🧠:Okay, so I need to figure out the smallest possible number of children in the Pascal family where each child has at least 2 brothers and at least 1 sister. Let me start by understanding the problem.First, the question says "each child has at least 2 brothers and at least 1 sister." That means, for every child in the family, no matter if they're a boy or a girl, they must have at least two brothers and one sister. Wait, but hold on—if a child is a girl, her brothers would be the male children, and her sisters would be the other female children. Similarly, if a child is a boy, his brothers are the other males, and his sisters are the females. So the family must have enough boys and girls such that each child, regardless of their gender, meets these conditions.The question is asking for the smallest total number of children. The options are from 3 up to 7. Let's start by considering the minimum numbers required.First, let's think about the brothers. Each child needs at least 2 brothers. If a child is a boy, then his brothers are the other boys in the family. So, if a boy has at least 2 brothers, that means there must be at least 3 boys in total (since the boy himself plus 2 brothers makes 3). Similarly, if a child is a girl, she also needs at least 2 brothers. So, regardless of the girl's count, there must be at least 2 boys in the family. But wait, if the girl needs at least 2 brothers, that means the family must have at least 2 boys. But if a boy needs at least 2 brothers, the family must have at least 3 boys. Therefore, the number of boys must be at least 3.Now, moving on to sisters. Each child needs at least 1 sister. If a child is a girl, her sisters are the other girls. So, if a girl is to have at least 1 sister, there must be at least 2 girls in the family. Similarly, if a child is a boy, he needs at least 1 sister, meaning the family must have at least 1 girl. But here's the catch: a girl needs at least 1 sister, so there must be at least 2 girls. But a boy just needs at least 1 sister, so the family could have 1 girl. But since the family must satisfy all children's conditions, the girls require at least 2 girls. Therefore, the number of girls must be at least 2.Putting this together: boys >= 3 and girls >= 2. So the total number of children would be at least 3 + 2 = 5. But wait, the answer options include 5 (option C), but let me verify if that's sufficient.Let me check with 3 boys and 2 girls. Let's see if each child meets the requirements.First, take a boy. Each boy has 2 brothers (since there are 3 boys total, each boy has 2 brothers) and at least 1 sister (there are 2 girls). So each boy is okay.Now, take a girl. Each girl has 3 brothers (since there are 3 boys) and 1 sister (since there are 2 girls). So each girl has 3 brothers (which is more than the required 2) and 1 sister (which meets the requirement). So in this case, with 3 boys and 2 girls (total 5 children), all conditions are satisfied.But wait, hold on. The problem is presented as each child having "at least" 2 brothers and "at least" 1 sister. So if 3 boys and 2 girls work, then 5 is possible. But let me check the answer options. Option C is 5, which would be the answer. However, before confirming, let me check if perhaps a smaller number is possible. The options given start at 3. Let's see if 4 is possible.Suppose we try with 4 children. Let's see if we can arrange boys and girls such that each child has at least 2 brothers and 1 sister.Let's try different combinations:Case 1: 3 boys and 1 girl.Total children: 4.Check each child:For a boy: brothers = 2 (since total boys are 3, so each boy has 2 brothers). Sisters = 1 (the girl). So each boy has 2 brothers and 1 sister. That's okay.For the girl: brothers = 3 (all the boys), which is more than 2, so okay. Sisters = 0, but she needs at least 1 sister. So this fails because the girl has no sisters. Therefore, this combination doesn't work.Case 2: 2 boys and 2 girls.Total children: 4.Check each child:For a boy: brothers = 1 (since there are 2 boys total), which is less than the required 2. So this fails. Each boy needs at least 2 brothers, but here they only have 1. So this doesn't work.For a girl: brothers = 2 (which meets the required 2), sisters = 1 (since there are 2 girls). So girls are okay. But the boys are not. So this combination fails.Case 3: 4 boys and 0 girls. Obviously, the girls requirement isn't met for any child. So that's invalid.Case 4: 1 boy and 3 girls.For the boy: brothers = 0 (needs at least 2), so invalid.Girls: each has 1 brother (needs at least 2) so also invalid.So in all cases with 4 children, it's impossible. Therefore, 4 is not possible.What about 3 children? Let's check:Possible combinations:1 boy and 2 girls.Boy has 0 brothers (needs at least 2) → invalid.Girls each have 1 brother (needs at least 2) → invalid.2 boys and 1 girl.Each boy has 1 brother (needs 2) → invalid.Girl has 2 brothers (okay), but no sisters (needs at least 1) → invalid.3 boys and 0 girls: no sisters for anyone → invalid.3 girls: no brothers → invalid.So 3 is impossible.Thus, the next number is 5. As we saw earlier, 3 boys and 2 girls works. Each child has the required number of brothers and sisters. So 5 is possible. But wait, before confirming, let me check if there's another combination with 5 children that might not work but perhaps a different one does. But 3 boys and 2 girls is the minimal based on the earlier reasoning.Wait, but what if we have 4 boys and 1 girl. Let's see:Total children 5.For the girl: brothers = 4 (which is more than 2), sisters = 0 (needs at least 1) → invalid.For boys: each has 3 brothers and 1 sister. The sister part is okay, but the girl's condition isn't met. So invalid.Similarly, 1 boy and 4 girls:Boy has 0 brothers → invalid.Girls each have 1 brother (needs 2) → invalid.So the only viable 5-child family is 3 boys and 2 girls. Which works. Therefore, the answer should be 5, which is option C.But hold on, the problem says "the Pascal family," so maybe there's a trick here. Wait, the problem states "each child has at least 2 brothers and at least 1 sister." So regardless of gender, each child must have at least two brothers and at least one sister. So if a child is a girl, she needs two brothers and one sister. If a child is a boy, he needs two brothers and one sister.In the case of 3 boys and 2 girls:Each boy has 2 brothers (since there are 3 boys) and 2 sisters (the two girls). Wait, wait, hold on: no, each boy would have 2 brothers (other boys) and 2 sisters (the girls). So the brothers part is okay, and the sisters part is more than required.Each girl has 3 brothers (the three boys) and 1 sister (the other girl). So sisters: 1, which meets the requirement. Brothers: 3, which is more than required. So yes, that works.Wait, but in the girl's case, she has 1 sister. That meets the "at least 1 sister" condition. The brothers are 3, which is more than the required 2. So all conditions are satisfied.Therefore, 5 children is possible. And since 4 is not possible, as checked earlier, 5 is the minimal number.But let me just think again: is there any other combination with 5 children that could work? Suppose 4 boys and 1 girl: the girl has 0 sisters, which is invalid. 2 boys and 3 girls: each boy has 1 brother (insufficient), so invalid. 3 boys and 2 girls is the only valid one.Yes, so the answer should be 5, option C.

✅:The problem requires determining the smallest number of children in a family where each child has at least 2 brothers and at least 1 sister. Here's the systematic breakdown:1. Key Constraints: - Brothers: Every child must have ≥2 brothers. If a child is a boy, this means there must be ≥3 boys (since a boy needs 2 brothers). If a child is a girl, the family must have ≥2 boys. - Sisters: Every child must have ≥1 sister. If a child is a girl, this means there must be ≥2 girls. If a child is a boy, the family must have ≥1 girl.2. Minimum Requirements: - Boys: At least 3 (to satisfy the brother condition for boys). - Girls: At least 2 (to satisfy the sister condition for girls).3. Testing Combinations: - 3 boys and 2 girls (Total = 5): - Boys: Each has 2 brothers (3 boys total) and 2 sisters (meets/exceeds requirements). - Girls: Each has 3 brothers (≥2 required) and 1 sister (exactly meets requirement). - All conditions are satisfied. - Smaller Numbers (3, 4): - 3 children: Impossible (either insufficient brothers or sisters). - 4 children: All combinations (e.g., 3 boys + 1 girl, 2 boys + 2 girls) fail either the brother or sister requirement.4. Conclusion: The smallest valid configuration is 3 boys and 2 girls, totaling 5 children.Answer: boxed{C}

🔑:1. Determine the minimum number of boys: - Each child has at least 2 brothers. - This means each boy must have at least 2 other brothers. - Thus, there must be at least 3 boys in the family (since having at least 2 brothers requires 3 boys in total). Let's denote the number of boys as ( b ). [ b geq 3 ]2. Determine the minimum number of girls: - Each child has at least 1 sister. - This means each girl must have at least 1 other sister. - Thus, there must be at least 2 girls in the family (since having at least 1 sister requires 2 girls in total). Let's denote the number of girls as ( g ). [ g geq 2 ]3. Calculate the smallest possible total number of children: - The total number of children in the family would be the sum of the boys and the girls: [ text{Total Children} = b + g ] - From the previous steps, we know ( b geq 3 ) and ( g geq 2 ). Therefore: [ text{Minimum Total Children} = 3 + 2 = 5 ]4. Conclusion: - Verifying: In a family with 3 boys and 2 girls: - Each boy has 2 brothers and 2 sisters. - Each girl has at least 1 sister and 3 brothers. - Thus, the requirements are met with a total of 5 children. [ boxed{C} ]

🧠:Okay, so I need to find the intersection points of these two circles. Let me start by writing down the equations again to make sure I have them right. The first circle is given by ((x - 2)^2 + (y - 10)^2 = 50). The second circle is (x^2 + y^2 + 2(x - y) - 18 = 0). Hmm, maybe I should rewrite the second equation in standard form to better understand its center and radius. Starting with the second equation: (x^2 + y^2 + 2x - 2y - 18 = 0). To convert this into the standard form of a circle ((x - h)^2 + (y - k)^2 = r^2), I need to complete the square for both the x and y terms. For the x terms: (x^2 + 2x). Completing the square here would require adding ((2/2)^2 = 1). For the y terms: (y^2 - 2y). Completing the square here would need adding ((-2/2)^2 = 1). But since I'm adding these to both sides, let me adjust the equation accordingly. So, starting with (x^2 + 2x + y^2 - 2y - 18 = 0). Adding 1 and 1 to complete the squares: (x^2 + 2x + 1 + y^2 - 2y + 1 = 18 + 1 + 1)This simplifies to ((x + 1)^2 + (y - 1)^2 = 20). Wait, let me check that. Original equation: (x^2 + y^2 + 2x - 2y - 18 = 0). Moving the constant term to the right side: (x^2 + 2x + y^2 - 2y = 18). Completing the square for x: (x^2 + 2x + 1 = (x + 1)^2). Completing the square for y: (y^2 - 2y + 1 = (y - 1)^2). So, adding those 1s to both sides: Left side becomes ((x + 1)^2 + (y - 1)^2). Right side becomes 18 + 1 + 1 = 20. So yes, the second circle is ((x + 1)^2 + (y - 1)^2 = 20). So now we have two circles:1. Center at (2, 10) with radius √50 (which is 5√2)2. Center at (-1, 1) with radius √20 (which is 2√5)Now, to find their intersection points, I can solve these two equations simultaneously. The standard method is to subtract one equation from the other to eliminate the quadratic terms, leading to a linear equation which can be solved for one variable in terms of the other. Then substitute back into one of the original equations. Let me expand both equations first to make subtraction easier. First circle: ((x - 2)^2 + (y - 10)^2 = 50)Expanding this:(x^2 - 4x + 4 + y^2 - 20y + 100 = 50)Simplify:(x^2 + y^2 - 4x - 20y + 104 = 50)Then subtract 50 from both sides:(x^2 + y^2 - 4x - 20y + 54 = 0) ... (1)Second circle already in expanded form is:(x^2 + y^2 + 2x - 2y - 18 = 0) ... (2)Now, subtract equation (2) from equation (1):(1) - (2): (x^2 + y^2 - 4x - 20y + 54) - (x^2 + y^2 + 2x - 2y - 18) = 0 - 0Simplify term by term:x^2 - x^2 = 0y^2 - y^2 = 0-4x - 2x = -6x-20y - (-2y) = -20y + 2y = -18y54 - (-18) = 54 + 18 = 72So the result is:-6x - 18y + 72 = 0Let me simplify this equation. Divide all terms by -6:x + 3y - 12 = 0So, x + 3y = 12. Therefore, x = 12 - 3y.Okay, so that's the linear equation we get by subtracting the two circles. Now, we can substitute this expression for x into one of the original equations to solve for y. Let's pick the second circle's equation because the numbers might be smaller. Second circle's equation: (x^2 + y^2 + 2x - 2y - 18 = 0)Substitute x = 12 - 3y into this:(12 - 3y)^2 + y^2 + 2(12 - 3y) - 2y - 18 = 0First, expand (12 - 3y)^2:12^2 - 2*12*3y + (3y)^2 = 144 - 72y + 9y^2So the equation becomes:144 - 72y + 9y^2 + y^2 + 24 - 6y - 2y - 18 = 0Combine like terms:First, the y^2 terms: 9y^2 + y^2 = 10y^2Then the y terms: -72y -6y -2y = -80yConstant terms: 144 + 24 - 18 = 144 + 6 = 150So now, the equation is:10y^2 - 80y + 150 = 0Let me divide all terms by 10 to simplify:y^2 - 8y + 15 = 0This is a quadratic equation in y. Let's solve it using the quadratic formula or factoring. Looking at y^2 -8y +15, can we factor it?Looking for two numbers that multiply to 15 and add up to -8. Those numbers are -3 and -5.So, (y - 3)(y - 5) = 0Therefore, y = 3 or y = 5.Now, substitute these y-values back into the equation x = 12 - 3y to find the corresponding x-values.First, y = 3:x = 12 - 3*3 = 12 - 9 = 3So one point is (3, 3)Second, y = 5:x = 12 - 3*5 = 12 - 15 = -3So the other point is (-3, 5)Therefore, the intersection points are (3, 3) and (-3, 5). But let me verify these points by plugging them back into both original equations to make sure they satisfy both.First, check (3, 3):First circle: (3 - 2)^2 + (3 - 10)^2 = 1 + 49 = 50. Correct.Second circle: 3^2 + 3^2 + 2*(3 - 3) -18 = 9 + 9 + 0 -18 = 0. Correct.Second, check (-3, 5):First circle: (-3 - 2)^2 + (5 - 10)^2 = (-5)^2 + (-5)^2 = 25 + 25 = 50. Correct.Second circle: (-3)^2 + 5^2 + 2*(-3 -5) -18 = 9 + 25 + 2*(-8) -18 = 34 -16 -18 = 0. Correct.So both points satisfy both equations. Therefore, the coordinates are (3, 3) and (-3, 5). But just to make sure I didn't make a mistake in the algebra, let me go through the steps again quickly.Starting with subtracting the two equations to eliminate the quadratic terms. Got x + 3y = 12. Then substituted into the second equation, expanded, combined terms correctly. Quadratic equation led to y = 3 and y = 5. Plugging back gives x = 3 and x = -3. Checked both points in both equations. All steps seem correct. I think that's thorough. So the final answer should be those two points.

✅:The coordinates of the intersection points of the given circles are (boxed{(3, 3)}) and (boxed{(-3, 5)}).Step-by-Step Explanation:1. Rewrite the Second Circle's Equation: Start with the equation (x^2 + y^2 + 2(x - y) - 18 = 0). Expand and complete the square: (x^2 + 2x + y^2 - 2y = 18) ((x + 1)^2 + (y - 1)^2 = 20). 2. Equations of the Circles: - First circle: ((x - 2)^2 + (y - 10)^2 = 50) - Second circle: ((x + 1)^2 + (y - 1)^2 = 20)3. Expand Both Equations: - First circle expanded: (x^2 + y^2 - 4x - 20y + 54 = 0) - Second circle expanded: (x^2 + y^2 + 2x - 2y - 18 = 0)4. Subtract the Equations to Eliminate Quadratic Terms: [ (x^2 + y^2 - 4x - 20y + 54) - (x^2 + y^2 + 2x - 2y - 18) = 0 ] Simplifies to: (-6x - 18y + 72 = 0) Divide by -6: (x + 3y = 12) ⇒ (x = 12 - 3y)5. Substitute (x = 12 - 3y) into the Second Circle's Equation: [ (12 - 3y)^2 + y^2 + 2(12 - 3y) - 2y - 18 = 0 ] Simplify to: (10y^2 - 80y + 150 = 0) Divide by 10: (y^2 - 8y + 15 = 0) Factor: ((y - 3)(y - 5) = 0) ⇒ (y = 3) or (y = 5)6. Find Corresponding (x)-values: - For (y = 3): (x = 12 - 3(3) = 3) - For (y = 5): (x = 12 - 3(5) = -3)7. Verify Solutions: Both points ((3, 3)) and ((-3, 5)) satisfy the original equations.Final Answer: The intersection points are (boxed{(3, 3)}) and (boxed{(-3, 5)}).

🔑:Let us find the coordinates of the points of intersection ( A(x_1, y_1) ) and ( B(x_2, y_2) ) of the given circles by solving the system of equations:[begin{cases}(x-2)^2 + (y-10)^2 = 50 x^2 + y^2 + 2(x-y) - 18 = 0 end{cases}]# Step 1: Expand the equationsStart by expanding the equations:1. Expand ((x-2)^2 + (y-10)^2 = 50):[(x-2)^2 + (y-10)^2 = 50 Rightarrow (x^2 - 4x + 4) + (y^2 - 20y + 100) = 50 Rightarrow x^2 + y^2 - 4x - 20y + 104 = 50 Rightarrow x^2 + y^2 - 4x - 20y + 54 = 0]2. Expand (x^2 + y^2 + 2(x-y) - 18 = 0):[x^2 + y^2 + 2x - 2y - 18 = 0]Thus, the equations become:[begin{cases}x^2 + y^2 - 4x - 20y + 54 = 0 x^2 + y^2 + 2x - 2y - 18 = 0 end{cases}]# Step 2: Subtract the second equation from the firstSubtract the second equation from first:[(x^2 + y^2 - 4x - 20y + 54) - (x^2 + y^2 + 2x - 2y - 18) = 0 Rightarrow x^2 + y^2 - 4x - 20y + 54 - x^2 - y^2 - 2x + 2y + 18 = 0 Rightarrow -6x - 18y + 72 = 0 Rightarrow -6x - 18y = -72 Rightarrow 2x + 6y = 24 Rightarrow x + 3y = 12 Rightarrow x = 12 - 3y]# Step 3: Substitute (x = 12 - 3y) back into one of the original equationsSubstitute (x = 12 - 3y) into (x^2 + y^2 + 2x - 2y - 18 = 0):[(12 - 3y)^2 + y^2 + 2(12 - 3y) - 2y - 18 = 0 Rightarrow (144 - 72y + 9y^2) + y^2 + 24 - 6y - 2y - 18 = 0 Rightarrow 10y^2 - 80y + 150 = 0 Rightarrow 5y^2 - 40y + 75 = 0]# Step 4: Solve the quadratic equationSolve the quadratic equation (5y^2 - 40y + 75 = 0):[y = frac{-b pm sqrt{b^2 - 4ac}}{2a} Rightarrow y = frac{40 pm sqrt{(-40)^2 - 4 cdot 5 cdot 75}}{2 cdot 5} Rightarrow y = frac{40 pm sqrt{1600 - 1500}}{10} Rightarrow y = frac{40 pm sqrt{100}}{10} Rightarrow y = frac{40 pm 10}{10} Rightarrow y = frac{50}{10} = 5 quad text{or} quad y = frac{30}{10} = 3]# Step 5: Find corresponding ( x ) valuesPlugging back ( y = 5 ) and ( y = 3 ) into ( x = 12 - 3y ):- For ( y = 5 ):[x = 12 - 3 cdot 5 = 12 - 15 = -3]- For ( y = 3 ):[x = 12 - 3 cdot 3 = 12 - 9 = 3]# ConclusionThe points of intersection are:[(3, 3) quad text{and} quad (-3, 5)](boxed{(3, 3), (-3, 5)})

🧠:Okay, let's see. I need to find the length of a crocodile in meters using these Japanese units called ken and shaku. The problem says two different things about the length from head to tail and tail to head. Hmm, maybe it's a play on words or a riddle? Let me read it again carefully."The length of a crocodile from head to tail is three times smaller than ten ken, and from tail to head it is three ken and two shaku. It is known that one shaku is equal to 30 cm. Find the length of the crocodile in meters."Wait, so from head to tail it's three times smaller than ten ken. That phrasing is a bit confusing. If something is three times smaller than ten ken, does that mean it's ten ken divided by three? Because if something is three times smaller, it's like original size divided by (3 + 1)? Wait, no, actually, when people say "three times smaller," sometimes they mean dividing by three. But this can be ambiguous. Let's clarify.Suppose the length from head to tail is three times smaller than ten ken. If "three times smaller" means that ten ken is three times the length, then the length would be ten ken divided by three. For example, if something is three times as big as X, then X is three times smaller than that. So maybe the length is (10 ken)/3. Alternatively, if "three times smaller" is taken as subtracting three times the length from the original, but that doesn't make much sense here.Then from tail to head, it's three ken and two shaku. Wait, tail to head is the same as head to tail, right? Unless there's a trick here. Maybe the problem is trying to say that when measured in one direction it's three times smaller than ten ken, and in the other direction it's three ken and two shaku. But that would mean both measurements are the same length, but expressed differently. So, maybe the length is both (10 ken)/3 and 3 ken + 2 shaku. Therefore, these two expressions are equal, and we can solve for ken, then find the length in ken and shaku, then convert to meters using the given shaku to cm conversion.Yes, that makes sense. The problem is giving the same length in two different units or expressions, so setting them equal to find the value of ken in terms of shaku, then converting to meters.Let me write down the equations.Let L be the length of the crocodile.First statement: L is three times smaller than ten ken. So, L = (10 ken) / 3.Second statement: L is three ken and two shaku. So, L = 3 ken + 2 shaku.Since both equal L, set them equal:(10 ken)/3 = 3 ken + 2 shaku.Now, we need to relate ken and shaku. The problem tells us that 1 shaku = 30 cm. But we need to know how many shaku are in a ken. I think ken is a larger unit. Let me check if there's a standard conversion between ken and shaku. Wait, in Japanese units, 1 ken is equal to 6 shaku. Let me verify this. Yes, historically, 1 ken (間) is 6 shaku (尺). So, 1 ken = 6 shaku. Therefore, 1 shaku = 1/6 ken.But maybe the problem expects us to figure that out? Wait, the problem only gives the conversion for shaku to cm. So perhaps we need to define ken in terms of shaku. Let me think.If we don't know the conversion between ken and shaku, but have two equations involving ken and shaku, can we solve for ken in terms of shaku? Let's see.We have:(10/3) ken = 3 ken + 2 shaku.Let me rearrange this equation:(10/3) ken - 3 ken = 2 shaku.Convert 3 ken to thirds: 3 ken = 9/3 ken.So,(10/3 - 9/3) ken = 2 shaku.(1/3) ken = 2 shaku.Therefore,1 ken = 6 shaku.So that's where the 1 ken = 6 shaku comes from. The problem didn't give us that directly, but by setting the two expressions for L equal, we can derive that 1 ken is 6 shaku. That's clever.So, now that we know 1 ken = 6 shaku, we can express everything in shaku or in ken. Let's proceed.We need to find L in meters. Let's first find L in ken or shaku and then convert.First, let's express L in shaku. Since 1 ken = 6 shaku, then:From the second equation, L = 3 ken + 2 shaku = 3*6 shaku + 2 shaku = 18 shaku + 2 shaku = 20 shaku.Alternatively, from the first equation, L = (10 ken)/3. Since 1 ken = 6 shaku, then:L = (10 * 6 shaku)/3 = 60 shaku /3 = 20 shaku.So both equations give L = 20 shaku, which is consistent. Good, that checks out.Now, since 1 shaku = 30 cm, then 20 shaku = 20 * 30 cm = 600 cm. Convert cm to meters: 600 cm = 6 meters.Therefore, the crocodile is 6 meters long.Wait, that seems straightforward once you set up the equations. Let me recap to make sure I didn't skip any steps.1. Interpreted "three times smaller than ten ken" as (10 ken)/3.2. Set that equal to "three ken and two shaku" which is 3 ken + 2 shaku.3. Derived that 1 ken = 6 shaku through the equation (10/3) ken = 3 ken + 2 shaku.4. Converted both expressions for L to shaku to confirm they both equal 20 shaku.5. Converted shaku to cm (30 cm each) giving 600 cm, which is 6 meters.Yes, that seems solid. I don't see any mistakes here. The key was recognizing that "three times smaller" likely meant dividing by three and that the two different expressions for the length had to be set equal to solve for the relationship between ken and shaku. Once that was established, converting to meters was straightforward.Another way someone might get confused is if they misinterpret "three times smaller." If they thought it meant subtracting three times the length from ten ken, that would lead to a different (and incorrect) equation. But given the context and the fact that setting them equal gives a logical answer, the division interpretation is correct.Also, verifying that both expressions give the same length in shaku adds confidence to the solution. So the final answer is 6 meters.

✅:The length of the crocodile can be determined by solving the given equations and converting units systematically:1. Interpret the statements: - From head to tail: The length is three times smaller than ten ken, so ( L = frac{10 , text{ken}}{3} ). - From tail to head: The length is ( 3 , text{ken} + 2 , text{shaku} ), so ( L = 3 , text{ken} + 2 , text{shaku} ).2. Set the equations equal to solve for ken in terms of shaku: [ frac{10}{3} , text{ken} = 3 , text{ken} + 2 , text{shaku} ] Rearranging gives: [ frac{1}{3} , text{ken} = 2 , text{shaku} quad Rightarrow quad 1 , text{ken} = 6 , text{shaku}. ]3. Convert the length to shaku: - Substitute ( 1 , text{ken} = 6 , text{shaku} ) into ( L = 3 , text{ken} + 2 , text{shaku} ): [ L = 3 times 6 , text{shaku} + 2 , text{shaku} = 20 , text{shaku}. ]4. Convert shaku to meters: - Given ( 1 , text{shaku} = 30 , text{cm} ), the length in meters is: [ 20 , text{shaku} times 30 , text{cm/shaku} = 600 , text{cm} = 6 , text{m}. ]Final Answer: The length of the crocodile is (boxed{6}) meters.

🔑:Let us denote the length of a crocodile from head to tail as ( L ) in ken (Japanese unit of length).1. According to the problem, the length of the crocodile from head to tail is three times less than ten ken. Mathematically, this can be represented as: [ 3L = 10 , text{ken} ] Solving for ( L ): [ L = frac{10 , text{ken}}{3} ]2. The length of the crocodile from tail to head is given as (3 , text{ken} + 2 , text{syaku}). Given (1 , text{syaku} = 30 , text{cm}), we convert syaku to ken. Knowing that (1 , text{ken} = 6 , text{syaku}), we can write: [ 3 , text{ken} + 2 , text{syaku} = 3 , text{ken} + frac{2}{6} , text{ken} = 3 , text{ken} + frac{1}{3} , text{ken} = frac{9}{3} , text{ken} + frac{1}{3} , text{ken} = frac{10}{3} , text{ken} ]3. As the two lengths (from head to tail and from tail to head) must represent the same quantity, we equate them: [ L = 3 , text{ken} + 2 , text{syaku} = frac{10 , text{ken}}{3} ]4. Now, we convert the total length in ken to centimeters. The length from head to tail ( L ) is (frac{10 , text{ken}}{3}). We know that (1 , text{ken} = 180 , text{cm}), hence: [ L = frac{10 , text{ken}}{3} = frac{10 times 180 , text{cm}}{3} = 600 , text{cm} ]5. To convert this length to meters, we use the conversion (1 , text{m} = 100 , text{cm}): [ L = frac{600 , text{cm}}{100 , text{cm/m}} = 6 , text{m} ] ConclusionThe length of the crocodile in meters is:[boxed{6 , text{m}}]

🧠:Okay, so I need to prove that in a parallelogram ABCD, with point P on side AD such that the ratio AP to AD is 1:n, the point Q where lines AC and BP intersect divides AC in the ratio AQ:AC = 1:(n+1). Hmm. Let me start by visualizing the problem.First, let me sketch a parallelogram ABCD. In a parallelogram, opposite sides are equal and parallel. So AB is parallel to CD, AD is parallel to BC. Let's label the vertices in order: A at the bottom left, B at the bottom right, C at the top right, D at the top left. So AD is the left side, and AB is the bottom side.Point P is on AD such that AP:AD = 1:n. So if I let AD be some length, say, n units, then AP would be 1 unit, and PD would be (n-1) units. Wait, but the ratio is 1:n, so maybe it's better to think of AD as length n, making AP = 1. Hmm, actually, ratios can be scaled. The exact lengths might not matter, but the ratios do.Since it's a parallelogram, diagonals bisect each other. Wait, but here the intersection is not the midpoint necessarily. AC is a diagonal, and BP is a line from B to P. Their intersection is Q, and we need to find AQ:AC.Maybe coordinate geometry would help here. Assign coordinates to the points and find equations of lines AC and BP, then find their intersection Q, then compute the ratio.Let's set coordinate system with point A at (0,0). Since it's a parallelogram, let's let AB be along the x-axis and AD along the y-axis. Wait, but in a general parallelogram, sides are not necessarily axis-aligned. Hmm, maybe to simplify, I can assign coordinates such that point A is (0,0), point B is (b,0), point D is (0,d), so point C would be (b,d) because in a parallelogram, opposite sides are equal and parallel. So vector AB is (b,0), vector AD is (0,d), so coordinates are:A: (0,0)B: (b,0)D: (0,d)C: (b,d)Then side AD is from (0,0) to (0,d). Point P is on AD such that AP:AD = 1:n. So AP is (1/n)*AD. Since AD has length d (if we consider the coordinates), but in terms of the ratio, maybe it's better to parametrize. Let's let AD be from (0,0) to (0,1) for simplicity, so scaling appropriately. Wait, maybe even better to set AD as length n, so that AP is 1. But perhaps coordinates can be set without loss of generality.Wait, maybe using vectors would be helpful here. Alternatively, mass point geometry. But let's try coordinate geometry first.Let me assign coordinates:Let me set point A at (0,0). Since ABCD is a parallelogram, let me let AB be along the x-axis. Let’s suppose AB has length ‘a’ and AD has length ‘b’. But since ratios are important, maybe setting specific values would make it easier. Let’s choose coordinates such that:- A is at (0, 0)- B is at (1, 0) (so AB has length 1)- D is at (0, n) (so AD has length n, making AP:AD = 1:n mean that AP = 1, so P is at (0,1))Wait, but AP:AD = 1:n. If AD is length n, then AP would be 1, so P is 1 unit up from A on AD. So if AD is from (0,0) to (0,n), then P is at (0,1). Then D is at (0,n). Then point C, being the opposite vertex, is at (1, n) because AB is (1,0), so adding vector AB to point D(0,n) gives C(1,n). Hmm, that makes ABCD a parallelogram with sides AB=1, AD=n.But actually, in a general parallelogram, sides don't have to be axis-aligned, but if we assign coordinates this way, it should still hold. So coordinates:A: (0,0)B: (1,0)D: (0,n)C: (1,n)P: (0,1) because AP=1, AD=n, so ratio 1:n.Now, find the intersection Q of AC and BP.First, let's find equations for lines AC and BP.Line AC connects A(0,0) and C(1,n). The parametric equations for AC can be written as:x = ty = n twhere t ranges from 0 to 1.Line BP connects B(1,0) and P(0,1). Let's find its equation.The slope of BP is (1 - 0)/(0 - 1) = -1. So the equation is y - 0 = -1(x - 1), which simplifies to y = -x + 1.Parametric equations for BP can be written as starting at B(1,0) and going to P(0,1). Let parameter s go from 0 to 1:x = 1 - sy = 0 + sSo x = 1 - s, y = s.Now, find intersection Q of AC and BP.Set the coordinates equal:From AC: x = t, y = n tFrom BP: x = 1 - s, y = sSo set t = 1 - s and n t = s.Substitute t = 1 - s into n t = s:n(1 - s) = sn - n s = sn = s + n sn = s(1 + n)Therefore, s = n / (1 + n)Then, t = 1 - s = 1 - [n / (n + 1)] = (n + 1 - n)/ (n +1 ) = 1/(n +1 )Therefore, point Q has coordinates (t, n t) = (1/(n+1), n/(n+1))Now, compute AQ:AC.Vector AC goes from (0,0) to (1,n), so its length is sqrt(1 + n²), but since we are interested in the ratio along the segment, not the actual distance, we can use the parameter t. Since Q corresponds to t = 1/(n+1), the ratio AQ:AC is t : 1, which is 1/(n+1) : 1, so 1 : (n+1). Hence, AQ:AC = 1:(n+1), which is what was to be proven.Alternatively, since in parametric terms, AQ is the segment from A to Q, which is t = 1/(n+1), so the ratio AQ:AC is 1/(n+1) : 1, which simplifies to 1:(n+1).Wait, that seems straightforward. Let me check if my coordinate assignment affected the generality. I set AB to length 1 and AD to length n, but in reality, the parallelogram could have any dimensions. However, since the ratios are preserved under scaling, the result should hold generally.Alternatively, if I use vectors, let's see.Let’s consider vectors with A at the origin.Let vector AB = b, vector AD = d.Then point B is at b, D is at d, C is at b + d.Point P is on AD such that AP:AD = 1:n, so vector AP = (1/n) AD = (1/n)d. Therefore, P is at (1/n)d.Line BP connects B(b) and P((1/n)d). The parametric equation for BP is b + t[(1/n)d - b], where t ∈ [0,1].Line AC connects A(0) and C(b + d). Parametric equation: s(b + d), s ∈ [0,1].Find intersection Q. So set b + t[(1/n)d - b] = s(b + d).Solve for s and t.Expanding the left side: b + t*(1/n d - b) = b - tb + (t/n)dRight side: sb + sdEquate coefficients of b and d:For b: 1 - t = sFor d: t/n = sTherefore, from the d component: s = t/nFrom the b component: 1 - t = t/nSolve for t:1 = t + t/n = t(1 + 1/n) = t(n + 1)/nThus, t = n / (n + 1)Then, s = t/n = (n / (n +1 )) / n = 1/(n +1 )Therefore, the parameter s along AC is 1/(n +1 ), so AQ:AC = s :1 = 1:(n+1 ). Hence proved.This vector approach also leads to the same conclusion, confirming the result is general and not dependent on specific coordinate choices.Alternatively, using mass point geometry. Let's see.In mass point, since ABCD is a parallelogram, so diagonals bisect each other, but here we have a different intersection. Let me consider the lines BP and AC.We need to find the ratio AQ:QC. Wait, AQ:AC = 1:(n+1) implies AQ:QC = 1:n.Let’s assign masses to points A and C such that the mass at A times AQ = mass at C times QC. If AQ:QC = 1:n, then mass at A is n and mass at C is 1, making the mass at Q equal to n +1.But how does point P come into play? Since Q is the intersection of BP and AC, maybe we can assign masses at B and P such that the masses are in the ratio BP:PD or something.Wait, mass point might require setting up the masses appropriately. Let's think:Along AD, point P divides AD into AP:PD = 1:(n-1). Wait, AP:AD =1:n, so AP:PD =1:(n-1). So if we assign masses at A and D such that mass at A * AP = mass at D * PD. So mass at A / mass at D = PD / AP = (n-1)/1. So mass at A is n-1, mass at D is 1. Then the mass at P is mass at A + mass at D = (n -1 ) +1 = n.Now, looking at line BP. Point B is on AB, but we need to see how masses at B and P affect the intersection Q on AC.But perhaps this is getting complicated. Alternatively, using coordinate geometry as before was more straightforward.Another approach: use Menelaus' theorem on triangle ABC with transversal BPQ.Wait, Menelaus' theorem states that for a triangle, if a line crosses the three sides (or their extensions), the product of the segment ratios is equal to 1.But let's see. Let's consider triangle ABC. The line BP intersects AC at Q and BC at... Wait, BP starts at B, goes to P, which is on AD. Since ABCD is a parallelogram, AD is congruent to BC. Wait, maybe this is not the right triangle.Alternatively, consider triangle ABD. Wait, not sure.Alternatively, use Menelaus on triangle ACD with transversal BPQ. Hmm, not sure.Alternatively, coordinate geometry was the most straightforward.Alternatively, use vectors as done before.Alternatively, area ratios.Since ABCD is a parallelogram, area of ABC is half the area of ABCD.Let’s consider the areas to find the ratio AQ:AC.Let me denote the area of the parallelogram as S.Since AP:AD =1:n, then the height from P to AB is 1/n times the height of the parallelogram.But maybe this approach is more involved. Let me think.If we consider triangles. Line BP intersects AC at Q. We need the ratio AQ:AC.Consider triangle ABP. Wait, point Q is inside the parallelogram, intersection of BP and AC.Alternatively, use coordinates. Since I already did that and it worked, maybe that's the simplest.Alternatively, parametric approach.Yes, so both coordinate and vector methods give the same result, so the ratio is indeed 1:(n+1). I think this is sufficient. Let me just verify once again.If in coordinates, with A(0,0), B(1,0), D(0,n), C(1,n), P(0,1). Line AC is from (0,0) to (1,n), parametric equations x = t, y = nt.Line BP is from (1,0) to (0,1), parametric equations x=1-s, y=s.Intersection when t =1 -s and nt = s.Substitute t =1 -s into nt =s: n(1 -s)=s ⇒ n - ns = s ⇒ n = s(n +1 ) ⇒ s= n/(n +1 )Then t =1 -n/(n +1 )=1/(n +1 )Therefore, Q is at (1/(n+1), n/(n+1)), so along AC, which goes from (0,0) to (1,n), the ratio AQ:AC is the parameter t, which is 1/(n+1), hence 1:(n+1).Yes, this checks out.So, after exploring coordinate geometry, vector methods, and considering alternative approaches like mass point or Menelaus', the coordinate approach provided a clear and direct proof. The key was setting up coordinates to model the problem, deriving the equations of the relevant lines, solving for their intersection, and then computing the ratio based on the parameter along the diagonal AC. This systematic approach confirms the desired ratio holds true.

✅:To prove that ( AQ:AC = 1:(n+1) ) in parallelogram ( ABCD ) with point ( P ) on ( AD ) such that ( AP:AD = 1:n ), follow these steps:1. Coordinate Setup: - Place ( A ) at the origin: ( A(0,0) ). - Let ( B ) be ( (1,0) ), ( D ) be ( (0,n) ), and ( C ) be ( (1,n) ). - Point ( P ) divides ( AD ) in the ratio ( 1:n ), so ( P(0,1) ).2. Equations of Lines: - Diagonal ( AC ): Parametric form: ( x = t ), ( y = nt ) where ( t in [0,1] ). - Line ( BP ): Connects ( B(1,0) ) and ( P(0,1) ). Slope = (-1), so equation: ( y = -x + 1 ). Parametric form: ( x = 1 - s ), ( y = s ) where ( s in [0,1] ).3. Find Intersection ( Q ): - Set coordinates equal: [ t = 1 - s quad text{and} quad nt = s. ] - Substitute ( t = 1 - s ) into ( nt = s ): [ n(1 - s) = s implies n = s(n + 1) implies s = frac{n}{n+1}. ] - Then, ( t = 1 - frac{n}{n+1} = frac{1}{n+1} ).4. Determine Ratio ( AQ:AC ): - The parameter ( t = frac{1}{n+1} ) along ( AC ) corresponds to ( Q ). - Thus, ( AQ:AC = frac{1}{n+1}:1 = 1:(n+1) ).Conclusion: By coordinate analysis, the intersection point ( Q ) divides ( AC ) in the ratio ( 1:(n+1) ), proving ( AQ:AC = 1:(n+1) ).[boxed{AQ:AC = frac{1}{n+1}}]

🔑:1. We start by considering the properties and characteristics of the parallelogram (ABCD). In a parallelogram, opposite sides are equal and parallel, and diagonals bisect each other. We also know that point (P) is taken on side (AD) such that: [ frac{AP}{AD} = frac{1}{n}. ]2. Next, let’s denote the coordinates of the points in the parallelogram to make the problem more tangible: - Let (A) be the origin, i.e., (A = (0, 0)). - Let (B = (a, b)), (C = (a+c, b+d)), and (D = (c, d)).3. Given that (P) divides (AD) in the ratio (1:n), the coordinates of (P) can be written as: [ P = left(frac{c}{n+1}, frac{d}{n+1}right). ]4. Now, we identify the intersection point (Q) of the lines (AC) and (BP). The parametric equations for line (AC) are: [ AC: left(x, frac{d}{c} x right), ] since (A) is the origin and (C = (a+c, b+d)).5. The parametric equations for line (BP), assuming (B = (a, b)) and (P = left(frac{c}{n+1}, frac{d}{n+1} right)), are: [ BP: left((1-t)a+tfrac{c}{n+1}, (1-t) b + t frac{d}{n+1} right), ] where (t) is the parameter.6. We solve for the intersection point (Q). Setting the (x) coordinates equal, we get: [ x = (1-t)a + t frac{c}{n+1}, ] and setting the (y) coordinates equal, we have: [ frac{d}{c} x = (1-t)b + t frac{d}{n+1}. ]7. To find the specific coordinates of (Q), we solve these simultaneous equations. Given (Q) is the intersection, then (Q(x_0, y_0)) must satisfy both parametric equations: [ x_0 = (1-t)a + t frac{c}{n+1}, quad text{and} quad y_0 = frac{d}{c} x_0. ]8. Due to symmetry and similarity (triangle AQP sim triangle CQB), we infer the ratio: [ frac{AQ}{QC} = frac{AP}{BC} = frac{1}{n}. ]9. Since (AQ + QC = AC), we write: [ frac{AQ}{QC} = frac{1}{n} implies m cdot frac{AQ}{QC} = 1 + frac{1}{n} = frac{n+1}{n} = m. ]10. Let’s denote (AQ = x) then (QC = nx). Thus, we have: [ AC = AQ + QC = x + nx = (n+1)x, ] implying that: [ x = frac{AC}{n+1}. ]11. Therefore, we conclude: [ frac{AQ}{AC} = frac{x}{(n+1)x} = frac{1}{n+1}. ] Conclusion:[boxed{ frac{ AQ }{ AC } = frac{ 1 }{ n+1 } }.]

🧠:Okay, let's try to figure out this brick weight problem. So the question is: "A brick weighs 2 kg plus half of the brick. How much does the brick weigh?" Hmm, that's a bit confusing at first glance. Let me break it down step by step.First, the problem says the brick's weight is equal to 2 kg plus half of the brick. Wait, that seems a bit circular because the brick's weight is defined in terms of itself. Maybe I need to set up an equation to represent this relationship.Let me denote the weight of the brick as B kg. According to the problem, B is equal to 2 kg plus half of B. So translating that into an equation, it would be:B = 2 + (1/2)BNow, I need to solve for B here. Let me subtract (1/2)B from both sides to get the terms involving B on one side. That would give:B - (1/2)B = 2Simplifying the left side, B minus half of B is the same as (1 - 1/2)B, which is (1/2)B. So now the equation is:(1/2)B = 2To solve for B, I can multiply both sides by 2. That cancels out the 1/2 on the left side:B = 2 * 2B = 4 kgWait, so the brick weighs 4 kg? Let me check if that makes sense. If the brick is 4 kg, then half of the brick would be 2 kg. Adding that to the 2 kg mentioned in the problem gives 2 kg + 2 kg = 4 kg, which matches the total weight. Okay, that seems to work.But let me think again to make sure I didn't make a mistake. Sometimes these self-referential problems can be tricky. So the equation was set up correctly, right? The total weight B is 2 kg plus half of B. Solving that step by step gives B=4. And plugging it back in, it checks out. Alternatively, maybe there's another way to approach this without algebra. If I consider that the brick's weight is composed of two parts: 2 kg and half its weight. So that means the 2 kg must be the other half of the brick. Because if half the brick is part of the total weight, then the remaining half must be the 2 kg. Therefore, the whole brick would be twice that, which is 4 kg. Yeah, that makes sense too.So both methods—algebraic and logical reasoning—lead to the same answer. I think that's solid. No contradictions here. The key was recognizing that the 2 kg represents half of the brick's weight, so doubling it gives the total weight.

✅:The problem states that a brick weighs 2 kg plus half of the brick. To solve this, we define the brick's total weight as ( B ) kg. The equation derived from the problem is:[B = 2 + frac{1}{2}B]Subtract (frac{1}{2}B) from both sides to isolate terms involving ( B ):[B - frac{1}{2}B = 2 quad Rightarrow quad frac{1}{2}B = 2]Multiply both sides by 2 to solve for ( B ):[B = 4 , text{kg}]Verification: If the brick weighs 4 kg, half of its weight is ( frac{4}{2} = 2 ) kg. Adding this to the given 2 kg yields ( 2 + 2 = 4 ) kg, which matches the total weight. Thus, the solution is consistent.Final Answer: The brick weighs (boxed{4}) kilograms.

🔑:Let the weight of the brick be ( x ) kilograms. According to the problem statement, the brick weighs 2 kg plus half of its own weight. This can be expressed as the equation:[x = 2 + frac{x}{2}]1. Begin by eliminating the fraction. Multiply all terms in the equation by 2:[2x = 4 + x]2. Next, isolate ( x ) by subtracting ( x ) from both sides of the equation:[2x - x = 4]3. Simplify the left-hand side:[x = 4]Conclusion: Therefore, the weight of the brick is:[boxed{4 text{ kg}}]

🧠:Okay, so there's this chess tournament question. Let me try to figure it out step by step. First, let me understand the problem.We have a round-robin tournament with 9 boys and 3 girls. Each player plays against every other player exactly once. The scoring is standard: 1 point for a win, 0.5 for a draw, and 0 for a loss. The question is asking if it's possible that the total points scored by all the boys combined equal the total points scored by all the girls combined.Hmm. Let me break it down. In a round-robin tournament, the total number of games played is C(n, 2), where n is the total number of players. Here, there are 9 boys + 3 girls = 12 players. So total games are C(12, 2) = 66 games. Each game contributes 1 point to the total pool (since a win gives 1 point total, and a draw splits 1 point between two players). So the total points scored by all players combined is 66 * 1 = 66 points.Now, the problem asks if the boys' total points can equal the girls' total points. That would mean each group scores 33 points. So, can the girls (3 players) score 33 points in total? And the boys (9 players) also score 33 points?Wait, but let me check. The girls are 3 players. Each plays against every other player, which includes other girls and boys. So each girl plays 11 games (since there are 12 players total, subtract 1 for herself). Similarly, each boy plays 11 games.But when considering the total points scored by girls, we need to sum all the points they earn in all their games. Similarly for the boys. The key is to see if the sum of the girls' points can be 33 and the boys' points 33 as well.But wait, 3 girls scoring 33 points in total. Let's see: if each of the 3 girls played 11 games, their total number of games is 3*11 = 33 games. However, each game between two girls is counted twice here (once for each girl), so actually, the games among the girls themselves are C(3, 2) = 3 games. So the total games involving girls are:- Games among girls: 3 games.- Games between girls and boys: 3 girls * 9 boys = 27 games.So total games involving girls: 3 + 27 = 30 games. Each game contributes 1 point, so the maximum points girls can earn is 30 points. But the total points in the tournament is 66, so girls can't have 33 points because their maximum possible is 30. Therefore, it's impossible for the girls to have 33 points, so the boys can't have 33 either. Hence, the answer is no, it cannot happen.Wait, but let me verify again. Maybe I made a mistake in calculating the maximum points girls can earn. Let's re-examine.Each girl plays 11 games. The 3 girls play 3*11 = 33 games, but this counts each game between two girls twice. So actual number of games among girls is 3, contributing 3 points (since each game is 1 point). The games between girls and boys are 3*9 = 27, but each of these is a game between a girl and a boy, so 27 games, contributing 27 points. So total points possible for girls are points from intra-girl games (max 3 points if all girls win their games against other girls) plus points from games against boys. But in games against boys, the girls can earn at most 27 points if they win all those games. However, since each game between a girl and a boy is played between one girl and one boy, the points for that game are split between them. So if a girl wins against a boy, she gets 1 point, the boy gets 0. If they draw, each gets 0.5. So the maximum points girls can get from games against boys is 27 (all wins), and from intra-girl games 3 (all wins). So total maximum possible for girls is 30 points. Therefore, girls cannot reach 33 points. Since the total points in the tournament is 66, if girls can't reach 33, then boys can't have 33 either. Therefore, the total points can't be equal.Wait, but maybe there's a way where some of the games between boys and girls result in draws, but that doesn't change the total points allocated. The total points from all games is 66, which is fixed. If girls can't get 33, then boys can't either. But girls can only get up to 30, so the maximum possible for girls is 30, meaning boys would have at least 36 points. Therefore, it's impossible for the girls to have 33, so the answer is no.But let me check another angle. Maybe the initial assumption that girls can only get 30 points is correct. Let's confirm:Total games in the tournament: 66. Each game contributes 1 point, so total points is 66. Now, consider the games:1. Games among boys: C(9, 2) = 36 games. These games contribute 36 points, all to boys.2. Games among girls: C(3, 2) = 3 games. These contribute 3 points, all to girls.3. Games between boys and girls: 9*3 = 27 games. These contribute 27 points, split between boys and girls.So total points for boys would be points from intra-boy games (36 points) plus points from boy-girl games (let's call this B). Similarly, total points for girls would be points from intra-girl games (3 points) plus points from girl-boy games (G). Since in the boy-girl games, each game gives B + G = 27 points. So boys get B points, girls get G points, with B + G = 27.Therefore, total boys' points = 36 + B.Total girls' points = 3 + G.But since B + G = 27, then total girls' points = 3 + (27 - B) = 30 - B.We want boys' total = girls' total:36 + B = 30 - BSo 36 + B = 30 - B → 2B = -6 → B = -3.But B can't be negative because points can't be negative. This is impossible, so there's no solution. Hence, the total points of boys and girls cannot be equal.Therefore, the answer is no, it cannot happen.

✅:In a round-robin tournament with 9 boys and 3 girls (12 players total), each game contributes 1 point to the total pool. The total number of games is ( binom{12}{2} = 66 ), so the total points scored by all players is 66 points. For the boys' and girls' total points to be equal, each group would need to score 33 points. However, this is impossible due to the structure of the games:1. Intra-boy games: ( binom{9}{2} = 36 ) games, contributing 36 points exclusively to boys.2. Intra-girl games: ( binom{3}{2} = 3 ) games, contributing 3 points exclusively to girls.3. Boy-girl games: ( 9 times 3 = 27 ) games, where points are split between boys and girls. Let ( B ) be the boys' points and ( G ) the girls' points from these games. Here, ( B + G = 27 ).The total points for boys and girls are:- Boys: ( 36 + B ),- Girls: ( 3 + G = 3 + (27 - B) = 30 - B ).Setting these equal:[36 + B = 30 - B implies 2B = -6 implies B = -3,]which is impossible since points cannot be negative. Thus, the total points scored by all boys and all girls cannot be equal.Final Answer: boxed{No}

🔑:Let us analyze the situation of the chess tournament thoroughly.1. Total Number of Players: There are 12 players in total, consisting of 9 boys and 3 girls.2. Total Matches: Each player plays with every other player exactly once. Therefore, the total number of matches is calculated using the combination formula ( binom{12}{2} ): [ binom{12}{2} = frac{12 times 11}{2} = 66 ]3. Total Points Distributed: Each match will contribute 1 point to the total score (since a win gives 1 point and a draw gives 0.5 points to each player, summing up to 1 point in total). Hence, the total number of points distributed among all players is 66.4. Points Scored by Boys Among Themselves: Since there are 9 boys, and each boy plays a match with each other boy once, the total number of matches among boys is: [ binom{9}{2} = frac{9 times 8}{2} = 36 ] Therefore, the total points distributed in matches among boys is 36 points.5. Points Scored by Girls Among Themselves: Similarly, the total number of matches among the 3 girls is: [ binom{3}{2} = frac{3 times 2}{2} = 3 ] Hence, the total points distributed in matches among girls is 3 points.6. Points in Matches Between Boys and Girls: Each boy plays with each girl exactly once, resulting in (since there are 9 boys and 3 girls): [ 9 times 3 = 27 text{ matches} ] Thus, the total points distributed in matches between boys and girls is 27 points.7. Summing Up the Distributed Points: Summarizing the points distributed: [ 36 text{ (points among boys)} + 3 text{ (points among girls)} + 27 text{ (points between boys and girls)} = 66 text{ points} ]8. Analyzing If Boys’ Points Can Equal Girls’ Points: Let's denote the total points scored by all boys as ( P_B ) and by all girls as ( P_G ). According to the analysis above, the points gained by boys just from playing among themselves already amounts to 36 points. Now, consider the points from matches between boys and girls. Since every point in those matches also contributes to the total points distribution among girls, we notice: - Points boys could gain in matches against girls. - Points girls could gain in matches against boys. Since the total points for boys and girls must be equal to 66 points, and already 36 points are distributed among boys from their matches, the remaining 30 points must be divided between the matches with girls and boys. - If boys can score a total of x points in matches against girls, it implies girls will score ( 27 - x ) points. Therefore, summing up the boys' total points: [ P_B = 36 + x ] And girls' total points: [ P_G = 3 + (27 - x) = 30 - x + 3 = 30 - x ] Setting ( P_B = P_G ): [ 36 + x = 30 - x ] Solving for ( x ): [ 2x = -6 implies x = -3 ] Which is impossible as points cannot be negative. Therefore, the total points scored by boys cannot be equal to the total points scored by girls. Conclusion:[boxed{text{No, it is not possible.}}]