formed by three vertices are right-angled. So if I arrange four points as a rectangle (including square), all triangles are right-angled, so no obtuse triangles. So four points work.But can we have five points? Let's think.Suppose we try to place five points in the plane such that every triangle formed is acute or right-angled. How to approach this? Maybe try to place points and check for obtuse angles.Alternatively, maybe there's a known result. I recall that in the plane, the maximum number of points with no three forming an obtuse triangle is four. Wait, but I need to verify this.Alternatively, maybe five points can be arranged. Let me try.Suppose I take a regular pentagon. In a regular pentagon, all sides and angles are equal. Let's check the triangles formed by three vertices. In a regular pentagon, each internal angle is 108 degrees, which is obtuse. But that's the angle inside the pentagon. However, the triangles formed by three vertices may have angles that are obtuse. Let's check.Take three consecutive vertices of a regular pentagon: A, B, C. The angle at B is 108 degrees, which is obtuse. So triangle ABC would have an obtuse angle. Therefore, a regular pentagon is bad because it creates obtuse triangles. So regular pentagon is out.Alternatively, maybe arrange five points in a different configuration. Maybe four points forming a square and the fifth point somewhere. Let's try.Suppose we have a square ABCD and a fifth point E placed at the center of the square. Let's see if this creates any obtuse triangles.Take triangle ABE. The center E is equidistant from all four vertices. In a square with side length s, the distance from center to vertices is s√2/2. So AE = BE = CE = DE = s√2/2. The sides AB = s.So in triangle ABE, sides are AB = s, AE = BE = s√2/2. Let's check the angles. The longest side is AB = s. Let's compute the angle at E. Using Law of Cosines:AB² = AE² + BE² - 2*AE*BE*cosθs² = (s²/2) + (s²/2) - 2*(s√2/2)*(s√2/2)*cosθs² = s² - 2*(s²/2)*cosθs² = s² - s² cosθSubtract s²:0 = -s² cosθ => cosθ = 0 => θ = 90 degrees.So angle at E is 90 degrees. So triangle ABE is right-angled. That's okay. What about triangle ADE? AD is the diagonal of the square, length s√2. AE and DE are both s√2/2. So in triangle ADE, sides are AD = s√2, AE = DE = s√2/2. The longest side is AD. Let's compute the angle at A. Using Law of Cosines:AD² = AE² + DE² - 2*AE*DE*cosφ(s√2)² = (s√2/2)² + (s√2/2)² - 2*(s√2/2)*(s√2/2)*cosφ2s² = (s²/2) + (s²/2) - 2*(s²/2)*cosφ2s² = s² - s² cosφSubtract s²:s² = -s² cosφ => cosφ = -1 => φ = 180 degrees.Wait, that can't be right. Wait, maybe I made a mistake here. Wait, triangle ADE has sides AD, AE, DE. AD is the diagonal of the square, length s√2. AE and DE are both s√2/2. So triangle ADE has sides s√2, s√2/2, s√2/2. So actually, triangle ADE is an isosceles triangle with two sides equal to s√2/2 and the base s√2. Wait, but such a triangle would have the base longer than the other two sides, so the angle opposite the base would be the largest angle.But according to the calculation above, the angle at A (which is between sides AE and AD) would be 180 degrees, which is impossible because the points are not collinear. Wait, but in reality, point E is the center of the square, so points A, D, E are not collinear. Therefore, my calculation must have an error.Wait, let's recast the triangle. Let me label the square with coordinates for clarity. Let’s assign coordinates: let’s place the square ABCD with A at (0,0), B at (1,0), C at (1,1), D at (0,1), and E at (0.5,0.5).Then, triangle ADE would have points A(0,0), D(0,1), E(0.5,0.5). Let's compute the lengths:AD: distance from A to D is 1 (vertical side).AE: distance from A to E is sqrt((0.5)^2 + (0.5)^2) = sqrt(0.25 + 0.25) = sqrt(0.5) ≈ 0.707.DE: distance from D to E is sqrt((0.5 - 0)^2 + (0.5 - 1)^2) = sqrt(0.25 + 0.25) = sqrt(0.5) ≈ 0.707.So triangle ADE has sides AD=1, AE=DE≈0.707. Therefore, the sides are 1, ~0.707, ~0.707. So the longest side is AD=1. Let's compute the angle at A.Using Law of Cosines:AD² = AE² + DE² - 2*AE*DE*cosφ1² = 0.5 + 0.5 - 2*(sqrt(0.5))*(sqrt(0.5))*cosφ1 = 1 - 2*(0.5)*cosφ1 = 1 - cosφSubtract 1:0 = -cosφ => cosφ = 0 => φ = 90 degrees.Wait, so the angle at A is 90 degrees. So triangle ADE is right-angled at A. Hmm, so that's okay. Then, where did I go wrong earlier? Because in the previous calculation, when I considered the sides as s√2, s√2/2, s√2/2, which in the coordinate system above, AD is length 1 (s=1), so s√2 would be √2, but in reality, AD is 1. So my mistake was in scaling. So in the coordinate system, the square has side length 1, so AE and DE are sqrt(0.5), not s√2/2 (which would be sqrt(0.5) if s=1). So the calculation is correct, and angle at A is 90 degrees. So triangle ADE is right-angled, not obtuse.So, in this configuration with five points (square plus center), are there any obtuse triangles?Let me check another triangle, say, B, C, E. Points B(1,0), C(1,1), E(0.5,0.5). Compute the angles.First, compute the sides:BC: length 1 (vertical side).BE: distance from B to E: sqrt((0.5)^2 + (0.5)^2) = sqrt(0.5) ≈ 0.707.CE: distance from C to E: same as BE ≈ 0.707.So triangle BCE has sides BC=1, BE=CE≈0.707. The longest side is BC=1. Let's compute the angle at B.Law of Cosines:BC² = BE² + CE² - 2*BE*CE*cosθ1² = 0.5 + 0.5 - 2*(0.5)*cosθ1 = 1 - cosθ => cosθ = 0 => θ = 90 degrees. So right-angled at B.Another triangle: A, E, C. Points A(0,0), E(0.5,0.5), C(1,1). Let's compute the angles.Sides:AC: diagonal of the square, length sqrt(2) ≈ 1.414.AE: sqrt(0.5) ≈ 0.707.EC: sqrt(0.5) ≈ 0.707.So triangle AEC has sides AC≈1.414, AE≈0.707, EC≈0.707. The longest side is AC. Let's compute the angle at A.Law of Cosines:AC² = AE² + EC² - 2*AE*EC*cosφ(1.414)^2 ≈ 2 ≈ 0.5 + 0.5 - 2*0.5*cosφ2 ≈ 1 - cosφ => cosφ ≈ -1 => φ ≈ 180 degrees.Wait, that can't be. Points A, E, C are colinear? Wait, in the coordinate system, A(0,0), E(0.5,0.5), C(1,1). These three points are colinear on the line y = x. Therefore, triangle AEC is degenerate, a straight line. But the problem states that no three points are colinear. Therefore, placing point E at the center of the square causes three points (A, E, C) to be colinear, which violates the problem's condition. Therefore, this configuration is invalid.Ah, right! The problem says no three points are colinear, so we cannot have point E lying on the diagonal AC. Therefore, placing the fifth point at the center of the square is invalid. So we need to place the fifth point somewhere else not on any diagonal or side.Let me try another configuration. Let's take a square ABCD and place the fifth point E slightly off-center, not on any of the diagonals or sides. Let's say E is at (0.5, 0.5 + ε) where ε is a small positive number. Then, points A, E, C are not colinear because E is not on the diagonal. Let's check triangles involving E.Take triangle ABE. A(0,0), B(1,0), E(0.5, 0.5 + ε). Let's compute the sides.AB = 1.AE = sqrt((0.5)^2 + (0.5 + ε)^2).BE = sqrt((0.5 - 1)^2 + (0.5 + ε - 0)^2) = sqrt(0.25 + (0.5 + ε)^2).Compute the angles in triangle ABE. The sides are AB=1, AE, BE. Let's check the largest angle. If ε is very small, say ε approaching 0, then AE and BE approach sqrt(0.25 + 0.25) = sqrt(0.5) ≈ 0.707. But as ε increases, AE and BE increase slightly. The longest side would still be AB=1. Let's compute the angle at E.Using Law of Cosines:AB² = AE² + BE² - 2*AE*BE*cosθ1² = AE² + BE² - 2*AE*BE*cosθBut as ε approaches 0, AE and BE approach sqrt(0.5). Let's plug in ε = 0 for approximation:1 ≈ 0.5 + 0.5 - 2*(0.5)*cosθ1 ≈ 1 - cosθ => cosθ ≈ 0 => θ ≈ 90 degrees.But with ε > 0, AE and BE are slightly larger. Let's compute numerically for ε = 0.1.AE = sqrt(0.25 + (0.6)^2) = sqrt(0.25 + 0.36) = sqrt(0.61) ≈ 0.781.BE = sqrt(0.25 + (0.6)^2) = same as AE ≈ 0.781.Then,1 ≈ 0.61 + 0.61 - 2*0.781*0.781*cosθ1 ≈ 1.22 - 2*(0.61)*cosθWait, 2*0.781*0.781 ≈ 2*0.61 ≈ 1.22So,1 ≈ 1.22 - 1.22 cosθThen,1 - 1.22 ≈ -1.22 cosθ-0.22 ≈ -1.22 cosθcosθ ≈ 0.22 / 1.22 ≈ 0.18θ ≈ 80 degrees. So the angle at E is acute.Similarly, other angles in triangle ABE can be computed. Let's check angle at A.Using Law of Cosines:BE² = AB² + AE² - 2*AB*AE*cosφ0.61 ≈ 1 + 0.61 - 2*1*0.781*cosφ0.61 ≈ 1.61 - 1.562*cosφSubtract 1.61:-1 ≈ -1.562*cosφcosφ ≈ 0.64φ ≈ 50 degrees. Acute. Similarly, angle at B is same as angle at A, so 50 degrees. So triangle ABE is acute.Similarly, other triangles might be okay. But we need to check all possible triangles involving E. Let's check triangle ADE. A(0,0), D(0,1), E(0.5,0.6). Compute the sides:AD = 1.AE = sqrt(0.5² + 0.6²) ≈ sqrt(0.25 + 0.36) ≈ sqrt(0.61) ≈ 0.781.DE = sqrt((0.5 - 0)^2 + (0.6 - 1)^2) = sqrt(0.25 + 0.16) = sqrt(0.41) ≈ 0.640.So triangle ADE has sides 1, 0.781, 0.640. The longest side is AD=1. Let's compute the angle at A.Law of Cosines:AD² = AE² + DE² - 2*AE*DE*cosθ1 = 0.61 + 0.41 - 2*0.781*0.640*cosθ1 = 1.02 - 2*0.500*cosθWait, 0.781*0.640 ≈ 0.500.So,1 = 1.02 - 1.000*cosθ1 - 1.02 = -cosθ-0.02 = -cosθcosθ = 0.02 => θ ≈ 88.8 degrees. That's acute (since less than 90). So angle at A is 88.8 degrees, acute.Angle at D: Let's use Law of Cosines:AE² = AD² + DE² - 2*AD*DE*cosφ0.61 = 1 + 0.41 - 2*1*0.640*cosφ0.61 = 1.41 - 1.28*cosφ0.61 - 1.41 = -1.28*cosφ-0.8 = -1.28*cosφcosφ = 0.8 / 1.28 ≈ 0.625φ ≈ 51.3 degrees. Acute.Angle at E: 180 - 88.8 - 51.3 ≈ 39.9 degrees. Acute. So triangle ADE is acute.Similarly, let's check triangle BCE. Points B(1,0), C(1,1), E(0.5,0.6). Sides:BC = 1.BE ≈ 0.781 (same as before).CE = sqrt((1 - 0.5)^2 + (1 - 0.6)^2) = sqrt(0.25 + 0.16) = sqrt(0.41) ≈ 0.640.So triangle BCE has sides 1, 0.781, 0.640. Similar to triangle ADE. The angle at B:Law of Cosines:BC² = BE² + CE² - 2*BE*CE*cosθ1 = 0.61 + 0.41 - 2*0.781*0.640*cosθSame as before, cosθ ≈ 0.02 => θ ≈ 88.8 degrees. Acute.Another triangle to check is CDE. C(1,1), D(0,1), E(0.5,0.6). Sides:CD = 1.CE ≈ 0.640.DE ≈ 0.640 (distance from D(0,1) to E(0.5,0.6): sqrt(0.25 + 0.16) = sqrt(0.41) ≈ 0.640).So triangle CDE has sides 1, 0.640, 0.640. The longest side is CD=1. Compute angle at C:Law of Cosines:CD² = CE² + DE² - 2*CE*DE*cosφ1 = 0.41 + 0.41 - 2*0.640*0.640*cosφ1 = 0.82 - 2*0.4096*cosφ1 = 0.82 - 0.8192*cosφ1 - 0.82 = -0.8192*cosφ0.18 = -0.8192*cosφcosφ ≈ -0.18 / 0.8192 ≈ -0.22φ ≈ 102.6 degrees. Oh, that's obtuse! So angle at C is obtuse. Therefore, triangle CDE is obtuse, which violates the condition. So even by moving the fifth point slightly off the center, we can end up with an obtuse triangle.Hmm, so placing the fifth point near the center but not exactly at the center still can result in an obtuse triangle when combined with other points. Therefore, maybe adding a fifth point is impossible?Alternatively, maybe there's another configuration where five points don't form an obtuse triangle. Let me think.Another approach: in the plane, the maximum number of points such that all triangles are non-obtuse is 5? Or is it 4?Wait, I recall that in the plane, the maximum is 5. Wait, but how? Let me check.Wait, according to some references, the problem is known. For example, the Erdős–Furedi problem on acute sets. But wait, there might be a difference between acute and non-obtuse. Acute triangles require all angles less than 90 degrees, whereas non-obtuse allows right angles.Wait, the question here is "none of the triangles formed by these points have an obtuse angle". So triangles can be acute or right-angled. So the problem is about non-obtuse sets, not necessarily all acute. So allowing right angles.In the plane, the maximum number of points with no three forming an obtuse triangle is 5. Wait, according to some sources, but I need to verify.Wait, here's a thought. In 1952, Gerriets and Poole conjectured that the maximum number is 5, and in 1967, Erdos and Furedi proved that in n-dimensional space, the maximum size is 2^n, but for acute angles, not sure. Wait, maybe different problem.Alternatively, maybe the maximum is 5. For example, arranging points as a regular pentagon, but we saw that regular pentagons have obtuse triangles. So that's no good.Alternatively, maybe arrange the five points as a convex quadrilateral with a point inside. Wait, but then triangles formed with the interior point might have obtuse angles.Alternatively, if we take four points as a square and fifth point very close to one of the vertices but not exactly on it. Wait, but then triangles formed with the nearby vertex might have obtuse angles.Alternatively, arranging the five points as a regular pentagon but perturbed slightly to avoid obtuse angles. However, this seems difficult.Alternatively, arrange the five points such that all mutual distances are equal, forming a regular pentagon. But as we saw, regular pentagons have triangles with obtuse angles.Alternatively, maybe arrange the five points in a circle such that every arc between consecutive points is less than 90 degrees. Wait, if you have five points on a circle, each arc is 72 degrees. Then, the central angle between any two points is 72 degrees. However, inscribed angles subtended by those arcs would be half, 36 degrees. But triangles formed by three points on the circle can have angles that are combinations of these.Wait, let's take three points on a circle. The angles in the triangle depend on the arcs between them. The largest angle in a triangle inscribed in a circle is equal to half the measure of the largest arc opposite it. So if all arcs are 72 degrees, then the largest arc between any three points is at most 144 degrees (if you skip one point). Wait, with five points on a circle, the arcs are 72 degrees each. If you take three points, the arcs between them could be 72, 72, and 144 degrees (if you skip one point). Then, the inscribed angle opposite the 144-degree arc would be 72 degrees. So all angles in the triangle would be 72, 72, and 36 degrees. Wait, no. Wait, inscribed angle is half the measure of the arc. So if there's an arc of 144 degrees, the inscribed angle opposite it would be 72 degrees. The other two arcs would be 72 degrees each, leading to inscribed angles of 36 degrees each. Therefore, the triangle would have angles 72, 36, 36 degrees. All acute. Wait, but is that correct?Wait, let's take three points A, B, C on a circle with five equally spaced points. If the arcs AB, BC, and CA are 72°, 72°, and 144°, then angle at B (opposite arc AC) is half of 144°, which is 72°. Angles at A and C would each be half of 72°, which is 36°. So triangle ABC has angles 72°, 36°, 36°, all acute. So in this case, the triangle is acute. Wait, but if we take three points with arcs 144°, 72°, 72°, the triangle is still acute.Wait, but if we take three points where two of the arcs are larger. For example, with five points on a circle, the maximum arc between two points is 144° (skipping one point). If we take three consecutive points, the arcs between them are 72°, 72°, and 144°. Wait, no. If the five points are equally spaced on the circle, then between any two consecutive points is 72°. If you take three consecutive points, the arcs between them are 72°, 72°, and the remaining arc is 360 - 72*2 = 216°, but wait, that can't be. Wait, no, three points on the circle divide the circle into three arcs. If the points are equally spaced with five points, then taking three consecutive points A, B, C, the arcs AB and BC are each 72°, and the arc CA is 360 - 72*2 = 216°, but in a five-point circle, the arcs between consecutive points are 72°, so three points would have arcs of 72°, 72°, and 216°? Wait, that can't be. Wait, no.Wait, let's label the five points as A0, A1, A2, A3, A4 arranged around a circle. The arc between A0 and A1 is 72°, A1 to A2 is 72°, etc. If we take three points A0, A1, A2, then the arcs between them are 72°, 72°, and 288° (since from A2 back to A0 is 3 intervals of 72°, totaling 216°? Wait, no. Wait, the circumference is 360°, five points, each arc is 72°. If we take three consecutive points, A0, A1, A2, then the arcs are A0A1=72°, A1A2=72°, and A2A0=360 - 72*2 = 216°. Wait, yes. So the arcs are 72°, 72°, 216°. Therefore, the inscribed angles would be half of the opposing arcs. So in triangle A0A1A2:Angle at A1 is opposite arc A0A2, which is 144° (since from A0 to A2 is two arcs of 72°). Wait, no. Wait, arc A0A2 is 144°, so the inscribed angle at A1 is half of that, 72°. Similarly, angle at A0 is opposite arc A1A2=72°, so angle is 36°. Angle at A2 is opposite arc A0A1=72°, so 36°. Therefore, triangle A0A1A2 has angles 72°, 36°, 36°, all acute.But if we take three points that are not consecutive. For example, A0, A2, A4. Then, the arcs between them are each 144°, since skipping one point. So each arc is 144°, leading to inscribed angles of 72°, making the triangle equiangular with all angles 72°, which is acute. Wait, but in reality, if all arcs are 144°, the triangle would be inscribed in a circle with each angle equal to 72°, so an acute triangle.Wait, so if all triangles formed by three points on a regular pentagon are acute? That contradicts my earlier thought that consecutive three points would form a triangle with a 72° angle. Wait, but 72° is still acute. So all triangles on a regular pentagon are acute? Then, a regular pentagon would be a valid configuration for five points with no obtuse triangles. But that contradicts my initial thought when I considered three consecutive points forming a triangle with an angle of 108°, but that was a mistake.Wait, in the regular pentagon, the internal angle at each vertex is 108°, which is the angle inside the pentagon. However, the angles of the triangles formed by three vertices are different. For example, taking three consecutive vertices A, B, C, the triangle ABC has angles that are different from the internal angles of the pentagon. As calculated earlier, in triangle ABC, the angle at B is 72°, and the other angles are 36°, which are all acute. Therefore, maybe a regular pentagon is a valid configuration with five points where all triangles are acute.Wait, let's verify this again with coordinates. Let's take a regular pentagon inscribed in a unit circle. The coordinates of the vertices can be given by (cos(72°k), sin(72°k)) for k=0,1,2,3,4.Let's take three consecutive points: A(1,0), B(cos 72°, sin 72°), C(cos 144°, sin 144°). Let's compute the angles in triangle ABC.First, compute the lengths of the sides:AB: distance between A and B.Using coordinates:A(1,0), B(cos 72°, sin 72°). The distance AB is sqrt[(cos72 - 1)^2 + (sin72)^2].Similarly, BC: distance between B and C.C(cos144°, sin144°). Distance BC = sqrt[(cos144 - cos72)^2 + (sin144 - sin72)^2].AC: distance between A and C.Distance AC = sqrt[(cos144 - 1)^2 + (sin144)^2].Alternatively, since all points are on the unit circle, the chord length between two points is 2*sin(θ/2), where θ is the central angle between them.Central angle between A and B is 72°, so AB = 2*sin(36°) ≈ 1.1756.Between B and C is also 72°, so BC ≈ 1.1756.Between A and C is 144°, so AC = 2*sin(72°) ≈ 1.9021.So triangle ABC has sides ≈1.1756, ≈1.1756, ≈1.9021.Let's compute the angles using Law of Cosines.Angle at A (opposite BC):cosφ = (AB² + AC² - BC²)/(2*AB*AC)Wait, but BC is opposite angle at A. Wait, no. Wait, angle at A is opposite side BC.Law of Cosines:BC² = AB² + AC² - 2*AB*AC*cosφSo,cosφ = (AB² + AC² - BC²)/(2*AB*AC)Plugging the values:AB² ≈ (1.1756)^2 ≈ 1.381AC² ≈ (1.9021)^2 ≈ 3.617BC² ≈ (1.1756)^2 ≈ 1.381cosφ = (1.381 + 3.617 - 1.381)/(2*1.1756*1.9021)cosφ = (3.617)/(4.472) ≈ 0.809Thus, φ ≈ 36°, which is acute.Angle at B (opposite side AC):Similarly,AC² = AB² + BC² - 2*AB*BC*cosθSo,cosθ = (AB² + BC² - AC²)/(2*AB*BC)cosθ = (1.381 + 1.381 - 3.617)/(2*1.1756*1.1756)cosθ = (-0.855)/(2.762) ≈ -0.309Thus, θ ≈ 108°, which is obtuse. Wait, so angle at B is 108°, which is obtuse. That contradicts the earlier assertion. So this triangle has an obtuse angle. Therefore, the regular pentagon does have triangles with obtuse angles. Therefore, a regular pentagon is invalid.Therefore, my previous calculation was wrong. The angle at B is 108°, obtuse. So the regular pentagon does not work.Therefore, this suggests that five points in a plane cannot be arranged without forming an obtuse triangle. So maybe the maximum is four points, as in the square. But wait, earlier I saw that in a square all triangles are right-angled, which are allowed. So four points work. But five points cannot be arranged without an obtuse triangle.But how to confirm this? Maybe it's a known result. From what I recall, in the plane, the maximum number of points with all triangles being non-obtuse (i.e., right or acute) is five. Wait, but our calculation shows that regular pentagon fails. Maybe there's a different configuration.Wait, perhaps a different configuration. For example, a convex quadrilateral with a point inside arranged such that all triangles are acute. But I think adding a fifth point inside a convex quadrilateral would result in some triangles being obtuse.Alternatively, consider the vertices of a rectangle with an additional point forming a pyramid-like structure in 3D, but we are restricted to the plane. So maybe no.Alternatively, according to some references, in the plane, the maximum number of points with all triangles acute is five. However, the problem here allows right angles. So maybe it's possible to have more points.Wait, but actually, if you allow right angles, then the square has four points with all triangles right-angled. If we can add another point without creating an obtuse triangle, maybe five points. But previous attempts showed that adding a fifth point tends to create an obtuse triangle. Alternatively, consider the vertices of a square and the center. But we saw that this creates colinear points, which is invalid. If we move the center slightly, we still get obtuse triangles. So maybe five points are impossible.According to some research, the problem of placing points in the plane with no three forming an obtuse triangle is a known problem. According to "Obtuse Triangles and the Euclidean Ball" by Erdos and Furedi, but I don't recall the exact result. However, according to the "no three points form an obtuse triangle" problem, it's known that the maximum number is 5. Wait, but I need to check.Wait, here's a source: "Acute Sets" by Erdős and Füredi. They show that in n-dimensional space, the maximum size is exponential in n. For the plane, they show you can have at least 5 points, but it might be possible to have more. Wait, but perhaps not.Wait, another approach: in a non-obtuse set (allowing right angles), the maximum in the plane is 5. For example, the vertices of a regular pentagon do not work, but perhaps another configuration. Alternatively, if you take the vertices of a square and add a point close to the center but not exactly at the center. But as shown earlier, this can lead to an obtuse triangle.Alternatively, take four points forming a square and fifth point forming a convex quadrilateral. Wait, but fifth point would make a convex pentagon, which might have triangles with obtuse angles.Alternatively, in 1995, Gerriets and Poole proved that the maximum number is five for non-obtuse sets in the plane. So that's possible. But how?Wait, let's think of the following configuration: take a regular pentagon and perturb it slightly to make all triangles acute. But if the regular pentagon already has obtuse triangles, perturbing slightly might not help.Alternatively, take points on a circle with arcs less than 90 degrees. If five points are placed on a circle such that each arc between consecutive points is less than 90 degrees, then any three points would form a triangle with all angles acute. Because the largest arc between any two points would be less than 180 degrees, and the inscribed angle would be less than 90 degrees. Wait, but with five points on a circle, each arc is 72 degrees. Wait, inscribed angles would be 36 degrees. However, if three points are selected such that the arcs between them sum up to more than 180 degrees, the opposite angle would be more than 90 degrees. Wait, no. Wait, the inscribed angle is half the measure of the arc. So if three points A, B, C are on a circle such that the arc AC is greater than 180 degrees, then angle at B would be greater than 90 degrees. So to prevent this, all arcs between consecutive points must be less than 90 degrees. But with five points on a circle, how?If you want each arc between consecutive points to be less than 90 degrees, then the total circumference would be less than 5*90 = 450 degrees, but a circle is 360 degrees. So impossible. Therefore, it's impossible to place five points on a circle such that each arc is less than 90 degrees. Therefore, there must be at least one arc of at least 72 degrees (360/5). So inscribed angles of 36 degrees. But if three points are such that the largest arc between two of them is 144 degrees (skipping two points), then the inscribed angle would be 72 degrees. So triangle formed by those three points would have a 72-degree angle, which is acute. Wait, but if three points are such that the largest arc is 216 degrees (three arcs), then inscribed angle is 108 degrees, which is obtuse. Wait, but how?If you have five points on a circle, the maximum arc between any two points is 144 degrees (skipping one point). So inscribed angle is 72 degrees. So no angle in any triangle would exceed 72 degrees. Wait, but if you take three points with two arcs of 144 degrees and one arc of 72 degrees, then the inscribed angles would be 72, 36, 36. All acute. So maybe all triangles would be acute. Wait, but if the three points are arranged such that the arcs are 144°, 144°, and 72°, the inscribed angles would be 72°, 72°, and 36°, summing to 180. Wait, 72+72+36=180. So angles are 72°, 72°, 36°, all acute. Therefore, maybe five points equally spaced on a circle (regular pentagon) actually form only acute triangles? But earlier calculation showed that taking three consecutive points leads to a triangle with a 108° angle. Wait, this is a confusion between internal angles of the pentagon and the angles of the triangles formed by three vertices.Wait, let's clarify. In a regular pentagon, the internal angle at each vertex is 108°, but when you form a triangle by three vertices, the angles of that triangle are different. Let's take three consecutive vertices of a regular pentagon: A, B, C. The arcs between them are 72°, 72°, and 144° (since from C back to A is 144°). The inscribed angles in the triangle would be half of the opposing arcs. Therefore, angle at B is half of 144°, which is 72°, angles at A and C are half of 72°, which is 36°. So triangle ABC has angles 72°, 36°, 36°, all acute. Similarly, taking three points with arcs 144°, 72°, 144°, the angles would be 72°, 72°, 36°, still all acute. Therefore, all triangles formed by three vertices of a regular pentagon are acute. Wait, but earlier, when I took three points A0, A2, A4 in a regular pentagon, the central angles were each 144°, leading to inscribed angles of 72°, forming an equilateral triangle (if regular), but in reality, in the regular pentagon, the three points A0, A2, A4 form a triangle with each side equal to the chord length of 144°, which is 2*sin(72°) ≈ 1.902. The angles in such a triangle would all be 72°, as the central angles are 144°, so the inscribed angles are 72°. Therefore, all angles are acute. So maybe the regular pentagon is actually a valid configuration with five points where all triangles are acute. Then why did I get a 108° angle earlier?Wait, perhaps I made a mistake in labeling. Let me take specific coordinates. Let's consider a regular pentagon inscribed in a unit circle:Points:A0: (1, 0)A1: (cos 72°, sin 72°) ≈ (0.3090, 0.9511)A2: (cos 144°, sin 144°) ≈ (-0.8090, 0.5878)A3: (cos 216°, sin 216°) ≈ (-0.8090, -0.5878)A4: (cos 288°, sin 288°) ≈ (0.3090, -0.9511)Take three consecutive points: A0, A1, A2.Compute the angles in triangle A0A1A2.First, compute the vectors:A0A1: from A0 to A1: (-0.6910, 0.9511)A0A2: from A0 to A2: (-1.8090, 0.5878)A1A2: from A1 to A2: (-1.1180, -0.3633)Now, compute the angles using the dot product.Angle at A0:Vectors A0A1 and A0A2.Dot product: (-0.6910)(-1.8090) + (0.9511)(0.5878) ≈ 1.250 + 0.559 ≈ 1.809Magnitude of A0A1: sqrt(0.6910² + 0.9511²) ≈ sqrt(0.477 + 0.904) ≈ sqrt(1.381) ≈ 1.1756Magnitude of A0A2: sqrt(1.8090² + 0.5878²) ≈ sqrt(3.272 + 0.345) ≈ sqrt(3.617) ≈ 1.902So cosθ = 1.809 / (1.1756 * 1.902) ≈ 1.809 / 2.236 ≈ 0.809θ ≈ 36°, which matches the previous calculation.Angle at A1:Vectors A1A0 and A1A2.A1A0: (0.6910, -0.9511)A1A2: (-1.1180, -0.3633)Dot product: (0.6910)(-1.1180) + (-0.9511)(-0.3633) ≈ -0.773 + 0.345 ≈ -0.428Magnitudes:|A1A0| ≈ 1.1756|A1A2| ≈ sqrt(1.1180² + 0.3633²) ≈ sqrt(1.250 + 0.132) ≈ sqrt(1.382) ≈ 1.1756cosφ = -0.428 / (1.1756 * 1.1756) ≈ -0.428 / 1.381 ≈ -0.310φ ≈ 108°, which is obtuse. So here's the mistake. Wait, the angle at A1 is 108°, which is obtuse. Therefore, triangle A0A1A2 has an obtuse angle. Therefore, the regular pentagon does contain obtuse triangles. Therefore, my initial calculation was correct, but my later reasoning was wrong. Therefore, regular pentagons cannot be used.Therefore, the conclusion is that in the plane, you cannot have five points without an obtuse triangle, so the maximum is four points arranged in a square, where all triangles are right-angled.But wait, is there another configuration of four points where all triangles are acute? For example, a regular tetrahedron projected onto a plane, but that would likely have obtuse triangles.Alternatively, a convex quadrilateral with all triangles acute. Is that possible?Yes, there are convex quadrilaterals called acute quadrilaterals, where all four triangles formed by their three vertices are acute. For example, a square is not acute since all triangles are right-angled. But if you slightly deform the square into a convex quadrilateral where all angles are acute and the diagonals are such that all triangles are acute. However, constructing such a quadrilateral might be possible, but even if so, adding a fifth point would still be problematic.But according to the problem, we need the maximum number of points such that no triangle has an obtuse angle. If we can find a convex quadrilateral with all triangles acute, then four points. Otherwise, if the square is the only quadrilateral where all triangles are right-angled, then the maximum is four.But I think there are convex quadrilaterals with all triangles acute. For example, a kite shape with two pairs of adjacent sides equal, but angles all acute. But need to verify.Let's take a convex quadrilateral with vertices A, B, C, D such that all four triangles ABC, ABD, ACD, BCD are acute.For example, take a rhombus with acute angles. A rhombus has all sides equal, and opposite angles equal. If the angles are acute (less than 90°), then the triangles formed by the diagonals would be acute.Wait, in a rhombus with acute angles, the diagonals are not equal. The diagonals split the angles into two equal parts. For example, in a rhombus with angles 80° and 100°, the triangles formed by the diagonals would have angles of 40°, 50°, 90°. Wait, no. Let's compute.In a rhombus, the diagonals bisect the angles. So if the rhombus has angles 80° and 100°, each half-angle at the vertex would be 40° and 50°. The triangles formed by the diagonals would have angles 40°, 50°, 90°, which is right-angled. Therefore, the triangles formed are right-angled. So rhombus is not good.Alternatively, take a convex quadrilateral where all angles are acute and the diagonals are such that all resulting triangles are acute. This requires careful construction.For example, take a convex quadrilateral ABCD with all angles less than 90°, and both diagonals AC and BD are such that all four triangles are acute. This might be possible but requires specific lengths.Alternatively, according to some geometrical references, it's possible to have a convex quadrilateral with all four triangles acute. For example, see "Acute Quadrilaterals" (but I don't have the reference here). If such a quadrilateral exists, then four points can be placed in the plane with all triangles acute. However, adding a fifth point would still be problematic, as shown before.But since the problem allows right-angled triangles, perhaps the square with four points is the safest answer, as any fifth point seems to introduce an obtuse triangle. Therefore, the maximum number is four for part a).Now, moving to part b) in space. What is the maximum number of points in space such that no triangle formed by any three points has an obtuse angle.In three dimensions, we can potentially place more points without forming obtuse triangles. For example, the vertices of a regular simplex (tetrahedron) have all triangles acute. In a regular tetrahedron, all faces are equilateral triangles, so all angles are 60 degrees. However, when considering all possible triangles formed by any three vertices, not just the faces. Wait, in a regular tetrahedron, there are four vertices. Any three vertices form an equilateral triangle, which is acute. So four points in space can be placed as a regular tetrahedron with all triangles acute.Can we add a fifth point? Let's think.If we try to add a fifth point such that all triangles formed with the existing four points are acute. This is challenging. The regular simplex in 3D is the tetrahedron with four points. The next possible configuration might be the vertices of a regular octahedron. Let's check.A regular octahedron has six vertices. Each face is an equilateral triangle. However, when considering triangles formed by any three vertices, there are also non-face triangles. For example, take two opposite vertices and a third one. The triangle formed by these three points might have obtuse angles.Wait, take a regular octahedron with vertices at (±1,0,0), (0,±1,0), (0,0,±1). Let's take three vertices: (1,0,0), (-1,0,0), (0,1,0). The triangle formed by these points has sides of length 2 (between (1,0,0) and (-1,0,0)), sqrt(2) (between (1,0,0) and (0,1,0)), and sqrt(2) (between (-1,0,0) and (0,1,0)). So the sides are 2, sqrt(2), sqrt(2). The longest side is 2. Let's check the angles.Using Law of Cosines for the angle opposite the longest side (2):cosθ = (sqrt(2)^2 + sqrt(2)^2 - 2^2)/(2*sqrt(2)*sqrt(2)) = (2 + 2 - 4)/ (2*2) = 0/4 = 0 => θ = 90 degrees. So the triangle has a right angle. Therefore, the regular octahedron has right-angled triangles, which are allowed in the problem, but if we require no obtuse triangles, right angles are acceptable. However, in the octahedron, there are also triangles with obtuse angles.Wait, take three vertices: (1,0,0), (0,1,0), (0,0,1). The triangle formed by these points has all sides equal to sqrt(2). So it's an equilateral triangle, all angles 60 degrees. Acute.Another triangle: (1,0,0), (0,1,0), (0,0,1). Wait, same as above. What about (1,0,0), (0,1,0), (-1,0,0). As before, right-angled. Another triangle: (1,0,0), (0,1,0), (0,-1,0). This is a square in the xy-plane, but the triangle formed by these three points would have sides 2, sqrt(2), sqrt(2). Again, right-angled.Wait, but the problem states that no triangle should have an obtuse angle. Since right angles are allowed, the octahedron might be acceptable. But wait, in the regular octahedron, are there any obtuse triangles? Let's check.Take three vertices: (1,0,0), (0,1,0), (0,0,1). Equilateral triangle, all angles acute. Take three vertices: (1,0,0), (0,1,0), (0,0,0). Wait, but (0,0,0) is not a vertex of the octahedron. The octahedron has vertices at (±1,0,0), etc. So if we take three vertices that are not coplanar. Wait, actually, all vertices of the octahedron are either on the axes or origin? No, the regular octahedron has vertices at (±1,0,0), (0,±1,0), (0,0,±1). So no three of them are colinear, and any three either form an equilateral triangle (if they are adjacent) or a right-angled triangle (if they include two opposites and one other).Wait, in the octahedron, any three vertices either form an equilateral triangle (on a face) or a right-angled triangle (if they include two opposite vertices). So there are no obtuse triangles in a regular octahedron. All triangles are either right-angled or equilateral. Therefore, the regular octahedron has six vertices with all triangles being either right-angled or acute. Therefore, six points in space can be placed without any obtuse triangles.But wait, earlier example with three vertices (1,0,0), (-1,0,0), (0,1,0) gives a right-angled triangle, which is allowed. But what about another triangle: (1,0,0), (0,1,0), (0,0,1). This is an equilateral triangle. All angles acute.Is there any triangle in the octahedron with an obtuse angle? Let's check.Take three vertices not on the same face. For example, (1,0,0), (0,1,0), (0,0,1). As before, equilateral. Another example: (1,0,0), (0,1,0), (-1,0,0). This gives a right-angled triangle. Another example: (1,0,0), (0,1,0), (0,-1,0). Right-angled triangle. What about (1,0,0), (0,1,0), (0,0.5,0.5). Wait, but (0,0.5,0.5) is not a vertex of the octahedron.Thus, all triangles formed by three vertices of the regular octahedron are either right-angled or equilateral. So no obtuse triangles. Therefore, six points in space can be placed without any obtuse triangles.But can we have more than six points? For example, a cube. A cube has eight vertices. However, in a cube, there are triangles with obtuse angles. For example, consider three vertices: (0,0,0), (1,0,0), (0,1,1). The triangle formed by these points would have sides of length 1, sqrt(2), sqrt(3). The largest side is sqrt(3), so the angle opposite to it would be:cosθ = (1^2 + (√2)^2 - (√3)^2)/(2*1*√2) = (1 + 2 - 3)/(2√2) = 0/2√2 = 0 => θ = 90 degrees. So right-angled. But another triangle: (0,0,0), (1,1,0), (1,0,1). Compute sides:Between (0,0,0) and (1,1,0): sqrt(2).Between (0,0,0) and (1,0,1): sqrt(2).Between (1,1,0) and (1,0,1): sqrt((0)^2 + (1)^2 + (1)^2) = sqrt(2).So equilateral triangle. All angles 60 degrees. Acute.But in a cube, other triangles can have obtuse angles. For example, take three vertices: (0,0,0), (1,0,0), (1,1,1). The sides are 1, sqrt(3), sqrt( (1)^2 + (1)^2 + (1)^2 ) = sqrt(3). Wait, no. (0,0,0) to (1,0,0) is 1. (0,0,0) to (1,1,1) is sqrt(3). (1,0,0) to (1,1,1) is sqrt(0 + 1 + 1) = sqrt(2). So sides are 1, sqrt(2), sqrt(3). The largest side is sqrt(3). Let's compute the angle opposite to sqrt(3):cosθ = (1^2 + (sqrt(2))^2 - (sqrt(3))^2)/(2*1*sqrt(2)) = (1 + 2 - 3)/(2*sqrt(2)) = 0/(2.828) = 0 => θ = 90 degrees. So right-angled again. Hmm.Wait, maybe the cube does not have obtuse triangles. Wait, let's check another triangle. Take (0,0,0), (1,1,0), (1,0,1). As before, sides are sqrt(2), sqrt(2), sqrt(2). Equilateral. All angles acute.Another triangle: (0,0,0), (1,0,0), (0,1,0). Right-angled. Another triangle: (0,0,0), (1,1,1), (0,1,1). Sides from (0,0,0) to (1,1,1): sqrt(3). From (0,0,0) to (0,1,1): sqrt(2). From (1,1,1) to (0,1,1): 1. So sides are 1, sqrt(2), sqrt(3). Again, right-angled.Wait, so maybe all triangles in the cube are either right-angled or acute. If that's the case, then the cube's eight vertices would satisfy the condition. But I need to check if there exists any obtuse triangle in the cube.Take three vertices that form a face diagonal and a space diagonal. For example, (0,0,0), (1,1,0), (1,1,1). The sides are:(0,0,0) to (1,1,0): sqrt(2).(0,0,0) to (1,1,1): sqrt(3).(1,1,0) to (1,1,1): 1.So sides are 1, sqrt(2), sqrt(3). Angle opposite sqrt(3) is 90 degrees.Another triangle: (0,0,0), (1,0,1), (0,1,1). Sides are sqrt(2), sqrt(2), sqrt(2). Equilateral.Wait, seems like all triangles in the cube are either right-angled or equilateral. Therefore, no obtuse triangles. Therefore, eight points in space (cube's vertices) satisfy the condition.But this contradicts my initial thought. Wait, perhaps there are obtuse triangles in the cube.Take three vertices: (0,0,0), (1,0,0), (1,0,1). This forms a right-angled triangle with sides 1,1,sqrt(2). Right angle at (1,0,0).Another example: (0,0,0), (1,1,0), (1,0,1). As before, sides are sqrt(2), sqrt(2), sqrt(2). Equilateral.Hmm. Let's check another triangle: (0,0,0), (1/2,1/2,1/2), (1,1,1). Wait, but (1/2,1/2,1/2) is not a vertex of the cube.If we only consider cube vertices, any three vertices form a triangle with sides that are either edges, face diagonals, or space diagonals. The possible triangles are:1. Three edges meeting at a corner: forms a right-angled triangle.2. Two edges and a face diagonal: right-angled triangle.3. One edge, one face diagonal, one space diagonal: right-angled triangle.4. Three face diagonals on the same face: equilateral triangle (but in cube, face diagonals are sqrt(2), but three such diagonals would form a larger triangle? Wait, no, three face diagonals on the same face would form a triangle with sides sqrt(2), sqrt(2), 2, which is right-angled.5. Three space diagonals: forming an equilateral triangle with sides sqrt(3).Wait, but in a cube, the space diagonals are from (0,0,0) to (1,1,1), etc. Taking three space diagonals, for example, (0,0,0) to (1,1,1), (1,1,0) to (0,0,1), etc. Wait, no, three vertices that are connected by space diagonals. Wait, no, each space diagonal connects two vertices. So three space diagonals would require six distinct vertices. So three vertices can't all be connected by space diagonals.Therefore, in the cube, all triangles formed by three vertices are either right-angled or equilateral. Therefore, no obtuse triangles. Therefore, eight points can be placed in space without any obtuse triangles.But is this true? Let me verify with a specific example.Take three vertices: (0,0,0), (1,1,0), (1,0,1). The distances are:From (0,0,0) to (1,1,0): sqrt(2).From (0,0,0) to (1,0,1): sqrt(2).From (1,1,0) to (1,0,1): sqrt((0)^2 + (1)^2 + (1)^2) = sqrt(2).So equilateral triangle, all angles 60 degrees.Another example: (0,0,0), (1,0,0), (1,1,1). Sides:From (0,0,0) to (1,0,0): 1.From (0,0,0) to (1,1,1): sqrt(3).From (1,0,0) to (1,1,1): sqrt(0 + 1 + 1) = sqrt(2).So sides are 1, sqrt(2), sqrt(3). The angle opposite sqrt(3) is 90 degrees.Another example: (0,0,0), (0,1,1), (1,0,1). Sides:From (0,0,0) to (0,1,1): sqrt(0 + 1 + 1) = sqrt(2).From (0,0,0) to (1,0,1): sqrt(2).From (0,1,1) to (1,0,1): sqrt(1 + 1 + 0) = sqrt(2). So equilateral.Another example: (0,0,0), (1,0,1), (1,1,0). As above, equilateral.Therefore, it seems that in the cube, all triangles are either right-angled or equilateral. Hence, no obtuse triangles. Therefore, eight points can be placed in space.But wait, this seems too good. Is there a configuration with more than eight points?If we consider the hypercube in 3D, which is the cube, we have eight vertices. What about other polyhedrons? For example, the regular icosahedron has 12 vertices. However, in an icosahedron, triangles are all equilateral, so all acute. However, when considering any three vertices, not just the faces. Wait, in a regular icosahedron, edges are all equal, but the distances between non-adjacent vertices vary. Some triangles formed by non-face vertices might have obtuse angles.For example, in a regular icosahedron, the distance between two vertices not connected by an edge is longer. Let's consider three vertices where two are adjacent, and the third is not. The triangle formed may have an obtuse angle.Alternatively, perhaps in the regular icosahedron, all triangles are acute. I need to check.The regular icosahedron has 12 vertices, 30 edges, and 20 triangular faces. Each face is equilateral. The question is whether any three vertices not forming a face have an obtuse triangle.According to some references, in a regular icosahedron, the dihedral angles are approx 138°, but the angles within the triangles would depend on the distances.Take three vertices where two are adjacent, and the third is across the icosahedron. For example, take a vertex A, its neighbor B, and a vertex C opposite to A. The distances AB is the edge length, AC is the distance across the icosahedron. The triangle ABC would have sides AB, AC, and BC. If AC is longer than AB, the angle at B might be obtuse.However, calculating exact angles requires specific coordinates. A regular icosahedron can be embedded with vertices at (0, ±1, ±φ), (±1, ±φ, 0), (±φ, 0, ±1), where φ is the golden ratio (~1.618). Let's pick three points:A(0, 1, φ), B(0, 1, -φ), and C(0, -1, φ). Wait, these points might not form a triangle with obtuse angles, but it's complicated.Alternatively, take three vertices: A(0, 1, φ), B(1, φ, 0), and C(φ, 0, 1). The distances between these points can be calculated, but it's tedious. Given the complexity, it's possible that some triangles in the regular icosahedron have obtuse angles, so it might not satisfy the problem's condition.Therefore, perhaps the cube with eight vertices is the maximum. But according to some research, in three dimensions, the maximum number of points with all triangles acute or right-angled is eight (cube), but I need to confirm.Wait, another thought: the hypercube in n dimensions has 2^n vertices, and in 3D, it's the cube. If in each dimension, the maximum is 2^n, then in 3D, it's eight. However, allowing right angles, maybe you can have more points.Alternatively, according to some references, the maximum number in three-dimensional space is eight, achieved by the cube vertices. However, if we can find a configuration with more than eight points, that would be better.Another approach: in 3D space, the maximum number of points such that every triangle is non-obtuse is 8. This is because you can place them at the vertices of a cube, and as shown, all triangles are either right-angled or equilateral. Adding a ninth point would necessarily create an obtuse triangle.But how to be sure? Suppose we try to add a ninth point. Where can we place it such that all triangles with existing eight points are non-obtuse. For example, adding a point at the center of the cube. Let's check.Point E at (0.5,0.5,0.5). Now, forming triangles with E and two cube vertices. For example, take E and two adjacent vertices: (0,0,0), (1,0,0), E. The sides are 1, sqrt(0.5), sqrt(0.5). The longest side is 1. Compute the angle at E.Law of Cosines:1² = 0.5 + 0.5 - 2*0.5*cosθ1 = 1 - cosθ => cosθ = 0 => θ = 90 degrees. So right-angled.Another triangle: E, (0,0,0), (1,1,1). The distance from E to (0,0,0) is sqrt(0.25*3) = sqrt(0.75) ≈ 0.866. Distance from E to (1,1,1) is same. Distance from (0,0,0) to (1,1,1) is sqrt(3) ≈ 1.732. The triangle has sides ≈0.866, ≈0.866, ≈1.732. The angle at E:Law of Cosines:(1.732)² ≈ 3 ≈ 0.75 + 0.75 - 2*0.75*cosφ3 ≈ 1.5 - 1.5 cosφ1.5 ≈ -1.5 cosφ => cosφ ≈ -1 => φ ≈ 180°, which is impossible. Wait, but points E, (0,0,0), (1,1,1) are colinear? Let's check. The line from (0,0,0) to (1,1,1) passes through E(0.5,0.5,0.5). Therefore, these three points are colinear, forming a degenerate triangle. Therefore, violating the problem's condition of no three points being colinear. Therefore, E cannot be placed at the center.Instead, place the ninth point near the center but not exactly. Let's say E(0.5,0.5,0.5 + ε). Now, forming triangle with (0,0,0), (1,1,1), and E. The distance from E to (0,0,0) is sqrt(0.25 + 0.25 + (0.5 + ε)^2). Similarly, distance from E to (1,1,1) is sqrt(0.25 + 0.25 + (0.5 - ε)^2). The distance between (0,0,0) and (1,1,1) is sqrt(3). Now, compute angles.But even with ε ≠ 0, the triangle may have an obtuse angle. For example, if ε is small, the point E is almost along the line, creating a very elongated triangle. The angle at E would be close to 180°, making it obtuse. Therefore, adding a ninth point is problematic.Therefore, it seems that eight points (cube vertices) is the maximum in three-dimensional space. Hence, the answer is:a) In the plane: 4 points (square vertices).b) In space: 8 points (cube vertices).But I need to verify if there's a configuration with more points in space.Another idea: the regular simplex in 3D is a tetrahedron with four points. The cross polytope (octahedron) has six points. The cube has eight points. Maybe the cube is the maximum.According to some mathematical results, in n-dimensional space, the maximum number of points such that all angles are non-obtuse is 2^n. In three dimensions, 2^3 = 8, which matches the cube. Therefore, the answer for part b) is eight.Therefore, the final answers are:a) 4b) 8Final Answera) boxed{4}b) boxed{8}

✅:To determine the maximum number of points that can be placed in a plane and in space such that none of the triangles formed by these points have an obtuse angle, we need to consider configurations where all triangles are either acute or right-angled. Part a) In the Plane1. Square Configuration: - Placing four points at the vertices of a square results in all triangles formed being right-angled. This configuration satisfies the condition as right angles are allowed. - Adding a fifth point introduces the risk of forming an obtuse triangle. For example, placing a point near the center of the square or perturbing a regular pentagon results in at least one obtuse triangle. - The regular pentagon and other configurations of five points were shown to have obtuse triangles, confirming that five points cannot be placed without forming an obtuse triangle.Conclusion: The maximum number of points in the plane is 4, forming a square. Part b) In Space1. Cube Configuration: - Placing eight points at the vertices of a cube results in all triangles formed being either right-angled or equilateral. This configuration satisfies the condition as right angles and acute angles are allowed. - The regular octahedron (six points) was considered, but the cube configuration allows more points without forming obtuse triangles. - Adding a ninth point introduces the risk of forming an obtuse triangle, especially when near the center of the cube, leading to degenerate or elongated triangles.Conclusion: The maximum number of points in space is 8, forming a cube. Final Answersa) boxed{4}b) boxed{8}

🔑:To determine the maximum number of points that can be placed on a plane (or in space) such that no triangle formed by these points is obtuse-angled, we need to consider both the conditions given and some fundamental geometric principles.# Part (a): On the PlaneLet's denote ( A_1, A_2, ldots, A_n ) as our ( n ) points on the plane.1. Observation: Any set of points satisfying the given condition must lie on the boundary of their convex hull. This is because any point inside the convex hull will form triangles that include obtuse angles with points on the boundary.2. Convex Polygon Angle Sum: If the points form a convex ( n )-sided polygon, the sum of the interior angles is ( 180^circ (n - 2) ). [ sum_{i=1}^n text{Angle}(A_i) = 180^circ (n - 2) ]3. Angle Condition: For no triangle to be obtuse, each angle of our convex polygon must be at most ( 90^circ ). Therefore, [ n times 90^circ geq 180^circ (n - 2) ]4. Solving the Inequality: Simplify the inequality: [ 90n geq 180(n - 2) ] [ 90n geq 180n - 360 ] [ 360 geq 90n ] [ n leq 4 ]5. Conclusion on the Plane: Thus, the maximum number of points that can be placed on a plane under the given condition is ( 4 ). An example of such an arrangement is the vertices of a square. [ boxed{4} ]# Part (b): In SpaceNow we consider the points in three-dimensional space.1. 3D Convex Hull Volume: Any point inside a convex solid in 3D space will form obtuse triangles with the boundary points. Therefore, the points must all lie on the boundary of the convex hull.2. Polyhedron with at most Right Angles: Consider a convex polyhedron ( V ). We construct ( V_i ) from ( V ) by translating ( V ) along vectors ( overrightarrow{A_1A_i} ). Due to the translation, the polyhedra ( V_i ) will intersect ( V ) only on the boundary.3. Volume Calculation: Let ( v ) be the volume of the original polyhedron ( V ). Since translations do not affect the volume and creating copies through translation, the minimum enveloping volume (convex hull) ( V' ) will have volume ( 8v ) (due to homothety of factor 2 centered at one vertex).4. Volume Inequality: Given the volume relationships and the non-overlapping interiors of ( V, V_1, ldots, V_n ), [ n times v leq 8v ] [ n leq 8 ]5. Conclusion in Space: Hence, the maximum number of points satisfying the condition in space is ( 8 ). An example of such an arrangement is the vertices of a cube. [ boxed{8} ]Conclusion:The maximum number of points where no triangle is obtuse is:- on the plane: ( 4 ) points- in space: ( 8 ) points.[boxed{a) 4 points, b) 8 points}]

🧠:Okay, so I need to figure out for which values of k the chances of A and B winning are equal when they randomly choose k numbers from the first 100 positive integers. The game is such that if the sum of the chosen numbers is even, A wins; otherwise, B wins. Hmm, interesting. Let me start breaking this down step by step.First, let's understand the problem. We have numbers from 1 to 100, which includes 50 even numbers and 50 odd numbers. When we pick k numbers randomly, the sum's parity (even or odd) depends on the number of odd numbers in the selected set. Because adding an even number doesn't change the parity, but adding an odd number flips it. So, if there's an even number of odd numbers in the selected k numbers, the sum is even; otherwise, it's odd. Therefore, the problem reduces to determining when the probability of selecting an even number of odd numbers is equal to the probability of selecting an odd number of odd numbers. That is, when is the probability that the number of odd numbers in the sample is even equal to the probability that it's odd?So, restating the problem: in a set of 100 numbers with 50 evens and 50 odds, when selecting a subset of size k, for which k is the probability that the subset contains an even number of odds equal to the probability that it contains an odd number of odds?Hmm, this seems like a classic combinatorial probability problem. Let me recall some related concepts. The number of ways to choose k numbers with exactly m odd numbers is C(50, m) * C(50, k - m), since we have to choose m odd numbers from the 50 available and k - m even numbers from the 50 evens. The total number of possible subsets is C(100, k). Therefore, the probability that the sum is even (i.e., even number of odds) is the sum over all even m of [C(50, m) * C(50, k - m)] divided by C(100, k). Similarly, the probability the sum is odd is the same but summing over odd m. We need these two probabilities to be equal.Therefore, the condition is:Sum_{m even} C(50, m) * C(50, k - m) = Sum_{m odd} C(50, m) * C(50, k - m)Which implies that the sum over all m of (-1)^m C(50, m) * C(50, k - m) = 0. Because if you subtract the two sums, you get Sum_{m} [1 - (-1)^m] C(50, m) C(50, k - m) = 0, but actually, more straightforwardly, if even sum equals odd sum, then their difference is zero. The difference can be written as Sum_{m} (-1)^m C(50, m) C(50, k - m) = 0.Alternatively, maybe using generating functions could help here. Let me recall that generating functions can be useful for such parity problems. For example, the generating function for selecting even numbers is (1 + x)^50, and similarly for odd numbers. But since we have two groups: 50 even numbers and 50 odd numbers. Wait, but in terms of generating functions, the generating function for choosing from the evens would be (1 + x)^50 (since each even can be chosen or not), and similarly for the odds, but perhaps with a twist. Wait, actually, since the evens don't affect the parity of the sum, and the odds do. Hmm, but perhaps considering the generating function where we track the parity.Alternatively, consider that the generating function for the number of subsets with even sum is ( (1 + x)^{50} + (1 - x)^{50} ) / 2, and similarly for odd sums it's ( (1 + x)^{50} - (1 - x)^{50} ) / 2. But we also have to consider that we are choosing from both evens and odds. Wait, perhaps not exactly. Let me think.Wait, actually, the total generating function for the entire set (100 numbers, 50 evens and 50 odds) would be the product of the generating functions for evens and odds. Since choosing an even number doesn't affect the parity, their generating function is (1 + x)^{50}. Choosing an odd number does affect the parity, so their generating function is also (1 + x)^{50}, but when considering parity, we can track it with a variable, say, y. Wait, perhaps using a two-variable generating function where one variable tracks the number of elements and the other tracks the parity. Let me recall that technique.Alternatively, use a substitution. Let me recall that in generating functions, if we want to consider the parity, we can substitute x with 1 and -1. Specifically, the sum over subsets of size k with even sum is equal to [f(1) + f(-1)] / 2, where f(x) is the generating function. Similarly, the sum over subsets with odd sum is [f(1) - f(-1)] / 2. Therefore, for the total generating function f(x) = (1 + x)^{100}, the number of subsets with even sum is [ (1 + 1)^{100} + (1 - 1)^{100} ] / 2 = [2^{100} + 0]/2 = 2^{99}. Similarly, the number of subsets with odd sum is also 2^{99}, which would mean that for the entire set of 100 numbers, the number of even and odd subsets is equal. Wait, but this seems contradictory, because in reality, the parity of the sum when selecting subsets from numbers 1 to 100. Wait, but if all numbers are considered, then indeed, each number can be either included or not, and the total number of subsets is 2^{100}. For each subset, the sum's parity is equally likely? That would be the case if the numbers are such that each bit is equally likely, but since numbers can be even or odd, maybe there's a bias. Wait, but in this problem, the numbers are fixed: 50 evens and 50 odds. So maybe when selecting a random subset, the parity of the sum depends on the number of odd numbers selected.But wait, the entire set of numbers includes both even and odd numbers. So the parity of the sum is determined by the number of odd numbers in the subset. Because each even number contributes 0 to the parity, and each odd number contributes 1. Therefore, the sum's parity is equal to the parity of the number of odd numbers in the subset. Therefore, the problem reduces to: when selecting k numbers from 50 evens and 50 odds, what is the probability that the number of odd numbers selected is even? We need this probability to be 1/2, equal to the probability of it being odd.Therefore, the question becomes: for which k is the number of subsets of size k with an even number of odd elements equal to the number with an odd number of odd elements?In combinatorics, this is equivalent to asking when the number of such subsets is equal. So, let me denote E(k) as the number of ways to choose k numbers with an even number of odds, and O(k) as the number with an odd number of odds. Then, we need E(k) = O(k). Since the total number of subsets is C(100, k), this would imply that E(k) = O(k) = C(100, k)/2.To find such k, we need to analyze E(k) and O(k). Let me recall that E(k) = Sum_{m even} C(50, m) * C(50, k - m), and O(k) = Sum_{m odd} C(50, m) * C(50, k - m). Then, E(k) - O(k) = Sum_{m} (-1)^m C(50, m) * C(50, k - m). So, setting E(k) - O(k) = 0 is equivalent to Sum_{m} (-1)^m C(50, m) * C(50, k - m) = 0.Alternatively, using generating functions. Let me recall that the generating function for E(k) - O(k) is given by:Sum_{k} [E(k) - O(k)] x^k = [C(50,0) - C(50,1)x + C(50,2)x^2 - ... + (-1)^{50}C(50,50)x^{50}] * [C(50,0)x^0 + C(50,1)x^1 + ... + C(50,50)x^{50}]Wait, no. Wait, actually, the generating function for E(k) is ( (1 + x)^{50} + (1 - x)^{50} ) / 2 * (1 + x)^{50}, and similarly for O(k) it's ( (1 + x)^{50} - (1 - x)^{50} ) / 2 * (1 + x)^{50}. Wait, maybe not. Let me think again.Wait, the number of ways to choose even number of odds is Sum_{m even} C(50, m) * C(50, k - m). So, the generating function would be [Sum_{m even} C(50, m) x^m] * [Sum_{n} C(50, n) x^n]. Similarly for the odd. The generating function for even m is ( (1 + x)^{50} + (1 - x)^{50} ) / 2, and for odd m is ( (1 + x)^{50} - (1 - x)^{50} ) / 2. Therefore, the generating function for E(k) is [(1 + x)^{50} + (1 - x)^{50}]/2 * (1 + x)^{50}, and for O(k) it's [(1 + x)^{50} - (1 - x)^{50}]/2 * (1 + x)^{50}. Therefore, the difference E(k) - O(k) would be [(1 + x)^{50} + (1 - x)^{50}]/2 * (1 + x)^{50} - [(1 + x)^{50} - (1 - x)^{50}]/2 * (1 + x)^{50} = [ (1 - x)^{50} ] * (1 + x)^{50}.Therefore, E(k) - O(k) = [ (1 + x)^{50} * (1 - x)^{50} ] evaluated at x^k coefficient. But (1 + x)^50*(1 - x)^50 = (1 - x^2)^50. Therefore, the difference E(k) - O(k) is equal to the coefficient of x^k in (1 - x^2)^50. Therefore, E(k) - O(k) = (-1)^{k/2} C(50, k/2) if k is even, and 0 otherwise. Wait, let me check.Wait, (1 - x^2)^50 expands as Sum_{m=0}^{50} C(50, m) (-1)^m x^{2m}. Therefore, the coefficient of x^k is (-1)^{m} C(50, m) if k = 2m, and 0 otherwise. Therefore, E(k) - O(k) is 0 when k is odd, and (-1)^m C(50, m) when k = 2m. Therefore, for even k, E(k) - O(k) = (-1)^{k/2} C(50, k/2). Therefore, in order to have E(k) = O(k), we need E(k) - O(k) = 0. Therefore, when is (-1)^{k/2} C(50, k/2) = 0? But C(50, k/2) is only zero when k/2 > 50, but k is at most 100. However, (-1)^{k/2} times C(50, k/2) can only be zero if C(50, k/2) is zero, which is not the case for k up to 100. Wait, but for even k, the difference is non-zero, and for odd k, the difference is zero. Wait, this seems conflicting with the earlier conclusion.Wait, hold on. If k is odd, then E(k) - O(k) = 0? But according to the expansion above, the coefficient of x^k in (1 - x^2)^50 is zero when k is odd, so yes. Therefore, for odd k, E(k) - O(k) = 0, which would mean that E(k) = O(k) for all odd k. But that can't be right, can it? If k is odd, then choosing k numbers, the number of odd numbers can be even or odd, but is it symmetric?Wait, let me take a small example. Suppose we have 2 even numbers and 2 odd numbers, total 4 numbers. Let k = 1. Then, the number of subsets with even sum (even number of odds) is choosing 0 odds and 1 even: C(2,0)*C(2,1) = 2. The number of subsets with odd sum is choosing 1 odd and 0 evens: C(2,1)*C(2,0) = 2. So, E(1) = O(1) = 2. Wait, so in this case, with k odd, E(k) = O(k). But in the original problem, if we have 50 evens and 50 odds, and k is odd, then according to the generating function, E(k) - O(k) = 0, which would mean E(k) = O(k). But in the small example, this holds. So, indeed, for odd k, the number of subsets with even sum equals the number with odd sum. Hmm, that's interesting.But in the original problem, the user is asking for which k the chances are equal. So, according to this, all odd k would satisfy E(k) = O(k). But that seems counterintuitive. Wait, let's take another example. Suppose we have 1 even and 1 odd number. Total numbers = 2. Let k = 1. Then, subsets of size 1: 1 even or 1 odd. Sum even: 1 subset (even), sum odd: 1 subset (odd). So, equal. Similarly, if k = 1 in this case, it's equal. But if we have 2 evens and 2 odds, k=1, as above, also equal. Wait, so maybe for any k, if the total number of evens and odds are equal, then for odd k, the probabilities are equal? Is that a general result?Wait, but let's take a different example. Suppose we have 3 evens and 1 odd. So total 4 numbers. Let k=1. Then, E(1) = number of subsets with 0 odds and 1 even: C(3,1) = 3. O(1) = subsets with 1 odd and 0 evens: C(1,1) = 1. So 3 vs 1, not equal. Therefore, in this case, with unequal number of evens and odds, for k odd, E(k) ≠ O(k). But in the previous examples where the number of evens and odds are equal, for k odd, E(k) = O(k). So perhaps when the number of evens and odds are equal, then for any odd k, E(k) = O(k). That seems to be the case.Wait, but let's verify with another example where evens and odds are equal. Suppose 2 evens and 2 odds, k=3. Then, E(3) is the number of subsets of size 3 with even number of odds (i.e., 0 or 2 odds). 0 odds: C(2,3) which is 0. 2 odds: C(2,2)*C(2,1) = 1*2=2. So E(3)=0 + 2 = 2. O(3) is subsets with 1 or 3 odds. 1 odd: C(2,1)*C(2,2)=2*1=2. 3 odds: C(2,3)*C(2,0)=0. So O(3)=2+0=2. Therefore, E(3)=O(3). So, even when k is odd, E(k)=O(k).Another example: 4 evens and 4 odds, k=5. Hmm, let's see. E(5) would be subsets with 0, 2, 4 odds. O(5) would be 1,3,5 odds. Let's compute E(5): C(4,0)*C(4,5) + C(4,2)*C(4,3) + C(4,4)*C(4,1). But C(4,5)=0, so first term 0. Second term: 6 * 4 =24. Third term:1 *4=4. So E(5)=24+4=28. O(5): C(4,1)*C(4,4) + C(4,3)*C(4,2) + C(4,5)*C(4,0). C(4,1)*C(4,4)=4*1=4. C(4,3)*C(4,2)=4*6=24. C(4,5)=0. So O(5)=4+24=28. So E(5)=O(5)=28. Hence, again equal.Therefore, when the number of evens and odds are equal, then for any odd k, E(k)=O(k). Therefore, in the original problem, since there are 50 evens and 50 odds, all odd k from 1 to 99 would satisfy E(k)=O(k). Therefore, the chances of A and B winning are equal for all odd k.But wait, the answer can't be all odd k. Let me check k=0. If k=0, then the sum is 0, which is even, so A always wins. So, k=0 is not in the solution set. Similarly, k=100: the sum is the sum of all numbers from 1 to 100. The sum of 1 to 100 is (100*101)/2=5050, which is even. Therefore, when k=100, the sum is even, so A always wins. Therefore, k=100 is not in the solution set. So the solution set is all odd k from 1 to 99. But wait, does this hold?Wait, let me take k=50, which is even. Then, according to our previous analysis, E(50) - O(50) = (-1)^{25} C(50,25). Which is non-zero, so E(50) ≠ O(50). So for even k, E(k) ≠ O(k). For odd k, E(k) - O(k)=0, so E(k)=O(k). Therefore, the answer is all odd k. But in the problem statement, it's stated that k numbers are chosen from the first 100 positive integers. So k can be from 1 to 100. Therefore, all odd k from 1 to 99 inclusive. But wait, the problem says "random k numbers are chosen", so k is a fixed number, and the question is for which k the probabilities are equal. So according to our analysis, all odd k. However, I should verify this with an example.Take k=1. Choosing 1 number: the probability that it's even is 50/100=1/2, odd is 1/2. The sum is even if the number is even, odd if the number is odd. So A and B have equal chances. Similarly, k=3: the probability that the number of odd numbers is even (0 or 2) vs odd (1 or 3). But wait, with k=3, the number of odd numbers can be 0,1,2,3. The total number of subsets: C(100,3). Number of subsets with 0 odds: C(50,0)*C(50,3). With 2 odds: C(50,2)*C(50,1). So E(3)=C(50,0)*C(50,3) + C(50,2)*C(50,1). O(3)=C(50,1)*C(50,2) + C(50,3)*C(50,0). Which is symmetric. Therefore, E(3)=O(3). So yes, equal.Similarly, for k=99. The number of subsets of size 99 is equal to the number of subsets of size 1 (since choosing 99 elements is equivalent to leaving out 1). Therefore, the parity of the number of odds in 99 elements is the same as the parity of the number of evens in the 1 element left out. Since the total number of odds is 50, the number of odds in the subset of 99 is 50 - (number of odds left out). If the left-out element is even, then the number of odds in the subset is 50 (even), if left-out is odd, then 49 (odd). Therefore, the number of subsets with even sum is equal to the number of left-out evens (50), and subsets with odd sum is equal to the number of left-out odds (50). Therefore, equal. So yes, k=99 also works.Therefore, it seems that all odd k from 1 to 99 inclusive result in equal probabilities for A and B. However, let me check with k=50. Wait, k=50 is even. Then, according to our generating function analysis, E(50) - O(50) = (-1)^{25} C(50,25). Which is negative of C(50,25). Therefore, E(50) = O(50) - C(50,25). Therefore, E(50) < O(50). So their probabilities are not equal. Therefore, only odd k satisfy the condition.Wait, but the problem states that the numbers are chosen from the first 100 positive integers. The first 100 positive integers include 50 even and 50 odd numbers. So, our analysis holds. Therefore, the answer should be all odd integers k between 1 and 99 inclusive. But let me check a specific case where k is even. For example, k=2. Let's compute E(2) and O(2). E(2) is the number of subsets with 0 or 2 odds. 0 odds: C(50,0)*C(50,2). 2 odds: C(50,2)*C(50,0). So E(2)=2*C(50,2). O(2) is subsets with 1 odd: C(50,1)*C(50,1). So E(2)=2*C(50,2)=2*(1225)=2450. O(2)=50*50=2500. So 2450 vs 2500, not equal. Therefore, for k=2, probabilities are not equal. Thus, only odd k satisfy the condition.Therefore, the conclusion is that for all odd k from 1 to 99 inclusive, the chances of A and B are equal. Therefore, the answer is all odd integers k between 1 and 99. But the problem is presented as "random k numbers are chosen from the first 100 positive integers". So, k can be 1 to 100, inclusive. Hence, the answer is all odd k in 1 ≤ k ≤ 99.But wait, let me confirm with k=1 and k=99. For k=1, as we saw, it's 50/50. For k=99, since choosing 99 elements is the same as leaving out 1, and the parity of the sum of 99 elements is the same as the parity of the left-out element (since sum_total = sum_subset + sum_leftout). Since the total sum is even (sum from 1 to 100 is 5050), then sum_subset is even if and only if sum_leftout is even. Therefore, the parity of the subset sum is equal to the parity of the left-out element. Therefore, the probability that the subset sum is even is equal to the probability that the left-out element is even, which is 50/100=1/2. Therefore, same as k=1. Therefore, k=99 also works. Similarly, for k=3, as we saw earlier, it works.Therefore, all odd k. Hence, the answer is all odd integers k from 1 to 99 inclusive. But the problem asks "for which values of k are the winning chances of A and B equal?" So, the answer is all odd k between 1 and 99. But to write this in mathematical terms, it's k ≡ 1 mod 2, 1 ≤ k ≤ 99. Or, in boxed notation, the values of k are all odd integers from 1 to 99. But in the problem statement, it might require a specific form. Let me see if there's another way. Sometimes in such problems, when the number of elements is even, the answer is all odd k. Here, since there are 50 even and 50 odd numbers, the symmetry leads to all odd k.Alternatively, is there a case where even k could also work? For example, if the number of odd and even numbers were unequal, but in this problem, they are equal. Therefore, the symmetry when k is odd allows for equal probabilities. Hence, the answer is all odd k.But let me check another even k, say k=4. E(4) = C(50,0)C(50,4) + C(50,2)C(50,2) + C(50,4)C(50,0) = C(50,4) + [C(50,2)]^2 + C(50,4). Similarly, O(4) = C(50,1)C(50,3) + C(50,3)C(50,1) = 2*C(50,1)C(50,3). Calculating these would take time, but intuitively, since the difference E(k) - O(k) for even k is non-zero, as per the generating function, which is (1 - x^2)^50, the coefficient for x^k is non-zero for even k, hence E(k) ≠ O(k). Therefore, only odd k satisfy E(k) = O(k).Therefore, the answer is all odd integers k between 1 and 99 inclusive. Thus, the final answer is that the winning chances are equal for all odd values of k from 1 to 99. Therefore, in boxed notation, we can write this as all integers k where k is odd, i.e., k ≡ 1 mod 2, but since the problem is in the context of a competition, likely expecting the answer as "all odd integers between 1 and 99" or in a more compact form.But let me recall that in some similar problems, when the number of elements is even and the number of odd and even elements are equal, the probabilities are equal when k is odd. This is because selecting an odd number of elements, the parity is equally likely due to the symmetry between even and odd numbers. Hence, the answer is all odd k. Therefore, the box answer is boxed{k text{ is odd}} or in specific terms, boxed{k text{ odd}}.However, depending on the problem's expectations, sometimes specific bounds are required. Since k is chosen from 1 to 100, and k=100 is even (sum even), and k=0 is excluded, the odd k are from 1 to 99. Hence, specifying all odd integers between 1 and 99 inclusive. But in the problem statement, it just says "the first 100 positive integers", and k is any number chosen, so k can be from 1 to 100. But since k=100 gives a certain win to A, it's excluded. Hence, the answer is all odd k from 1 to 99. However, in mathematical terms, it's more elegant to say "all odd k", but given the range, specifying 1 ≤ k ≤ 99 and k odd.But the problem might expect the answer in a specific form. Let me check again the problem statement:"For which values of ( k ) are the winning chances of ( A ) and ( B ) equal?"So, the answer is all odd integers k between 1 and 99 inclusive. In LaTeX, this can be written as boxed{k text{ is odd}} but given the problem is likely expecting numerical values or a specific characterization. Alternatively, if the answer is all odd numbers, then that's the case. But let me recall that in the case of equal number of even and odd elements, the number of subsets with even size parity is equal to the number with odd parity when the subset size is odd. Therefore, the answer is all odd k. So, boxed as boxed{k text{ is odd}}.But in some contexts, especially competitions like Olympiads, the answer is expected to be given as boxed{all odd integers between 1 and 99} or similar. However, given the problem's phrasing, it's more likely that the answer is simply all odd k, so boxed{k text{ is odd}}.But to confirm, let me consider another approach. Let's use probability. The probability that the sum is even is equal to 1/2. For each element, the probability that it's selected is k/100. Wait, no, it's a random subset of size k, not each element being selected independently. Therefore, different approach.The number of odd numbers in a sample of size k from a population of 50 odd and 50 even is hypergeometrically distributed. The probability mass function is P(m) = C(50, m) * C(50, k - m) / C(100, k). We need Sum_{m even} P(m) = Sum_{m odd} P(m) = 1/2.In the hypergeometric distribution, when is the probability of even success equal to odd? There's a known result that for a hypergeometric distribution with parameters N, K, n, if K = N/2, then the probabilities of even and odd numbers are equal when n is odd. Wait, in our case, N=100, K=50 (number of successes, i.e., odd numbers), and n=k. So, if N is even, K = N/2, then the hypergeometric probabilities for even and odd numbers of successes are equal if and only if n is odd. This seems to be the case here.Therefore, this result directly tells us that when the population has an equal number of successes and failures (K = N/2), then the probability of getting an even number of successes is equal to the probability of getting an odd number of successes if and only if the sample size n is odd. Therefore, in our problem, since N=100, K=50, which is N/2, then the required condition is that the sample size k is odd. Therefore, the answer is all odd k.Hence, confirming our previous conclusion. Therefore, the answer is all odd integers k from 1 to 99 inclusive, which can be succinctly written as "all odd k" given the context. Therefore, boxed as boxed{k text{ is odd}}.Final AnswerThe winning chances of ( A ) and ( B ) are equal for all odd values of ( k ). Thus, the solution is boxed{k text{ is odd}}.

✅:To determine for which values of ( k ) the winning chances of ( A ) and ( B ) are equal when ( k ) numbers are chosen from the first 100 positive integers, we analyze the parity of the sum of the chosen numbers. The key steps are as follows:1. Parity Analysis: The sum of the chosen numbers is even if and only if the number of odd numbers in the chosen set is even. This reduces the problem to determining when the probability of selecting an even number of odd numbers is equal to the probability of selecting an odd number of odd numbers.2. Combinatorial Probability: We need the number of subsets with an even number of odd numbers to be equal to the number of subsets with an odd number of odd numbers. This can be expressed using generating functions or hypergeometric distribution properties.3. Generating Functions: The generating function for the difference between the number of subsets with even and odd sums is ((1 - x^2)^{50}). The coefficient of ( x^k ) in this expansion is zero for odd ( k ), indicating that the number of subsets with even and odd sums are equal for odd ( k ).4. Hypergeometric Distribution: When the population has an equal number of successes (odd numbers) and failures (even numbers), the probability of an even number of successes is equal to the probability of an odd number of successes if and only if the sample size ( k ) is odd.5. Verification: Examples and verification with small cases confirm that for odd ( k ), the probabilities are equal, while for even ( k ), they are not.Thus, the winning chances of ( A ) and ( B ) are equal for all odd values of ( k ). The final answer is:[boxed{k text{ is odd}}]

🔑:To determine the values of k for which players A and B have equal chances of winning, we need to carefully analyze the parity of the sum of k randomly selected numbers from the first 100 positive integers. This analysis involves understanding when the sum is even and when it is odd.# Step-by-Step Analysis:1. Even and Odd Sum Selection: Consider the first 100 positive integers {1, 2, ..., 100}. We need to select k numbers randomly and check the parity of their sum. For the sum to be even, the number of odd numbers selected must be even; otherwise, the sum will be odd.2. Handling Edge Cases with 1 and 100: Start by examining groups that include either 1 or 100, but not both. These groups can be paired such that one group’s sum is even and the other’s is odd. This method can be generalized: choose (k-1) numbers from the first 98 numbers and add either 1 or 100 as the k-th number. This ensures each pair has one even and one odd sum. 3. Generalizing for all Adjacent Pairs: Extend this idea to other adjacent number pairs like {2, 99}, {3, 98}, and so on up to {50, 51}. For each pair, consider (k-1) numbers selected from the remaining numbers in the set, combined with one of the pair, ensuring that within each group the sums are balanced between even and odd.4. Impact of Parity on k: When k is odd, there is an equal amount of even and odd sums for the combinations that include any particular number combination. The reasoning follows that pairing always includes an odd number at least once, leading to the sum-changing nature evenly distributed.5. Counting Non-Paired Selections for Even k: For the remaining selections that can’t be paired using the method above, these selections must include both elements from several pairs. All combinations involving exact pairs will result in `k` group sums being all even or all odd, based on whether k/2 is odd or even.6. Conclusion for Winning Odds: - Odd k: Any sum-parity balancing rests on the equal distribution and such balance will favor neither player. [ boxed{text{For odd } k, text{ both } A text{ and } B text{ have equal chance of winning.}} ] - Even k: Only those selections where exactly k/2 is even produce strictly even sums and exact k/2 odd remain odd: - If k/2 is even, sums become naturally even, and hence A wins. - If k/2 is odd, sums become odd, favoring B.7. Formalizing the Case with Combinations: Using combinatorial calculations, verify the counts associated with text{choose } k from 100: [ binom{100}{k} ] Different solutions using combinatorial differences can confirm: [ text{For } k text{ odd: } ] [ frac{1}{2}binom{100}{k} ] For the exact difference in cases: [ frac{1}{2}left( binom{100}{k} - binom{50}{k/2} right) ] Conclusion:The equal winning chances are thus specifically under the condition of odd selections (k being odd). For other values, evaluate the combinatorial counts remaining after structured pairing methodologies confirming imbalance in sum parity selections.[boxed{k text{ is odd }}]

🧠:Okay, so I need to prove that a good botanical guide can't have more than 50 plants. The definition of a good guide here is that any two plants differ by more than half of the characteristics. Each plant has 100 characteristics, each either present or absent. So, essentially, each plant can be represented as a binary vector of length 100, right? And the condition is that the Hamming distance between any two vectors is more than 50. Hamming distance is the number of positions where the bits differ, so if two plants differ in more than 50 characteristics, their Hamming distance is greater than 50.The problem is asking to show that the maximum number of such vectors (plants) is 50. Wait, but 50 seems a bit small. Wait, actually, no. Let me think. In coding theory, there are bounds on the maximum number of codewords given certain distances. Maybe this is related to the concept of error-correcting codes. Specifically, binary codes with certain minimum Hamming distances. There's the Singleton bound, Hamming bound, Gilbert-Varshamov bound, and the Elias-Bassalygo bound, but maybe the most relevant here is the Johnson bound or the Plotkin bound?Plotkin's theorem comes to mind. Let me recall. Plotkin's theorem states that if the minimum distance d of a code satisfies d > (n-1)/2 * k, then the number of codewords is bounded. Wait, maybe not exactly. Let me check the exact statement.Wait, Plotkin's theorem says that if the minimum distance d of a binary code of length n satisfies d > n/2, then the maximum number of codewords is floor(n/(2d - n)). Wait, is that right? Let me confirm. For example, if n=100, d=51, which is more than n/2=50. Then according to Plotkin's theorem, the maximum number of codewords is floor(100/(2*51 - 100)) = floor(100/(102 - 100)) = floor(100/2) = 50. Exactly! So this would directly give the result that the maximum number is 50. But since the problem states "more than half", so d >= 51, so Plotkin's theorem would apply here. Therefore, the maximum number is 50. Therefore, the answer is that you can't have more than 50 plants.But maybe the user expects a proof without invoking Plotkin's theorem, since this seems like a math competition problem, and maybe the solution is expected to use more elementary methods. So perhaps I need to construct a proof using linear algebra or combinatorial arguments.Alternatively, think about it this way: suppose there are N plants. Each plant is a vector in {0,1}^100. The condition is that for any two distinct vectors, their Hamming distance is at least 51. We need to show that N <= 50.Alternatively, maybe use the pigeonhole principle. Let's consider the sum of all pairwise distances. If each pair has distance at least 51, then the total sum is at least 51 * C(N,2). On the other hand, we can find an upper bound for the total sum of all pairwise distances.Alternatively, consider that for each characteristic, the number of pairs of plants that differ in that characteristic can be counted. For each characteristic, if k_i plants have the characteristic present, then the number of pairs differing in that characteristic is k_i*(N - k_i). Since each characteristic contributes 0 or 1 to the Hamming distance between two plants, the total sum of all pairwise Hamming distances is equal to the sum over all characteristics of k_i*(N - k_i).So total sum S = Σ_{i=1 to 100} k_i*(N - k_i).But each pairwise distance is at least 51, so S >= 51 * C(N,2) = 51 * N(N-1)/2.On the other hand, S is also equal to Σ_{i=1 to 100} k_i*(N - k_i). Let's compute this sum. Note that k_i*(N - k_i) is maximized when k_i = N/2, so each term is at most N^2/4. Therefore, the total sum S <= 100*(N^2/4) = 25N^2.Therefore, 25N^2 >= 51*N(N-1)/2.Multiply both sides by 2: 50N^2 >= 51N(N - 1)Divide both sides by N (assuming N > 0): 50N >= 51(N - 1)50N >= 51N - 5150N -51N >= -51- N >= -51Multiply both sides by -1 (reverse inequality):N <= 51Hmm, but that gives N <= 51, which is not tight enough. The actual answer should be N <= 50. So this approach gives a weaker result.Alternatively, maybe we can use a better bound on the sum S. Each term k_i*(N - k_i) can be rewritten as Nk_i - k_i^2. The sum over i would be N Σk_i - Σk_i^2.But maybe that's not helpful. Alternatively, note that if we use the Cauchy-Schwarz inequality. The sum Σk_i*(N - k_i) can be written as Σk_i*N - Σk_i^2 = N Σk_i - Σk_i^2.But perhaps Cauchy-Schwarz on Σk_i^2?Σk_i^2 >= (Σk_i)^2 / 100 by Cauchy-Schwarz.But let's see. Let’s denote T = Σk_i. Then Σk_i^2 >= T^2 / 100. Therefore, S = N*T - Σk_i^2 <= N*T - T^2 / 100.But then S <= N*T - T^2 /100. So we have:N*T - T^2 /100 >= 51*N(N-1)/2.But this seems complicated. Alternatively, maybe we can think of the average value of k_i.The average k_i is T /100. Let’s denote average k_i = μ. Then Σk_i = 100μ. Then Σk_i*(N - k_i) = N Σk_i - Σk_i^2 = 100Nμ - Σk_i^2.But Σk_i^2 >= 100μ^2 by Cauchy-Schwarz (or Jensen's inequality, since the function f(x) = x^2 is convex). So Σk_i^2 >= 100μ^2.Therefore, S = 100Nμ - Σk_i^2 <= 100Nμ - 100μ^2 = 100μ(N - μ).So S <= 100μ(N - μ). But also, S >= 51*N(N-1)/2.Therefore:100μ(N - μ) >= 51*N(N -1)/2Divide both sides by N:100μ(1 - μ/N) >= 51(N -1)/2But I'm not sure if this helps. Let's see. Let's suppose that μ = N/2, since the maximum of μ(N - μ) is at μ = N/2. Then S <= 100*(N/2)*(N - N/2) = 100*(N^2/4) = 25N^2, which is the same as before, leading to 25N^2 >= 51*N(N -1)/2, which gives N <=51.But the problem is that this approach is too lossy. We need a better bound.Alternatively, consider using linear algebra. Suppose we have N vectors in {0,1}^100 with pairwise Hamming distance >50. Let's consider these vectors as vectors in F_2^100. The inner product of two vectors u and v is equal to the number of coordinates where they agree minus the number where they disagree, modulo 2. But maybe the inner product isn't directly helpful here.Alternatively, consider the following: for any two distinct vectors u and v, the Hamming distance is >50, which is equivalent to saying that their dot product (mod 2) is less than or equal to 50 - 1, but I don't know.Wait, the Hamming distance between u and v is equal to the weight (number of 1s) of u + v (mod 2). So if the Hamming distance is >50, then the weight of u + v is >50. So each u + v has weight >50.Alternatively, consider the following approach: use the concept of antipodal pairs. For each vector u, consider its complement, where all bits are flipped. The complement of u would have a Hamming distance of 100 from u. But if the code (set of vectors) includes both u and its complement, then the distance between them is 100. But in our problem, the distance between any two distinct vectors must be >50, so including a vector and its complement is allowed. However, if the code has the property that no two vectors are too close, but how does that limit the size?Alternatively, think of the sphere packing bound. Each pair of vectors must be at least distance 51 apart, so the spheres of radius 25 around each codeword (in Hamming distance) do not overlap. The size of each sphere is the number of vectors within distance 25 from a codeword. The total number of vectors in all spheres must be at most 2^100. Therefore, N * V(n,25) <= 2^100, where V(n, r) is the volume of a Hamming ball of radius r in n dimensions. However, calculating V(100,25) is complicated, and this would give a bound on N, but I don't think this would give exactly 50. Moreover, sphere packing bounds are usually not tight.Alternatively, maybe use a probabilistic method? Not sure.Wait, perhaps another approach. Let me consider the vectors as elements of {0,1}^100. Suppose we have N vectors such that any two differ in more than 50 coordinates. Let's consider the sum over all pairs of the distance between them. As before, S >= 51 * C(N,2).But we also have S = Σ_{i=1}^100 k_i(N - k_i), where k_i is the number of vectors with a 1 in the i-th coordinate. Let's compute the average value of k_i(N - k_i). Since there are 100 coordinates, the average per coordinate is S / 100. So average k_i(N - k_i) >= [51 * C(N,2)] / 100.But since k_i(N - k_i) is maximized when k_i = N/2, giving N^2/4. The average can't exceed N^2/4, so:[51 * N(N - 1)/2] / 100 <= N^2/4Multiply both sides by 100:51 * N(N - 1)/2 <= 25N^2Multiply both sides by 2:51N(N - 1) <= 50N^251N^2 - 51N <= 50N^251N^2 -50N^2 -51N <=0N^2 -51N <=0N(N -51) <=0Which implies that N <=51 (since N>0). But again, this gives N <=51, which is not the desired result. So this approach is not sufficient.Wait, but the problem states "more than half", so the minimum distance is at least 51. So maybe the exact Plotkin bound is necessary here.Let me recall Plotkin's theorem precisely. The theorem states that if C is a code of length n over an alphabet of size q, with minimum distance d. If q*d > (q-1)*n, then |C| <= floor(q*d / (q*d - (q-1)*n)). Wait, maybe for binary codes, q=2. So if 2d >n, then |C| <= floor(2d / (2d -n)). In our case, d=51, n=100. Then 2d -n =102-100=2, so |C| <= floor(2*51 /2)=51. Wait, but 2d/(2d -n)=102/2=51. So floor(51)=51. But the theorem says that if 2d >n, then |C| <= floor(2d/(2d -n)). So in our case, |C| <=51. But the problem says that you can't have more than 50. There's a discrepancy here.Wait, maybe Plotkin's theorem has a different formulation. Let me check another source.Another version of Plotkin's theorem: Let C be a binary code of length n with minimum distance d. If d > n/2, then |C| <= floor( d / (2d -n) ). Wait, but in our case, d=51, n=100. So floor(51 / (102 -100))=floor(51/2)=25. Which is not matching. Hmm, maybe my recollection is off.Wait, perhaps the correct formula is |C| <= floor( (n*d) / (2d -n) ). Let me see. For example, if n=4, d=3, then 2d -n=6-4=2, so |C| <= (4*3)/2=6. But in reality, the maximum code size for n=4, d=3 is 4 (simplex code). So maybe this formula isn't correct.Alternatively, perhaps the theorem states that if d >n/2, then |C| <= floor( (d) / (2d -n) ). For n=100, d=51, floor(51/(102-100))=floor(51/2)=25. But that contradicts the previous thought.Wait, perhaps I need to refer to the actual Plotkin bound.Upon checking, Plotkin's theorem for binary codes states that if d > n/2, then the maximum number of codewords |C| satisfies |C| <= floor( d / (2d -n) ). But in our case, d=51, n=100, so 2d -n=102-100=2, so |C| <= floor(51/2)=25. But this is not correct, since we can have codes with more codewords. For example, the binary repetition code of length 4 with two codewords has distance 4, which is greater than 2, and |C|=2, but according to the formula, it would give floor(4/(8-4))=1, which is not matching. So maybe my application is wrong.Wait, maybe another version. Another source says:Theorem (Plotkin): If C is a code of length n with minimal distance d over an alphabet of size q, then if q*d > (q-1)*n, then |C| <= floor( q*d / (q*d - (q-1)*n) )For binary codes, q=2, so if 2d >n, then |C| <= floor( 2d / (2d -n) )In our case, d=51, n=100. Then 2d=102 >100, so |C| <= floor(102/(102-100))=floor(102/2)=51. So 51. But the problem states that no more than 50 plants can be described. So there's a conflict here. Is the Plotkin bound 51 or 50?Wait, maybe the exact bound is 50. Maybe when using Plotkin's theorem in some formulations, they require the distance to be at least (n+1)/2 when n is odd. But in our case, n=100 is even, so d>50 is 51. The formula gives 51, but the problem says 50. So why is there a difference?Wait, perhaps because Plotkin's theorem gives an upper bound, but sometimes it's not tight. For example, in even lengths, the maximum size might be lower. Let me check specific known codes.For example, in n=4, d=3. Plotkin gives |C| <= floor(6/(6-4))=floor(6/2)=3. But the maximum code is 4 (simplex code). Wait, no. Wait, in n=4, d=3, the maximum code is actually 4, since you can have all vectors with even weight, but distance 2. Wait, no. Wait, if you want minimal distance 3, the maximum code is 4, which is the simplex code. But according to Plotkin's bound, it's 3. So here Plotkin's bound is not tight. So maybe Plotkin's theorem gives an upper bound, but the actual maximum can be lower or higher? Wait, no, in that case, the code has size larger than Plotkin's bound, which would mean the bound is incorrect. But that's not possible. Wait, maybe I made a mistake.Wait, for n=4, d=3, according to Plotkin's theorem: If 2d >n, which is 6>4, so yes. Then |C| <= floor(6/(6-4))=3. But in reality, the maximum code with d=3 in n=4 is 4. So this contradicts. Therefore, my understanding of Plotkin's theorem must be wrong.Alternatively, maybe Plotkin's theorem has different conditions. Let me check again.Another source says: Plotkin's theorem states that for a code of length n with minimal distance d, if d >= (q-1)n/q, then |C| <= qd/(qd - (q-1)n). For binary codes, q=2, so if d >=n/2, then |C| <= 2d/(2d -n). However, if 2d -n divides 2d, then equality holds. Otherwise, the floor is taken. But in our case, d=51, n=100, 2d -n=2, so |C| <=51. But in reality, the maximum code size is 50. So where is the problem?Wait, maybe when d is exactly n/2 +1, then the Plotkin bound gives |C| <= floor(2d / (2d -n)) = floor( (n+2)/ (2)), since d =n/2 +1. For n even, this becomes floor( (n+2)/2 ) = n/2 +1. But in our case, n=100, d=51, so floor(102/2)=51. However, the actual maximum code size might be lower. Perhaps in even lengths, when d=n/2 +1, the maximum code size is n/2, not n/2 +1. So maybe there's an improvement to the Plotkin bound in specific cases.Alternatively, maybe the problem can be approached via orthogonality. Suppose we have N vectors in {0,1}^100 with pairwise Hamming distance >50. Let's convert these vectors to vectors in {1,-1}^100 by replacing 0 with 1 and 1 with -1. Then the inner product of two vectors u and v is equal to 100 - 2*d(u,v), where d(u,v) is the Hamming distance. Since d(u,v) >50, then the inner product u·v <100 - 2*51= -2. So all pairwise inner products are < -2.Now, consider the Gram matrix of these vectors. The Gram matrix G is an NxN matrix where G_{i,j} = u_i · u_j. The diagonal entries are 100, since each vector has norm squared 100. The off-diagonal entries are < -2. Now, we can use the fact that the Gram matrix must be positive semi-definite. Therefore, the matrix G must be positive semi-definite. Let's see if we can derive a contradiction for N >50.Let’s denote the all-ones vector as j. Then consider the quadratic form j^T G j. This is equal to Σ_{i,j} G_{i,j} = Σ_{i=1}^N Σ_{j=1}^N u_i · u_j = (Σ_{i=1}^N u_i) · (Σ_{j=1}^N u_j) = ||Σ_{i=1}^N u_i||^2 >=0.On the other hand, j^T G j = Σ_{i=1}^N G_{i,i} + Σ_{i≠j} G_{i,j} = N*100 + Σ_{i≠j} G_{i,j}.But each G_{i,j} for i≠j is < -2, so Σ_{i≠j} G_{i,j} < -2*N(N-1).Therefore, j^T G j < 100N - 2N(N -1).But since j^T G j >=0, we have 100N -2N(N -1) >0.Divide both sides by N (assuming N>0):100 -2(N -1) >0100 -2N +2 >0102 -2N >02N <102N <51Thus, N <=50. Which is exactly the result we need. Therefore, no more than 50 vectors can exist, proving that the maximum number of plants in a good guide is 50.This approach using linear algebra and the positive semi-definiteness of the Gram matrix seems to work. By converting the problem to vectors with inner products bounded above by -2, we derived an inequality that limits N to at most 50. This is a clean and elegant proof that doesn't rely on advanced coding theory, suitable for a math competition or textbook problem.

✅:To prove that no more than 50 plants can be described in a good guide, we model each plant as a binary vector in ({0,1}^{100}) and analyze their pairwise Hamming distances. A "good" guide requires that any two distinct vectors differ in more than 50 coordinates (Hamming distance (>50)).Key Steps:1. Vector Transformation: Convert binary vectors to ({-1,1}^{100}) by mapping (0 to 1) and (1 to -1). For two vectors (u) and (v), their inner product becomes: [ u cdot v = 100 - 2 cdot text{Hamming distance}(u,v). ] Since the Hamming distance (>50), the inner product satisfies: [ u cdot v < 100 - 2 cdot 51 = -2. ]2. Gram Matrix Analysis: Let (G) be the (N times N) Gram matrix where (G_{i,j} = u_i cdot u_j). The diagonal entries are (100), and off-diagonal entries are (<-2). The matrix (G) must be positive semi-definite (PSD), so for any vector (j = (1,1,dots,1)), [ j^T G j geq 0. ]3. Quadratic Form Bound: Compute the quadratic form: [ j^T G j = sum_{i=1}^N sum_{j=1}^N u_i cdot u_j = left| sum_{i=1}^N u_i right|^2 geq 0. ] Expanding the sum: [ j^T G j = N cdot 100 + sum_{i neq j} u_i cdot u_j < 100N - 2N(N-1). ] Combining with the PSD condition: [ 100N - 2N(N-1) > 0 implies 102 - 2N > 0 implies N < 51. ]Conclusion: Since (N) must be an integer, the maximum number of plants (N leq 50). Thus, a good botanical guide cannot describe more than 50 plants.Final Answer: boxed{50}

🔑:1. Let ( m ) be the number of plants described in a good botanical dictionary. We need to examine the total number of differences between all pairs of plants across all 100 characteristics.2. The number of pairs of plants is given by the binomial coefficient: [ binom{m}{2} = frac{m(m-1)}{2} ] Since each pair of plants differs by at least 51 characteristics, the total number of differences, ( S ), satisfies: [ S geq 51 times frac{m(m-1)}{2} ]3. Calculate ( S ) by considering each characteristic separately. Let ( m_i ) be the number of plants having the ( i )-th characteristic. Then, the number of pairs of plants that differ in the ( i )-th characteristic is: [ m_i times (m - m_i) ] Therefore, the total number of differences ( S ) can be expressed as: [ S = sum_{i=1}^{100} m_i times (m - m_i) ]4. Using the inequality ( m_i times (m - m_i) leq frac{m^2}{4} ) (since the function ( x(m - x) ) reaches its maximum value when ( x = frac{m}{2} )), we have: [ S leq 100 times frac{m^2}{4} = 25 m^2 ]5. Combining the inequalities, we get: [ 51 times frac{m(m-1)}{2} leq S leq 25 m^2 ] Simplifying the left-hand side: [ 51 times frac{m(m-1)}{2} = frac{51m^2 - 51m}{2} ] So, [ frac{51m^2 - 51m}{2} leq 25 m^2 ] 6. Multiplying through by 2 to clear the fraction: [ 51m^2 - 51m leq 50 m^2 ] [ m^2 - 51m leq 0 ]7. Factoring out ( m ): [ m(m - 51) leq 0 ] The inequality ( m(m - 51) leq 0 ) holds for ( 0 leq m leq 51 ). Since ( m ) represents the number of plants, it must be a positive integer. This means ( m le 51 ).8. To show that ( m neq 51 ), assume ( m = 51 ). Then the left side inequality: [ 51 cdot frac{51 cdot 50}{2} = 51 cdot 1275 = 65025 ] The other side is calculated as: [ 25 cdot 51^2 = 25 cdot 2601 = 65025 ] So the inequality becomes: [ 65025 leq S < 65025 ] Which leads to a contradiction, since ( S ) cannot be simultaneously less than and equal to 65025. Hence, [ m neq 51 ]Conclusion: Therefore, the number of plants ( m ) in a good botanical dictionary cannot exceed 50.[boxed{m leq 50}]

🧠:Okay, let's try to figure out this limit problem. The problem is asking for the limit as n approaches infinity of (1/n^5) multiplied by the sum of (5r^4 - 18r^2s^2 + 5s^4) where r and s are each from 1 to n. So, we need to evaluate:[lim_{n to infty} frac{1}{n^5} sum_{r=1}^n sum_{s=1}^n (5r^4 - 18r^2s^2 + 5s^4)]Hmm, first, I remember that when dealing with sums like this, especially when taking limits as n goes to infinity, it's often useful to approximate the sums by integrals. Because as n becomes very large, the sum can be approximated by the integral over a continuous variable. So maybe I can convert these sums into integrals.But before jumping into that, let me break down the expression inside the sum. The term is 5r^4 - 18r^2s^2 + 5s^4. Let's see if this can be simplified or factored in some way. Let me check:Is 5r^4 - 18r^2s^2 + 5s^4 a quadratic in terms of r^2 and s^2? Let's see:Let me write it as 5(r^4 + s^4) - 18r^2s^2. Hmm, that's 5(r^4 + s^4) - 18r^2s^2. Is there a way to factor this?Alternatively, maybe recognizing this as part of some expansion. Wait, maybe (r^2 - s^2)^2? Let's compute that:(r^2 - s^2)^2 = r^4 - 2r^2s^2 + s^4. Hmm, that's similar but the coefficients here are 1, -2, 1. Our expression is 5(r^4 + s^4) - 18r^2s^2. Let's see:If we take 5(r^4 + s^4) - 18r^2s^2 = 5r^4 + 5s^4 - 18r^2s^2. How does this compare to (r^2 - s^2)^2? If we multiply (r^2 - s^2)^2 by 5, we get 5r^4 - 10r^2s^2 + 5s^4. Then subtract 8r^2s^2 more: 5r^4 - 10r^2s^2 - 8r^2s^2 +5s^4 = 5r^4 -18r^2s^2 +5s^4. So yes! Therefore, the expression can be written as 5(r^2 - s^2)^2 - 8r^2s^2. Wait, no, let's check:Wait, if (r^2 - s^2)^2 = r^4 - 2r^2s^2 + s^4, then 5(r^2 - s^2)^2 = 5r^4 -10r^2s^2 +5s^4. Then, our original expression is 5r^4 -18r^2s^2 +5s^4, so that is 5(r^2 -s^2)^2 -8r^2s^2. So, yes, it can be expressed as such. But not sure if this helps, maybe?Alternatively, maybe not. Maybe it's easier to compute each term separately. Let's consider splitting the sum into three separate sums:Sum_{r,s} 5r^4 - Sum_{r,s}18r^2s^2 + Sum_{r,s}5s^4.So, that would be 5Sum_{r=1}^n Sum_{s=1}^n r^4 -18 Sum_{r=1}^n Sum_{s=1}^n r^2s^2 +5 Sum_{r=1}^n Sum_{s=1}^n s^4.Notice that the first and the third terms are similar. Let's compute each of them.First, Sum_{r=1}^n Sum_{s=1}^n r^4. Since r and s are independent variables here, the sum over s doesn't involve r. Therefore, this is equal to Sum_{r=1}^n [r^4 * Sum_{s=1}^n 1] ] = Sum_{r=1}^n [r^4 * n] = n * Sum_{r=1}^n r^4.Similarly, Sum_{r=1}^n Sum_{s=1}^n s^4 = n * Sum_{s=1}^n s^4. But since s is a dummy variable, this is also n * Sum_{r=1}^n r^4. Therefore, the first and third terms are both 5n times the sum of r^4 from 1 to n. So adding them together, we get 5n * Sum r^4 +5n * Sum r^4 =10n * Sum r^4.Then the middle term is -18 Sum_{r,s} r^2s^2. Since r and s are independent, this is -18 Sum_{r=1}^n r^2 * Sum_{s=1}^n s^2 = -18 [Sum r^2]^2.Therefore, putting it all together, the entire sum is 10n * Sum_{r=1}^n r^4 -18 [Sum_{r=1}^n r^2]^2.Therefore, the expression we need to evaluate the limit of is [10n * Sum r^4 -18 (Sum r^2)^2] / n^5.So, the problem reduces to evaluating the limit as n approaches infinity of [10n * Sum r^4 -18 (Sum r^2)^2] / n^5.Now, we need to find asymptotic expressions for Sum r^4 and Sum r^2 as n becomes large. I remember that the sum of k^m from k=1 to n is asymptotic to n^{m+1}/(m+1) + lower order terms. Specifically, for large n, Sum_{k=1}^n k^m ≈ n^{m+1}/(m+1) + n^m/2 + ... So, we can approximate Sum r^4 ≈ n^5/5 + n^4/2 + ... and Sum r^2 ≈ n^3/3 + n^2/2 + ... .Therefore, let's compute each term:First term: 10n * Sum r^4 ≈ 10n [n^5/5 + n^4/2 + ...] = 10n*(n^5/5) + 10n*(n^4/2) + ... = 10/5 n^6 + 10/2 n^5 + ... = 2n^6 +5n^5 + ... .Second term: -18 (Sum r^2)^2 ≈ -18 [n^3/3 + n^2/2 + ...]^2. Let's square that: [n^3/3]^2 + 2*(n^3/3)*(n^2/2) + ... = n^6/9 + (2*n^5/6) + ... = n^6/9 + n^5/3 + ... . Multiply by -18: -18*(n^6/9 +n^5/3 + ...) = -2n^6 -6n^5 + ... .Therefore, combining the two terms:First term + Second term ≈ 2n^6 +5n^5 -2n^6 -6n^5 = (2n^6 -2n^6) + (5n^5 -6n^5) = -n^5.Then, the entire numerator is approximately -n^5. Dividing by n^5 gives -1. Therefore, the limit is -1?Wait, but let's check the coefficients more carefully. Because when we approximate the sums, there might be more precise terms needed. Let's verify.First, let's recall the exact formula for Sum_{r=1}^n r^2 and Sum_{r=1}^n r^4.Sum_{r=1}^n r^2 = n(n + 1)(2n + 1)/6.Sum_{r=1}^n r^4 = n(n + 1)(2n + 1)(3n^2 + 3n -1)/30.These are exact formulas. Let's use these to compute asymptotic expansions.First, expand Sum r^2:n(n +1)(2n +1)/6 = n( n +1 )(2n +1 ) /6.Expanding term by term:n * (n +1) = n^2 +n.Multiply by (2n +1):(n^2 +n)(2n +1) = 2n^3 +n^2 +2n^2 +n = 2n^3 +3n^2 +n.Divide by 6: (2n^3 +3n^2 +n)/6 = (1/3)n^3 + (1/2)n^2 + (1/6)n.So Sum r^2 = (1/3)n^3 + (1/2)n^2 + (1/6)n.Similarly, Sum r^4 = n(n +1)(2n +1)(3n^2 +3n -1)/30.Let me expand this step by step. First, compute each factor:n(n +1) = n^2 +n.(2n +1)(3n^2 +3n -1):First multiply 2n*(3n^2 +3n -1) = 6n^3 +6n^2 -2n.Then add 1*(3n^2 +3n -1) = 3n^2 +3n -1.So total is 6n^3 +6n^2 -2n +3n^2 +3n -1 = 6n^3 +9n^2 +n -1.Therefore, the entire numerator is (n^2 +n)(6n^3 +9n^2 +n -1). Let's multiply this out:First, n^2*(6n^3 +9n^2 +n -1) =6n^5 +9n^4 +n^3 -n^2.Then, n*(6n^3 +9n^2 +n -1)=6n^4 +9n^3 +n^2 -n.Add them together:6n^5 +9n^4 +n^3 -n^2 +6n^4 +9n^3 +n^2 -n =6n^5 + (9n^4 +6n^4) + (n^3 +9n^3) + (-n^2 +n^2) + (-n) =6n^5 +15n^4 +10n^3 -n.Then, divide by 30:Sum r^4 = (6n^5 +15n^4 +10n^3 -n)/30 = (6/30)n^5 + (15/30)n^4 + (10/30)n^3 - (1/30)nSimplify:Sum r^4 = (1/5)n^5 + (1/2)n^4 + (1/3)n^3 - (1/30)n.Therefore, the asymptotic expansion for Sum r^4 is (1/5)n^5 + (1/2)n^4 + (1/3)n^3 + lower order terms.Similarly, Sum r^2 = (1/3)n^3 + (1/2)n^2 + (1/6)n.Therefore, let's compute each term in the numerator:First term:10n * Sum r^4 =10n*(1/5 n^5 +1/2 n^4 +1/3 n^3 -1/30 n )=10n*(1/5 n^5) +10n*(1/2 n^4) +10n*(1/3 n^3) +10n*(-1/30 n )=10*(1/5)n^6 +10*(1/2)n^5 +10*(1/3)n^4 +10*(-1/30)n^2=2n^6 +5n^5 + (10/3)n^4 - (1/3)n^2.Second term: -18*(Sum r^2)^2.First, compute (Sum r^2)^2:Sum r^2 = (1/3)n^3 + (1/2)n^2 + (1/6)n.So square this:[(1/3)n^3 + (1/2)n^2 + (1/6)n]^2.Let me expand this term by term:First, square each term:(1/3 n^3)^2 =1/9 n^6.(1/2 n^2)^2 =1/4 n^4.(1/6 n)^2 =1/36 n^2.Then cross terms:2*(1/3 n^3)*(1/2 n^2) =2*(1/6)n^5=1/3 n^5.2*(1/3 n^3)*(1/6 n)=2*(1/18)n^4=1/9 n^4.2*(1/2 n^2)*(1/6 n)=2*(1/12)n^3=1/6 n^3.So putting it all together:1/9 n^6 +1/3 n^5 + (1/4 +1/9)n^4 +1/6 n^3 +1/36 n^2.Compute coefficients:1/4 +1/9 = (9 +4)/36 =13/36.So,(Sum r^2)^2 =1/9 n^6 +1/3 n^5 +13/36 n^4 +1/6 n^3 +1/36 n^2.Multiply by -18:-18*(1/9 n^6 +1/3 n^5 +13/36 n^4 +1/6 n^3 +1/36 n^2 )= -18*(1/9 n^6) -18*(1/3 n^5) -18*(13/36 n^4) -18*(1/6 n^3) -18*(1/36 n^2 )= -2n^6 -6n^5 - (13/2)n^4 -3n^3 -0.5n^2.Therefore, combining the first term and the second term:First term:2n^6 +5n^5 + (10/3)n^4 - (1/3)n^2.Second term:-2n^6 -6n^5 - (13/2)n^4 -3n^3 -0.5n^2.Add them together:2n^6 -2n^6 =0.5n^5 -6n^5 = -n^5.(10/3)n^4 - (13/2)n^4 = (20/6 -39/6)n^4 = (-19/6)n^4.-3n^3.(-1/3)n^2 -0.5n^2 = (-1/3 -3/6)n^2 = (-1/3 -1/2)n^2 = (-5/6)n^2.Therefore, total numerator is:-n^5 - (19/6)n^4 -3n^3 - (5/6)n^2.Therefore, the entire expression is:[ -n^5 - (19/6)n^4 -3n^3 - (5/6)n^2 ] / n^5.Divide each term by n^5:= -1 - (19/6)(1/n) -3(1/n^2) - (5/6)(1/n^3).Now, take the limit as n approaches infinity:lim_{n→∞} [ -1 - (19/6)/n -3/n^2 - (5/6)/n^3 ] = -1 -0 -0 -0 = -1.So, the final answer should be -1.Wait, but let me verify once more. Because sometimes when approximating sums, higher-order terms might contribute, but in this case, we used the exact expressions for the sums, expanded them, and then subtracted. The leading terms canceled out (the 2n^6 -2n^6), and the next term was -n^5. Then when divided by n^5, the leading term is -1, and the rest vanish as n approaches infinity. So the limit is indeed -1.But let me check with another approach, maybe using integrals. Because sometimes converting the sum into integrals can be a quicker way for such limits.Given that the sum is over r and s from 1 to n, and we have the expression 5r^4 -18r^2s^2 +5s^4. Then, when divided by n^5, we can think of this as approximating a double integral over x and y from 0 to 1, where x = r/n and y = s/n. Then, as n approaches infinity, the sum becomes:Sum_{r,s} (5r^4 -18r^2s^2 +5s^4)/n^5 ≈ Sum_{r,s} [5(nx)^4 -18(nx)^2(ny)^2 +5(ny)^4]/n^5 = Sum_{r,s} [5n^4x^4 -18n^4x^2y^2 +5n^4y^4]/n^5 = Sum_{r,s} [5x^4 -18x^2y^2 +5y^4]/n.But since we are summing over r and s, each term is divided by n, and the sum over r and s would approximate a double integral over x and y from 0 to 1. The spacing between x and y is 1/n, so the sum becomes approximately n^2*(1/n^2)*integral over [0,1]x[0,1] of (5x^4 -18x^2y^2 +5y^4) dxdy. Wait, let me check:Wait, each term in the sum is (5r^4 -18r^2s^2 +5s^4)/n^5. Let me factor out n^4 from the numerator:= [5(r/n)^4 -18(r/n)^2(s/n)^2 +5(s/n)^4] * n^4 /n^5 = [5x^4 -18x^2y^2 +5y^4] * (1/n), where x = r/n, y = s/n.So each term is [5x^4 -18x^2y^2 +5y^4]*(1/n). Then, summing over all r and s from 1 to n is like integrating over x and y from 0 to 1 with step size 1/n. The sum would approximate the double integral multiplied by n^2 (number of terms) times (1/n). Wait, no:Wait, the sum Sum_{r=1}^n Sum_{s=1}^n f(r/n, s/n) * (1/n^2) approximates the double integral of f(x,y) over [0,1]x[0,1]. But in our case, each term is [5x^4 -18x^2y^2 +5y^4]*(1/n), and there are n^2 terms. So total sum is approximately n^2*(1/n) * integral. Wait, no, this seems confused.Let me be precise. When approximating a double sum by a double integral, the standard approach is:(1/n^2) Sum_{r=1}^n Sum_{s=1}^n f(r/n, s/n) ≈ ∫₀¹ ∫₀¹ f(x,y) dx dy.But in our case, the original expression is (1/n^5) Sum (5r^4 -18r^2s^2 +5s^4).Let me write this as (1/n^5) Sum_{r,s} [5r^4 +5s^4 -18r^2s^2].Let me separate the terms:5*(1/n^5)(Sum_{r=1}^n r^4 * Sum_{s=1}^n 1 + Sum_{s=1}^n s^4 * Sum_{r=1}^n 1 ) -18*(1/n^5) Sum_{r=1}^n r^2 * Sum_{s=1}^n s^2.Wait, this is similar to what I did before. So:First term:5*(1/n^5)(n Sum r^4 +n Sum s^4 ) =5*(2n Sum r^4)/n^5 =10 Sum r^4 /n^4.Second term:-18*(1/n^5)*(Sum r^2)^2 = -18 (Sum r^2)^2 /n^5.So the expression is 10*(Sum r^4 /n^5) -18*(Sum r^2)^2 /n^5.But Sum r^4 /n^5 ≈ (1/5)n^5 /n^5 =1/5 as n→∞.Wait, but actually, Sum r^4 ≈ (1/5)n^5, so Sum r^4 /n^5 ≈1/5. Similarly, Sum r^2 ≈ (1/3)n^3, so (Sum r^2)^2 ≈(1/9)n^6, then (Sum r^2)^2 /n^5 ≈(1/9)n^6 /n^5 =n/9. Which would go to infinity, but in our case, we have 10*(1/5) -18*(n/9) =2 -2n, which would go to negative infinity. But this contradicts the previous result of -1. So what's wrong here?Wait, this approach must be incorrect because when we use the leading term approximations, we need to consider the entire expression. Wait, actually, in the previous detailed calculation, the leading terms canceled out, leaving -n^5. So maybe when we use this integral approximation, we need to be careful.Wait, but perhaps converting the original sum into an integral isn't straightforward because the expression has cross terms. Wait, but let's see:Original expression:(1/n^5) Sum_{r,s} [5r^4 -18r^2s^2 +5s^4].Expressed in terms of x=r/n, y=s/n:Each term becomes 5(nx)^4 -18(nx)^2(ny)^2 +5(ny)^4 =5n^4x^4 -18n^4x^2y^2 +5n^4y^4.Then, the sum is over r and s from 1 to n, so each term is divided by n^5:[5n^4x^4 -18n^4x^2y^2 +5n^4y^4]/n^5 = [5x^4 -18x^2y^2 +5y^4]/n.Therefore, the entire sum is Sum_{r,s} [5x^4 -18x^2y^2 +5y^4]/n.But since x and y are spaced by 1/n, the sum is approximately a double integral over x and y from 0 to1, multiplied by n^2 (number of terms) times (1/n) (the term per sum):Integral approximation: n^2*(1/n)*∫₀¹∫₀¹ [5x^4 -18x^2y^2 +5y^4] dxdy =n* [∫₀¹∫₀¹ (5x^4 +5y^4 -18x^2y^2) dxdy].But as n approaches infinity, this would behave like n times the integral, which would go to infinity or negative infinity unless the integral is zero. However, in our previous detailed calculation, the limit was -1, which is finite. Therefore, there must be a mistake in this integral approach. Alternatively, maybe the integral approach is not suitable here because the cross terms lead to higher-order contributions?Wait, perhaps my initial approach using exact sums was correct, and the integral approach here is not appropriate because the leading terms in the sum cancel out, but when approximated via integral, we might not capture that cancellation. Let's compute the integral to check:Compute the integral ∫₀¹∫₀¹ [5x^4 +5y^4 -18x^2y^2] dxdy.First, integrate over x:∫₀¹ [5x^4 +5y^4 -18x^2y^2] dx =5∫x^4 dx +5y^4 ∫dx -18y^2 ∫x^2 dx.Compute each integral:5*(1/5) +5y^4*(1) -18y^2*(1/3) =1 +5y^4 -6y^2.Then integrate this over y from 0 to1:∫₀¹ [1 +5y^4 -6y^2] dy = ∫₀¹1 dy +5∫y^4 dy -6∫y^2 dy =1 +5*(1/5) -6*(1/3)=1 +1 -2=0.Therefore, the integral evaluates to zero. So according to the integral approximation, the expression tends to n*0=0, which contradicts the previous result of -1. But this is strange. So which one is correct?Wait, the problem arises because when approximating the sum by an integral, sometimes higher-order terms are necessary when the leading terms cancel out. In our case, the integral of the function is zero, which suggests that the leading term (which would be O(n)) cancels out, and the next term is O(1), which would give a finite limit. However, according to our exact calculation, the limit is -1, which is finite. Therefore, perhaps the integral approach needs to consider the next term beyond the leading order.In the method of approximating sums by integrals, when the leading term cancels, we need to consider the next term in the Euler-Maclaurin formula, which accounts for the correction terms. However, this can get complicated. Alternatively, since we already did an exact computation expanding the sums and found the limit to be -1, which is consistent, that's likely the correct answer. Therefore, the integral being zero suggests that the leading term cancels, but the exact computation shows that the next term gives a finite contribution. Therefore, the answer is indeed -1.Another way to check: use the exact expressions for the sums.From before:Sum r^2 = n(n +1)(2n +1)/6 ≈ (1/3)n^3 + (1/2)n^2 + (1/6)n.Sum r^4 =n(n +1)(2n +1)(3n^2 +3n -1)/30≈ (1/5)n^5 + (1/2)n^4 + (1/3)n^3 - (1/30)n.So, compute the numerator:10n * Sum r^4 -18*(Sum r^2)^2.Substitute the approximate expressions:10n*(1/5 n^5 +1/2 n^4 +1/3 n^3) -18*( (1/3 n^3 +1/2 n^2)^2 )=10n*(1/5 n^5) +10n*(1/2 n^4) +10n*(1/3 n^3) -18*(1/9 n^6 +1/3 n^5 +1/4 n^4 )=2n^6 +5n^5 + (10/3)n^4 -18*(1/9 n^6 +1/3 n^5 +1/4 n^4 )=2n^6 +5n^5 + (10/3)n^4 -2n^6 -6n^5 - (9/2)n^4.Then combine terms:(2n^6 -2n^6) + (5n^5 -6n^5) + (10/3 n^4 -9/2 n^4 )= -n^5 + (20/6 -27/6)n^4= -n^5 -7/6 n^4.Wait, this is different from the previous expansion where we had more terms. However, if we take more terms in the expansion of Sum r^4 and Sum r^2, maybe we get the -n^5 - (19/6)n^4 -... So perhaps truncating the expansion too early gives an incorrect coefficient, but the leading term -n^5 is correct.However, when divided by n^5, the leading term is -1, and the next term is -7/6n^4 /n^5 = -7/(6n), which goes to zero as n approaches infinity. Therefore, the limit is -1. Therefore, even with this approximate expansion, we get the same leading term.Thus, both methods confirm that the limit is -1. Therefore, the final answer is boxed{-1}.

✅:To evaluate the limit (lim_{n to infty} frac{1}{n^5} sum_{r=1}^n sum_{s=1}^n (5r^4 - 18r^2s^2 + 5s^4)), we start by breaking down the summation into individual terms:[sum_{r=1}^n sum_{s=1}^n (5r^4 - 18r^2s^2 + 5s^4) = 5 sum_{r=1}^n sum_{s=1}^n r^4 - 18 sum_{r=1}^n sum_{s=1}^n r^2s^2 + 5 sum_{r=1}^n sum_{s=1}^n s^4]Each term can be simplified using properties of summations. The first and third terms are:[5 sum_{r=1}^n sum_{s=1}^n r^4 = 5n sum_{r=1}^n r^4 quad text{and} quad 5 sum_{r=1}^n sum_{s=1}^n s^4 = 5n sum_{s=1}^n s^4]Since (sum_{r=1}^n r^4) and (sum_{s=1}^n s^4) are the same, we combine these terms:[10n sum_{r=1}^n r^4]The middle term is:[-18 sum_{r=1}^n sum_{s=1}^n r^2s^2 = -18 left( sum_{r=1}^n r^2 right) left( sum_{s=1}^n s^2 right) = -18 left( sum_{r=1}^n r^2 right)^2]Using the exact formulas for the sums of powers of integers:[sum_{r=1}^n r^2 = frac{n(n+1)(2n+1)}{6} approx frac{1}{3}n^3 + frac{1}{2}n^2 + frac{1}{6}n][sum_{r=1}^n r^4 = frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} approx frac{1}{5}n^5 + frac{1}{2}n^4 + frac{1}{3}n^3 - frac{1}{30}n]Substituting these approximations into the expression, we get:[10n left( frac{1}{5}n^5 + frac{1}{2}n^4 + frac{1}{3}n^3 right) - 18 left( frac{1}{3}n^3 + frac{1}{2}n^2 right)^2]Expanding and simplifying the terms:[10n left( frac{1}{5}n^5 right) + 10n left( frac{1}{2}n^4 right) + 10n left( frac{1}{3}n^3 right) - 18 left( frac{1}{9}n^6 + frac{1}{3}n^5 + frac{1}{4}n^4 right)][= 2n^6 + 5n^5 + frac{10}{3}n^4 - 2n^6 - 6n^5 - frac{9}{2}n^4]Combining like terms:[(2n^6 - 2n^6) + (5n^5 - 6n^5) + left( frac{10}{3}n^4 - frac{9}{2}n^4 right)][= -n^5 - frac{7}{6}n^4]Dividing by (n^5) and taking the limit as (n to infty):[lim_{n to infty} frac{-n^5 - frac{7}{6}n^4}{n^5} = lim_{n to infty} left( -1 - frac{7}{6n} right) = -1]Thus, the final answer is (boxed{-1}).

🔑:Given the problem:[ lim_{n to infty} frac{1}{n^5} sum (5r^4 - 18r^2s^2 + 5s^4) ]where the sum is over all ( r ) and ( s ) satisfying ( 0 < r, s leq n ).Let's break down and detail the solution step by step.1. Sum Components Evaluation: To evaluate: [ sum_{r=1}^{n} r^2 ] and [ sum_{r=1}^{n} r^4 ] we use known formulas for polynomial summation: - For ( r^2 ): [ sum_{r=1}^{n} r^2 = frac{n(n+1)(2n+1)}{6} ] - For ( r^4 ): [ sum_{r=1}^{n} r^4 = frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} ]2. Component Summation Approximations: Accurate approximations for large ( n ) for the sums are: - For ( sum_{r=1}^{n} r^2 ): [ sum_{r=1}^{n} r^2 = frac{n^3}{3} + frac{n^2}{2} + frac{n}{6} ] - For ( sum_{r=1}^{n} r^4 ): [ sum_{r=1}^{n} r^4 = frac{n^5}{5} + frac{n^4}{2} + O(n^3) ]3. Sum over ( r ) and ( s ) Components: For our problem, the expression includes sums of both ( r^4 ) and ( r^2s^2 ). Using additivity of sums and the fact that sums over independent variables can be squared, we have: - For ( r^2s^2 ): [ left( sum_{r=1}^{n} r^2 right) left( sum_{s=1}^{n} s^2 right) = left( frac{n^3}{3} + frac{n^2}{2} + frac{n}{6} right)^2 only<1> ]4. Expand the Given Sum: Substitute the approximations into the original sum: [ sum_{r=1}^{n} sum_{s=1}^{n} (5r^4 - 18r^2s^2 + 5s^4) ] This decomposes into: - Use the formulae above: [ 5 sum_{r=1}^{n} r^4 + 5 sum_{s=1}^{n} s^4 - 18 sum_{r=1}^{n} r^2 sum_{s=1}^{n} s^2 ] 5. Insert Summation Approximations: [ 5 left( frac{n^5}{5} + frac{n^4}{2} + O(n^3) right) + 5 left( frac{n^5}{5} + frac{n^4}{2} + O(n^3) right) - 18 left( frac{n^3}{3} + frac{n^2}{2} + frac{n}{6} right)^2 ]6. Simplify: Simplify each term: [ = 2 left( n^5 + frac{5n^4}{2} + O(n^3) right) - 2 left(frac{2n^6}{9} + frac{2n^5}{3} + O(n^4)right) ] 7. Combine and Factor: Combine like terms: [ n^6 + 5n^5 + O(n^4) - 2n^6 - 6n^5 + O(n^4) ] [ = -n^6 - n^5 + O(n^4) ]8. Limit Calculation: Divide the rounded sum by ( n^5 ) and take the limit as ( n to infty ): [ lim_{n to infty} frac{1}{n^5} left( -n^6 - n^5 + O(n^4) right) ] [ = lim_{n to infty} left( -n - 1 + frac{O(n^4)}{n^5} right) ]9. Conclusion: Since the higher-order terms ( frac{O(n^4)}{n^5} ) approach 0 faster than linearly, the limit reduces to: [ = -1 ] Therefore, the result is: [ boxed{-1} ]