🧠:[<think>Okay, let's try to figure out this problem. So, we have three mutually perpendicular planes α, β, and γ intersecting at point O. A point P has distances 3, 4, and 12 from these three planes respectively. We need to find the distance from P to O. Hmm, let's start by visualizing this setup.First, since the planes are mutually perpendicular and intersect at O, they probably form a coordinate system. In three-dimensional space, three mutually perpendicular planes intersecting at a point define a coordinate system where each pair of planes intersects along a coordinate axis. So, point O would be the origin, and the three planes could be the xy-plane, yz-plane, and xz-plane. But the problem doesn't specify which is which, so maybe it doesn't matter.Wait, the distances from P to each plane. In a coordinate system, the distance from a point to the xy-plane is the absolute value of its z-coordinate, right? Similarly, the distance to the yz-plane is the absolute value of the x-coordinate, and the distance to the xz-plane is the absolute value of the y-coordinate. So, if the three planes are like the coordinate planes, then the distances from P to each plane would correspond to the absolute values of its coordinates.So, if point P has distances 3, 4, and 12 from the three planes, then its coordinates would be (a, b, c) where |a| = 3, |b| = 4, and |c| = 12. Then, the distance from P to O (the origin) would be the square root of (a² + b² + c²). So, regardless of the signs of a, b, and c, the distance would be sqrt(3² + 4² + 12²). Let me check that.Calculating that: 3 squared is 9, 4 squared is 16, 12 squared is 144. Adding those up: 9 + 16 = 25, 25 + 144 = 169. The square root of 169 is 13. So, the distance from P to O is 13. Wait, is that it? It seems straightforward, but let me make sure I'm not missing anything.The problem says three mutually perpendicular planes intersecting at O. So, they form a coordinate system. The distance from a point to a plane in such a coordinate system is indeed the absolute value of the coordinate perpendicular to that plane. For example, the distance from (x, y, z) to the yz-plane (which is the x=0 plane) is |x|, right? Similarly, distance to the xz-plane (y=0) is |y|, and distance to the xy-plane (z=0) is |z|. So, if the distances are 3, 4, 12, then the coordinates would be (3,4,12) or (-3,4,12) or any combination of signs. But since distance is always positive, it's absolute value. Then the distance from P to O is sqrt(3² + 4² + 12²) = 13. That seems correct.But maybe I should think again. Are the three planes necessarily the coordinate planes? What if they are different? For example, maybe the planes are rotated or something. But the problem states that they are mutually perpendicular. In three-dimensional space, any three mutually perpendicular planes intersecting at a point can be considered as coordinate planes after an appropriate rotation or translation. Since they all intersect at O, there's no translation needed; they can be aligned with the standard coordinate planes through rotation. So, regardless of their original orientation, the coordinates can be transformed such that the distances correspond to the coordinates of P. Therefore, the answer should indeed be 13.Alternatively, another approach: if three planes are mutually perpendicular, they can be used to define a coordinate system with O as the origin. The distance from a point to each plane corresponds to the coordinate along the axis perpendicular to that plane. Hence, the coordinates of P would be those distances, and the distance from P to O is the norm of the vector from O to P, which is sqrt(3² + 4² + 12²) = 13. That makes sense.Wait, just to be thorough, let me recall that in 3D geometry, the distance from a point (x, y, z) to the plane ax + by + cz + d = 0 is |ax + by + cz + d| / sqrt(a² + b² + c²). In the case of coordinate planes, the equations are simpler. For example, the xy-plane is z = 0, so the distance from (x, y, z) to the xy-plane is |z|. Similarly, the distance to the yz-plane (x=0) is |x|, and distance to the xz-plane (y=0) is |y|. So yes, if the three planes are the coordinate planes, then the distances are indeed the absolute values of the coordinates.Since the problem states that the three planes are mutually perpendicular and intersect at O, we can safely assume that they form a coordinate system where the distance from P to each plane is the coordinate in that system. Therefore, regardless of the actual orientation of the planes, the distance from P to O can be computed using the Pythagorean theorem in three dimensions.So, 3, 4, and 12 are the distances, which are like the coordinates squared and summed. The square root of (9 + 16 + 144) = sqrt(169) = 13. Therefore, the answer is 13.I don't see any issues with this reasoning. The key insight is recognizing that three mutually perpendicular planes through a point create a coordinate system where distances to the planes correspond to coordinates. Then, applying the 3D distance formula gives the result. It seems straightforward once you make that connection.Final AnswerThe distance from point P to point O is boxed{13}.</think>Given three mutually perpendicular planes α, β, and γ intersecting at point O, we need to find the distance from point P to point O. The distances from point P to these planes are 3, 4, and 12 respectively.1. Understanding the Coordinate System: The three mutually perpendicular planes intersecting at point O form a coordinate system with O as the origin. The distances from point P to each plane correspond to the absolute values of the coordinates of P in this system.2. Coordinates of Point P: If the distances from P to the planes are 3, 4, and 12, then the coordinates of P (considering absolute values) are 3, 4, and 12.3. Distance Calculation: Using the distance formula in three-dimensional space, the distance from P to O is calculated as follows: [ text{Distance} = sqrt{3^2 + 4^2 + 12^2} = sqrt{9 + 16 + 144} = sqrt{169} = 13 ]Thus, the distance from point P to point O is (boxed{13}).,<think>Alright, so I need to find the distance from point P to point O, where O is the intersection point of three mutually perpendicular planes α, β, and γ. The distances from P to each of these planes are given as 3, 4, and 12. Hmm, let me think about how to approach this.First, I remember that when dealing with mutually perpendicular planes intersecting at a point, they essentially form a coordinate system. Each plane corresponds to one of the coordinate planes in a 3D Cartesian system. For example, in the standard coordinate system, the xy-plane, yz-plane, and xz-plane are mutually perpendicular and intersect at the origin. So, maybe I can model this problem using a coordinate system where point O is the origin, and the three planes α, β, and γ are the coordinate planes.If that's the case, then the distance from a point P to each of these coordinate planes would correspond to the absolute value of its coordinates. For instance, the distance from P to the yz-plane (which is the x=0 plane) is the absolute value of the x-coordinate of P. Similarly, the distance to the xz-plane (y=0) is the absolute y-coordinate, and the distance to the xy-plane (z=0) is the absolute z-coordinate. So, if the distances from P to planes α, β, γ are 3, 4, and 12, then the coordinates of P would be (3, 4, 12), right?Wait, but the problem doesn't specify which plane corresponds to which distance. However, since the planes are mutually perpendicular, it doesn't really matter which distance we assign to which coordinate axis because the final distance from P to O will be calculated using the Pythagorean theorem in three dimensions. The order shouldn't affect the result. So, regardless of how the distances 3, 4, and 12 are assigned to the x, y, or z coordinates, the distance from P to O will be the square root of (3² + 4² + 12²). Let me verify that.Calculating the squares: 3² is 9, 4² is 16, and 12² is 144. Adding them up: 9 + 16 = 25, then 25 + 144 = 169. The square root of 169 is 13. So, the distance from P to O should be 13 units. That seems straightforward. But let me make sure I didn't skip any steps or make wrong assumptions.The key assumption here is that the three mutually perpendicular planes form a coordinate system with O as the origin, and the distances from P to each plane correspond to the coordinates of P. In 3D geometry, the distance from a point to a plane is indeed the shortest distance, which is the perpendicular distance. Since the planes are mutually perpendicular, these distances align with the coordinate axes. Therefore, the coordinates of P relative to O are (3, 4, 12), and the distance from P to O is the magnitude of this vector, which is √(3² + 4² + 12²) = 13.Is there another way to approach this problem? Maybe using the formula for distance from a point to a plane? Let me recall that the distance from a point (x, y, z) to the plane ax + by + cz + d = 0 is |ax + by + cz + d| / √(a² + b² + c²). But in this case, the planes α, β, and γ are mutually perpendicular and intersect at O. If O is the origin, then the equations of the planes can be taken as the coordinate planes: x=0, y=0, and z=0. So, the distance from P to the x=0 plane is |x|, to the y=0 plane is |y|, and to the z=0 plane is |z|. Therefore, if the distances are 3, 4, 12, then the coordinates are (3,4,12) or any combination depending on the plane, but since the planes are just labels, the distance will still be √(3² +4² +12²) =13.Alternatively, if the planes are not the coordinate planes but still mutually perpendicular, would the formula still hold? Let's suppose the planes are given by different equations, but since they are mutually perpendicular, their normals are also perpendicular. The distance from P to each plane is still the absolute value of the signed distance along the respective normals. If we set up a coordinate system aligned with these planes, then it's equivalent to the standard coordinate system. Therefore, regardless of the specific orientation of the planes, as long as they are mutually perpendicular, the problem reduces to the coordinate system case, and the distance from P to O is the Euclidean norm of the vector formed by the distances to the planes.Therefore, the answer should be 13. Let me check once more: 3 squared is 9, 4 squared is 16, 12 squared is 144. 9 +16 is 25, 25 +144 is 169. Square root of 169 is 13. Yep, that's correct.Final AnswerThe distance from point P to point O is boxed{13}.</think>Given three mutually perpendicular planes α, β, and γ that intersect at point O, we need to find the distance from point P to point O. The distances from P to these planes are 3, 4, and 12 respectively.1. Modeling the Planes as Coordinate Planes: The three mutually perpendicular planes can be considered as the coordinate planes (x=0, y=0, z=0) intersecting at the origin O. The distances from point P to these planes correspond to the absolute values of the coordinates of P.2. Coordinates of Point P: The distances 3, 4, and 12 from the planes correspond to the coordinates of P. Therefore, the coordinates of P can be (3, 4, 12) or any permutation, but the order does not affect the final distance calculation.3. Distance Calculation: Using the Pythagorean theorem in three dimensions, the distance from P to O is calculated as: [ sqrt{3^2 + 4^2 + 12^2} = sqrt{9 + 16 + 144} = sqrt{169} = 13 ]Thus, the distance from point P to point O is (boxed{13}).]

🔑:According to the problem statement, the line segment OP can be seen as the diagonal of a rectangular prism (box), where the lengths of the three edges adjacent to O are 3, 4, and 12, respectively.To find the length of the diagonal (OP), we use the Pythagorean theorem in three dimensions. Specifically, if a, b, and c are the lengths of the edges of the rectangular prism, then the length of the diagonal d is given by d = sqrt{a^2 + b^2 + c^2} In this situation, we have a=3, b=4, and c=12. Substituting these values into the equation, we get OP = sqrt{3^2 + 4^2 + 12^2} OP = sqrt{9 + 16 + 144} OP = sqrt{169} OP = 13 Therefore, the distance from point P to point O is boxed{13}.

🧠:[<think>Okay, let's see. So, the problem is about two beds with strawberry bushes. The first bed has 5 times more bushes than the second one. Then, when they transplant 22 bushes from the first to the second, both beds end up with the same number of bushes. We need to find out how many bushes were on each bed originally.Hmm, let me start by trying to set up some variables. Let's call the number of bushes on the second bed something, maybe "x". Then, since the first bed has 5 times more, that would be 5x. Wait, wait. The wording here is important. It says "5 times more strawberry bushes on the first bed than on the second." In some cases, people might interpret "5 times more" as 5 times as many, which would be 5x, but others might think it's 5 times as many plus the original, making it 6x. Hmm, which is it here?Well, in standard math terminology, "5 times more" usually means 5 times as many, so 5x. But I know in some contexts, especially in everyday language, people might use "times more" to mean the original plus that multiple. But since this is a math problem, I think it's safer to go with 5x. Let me note that. So, second bed is x, first bed is 5x.Now, when 22 bushes are transplanted from the first to the second, the numbers become equal. So, after the transplant, the first bed has 5x - 22 bushes, and the second bed has x + 22 bushes. These two quantities are equal. So, we can set up an equation:5x - 22 = x + 22Let me write that down:5x - 22 = x + 22Now, solve for x. Subtract x from both sides:4x - 22 = 22Then, add 22 to both sides:4x = 44Divide both sides by 4:x = 11So, the second bed originally had 11 bushes, and the first bed had 5 times that, which is 55 bushes.Wait, let me check. If first bed is 55 and second is 11, moving 22 from first to second: 55 - 22 = 33, and 11 + 22 = 33. Yeah, they both become 33. That seems to work.But hold on, earlier I was thinking about whether "5 times more" is 5x or 6x. If the problem had meant "5 times as many," it should be 5x, but if "more" is emphasized, maybe it's 5x more, so total is x + 5x = 6x. Let's test that possibility too, just to be thorough.Suppose second bed is x, first bed is 6x (if "5 times more" means adding 5x to the original x). Then transplanting 22:First bed becomes 6x - 22, second becomes x + 22. Setting them equal:6x - 22 = x + 22Subtract x:5x - 22 = 22Add 22:5x = 44x = 8.8Hmm, that's 8.8 bushes, which is a fraction. But you can't have a fraction of a bush here, so that can't be right. Therefore, that interpretation must be incorrect. So, the correct interpretation is 5 times as many, so 5x. Therefore, the answer is first bed 55, second bed 11. So, the solution is correct.Alternatively, maybe the problem was translated from another language, and "5 times more" could have a different standard. But given that solving for 5x gives integer results and 6x gives fractional, and the problem is about bushes which are whole numbers, 5x is the correct way to interpret it. So, I think the answer is 55 and 11.But just to be 100% sure, let me re-express the problem again. The problem says "5 times more strawberry bushes on the first bed than on the second." If you have 5 times more, that is, the first bed = second bed + 5*(second bed) = 6*(second bed). But this is conflicting with the previous result. Wait, this is a common confusion. In some contexts, "times more" is used to mean "times as much." For example, if I have 10 apples and you have 5 times more, does that mean you have 50 or 60? It's ambiguous in everyday language. But in mathematics, typically, "times more" is considered the same as "times as many," which would be 5x. But in some literature, "more" might mean in addition to the original. So, this is a possible point of confusion.But in this problem, since the answer comes out to integer when we take 5x, but 6x gives a fraction, that might be a hint that the problem expects 5x. Also, given that when you take 5x and subtract 22, you get equal numbers. So, maybe the problem is intended to be 5x. Therefore, the answer is 55 and 11.Alternatively, maybe the problem is from a non-English source where "5 times more" is consistently used to mean 5x, not 6x. So, in this case, the answer is 55 and 11.Alternatively, perhaps the problem is designed to test this exact ambiguity. But given that math problems are usually precise, and the answer is an integer, the answer is 55 and 11.So, the conclusion is, the first bed had 55 bushes, the second bed had 11 bushes. After moving 22, both have 33. That works.Final AnswerThe first bed had boxed{55} bushes and the second bed had boxed{11} bushes.</think>Let ( x ) be the number of strawberry bushes on the second bed. According to the problem, the first bed has 5 times as many bushes as the second bed, which means the first bed has ( 5x ) bushes.When 22 bushes are transplanted from the first bed to the second bed, the number of bushes on each bed becomes the same. This gives us the equation:[5x - 22 = x + 22]Solving for ( x ):1. Subtract ( x ) from both sides: [ 4x - 22 = 22 ]2. Add 22 to both sides: [ 4x = 44 ]3. Divide both sides by 4: [ x = 11 ]Therefore, the second bed originally had ( x = 11 ) bushes, and the first bed had ( 5x = 5 times 11 = 55 ) bushes.After transplanting 22 bushes, the first bed has ( 55 - 22 = 33 ) bushes and the second bed has ( 11 + 22 = 33 ) bushes, confirming the solution is correct. Final AnswerThe first bed had boxed{55} bushes and the second bed had boxed{11} bushes.,<think>Okay, let's see. So the problem is about two beds with strawberry bushes. The first bed has 5 times as many bushes as the second one. Then, when they transplant 22 bushes from the first bed to the second, both beds end up with the same number of bushes. I need to find out how many bushes were on each bed originally. Hmm, sounds like algebra. Let me break it down step by step.First, let's define some variables to represent the number of bushes on each bed. Let me call the number of bushes on the second bed "x". Then, according to the problem, the first bed has 5 times as many, so that would be "5x". That makes sense because if the second bed has x, five times that is 5x. So initially, first bed: 5x, second bed: x.Then, the problem says that when 22 bushes are transplanted from the first bed to the second, the number of bushes on both beds becomes the same. So, after the transplant, the first bed will have 5x minus 22 bushes, and the second bed will have x plus 22 bushes. And these two quantities are equal. So I can set up an equation: 5x - 22 = x + 22. Then, solving for x should give me the original number on the second bed, and multiplying by 5 gives the first bed.Wait, let me check that again. If we start with 5x on the first bed and x on the second, moving 22 from first to second would subtract 22 from the first and add 22 to the second. So yes, first bed becomes 5x - 22, second becomes x + 22. And those are equal. So the equation is 5x - 22 = x + 22.Now, solving for x. Let's subtract x from both sides: 5x - x - 22 = 22. That simplifies to 4x - 22 = 22. Then, add 22 to both sides: 4x = 44. Divide both sides by 4: x = 11. So the second bed originally had 11 bushes, and the first bed had 5 times that, so 5 * 11 = 55. Therefore, originally, first bed: 55, second bed: 11. After moving 22, first bed: 55 - 22 = 33, second bed: 11 + 22 = 33. That checks out. Both have 33, which matches the problem statement. So the answer should be 55 on the first bed and 11 on the second.Wait, but let me just verify once more to make sure I didn't make any mistakes. Let's go through the problem again. "There were 5 times more strawberry bushes on the first bed than on the second." So if the second bed is x, first is 5x. Then, when 22 are moved, first bed becomes 5x - 22, second becomes x + 22. These are equal. Solving 5x - 22 = x + 22. Subtract x: 4x - 22 = 22. Add 22: 4x = 44. x = 11. So yes, that seems correct.Alternatively, maybe I can think of it another way. Let's say after transplantation, both have the same number. Let me call that number y. So after moving, first bed is y, second bed is y. Then, before moving, the first bed was y + 22, and the second bed was y - 22. But the original ratio was 5 times more on the first. So y + 22 = 5*(y - 22). Let's see if this gives the same answer.Expanding that equation: y + 22 = 5y - 110. Subtract y from both sides: 22 = 4y - 110. Add 110 to both sides: 132 = 4y. Then y = 132 / 4 = 33. So after the move, both beds have 33. Therefore, before moving, first bed had 33 + 22 = 55, second bed had 33 - 22 = 11. Which matches the previous answer. So that's another way to approach it and it gives the same result. That gives me more confidence.So whether I set up the equation based on initial quantities and the result after moving, or set up based on the final quantities and work backwards, I end up with the same numbers. So that must be correct.Alternatively, maybe think in terms of total bushes. The total number of bushes doesn't change when moving them. Originally, total bushes are 5x + x = 6x. After moving, each bed has the same number, so each has 3x. Wait, but 6x divided by 2 is 3x. But after moving, first bed is 5x - 22, second is x + 22. So 5x - 22 must equal 3x. Wait, that's another way to think about it.So 5x - 22 = 3x. Then, subtract 3x: 2x - 22 = 0. So 2x = 22. x = 11. Which is the same answer. Alternatively, since total bushes are 6x, each bed has 3x. So the first bed went from 5x to 3x, so they lost 2x bushes. According to the problem, they lost 22 bushes. Therefore, 2x = 22. x = 11. Exactly. So that's another way.So using total bushes and the fact that the total remains constant is another method. So multiple approaches confirm that x = 11. Therefore, the original numbers are 55 and 11. Moving 22 from first to second gives 33 each, which is equal. All methods check out.I think that's solid. So the answer is 55 on the first bed and 11 on the second.Final AnswerThe first bed had boxed{55} bushes and the second bed had boxed{11} bushes.</think>Let ( x ) be the number of strawberry bushes on the second bed. According to the problem, the first bed has 5 times as many bushes as the second bed, so the first bed has ( 5x ) bushes.When 22 bushes are transplanted from the first bed to the second bed, the number of bushes on each bed becomes the same. This gives us the equation:[5x - 22 = x + 22]Solving for ( x ):[5x - 22 = x + 22 5x - x = 22 + 22 4x = 44 x = 11]Thus, the number of bushes on the second bed is ( x = 11 ). The number of bushes on the first bed is ( 5x = 5 times 11 = 55 ).To verify, after transplanting 22 bushes:- The first bed has ( 55 - 22 = 33 ) bushes.- The second bed has ( 11 + 22 = 33 ) bushes.Both beds have the same number of bushes, confirming the solution is correct. Final AnswerThe first bed had boxed{55} bushes and the second bed had boxed{11} bushes.]

🔑:## 5. First method of solving.Let's build a graphical model of the problem's condition1) 22: 2=11 (bushes) - this is 1/5 of all the bushes (that were on the second bed);2) 11 times 5=55 (bushes) - this was on the first bed.Second method of solving (Nikolai Zherebtsov - student of School No. 46).1) 5+1=6 (parts) - in total;2) 6: 2=3 (parts) - this is how many parts each bed has;3) 5-3=2 (parts) - this is how many parts were transplanted from the first bed to the second;4) 22: 2=11 (bushes) - this is how many bushes are in one part (there were on the second bed)5) 11 times 5=55 (bushes) - this was on the first bed.Third method of solving (Vitaliy Semikin - student of School No. 48).1) 22+22=44 (bushes) - this is how many fewer bushes there are on the second bed compared to the first;2) 44: 4=11 (bushes) - this is how many bushes are in one part (there were on the second bed)3) 11 times 5=55 (bushes) - this was on the first bed.Answer: 11 bushes were on the second bed, 55 bushes were on the first.

🧠:[<think>Okay, let's try to figure out Cat's favorite number. So, according to the problem, Cat says that her favorite number is a two-digit positive integer that is the product of three distinct prime numbers. Claire then says she doesn't know the number yet, but she knows that among four numbers that might be Cat's favorite, you could start with any one of them, add a second, subtract a third, and get the fourth. Then Cat mentions her number isn't among those four, and now Claire knows the number. We need to figure out what that number is.First, let's break down the information step by step. 1. Two-digit number, product of three distinct primes. So, we need to list all two-digit numbers that are the product of three different primes. Let's call this set S.2. Claire doesn't know the number yet, but knows there are four numbers in consideration where starting with any one, adding a second, subtracting a third gives the fourth. This part is a bit tricky. Claire is referring to four numbers where for any permutation of starting with one, adding another, subtracting another, you get the fourth. That seems like some kind of relationship where the four numbers form a system where they can be combined in such a way. Maybe it's a set where the numbers have some additive and subtractive relationships?3. Cat says her number isn't among those four, and now Claire can deduce the number. So, after eliminating those four numbers, there's only one remaining possibility, which must be Cat's favorite number.Okay, let's start by listing all two-digit numbers that are products of three distinct primes. To do this, I need to find numbers where their prime factorization consists of exactly three different primes, and the product is between 10 and 99.First, let's recall that the smallest three distinct primes are 2, 3, and 5. Their product is 30. Then, the next combinations would be:2, 3, 7: 2*3*7=422, 3, 11: 662, 3, 13: 782, 3, 17: 102 (which is three digits, so too big)So moving to primes with 2, 5:2, 5, 7: 702, 5, 11: 110 (too big)Then 3, 5, 7: 105 (too big)Now using 2, 3, 5: 30Wait, maybe I need a systematic way. Let's list all possible combinations:Start with 2 as the smallest prime:2, 3, 5: 302, 3, 7: 422, 3, 11: 662, 3, 13: 782, 5, 7: 702, 5, 11: 110 (too big)2, 5, 13: 130 (too big)2, 7, 11: 154 (too big)So, that's 30, 42, 66, 78, 70Then starting with 3 as the smallest prime (excluding 2):3, 5, 7: 105 (too big)3, 5, 11: 165 (too big)So all others would be three digits. So the two-digit numbers that are products of three distinct primes are: 30, 42, 66, 70, 78.Wait, is that all? Let's check again.Wait, 2*3*5=302*3*7=422*3*11=662*3*13=782*5*7=70Is there another combination? Let's see:2*7*3=42 (already listed)3*5*7=105 (too big)So, yes, those are the only two-digit numbers. Wait, 105 is too big, so 30, 42, 66, 70, 78. So five numbers in total. Wait, but Claire refers to four numbers. Hmm.Wait, maybe I missed some. Let's check again.Wait, 2*5*7=702*5*3=302*5* anything higher would be over 100. 2*7* anything higher would also be over 100.So, 30, 42, 66, 70, 78. So five numbers. But Claire mentions four numbers. So perhaps one of these five is not considered? But Claire says "among four of the numbers that might be your favorite number," so maybe there are four numbers that satisfy the additive relationship.So, first, let's list the possible numbers: 30, 42, 66, 70, 78.Now, Claire knows that among four of these numbers, starting with any one, add a second, subtract a third, and get the fourth. So, for any permutation of the four numbers, there exists some combination where a + b - c = d.But Claire says, "you could start with any one of them, add a second, subtract a third, and get the fourth." Wait, maybe not for any permutation, but that the four numbers have the property that for each one, there exists some other three such that starting with it, adding one, subtracting another, gives the fourth. Hmm, the wording is a bit confusing.Alternatively, maybe Claire is saying that there are four numbers such that if you take any one of them as the starting point, you can add a second, subtract a third, and arrive at the fourth. So, for each of the four numbers, there's a way to combine the other three with addition and subtraction to get that number. That is, each number is expressible as a + b - c where a, b, c are the other three numbers. But this is a bit complex.Alternatively, maybe the four numbers form a system where if you take any one as a starting point, you can add one, subtract another, and get the fourth. So, for example, if we have numbers a, b, c, d, then:a + b - c = db + c - d = ac + d - a = bd + a - b = cOr something like that. But this is a bit of a stretch. Alternatively, maybe it's a set where the numbers are in an arithmetic progression or some other relation. Let me think.Alternatively, maybe the four numbers can be arranged in such a way that each is a linear combination of the others with coefficients +1, +1, -1. So, for four numbers w, x, y, z, we have:w = x + y - zx = y + z - wy = z + w - xz = w + x - yIf that's the case, then adding all four equations:w + x + y + z = (x + y - z) + (y + z - w) + (z + w - x) + (w + x - y)Simplify the right side:(x + y - z) + (y + z - w) + (z + w - x) + (w + x - y)Let's combine like terms:x cancels with -x, y cancels with -y, z cancels with -z, w cancels with -w. Wait, but actually:First term: x + y - zSecond term: y + z - wThird term: z + w - xFourth term: w + x - yAdding all together:x + y - z + y + z - w + z + w - x + w + x - yCombine like terms:x cancels with -x, y cancels with -y, z cancels with -z, w cancels with -w, but wait:Wait:x from first term and -x from third term: cancels.y from first term, y from second term, -y from fourth term: y + y - y = y.-z from first term, z from second term, z from third term: -z + z + z = z.-w from second term, w from third term, w from fourth term: -w + w + w = w.So total sum: y + z + w.But the left side was w + x + y + z. So:w + x + y + z = y + z + wTherefore, x = 0. But x is one of the numbers, which are two-digit positive integers, so this is impossible. Therefore, such a system of equations cannot hold unless x = 0, which is not possible here. Therefore, maybe my initial assumption is wrong.Alternatively, perhaps the four numbers are such that for any two of them, their sum minus another gives the fourth. But not sure. Let me try to see if any four numbers from the list 30, 42, 66, 70, 78 have such relationships.Let's check:Suppose we pick four numbers: 30, 42, 66, 78.Check if starting with any one, add another, subtract another, gives the fourth.Starting with 30: 30 + 42 - 66 = 6, not 78. 30 + 66 - 42 = 54, not 78. 30 + 78 - 42 = 66. Oh, here we go: 30 + 78 - 42 = 66. So 30 + 78 - 42 = 66. So that works if we start with 30, add 78, subtract 42, get 66.But the problem says "you could start with any one of them, add a second, subtract a third, and get the fourth." So, for each number, there must be a way to add one of the others and subtract another to get the fourth.So, let's check for each of the four numbers:For 30: 30 + a - b = c, where a, b, c are the other three numbers.30 + 42 - 66 = 6, not 78.30 + 66 - 42 = 54, not 78.30 + 78 - 42 = 66. So 30 + 78 - 42 = 66. So this gives 66, which is in the set.Wait, but the result should be the fourth number, which is 78? Wait, no. Wait, if we start with 30, add a second number, subtract a third, we should get the fourth. The four numbers are 30, 42, 66, 78.So starting with 30, if we add 42 and subtract 66, we get 6, not 78.If we add 66 and subtract 42, we get 54, not 78.If we add 42 and subtract 78, we get -6, not 66.Wait, maybe I have the wrong set. Let's check another combination.Take numbers 30, 42, 66, 70.Check if starting with any one:30 + 42 - 66 = 6 ≠ 7030 + 66 - 42 = 54 ≠ 7030 + 70 - 42 = 58 ≠ 6630 + 42 - 70 = -8 ≠ 66Doesn't seem to work.Another combination: 42, 66, 70, 78.Check starting with 42:42 + 66 - 70 = 38 ≠ 7842 + 70 - 66 = 46 ≠ 7842 + 78 - 66 = 54 ≠ 70Not working.Another combination: 30, 66, 70, 78.Start with 30:30 + 66 - 70 = 26 ≠ 7830 + 70 - 66 = 34 ≠ 7830 + 78 - 66 = 42, which is not in the set (the set here is 30, 66, 70, 78). So 42 is excluded, so no.Hmm. Maybe another approach. Maybe the four numbers form a quadruple where each number is the sum of two others minus the third. For example, in some sets of four numbers, each number can be expressed as the sum of two others minus the fourth. Let's see.Wait, for four numbers a, b, c, d, perhaps each number is equal to (sum of two others) minus the fourth. So:a = b + c - db = a + c - dc = a + b - dd = a + b - cBut this would imply that:From the first equation, a = b + c - dFrom the second, b = a + c - d. Substitute a from the first equation: b = (b + c - d) + c - d = b + 2c - 2dSubtract b from both sides: 0 = 2c - 2d => c = d. But all numbers are distinct, so this is impossible. Therefore, such a system is impossible unless numbers are not distinct, which they are. So maybe the relationships are different for each number. Maybe each number can be expressed as some combination, not necessarily the same structure.Alternatively, maybe it's like a,b,c,d where a + b = c + d, and similar relations. Let's check.Take 30, 42, 66, 78.30 + 78 = 10842 + 66 = 108So 30 + 78 = 42 + 66. That's an interesting relation. So maybe in this case, starting with any one, add another, subtract another gives the fourth. Let's check.Starting with 30: 30 + 42 - 66 = 6 ≠ 78. Wait, that doesn't work.But 30 + 78 - 42 = 66, which is another number in the set. Similarly, 42 + 66 - 30 = 78. 66 + 78 - 42 = 102, which is not in the set.Alternatively, using the fact that 30 + 78 = 42 + 66, then:If you start with 30, add 78, you get 108. Then subtract 42, you get 66. Or subtract 66, you get 42. Similarly, starting with 42, add 66, subtract 30, get 78. Starting with 66, add 42, subtract 30, get 78. Starting with 78, add 30, subtract 42, get 66. Hmm, so each number can be obtained by adding one and subtracting another. For example:30 + 78 - 42 = 6630 + 78 - 66 = 4242 + 66 - 30 = 7842 + 66 - 78 = 3066 + 42 - 30 = 7866 + 42 - 78 = 3078 + 30 - 42 = 6678 + 30 - 66 = 42So actually, if you take the two pairs that sum to 108 (30+78 and 42+66), then starting with any one number, adding its pair's counterpart, and subtracting the other number in the pair gives you the remaining number.Wait, so for example:Take 30. Its pair is 78. If you add 78 to 30, you get 108. Then subtract 42 (from the other pair), you get 66. Alternatively, subtract 66, you get 42. Similarly, take 42. Its pair is 66. Add 66 to 42, get 108. Subtract 30, get 78. Subtract 78, get 30. So in this way, each number can be used in such an equation. So if Claire is referring to these four numbers: 30, 42, 66, 78, then they have the property that starting with any one, add its pair, subtract another number, get the fourth. So Claire is saying that among four numbers (these four), you can perform this operation. Therefore, Claire narrows it down to these four, but Cat says her number isn't among them, leaving only 70 as the remaining possibility. Then Claire knows it's 70. Therefore, Cat's favorite number is 70.Wait, but let's check if 70 is indeed the remaining number. The possible numbers are 30, 42, 66, 70, 78. If Claire's four numbers are 30, 42, 66, 78, then the only one left is 70. So if Cat says her number isn't among those four, Claire can deduce it's 70. Therefore, the answer is 70.But let's verify that Claire actually could narrow it down. Suppose Claire, upon hearing that the number is a product of three distinct primes, considers all possibilities: 30, 42, 66, 70, 78. Then Claire knows that the number must be among four numbers (30, 42, 66, 78) which have that additive property. But Cat's number isn't among those four, so it must be 70. Therefore, 70 is the answer.But let me check again if Claire would have known. So Claire says, "I don't know your favorite number yet, but I do know that among four of the numbers that might be your favorite number, you could start with any one of them, add a second, subtract a third, and get the fourth!" So Claire is certain that the number is among these four, but she doesn't know which one. Then Cat says it's not among those four, so Claire deduces it's the remaining one, which is 70.Therefore, the answer should be 70. Let me check if there are other possibilities. Suppose there is another set of four numbers with that additive property, but looking at the list of possible numbers (30, 42, 66, 70, 78), the only set that has the additive relationship where each number can be obtained by adding two and subtracting another is 30, 42, 66, 78. The other number, 70, isn't part of such a quartet. Therefore, the answer is 70.Final Answerboxed{70}</think>Cat's favorite number is a two-digit positive integer that is the product of three distinct prime numbers. The possible numbers are 30, 42, 66, 70, and 78. Claire mentions that among four numbers, you can start with any one, add a second, subtract a third, and get the fourth. These four numbers must be 30, 42, 66, and 78 because they satisfy the relationship where each number can be obtained by adding two others and subtracting the fourth. For example:- (30 + 78 - 42 = 66)- (42 + 66 - 30 = 78)- (66 + 42 - 78 = 30)- (78 + 30 - 66 = 42)Cat states her favorite number is not among these four, leaving 70 as the only remaining possibility. Claire deduces the number must be 70.[boxed{70}],<think>Okay, let's try to figure out Cat's favorite number step by step. So, Cat says it's a two-digit positive integer that's the product of three distinct prime numbers. Claire then says she doesn't know the number yet, but knows that among four possible numbers, you can start with any one of them, add a second, subtract a third, and get the fourth. Then Cat says her number isn't among those four, and Claire figures it out. Hmm, this seems a bit complex, but let's break it down.First, let's list all two-digit numbers that are the product of three distinct primes. To do that, I need to recall that a product of three distinct primes will have exactly 8 divisors (since each prime can be either included or excluded, so 2^3 = 8). But actually, wait, for three distinct primes, the number of divisors is (1+1)(1+1)(1+1) = 8, right. So any number that's the product of three distinct primes will have exactly 8 divisors. So first step: list all two-digit numbers with exactly 8 divisors where each prime factor is distinct.Let me start by listing the primes under 50, since the number is two-digit. The primes less than 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. But since the number is two-digit, the product of three primes can't be too large. Let's find all possible combinations.Start with the smallest primes:2*3*5=302*3*7=422*3*11=662*3*13=782*3*17=102 (too big)So stop here for 2*3*p. Then 2*5*7=702*5*11=110 (too big)Next, 2*7*11=154 (too big)So moving to 3*5*7=105 (too big)What about 3*3*5=45, but that's not three distinct primes. So invalid.Wait, so starting with 2 as a prime, the possible combinations:2*3*5=302*3*7=422*3*11=662*3*13=782*5*7=70Then 3*5*7=105 (too big)So maybe those are the only ones. Wait, 2*5*7=70, 2*5*11=110 (too big). 2*7*11=154 (too big). 3*5*7=105 (too big). 3*5*2=30 (already listed). 3*5*7=105. So maybe the two-digit numbers are 30, 42, 66, 70, 78.Wait, let me check if I missed any.For primes starting with 2:2*3*5=302*3*7=422*3*11=662*3*13=782*5*7=702*5*11=110 (too big)2*5*13=130 (too big)Similarly, 2*7*11=154 (too big)So yes, 30, 42, 66, 70, 78.Now, primes starting with 3:3*5*7=105 (too big)3*5*2=30 (already listed)3*5*11=165 (too big)Similarly, all others will be over 100.So the possible numbers are 30, 42, 66, 70, and 78. Wait, but let me confirm each of these:30: 2*3*5=30. Correct.42: 2*3*7=42. Correct.66: 2*3*11=66. Correct.70: 2*5*7=70. Correct.78: 2*3*13=78. Correct.So these are the five two-digit numbers that are the product of three distinct primes. So Cat's favorite number is one of these: 30, 42, 66, 70, 78.Now, Claire says she doesn't know the number yet but knows that among four of the numbers that might be Cat's favorite number, you could start with any one of them, add a second, subtract a third, and get the fourth. So there's a set of four numbers where this relationship holds: starting with any one, add another, subtract another, and get the fourth. So for any a, b, c, d in the set, a + b - c = d? But the problem says "you could start with any one of them, add a second, subtract a third, and get the fourth". Hmm, maybe it's that for any permutation of the four numbers, one can be obtained by starting with another, adding another, subtracting another. Wait, that seems a bit unclear.Wait, the exact quote is: "you could start with any one of them, add a second, subtract a third, and get the fourth!" So for the four numbers, say w, x, y, z. Then for each of them, you can start with one, add another, subtract another, and get the fourth. So for example, starting with w, add x, subtract y, get z. But also starting with x, add y, subtract z, get w? Or starting with any one, add any other, subtract any other, and get the fourth? Hmm, this is a bit ambiguous.Alternatively, maybe the four numbers form a system where each can be expressed as a combination of the others through addition and subtraction. But the key is that Claire is referring to four numbers, and in those four numbers, such a relationship holds. So perhaps it's a set of four numbers where they form an additive quadruple in some way.Alternatively, maybe it's that for any two numbers in the set, their sum minus another number gives the fourth. But the wording says "start with any one of them, add a second, subtract a third, and get the fourth". So perhaps that for any choice of the starting number, you can find three others such that starting with the first, add a second, subtract a third, get the fourth. But if you can start with any one, then perhaps for each number in the set, there's some permutation where that number is the starting one, and the equation holds.Alternatively, maybe the four numbers are such that all four can be written in terms of each other via these operations. So maybe they form a closed system under these operations.Alternatively, maybe the four numbers are in an arithmetic progression? But with four numbers, but the operations here are addition and subtraction. Wait, but in an arithmetic progression, if you have four terms a, a+d, a+2d, a+3d, then you could have relationships like a + (a+d) - (a+2d) = a - d, which isn't in the set unless d=0, which is not possible. So maybe not arithmetic progression.Alternatively, maybe they form a modular arithmetic system? Hmm.Alternatively, think of it as four numbers where each is a linear combination of the others with coefficients +1, +1, -1.Wait, let's take an example. Suppose we have four numbers: a, b, c, d. Then, for each number, there's an equation:a = b + c - db = a + d - cc = a + d - bd = b + c - aBut this seems like a system of equations. Let's see if such a system is possible.If a = b + c - d, then rearranged: a - b - c + d = 0.Similarly, b = a + d - c => -a + b - d + c = 0.Similarly, c = a + d - b => -a + b + c - d = 0.d = b + c - a => a + d - b - c = 0.So each equation is a linear combination summing to zero. If we set up the system:1) a - b - c + d = 02) -a + b + c - d = 03) -a + b + c - d = 04) a + d - b - c = 0Wait, equations 2 and 3 are the same. Similarly, equation 1 and 4 are negatives of each other. If equation 1 holds, then equation 4 is equivalent to - (a - b - c + d) = 0, which would be the same. So essentially, there's only one unique equation here. So if we have four variables but only one equation, there are infinitely many solutions. But since we're dealing with specific numbers, maybe all four numbers satisfy this single equation a - b - c + d = 0. So a + d = b + c.So the condition is that for the four numbers, the sum of the first and fourth is equal to the sum of the second and third. If this is true for all permutations? Wait, no. Wait, if the four numbers can be arranged such that a + d = b + c, then such a relationship exists. So perhaps the four numbers form two pairs that add up to the same sum.For example, numbers 30, 42, 66, 78. Let's check their sums:30 + 78 = 10842 + 66 = 108So here, 30 and 78 add up to the same as 42 and 66. So if we have these four numbers, then indeed a + d = b + c. So in this case, starting with any one number, adding another, subtracting another, gives the fourth. Let's check.Take 30: 30 + 42 - 66 = 30 +42=72 -66=6, which is not 78. Wait, that doesn't work. Wait, maybe the arrangement is different. Wait, if the four numbers are 30, 42, 66, 78, and they satisfy 30 +78=42 +66=108, then if we take 30 +42 -66=6, which is not in the set. Hmm. Maybe the relationship is different.Alternatively, if we take 30 +66 -42=54, not in the set. 30 +78 -42=66. Ah, that works. So starting with 30, add 78, subtract 42, get 66. Similarly, starting with 30, add 66, subtract 42, get 54, which isn't in the set. Wait, but 30 +78 -42=66, which is in the set. Similarly, 30 +78 -66=42. 30 +42 -78= -6, which isn't in the set. Hmm.Wait, maybe the correct relationship is that given four numbers w, x, y, z, then w + x - y = z. So for each number, there exists some permutation where starting with w, add x, subtract y, get z. But in our example, 30 +78 -42=66, which is in the set. 30 +78 -66=42. 30 +42 -78= -6. 30 +42 -66=6. So only some combinations work.But Claire says that "you could start with any one of them, add a second, subtract a third, and get the fourth". So starting with any one, you can choose the other three in some order such that when you add one and subtract another, you get the fourth. Hmm. So for each number in the four, there exists some ordering of the other three such that starting with that number, adding one, subtracting another, gives the fourth.So for example, take 30. If we can find numbers x, y, z in the set such that 30 + x - y = z. Let's check:30 +42 -66=6, not in the set.30 +66 -42=54, not in the set.30 +78 -42=66, which is in the set. So 30 +78 -42=66. So yes, starting with 30, add 78, subtract 42, get 66.Similarly, starting with 42:42 +30 -78= -6, not in the set.42 +66 -30=78, which is in the set.42 +78 -30=90, not in the set.So 42 +66 -30=78. So that works.Starting with 66:66 +30 -42=54, not in the set.66 +42 -30=78, which is in the set.66 +78 -30=114, not in the set.So 66 +42 -30=78.Starting with 78:78 +30 -42=66, which is in the set.78 +42 -30=90, not in the set.78 +66 -30=114, not in the set.So 78 +30 -42=66.So in this case, each number can be used as the starting number with some combination. For 30: 30 +78 -42=66. For 42: 42 +66 -30=78. For 66: 66 +42 -30=78. For 78: 78 +30 -42=66. Wait, but in some cases, the same equation is used. So 30 and 78 both lead to 66, and 42 and 66 lead to 78. So maybe this satisfies Claire's statement? Because starting with any of the four, you can add one, subtract another, and get the fourth. However, in this case, each number has one such combination. So maybe these four numbers (30, 42, 66, 78) form the set Claire is talking about. Then Cat says her number isn't among those four. Since the possible numbers are 30, 42, 66, 70, 78. If Claire's four numbers are 30, 42, 66, 78, then the remaining number is 70. Then Claire says, "Now I know your favorite number!" So she must have deduced it's 70. Let's check if this makes sense.So Claire initially knew that the number was a product of three distinct primes (so among 30, 42, 66, 70, 78). Then she knew that it was among four numbers where this additive property holds, but after Cat says it's not among those four, Claire can deduce it's the remaining one, which is 70. Therefore, the favorite number would be 70. Let me confirm this.Wait, but wait. How did Claire know about the four numbers? Because Claire says, "I don't know your favorite number yet, but I do know that among four of the numbers that might be your favorite number, you could start with any one of them, add a second, subtract a third, and get the fourth!" So Claire must have some prior information. Wait, but how does Claire know about the four numbers? Maybe she knows that the number is in a set of four numbers with that additive property. But from the conversation, it seems that Claire deduced something based on Cat's statement. Wait, perhaps Claire is working with the same list we are, and upon hearing that it's a product of three distinct primes, she considers the possible numbers (30,42,66,70,78). Then she says she doesn't know the number yet, but knows that among four of them (so four numbers in the list), the additive property holds. Therefore, Claire must have figured out that the number is among a subset of four numbers with that property. But since Cat's number is not among those four, Claire can eliminate them and be left with 70.But let me verify if the four numbers 30,42,66,78 indeed have the additive property Claire mentions. Let's check if starting with any one, you can add a second, subtract a third, and get the fourth. So for each number:Starting with 30: 30 + 78 -42 = 66. Yes, 30 +78=108, 108-42=66.Starting with 42: 42 +66 -30 = 78. 42+66=108, 108-30=78.Starting with 66: 66 +42 -30 =78. Same as above.Starting with 78:78 +30 -42=66. Same as first.So yes, each of these four numbers can be obtained by starting with another, adding a third, and subtracting the fourth. So that works. So Claire is referring to these four numbers. Then Cat says her number isn't among those four, so it must be 70. Then Claire can conclude it's 70. Therefore, the answer is 70.Wait, but let me check if there are any other possible sets of four numbers with that property. The remaining numbers are 30,42,66,70,78. If the four numbers are 30,42,66,78, as we saw, they have that additive property. Is there another set of four numbers among these five that could have such a property? Let's check. If we take 30,42,70,78. Let's see if they can form such relationships.Starting with 30: 30 + something - something else. Let's see:30 +42 -70=2, not in the set.30 +70 -42=58, not in the set.30 +78 -42=66, which is not in this set.30 +78 -70=38, not in the set.30 +42 -78= -6, nope.30 +70 -78=22, nope.So no, 30 can't be used to get another number in the set. Similarly, starting with 42:42 +30 -70=2, nope.42 +70 -30=82, nope.42 +78 -30=90, nope.42 +78 -70=50, nope.Same with others. So this set doesn't work. Similarly, trying other combinations. What about 30,66,70,78.Starting with 30:30 +66 -70=26, nope.30 +78 -66=42, which is not in the set.30 +70 -66=34, nope.Nope. So no. Similarly, 42,66,70,78.Starting with 42:42 +66 -70=38, nope.42 +78 -70=50, nope.42 +70 -66=46, nope.No. So only the original set of 30,42,66,78 works. Therefore, Claire must be referring to those four numbers, leaving 70 as the only remaining possibility. Therefore, Cat's favorite number is 70. Let me confirm once more.Cat's number is a two-digit product of three distinct primes. Possible numbers: 30,42,66,70,78.Claire says she doesn't know the number but knows it's among four numbers with the additive property. The only such four numbers are 30,42,66,78. Therefore, if Cat's number is not among those, it must be 70. Then Claire can figure it out. So yes, 70 is the answer.Final Answerboxed{70}</think>Cat's favorite number is a two-digit positive integer that is the product of three distinct prime numbers. The possible numbers are 30, 42, 66, 70, and 78. Claire mentions that among four of these numbers, you can start with any one of them, add a second, subtract a third, and get the fourth. This implies a set of four numbers with a specific additive relationship. Checking the numbers, the set 30, 42, 66, and 78 form such a relationship:- Starting with 30: (30 + 78 - 42 = 66)- Starting with 42: (42 + 66 - 30 = 78)- Starting with 66: (66 + 42 - 30 = 78)- Starting with 78: (78 + 30 - 42 = 66)Since Cat's number is not among these four, the remaining number is 70. Claire deduces this, confirming Cat's favorite number is (boxed{70}).]

🔑:1. Identify the possible numbers: We need to find all two-digit positive integers that are the product of three distinct prime numbers. The prime numbers less than 10 are 2, 3, 5, and 7. We can form the following products: [ 2 cdot 3 cdot 5 = 30, quad 2 cdot 3 cdot 7 = 42, quad 2 cdot 3 cdot 11 = 66, quad 2 cdot 5 cdot 7 = 70, quad 2 cdot 3 cdot 13 = 78 ] Thus, the possible numbers are 30, 42, 66, 70, and 78.2. Analyze the condition given by Claire: Claire mentions that among four of the numbers, you could start with any one of them, add a second, subtract a third, and get the fourth. We need to check if this condition holds for any set of four numbers from the list.3. Check the condition for sets of four numbers: - Consider the set {30, 42, 66, 78}: [ 42 + 66 - 78 = 30 ] This satisfies the condition. Therefore, the set {30, 42, 66, 78} is a valid set.4. Determine Cat's favorite number: Cat's favorite number is not among the four numbers that satisfy Claire's condition. Therefore, Cat's favorite number must be the remaining number from the list, which is 70.The final answer is boxed{70}

🧠:[<think>Okay, let's try to figure out this problem step by step. So Johann is walking from the point (0, 2004) to a random lattice point inside the square with vertices (0,0), (0,99), (99,99), and (99,0). The question is asking for the probability that his path, including the endpoints, contains an even number of lattice points. First, I need to recall that a lattice point is a point where both coordinates are integers. So the starting point (0, 2004) is a lattice point, and the destination is another lattice point inside the square. The square's interior points would have coordinates from (1,1) up to (98,98), right? Because the boundaries are excluded, so the x and y coordinates can't be 0, 99, or anything beyond that. So the possible destinations are points (a, b) where 1 ≤ a ≤ 98 and 1 ≤ b ≤ 98. Now, the main question is about the number of lattice points on the path between (0, 2004) and some (a, b). The problem mentions that the path includes the endpoints, so we need to count both (0, 2004) and (a, b) when determining if the total is even. I remember that the number of lattice points on a line segment between two lattice points can be determined using the greatest common divisor (gcd) of the differences in coordinates. Specifically, the number of lattice points on the line segment from (x1, y1) to (x2, y2) is gcd(|x2 - x1|, |y2 - y1|) + 1. This includes both endpoints. So, if we compute the gcd of the horizontal and vertical distances between the two points and add 1, that gives the total number of lattice points on the path.Therefore, for Johann's path from (0, 2004) to (a, b), the horizontal distance is |a - 0| = a, and the vertical distance is |b - 2004| = 2004 - b (since b is between 1 and 98, so 2004 - b is positive). The number of lattice points on this path would be gcd(a, 2004 - b) + 1. Since we need the total number of lattice points to be even, we need gcd(a, 2004 - b) + 1 to be even. That implies that gcd(a, 2004 - b) must be odd. Because if gcd is odd, adding 1 makes it even. If gcd is even, adding 1 makes it odd. Therefore, the condition is that gcd(a, 2004 - b) is odd. So the problem reduces to finding the probability that gcd(a, 2004 - b) is odd, where a and b are integers between 1 and 98 inclusive. First, let's note that 2004 is an even number. Since 2004 is even, then 2004 - b will be even if and only if b is even. Because even minus even is even, and even minus odd is odd. So, if b is even, 2004 - b is even; if b is odd, 2004 - b is odd.Therefore, 2004 - b is even when b is even, and odd when b is odd. We need to compute gcd(a, 2004 - b). Let's consider two cases based on the parity of b:Case 1: b is even. Then 2004 - b is even. So the gcd(a, even number). For the gcd to be odd, both a and the even number (2004 - b) must have no factors of 2. But since 2004 - b is even, it has at least a factor of 2. Therefore, gcd(a, even number) can only be odd if a is also even, but since 2004 - b is even, their gcd will have at least a factor of 2. Wait, no. Wait, if both a and 2004 - b are even, then their gcd is at least 2. Therefore, gcd(a, 2004 - b) is even. If a is even and 2004 - b is even, then gcd is even. If a is odd and 2004 - b is even, then gcd(a, even) would be gcd(odd, even). Since a is odd and 2004 - b is even, the gcd can only be 1 if the odd number is coprime with the even number. Wait, but the gcd of an odd and even number is the same as the gcd of the odd number and half the even number. Wait, let me think. Suppose we have gcd(a, c) where a is odd and c is even. Then since a is odd, 2 is not a factor of a. Therefore, the gcd(a, c) must divide a and c. Since c is even, which is 2 * k, then gcd(a, c) = gcd(a, k). But since a is odd, gcd(a, k) must be gcd(a, k). So in this case, if a is odd and 2004 - b is even, then gcd(a, 2004 - b) = gcd(a, (2004 - b)/2). However, (2004 - b)/2 is an integer because 2004 - b is even. But since a is odd, this gcd could be 1 or some other odd number. Wait, but regardless of that, in this case, when a is odd and 2004 - b is even, the gcd(a, 2004 - b) can be odd or even? Wait, actually, since a is odd and 2004 - b is even, their gcd must be gcd(odd, even), which is gcd(odd, even). Since the even number has a factor of 2, but the odd number doesn't, so their gcd must be odd. Wait, no, that's not correct. Wait, for example, gcd(3, 6) is 3, which is odd. gcd(3, 4) is 1, which is odd. Wait, but in this case, even if the even number is a multiple of the odd number, the gcd would be the odd number. So yes, gcd(odd, even) is always odd. Because the even number can be written as 2 * k, and since the odd number has no factors of 2, the gcd(odd, 2k) is equal to gcd(odd, k). Which is the same as gcd(odd, k). Since odd and k can have some common divisors, but since odd is still, the gcd is odd. Therefore, in case a is odd and 2004 - b is even, the gcd(a, 2004 - b) is odd. If a is even and 2004 - b is even, then gcd(a, 2004 - b) is even. So to summarize:If b is even (i.e., 2004 - b is even):- If a is even, gcd(a, 2004 - b) is even.- If a is odd, gcd(a, 2004 - b) is odd.If b is odd (i.e., 2004 - b is odd):- Then gcd(a, 2004 - b) is gcd(a, odd number). Since 2004 - b is odd, regardless of a's parity. Now, since 2004 - b is odd, and a can be even or odd. Let's consider both cases:If a is even: Then gcd(even, odd) is gcd(odd, even). Wait, but we can just compute gcd(a, odd). Since a is even, and the other number is odd, their gcd must be odd (since the even number has factors of 2, which the odd number doesn't). Wait, no: gcd(even, odd) is equal to gcd(odd, even/2^k), where 2^k is the highest power of 2 dividing the even number. But since the odd number doesn't have any factors of 2, the gcd is gcd(odd, a/2^k). Which is an odd number. Wait, but let me verify with examples.For example, gcd(4, 3): 4 is even, 3 is odd. gcd(4,3) = 1, which is odd.gcd(6, 9): 6 is even, 9 is odd. gcd(6,9) = 3, which is odd.gcd(8, 5): 1, odd. So yes, gcd(even, odd) is odd. Similarly, if a is odd and 2004 - b is odd (since b is odd), then gcd(odd, odd) could be odd. So in this case, regardless of a's parity, when 2004 - b is odd, the gcd(a, 2004 - b) is the same as gcd(a, 2004 - b). Since both numbers could be odd or one is even. Wait, no: if 2004 - b is odd, then regardless of a's parity, gcd(a, 2004 - b) is equal to gcd(a, 2004 - b). If a is even, then as we saw, gcd(even, odd) is odd. If a is odd, gcd(odd, odd) can be odd or 1. So in either case, when 2004 - b is odd, the gcd(a, 2004 - b) is odd. Therefore, in the case when b is odd, regardless of a being even or odd, the gcd is odd. Therefore, putting it all together:The gcd(a, 2004 - b) is odd if and only if either:1. b is odd, or2. b is even and a is odd.Therefore, the total number of favorable pairs (a, b) is:Number of pairs where b is odd plus number of pairs where b is even and a is odd.Now, let's compute these quantities.First, the total number of possible destination points (a, b). Since a and b each range from 1 to 98 inclusive, there are 98 choices for a and 98 choices for b, so total 98*98 = 9604 points.Now, the number of pairs where b is odd: Since b ranges from 1 to 98. The number of odd b's is 49 (since 98/2 = 49). For each odd b, a can be any from 1 to 98, so 98 choices. So total is 49*98 = 4802.Next, the number of pairs where b is even and a is odd: The number of even b's is 98 - 49 = 49. For each even b, the number of odd a's is 49 (since a ranges from 1 to 98, half are odd). So total is 49*49 = 2401.Therefore, total favorable pairs are 4802 + 2401 = 7203.Therefore, the probability is 7203 / 9604.Simplify this fraction:Let's see if 7203 and 9604 have a common divisor. Let's check if 7 is a divisor:7203 ÷ 7 = 1029 (7*1000=7000, 7203-7000=203, 203 ÷7=29, so 1000+29=1029)9604 ÷7= 1372 (7*1000=7000, 9604-7000=2604; 2604 ÷7= 372, since 7*300=2100, 2604-2100=504; 504 ÷7=72, so 300+72=372. So 1000+372=1372)So dividing numerator and denominator by 7: 1029 / 1372.Check if 1029 and 1372 have a common divisor. Let's try 7 again:1029 ÷7=147 (7*140=980, 1029-980=49, 49 ÷7=7, so 140+7=147)1372 ÷7=196 (7*190=1330, 1372-1330=42, 42 ÷7=6, so 190+6=196)So now 147 / 196. Divide by 7 again:147 ÷7=21196 ÷7=2821/28. Divide by 7 again:21 ÷7=328 ÷7=4So 3/4. Therefore, 7203/9604 simplifies to 3/4.Wait, but let's verify the steps again because this seems too straightforward, but let me check.Original fraction: 7203 / 9604First divide by 7: 7203 ÷7=1029, 9604 ÷7=1372. Then 1029 ÷7=147, 1372 ÷7=196. Then 147 ÷7=21, 196 ÷7=28. Then 21 ÷7=3, 28 ÷7=4. So yes, 7203/9604 simplifies by dividing numerator and denominator by 7^3=343. 7*7*7=343. 7203 ÷343=21, 9604 ÷343=28. Then 21/28 simplifies to 3/4. So yes, the simplified probability is 3/4. Therefore, the probability is 3/4.But wait, let me cross-verify this with another approach to make sure.Alternative approach: Since the problem is about the gcd being odd, which is equivalent to both numbers being coprime with 2, i.e., both numbers being odd. Wait, no, because gcd(a, c) is odd if and only if the 2-adic valuations of a and c are both zero. In other words, both a and c are odd. Wait, no, that's not necessarily true. For example, gcd(3, 6) is 3, which is odd, but 6 is even. Wait, but in that case, even though 6 is even, the gcd is odd. So my previous analysis seems contradictory. Wait, but according to our previous reasoning, if c is even and a is odd, then gcd(a, c) is gcd(a, c/2^k), which can still be odd. So the presence of a single even number doesn't necessarily make the gcd even. Wait, but in the case where c is even and a is even, then the gcd is at least 2. So the gcd(a, c) is odd if and only if at least one of a or c is odd? Wait, no. Wait, let's take examples:gcd(3,6) = 3 (odd)gcd(3,4) = 1 (odd)gcd(6, 9) = 3 (odd)gcd(6, 8) = 2 (even)So when one is even and the other is odd, the gcd can be odd. When both are even, gcd is even. When both are odd, gcd can be odd or 1.Therefore, gcd(a, c) is odd if and only if at least one of a or c is odd? Wait, no, because if both are even, gcd is even. If both are odd, gcd can be odd or 1. If one is odd and one is even, gcd can be odd. Therefore, the gcd is odd unless both a and c are even. Therefore, the number of pairs where gcd(a, c) is odd is equal to total pairs minus pairs where both a and c are even. So maybe this is a simpler way to compute.Wait, let's see. Let c = 2004 - b. So since b ranges from 1 to 98, c ranges from 2004 - 98 = 1906 to 2004 -1 = 2003. But we might not need that.But if we consider that c = 2004 - b. Then when b is even, c is even, and when b is odd, c is odd. Therefore, if we let a and c be variables, then the pairs (a, c) where c = 2004 - b, so c ranges from 1906 to 2003 as above. But perhaps this complicates things. Alternatively, note that in the original problem, the pairs (a, b) with a and b from 1 to 98. Then, c = 2004 - b is determined by b. But the key is that c's parity is opposite to b's parity. So when b is even, c is even; when b is odd, c is odd. Therefore, we can model this as:The gcd(a, c) is odd if and only if either c is odd (i.e., b is odd), or c is even and a is odd. Because if c is even (i.e., b is even) and a is even, gcd(a, c) is even. So, as per before, the favorable cases are when either b is odd (so c is odd) or b is even and a is odd. Thus, total favorable pairs = (number of b odd)*(all a) + (number of b even)*(number of a odd). Which is exactly what we computed earlier.Therefore, the total favorable pairs are (49*98) + (49*49) = 49*(98 + 49) = 49*147 = 7203. Total pairs are 98*98=9604. Then 7203/9604=3/4 after simplifying. So that seems correct.Alternatively, since the probability that gcd(a, c) is odd is equal to 1 - probability that both a and c are even. Because if both a and c are even, gcd is even; otherwise, gcd is odd. So probability that gcd is odd = 1 - probability(a even and c even). Now, c = 2004 - b. So c is even when b is even, as before. Therefore, probability(a even and c even) = probability(a even and b even). Since a and b are independent (since the destination is a random lattice point, a and b are chosen independently), the probability that a is even and b is even is (number of even a / total a) * (number of even b / total b) = (49/98)*(49/98) = (1/2)*(1/2) = 1/4. Therefore, the probability that gcd(a, c) is odd is 1 - 1/4 = 3/4. Ah, this is a much simpler way to see it! So since the gcd(a, c) is odd unless both a and c are even, and c is even iff b is even. Therefore, the probability that both a and c are even is the probability that a is even and b is even, which is (1/2)(1/2)=1/4. Hence, the probability that gcd is odd is 1 - 1/4 = 3/4.Therefore, the answer is 3/4. This matches the previous result. So that gives me confidence that the answer is indeed 3/4. But let me just think through once more to check for any possible mistakes.Wait, in the alternative approach, we assumed that a and b are chosen uniformly and independently. But is that the case? The problem states that Johann is walking to a random lattice point in the interior of the square. The interior lattice points are those with coordinates (a, b) where 1 ≤ a, b ≤ 98. So there are 98x98 points, each equally likely. So yes, a and b are independent, each uniform over 1 to 98. Therefore, the probability that a is even is 49/98=1/2, same for b. Since the choices are independent, the probability that a is even and b is even is indeed 1/2 * 1/2 = 1/4. Therefore, the probability that gcd(a, 2004 - b) is odd is 1 - 1/4 = 3/4.Therefore, the answer is 3/4.But let me confirm with another example. Suppose we have a simpler case, say, starting from (0, 2) to a random interior point (a, b) in the square with vertices (0,0), (0,1), (1,1), (1,0). The interior points here would be just (1,1). Wait, no, the interior points would be those not on the boundary. If the square is from (0,0) to (1,1), the interior points would be empty. So maybe a different example.Let's take a square with vertices (0,0), (0,3), (3,3), (3,0). The interior points would be (1,1), (1,2), (2,1), (2,2). So 4 points. Suppose starting from (0, 4). Let's compute the probability that the path from (0,4) to each of these points has an even number of lattice points.Compute for each point:1. (1,1): The path is from (0,4) to (1,1). The differences are Δx=1, Δy=-3. gcd(1,3)=1. Number of lattice points: 1 + 1 = 2, which is even.2. (1,2): Δx=1, Δy=-2. gcd(1,2)=1. Number of lattice points: 1 + 1 = 2, even.3. (2,1): Δx=2, Δy=-3. gcd(2,3)=1. Number of lattice points: 1 + 1 = 2, even.4. (2,2): Δx=2, Δy=-2. gcd(2,2)=2. Number of lattice points: 2 +1=3, which is odd.So out of 4 points, 3 have even number of lattice points. So the probability is 3/4. Which matches the answer we obtained earlier. So this example also results in 3/4. Therefore, this gives more confidence that the answer is indeed 3/4.Therefore, I think the answer is 3/4.Final Answerboxed{dfrac{3}{4}}</think>Johann walks from the point ((0, 2004)) to a random lattice point inside the square with vertices ((0,0)), ((0,99)), ((99,99)), and ((99,0)). We need to find the probability that his path contains an even number of lattice points, including the endpoints.1. Number of Lattice Points: The number of lattice points on a line segment between ((x_1, y_1)) and ((x_2, y_2)) is given by (gcd(|x_2 - x_1|, |y_2 - y_1|) + 1). For the path from ((0, 2004)) to ((a, b)), this is (gcd(a, 2004 - b) + 1).2. Even Number of Lattice Points: For the total number of lattice points to be even, (gcd(a, 2004 - b) + 1) must be even, implying (gcd(a, 2004 - b)) must be odd.3. Parity Analysis: - (2004 - b) is even if (b) is even, and odd if (b) is odd. - If (2004 - b) is even, (gcd(a, 2004 - b)) is odd if (a) is odd. - If (2004 - b) is odd (i.e., (b) is odd), (gcd(a, 2004 - b)) is odd regardless of (a).4. Probability Calculation: - The total number of interior points is (98 times 98 = 9604). - The number of favorable pairs ((a, b)) where (gcd(a, 2004 - b)) is odd: - (b) is odd: (49 times 98 = 4802). - (b) is even and (a) is odd: (49 times 49 = 2401). - Total favorable pairs: (4802 + 2401 = 7203).5. Simplification: - The probability is (frac{7203}{9604}). - Simplifying this fraction gives (frac{3}{4}).Thus, the probability that Johann's path contains an even number of lattice points is (boxed{dfrac{3}{4}}).,<think>Okay, let me try to work through this problem step by step. So, the problem is about Johann walking from the point (0, 2004) to a random lattice point inside the square with vertices (0,0), (0,99), (99,99), and (99,0). We need to find the probability that his path, including the endpoints, contains an even number of lattice points.First, I need to understand the problem clearly. A lattice point is a point where both coordinates are integers. The square in question has vertices from (0,0) to (99,99), but the destination point is a random lattice point in the interior, meaning it's not on the boundary. So, the possible endpoints for Johann's walk are points (x, y) where x and y are integers between 1 and 98 inclusive. That gives us 98x98 = 9604 possible points, right?Johann is walking in a straight line from (0, 2004) to one of these interior points. The question is about the probability that the number of lattice points on his path (including both endpoints) is even. So, we need to count how many of these destination points result in a path with an even number of lattice points and then divide by the total number of possible destination points (9604) to get the probability.Now, I recall that the number of lattice points on a straight line between two points (x1, y1) and (x2, y2) is related to the greatest common divisor (gcd) of the differences in coordinates. Specifically, the number of lattice points on the line segment, including the endpoints, is gcd(delta_x, delta_y) + 1, where delta_x = |x2 - x1| and delta_y = |y2 - y1|. Is that right?Let me confirm. If two points are (0,0) and (2,2), the gcd of (2,2) is 2, so the number of lattice points should be 2 + 1 = 3. Indeed, the points are (0,0), (1,1), (2,2). That works. So yes, the formula is gcd(delta_x, delta_y) + 1.In our case, Johann is walking from (0, 2004) to (a, b), where (a, b) is an interior point (so 1 ≤ a, b ≤ 98). Then delta_x is |a - 0| = a, and delta_y is |b - 2004| = 2004 - b, since b is between 1 and 98, so 2004 - b is positive. Therefore, the number of lattice points on the path is gcd(a, 2004 - b) + 1.We need this number to be even. So, the condition is gcd(a, 2004 - b) + 1 ≡ 0 mod 2. Which simplifies to gcd(a, 2004 - b) ≡ 1 mod 2. Therefore, the gcd of a and (2004 - b) must be odd. So, the problem reduces to finding the number of pairs (a, b) where 1 ≤ a, b ≤ 98 and gcd(a, 2004 - b) is odd. Then, the probability is that number divided by 9604.So, first, let's analyze 2004 - b. Since b is between 1 and 98, 2004 - b is between 2004 - 98 = 1906 and 2004 - 1 = 2003. So, 1906 ≤ 2004 - b ≤ 2003.But perhaps more importantly, we need to look at the parity of gcd(a, 2004 - b). For the gcd to be odd, both numbers must be odd. Because if either number is even, then the gcd would be even (since 2 would be a common divisor). Wait, is that necessarily true?Wait, let's think. The gcd of two numbers is odd if and only if both numbers are odd. Because if at least one is even, then 2 divides both only if both are even. Wait, actually, if exactly one is even and the other is odd, then they have no common divisor of 2, so the gcd is odd. Wait, no. Wait, suppose a is even and 2004 - b is odd. Then, gcd(a, 2004 - b) would be gcd(even, odd) which is gcd(odd, even) which is gcd(odd, even). Since the even number is divisible by 2, but the odd number isn't, so the gcd must be odd. Similarly, if a is odd and 2004 - b is even, then gcd(odd, even) is also odd. However, if both are even, then gcd(a, 2004 - b) is at least 2, so even. If both are odd, then the gcd could be odd or even? Wait, no. Wait, if both are odd, their gcd can still be odd. For example, gcd(3,5)=1, which is odd. gcd(9, 15)=3, which is odd. So, in fact, the gcd of two odd numbers is odd, since all their divisors are odd.So, actually, the gcd of two numbers is odd if and only if both numbers are odd. Because if either number is even, then 2 could divide the gcd only if both are even. Wait, but if one is even and the other is odd, then 2 does not divide both, so the gcd is odd. Wait, that contradicts my earlier thought. Let me check with an example.Take a=2 (even) and c=3 (odd). gcd(2,3)=1, which is odd. Similarly, a=3 (odd) and c=4 (even). gcd(3,4)=1, which is odd. So, even if one is even and the other is odd, the gcd is odd. If both are even, gcd is even. If both are odd, gcd is odd. Wait, so actually:gcd(a, c) is odd if and only if at least one of a or c is odd? Wait, no. Let me check:If a is even and c is odd: gcd(even, odd) = gcd(odd, even) = gcd(odd, even). Let's use the Euclidean algorithm: gcd(2k, 2m+1). Since 2k and 2m+1 are coprime if k and 2m+1 are coprime. But the actual gcd is the same as gcd(2m+1, 2k). Since 2m+1 is odd, it cannot have 2 as a divisor, so gcd(2k, 2m+1) = gcd(k, 2m+1). If k and 2m+1 share a common divisor, which would have to be odd. So, the gcd is the same as gcd(k, 2m+1). Which could be odd or 1. But in any case, the gcd does not have 2 as a factor, so it is odd.Similarly, if a is even and c is even, then gcd(a, c) is at least 2. So, the gcd is even.If a and c are both odd, then gcd(a, c) can be odd. For example, gcd(3, 5)=1, gcd(9, 15)=3, which is odd.Therefore, the conclusion is: gcd(a, c) is odd if and only if at least one of a or c is odd. Wait, no. Wait, in the case where both are odd, gcd is odd. If one is even and the other is odd, gcd is odd. If both are even, gcd is even. Therefore, gcd(a, c) is odd if and only if at least one of a or c is odd? Wait, no. Wait, when both are odd, gcd is odd. When one is even and one is odd, gcd is still odd. Only when both are even, the gcd is even. Therefore, gcd(a, c) is odd if and only if it is not the case that both a and c are even. So, gcd(a, c) is odd ⇨ at least one of a or c is odd. That is equivalent to saying that it's not both even.Therefore, the number of pairs (a, b) where gcd(a, 2004 - b) is odd is equal to the number of pairs where at least one of a or (2004 - b) is odd. Since 2004 is even, 2004 - b is even if and only if b is even. Because 2004 is even, subtracting b (even or odd) will result in 2004 - b being even when b is even, and odd when b is odd. Therefore:2004 - b is even ⇨ b is even2004 - b is odd ⇨ b is oddSo, 2004 - b is even when b is even, and odd when b is odd. Therefore, the condition that gcd(a, 2004 - b) is odd is equivalent to "a is odd or b is odd" (since 2004 - b is odd iff b is odd). Therefore, the number of pairs (a, b) with gcd(a, 2004 - b) odd is equal to the number of pairs where a is odd or b is odd.Therefore, the problem reduces to finding the number of (a, b) pairs where a is odd or b is odd, with 1 ≤ a, b ≤ 98. Then, the probability is that number divided by 98*98.But wait, let me verify that. So, because 2004 - b is odd if and only if b is odd. So, the condition for gcd(a, 2004 - b) being odd is that at least one of a or (2004 - b) is odd. But since (2004 - b) is odd ⇨ b is odd, so the condition is equivalent to a is odd or b is odd. So, the count is the number of pairs where a is odd or b is odd.Therefore, the number of such pairs is equal to (number of a odd) * (number of b) + (number of a even) * (number of b odd) - (number of a odd) * (number of b odd). Wait, no. Wait, actually, using inclusion-exclusion principle: the number of pairs where a is odd or b is odd is equal to (number of a odd) * total b + total a * (number of b odd) - (number of a odd) * (number of b odd). But since a and b are independent variables from 1 to 98.Wait, more carefully: the total number of pairs where a is odd or b is odd is equal to:= (number of a odd) * 98 + 98 * (number of b odd) - (number of a odd) * (number of b odd)But that seems not quite right. Wait, inclusion-exclusion formula is:|A ∪ B| = |A| + |B| - |A ∩ B|Here, A is the set of pairs where a is odd, B is the set of pairs where b is odd. So, |A| = (number of a odd) * 98, |B| = 98 * (number of b odd), and |A ∩ B| = (number of a odd) * (number of b odd). Therefore,|A ∪ B| = (number of a odd) * 98 + 98 * (number of b odd) - (number of a odd)*(number of b odd)But since a and b are each from 1 to 98, which is 98 numbers. The number of odd numbers between 1 and 98 is 49 (since 98 is even). Similarly, the number of even numbers is 49 as well. Wait, 1 to 98: numbers 1,2,...,98. There are 98 numbers. Since 98 is even, half of them are odd and half are even. So, 49 odd and 49 even.Therefore, number of a odd = 49, number of b odd = 49.Therefore, |A| = 49*98, |B| = 98*49, |A ∩ B| = 49*49.Therefore, |A ∪ B| = 49*98 + 49*98 - 49*49 = 49*(98 + 98 - 49) = 49*(147) = 49*147.Calculate 49*147: 49*100=4900, 49*40=1960, 49*7=343; so 4900+1960=6860, 6860+343=7203. So, 7203 pairs where a is odd or b is odd.Therefore, the number of favorable pairs is 7203, total pairs is 98*98=9604. Therefore, the probability is 7203/9604.But let me confirm once again. Since the condition for gcd being odd is equivalent to a odd or b odd. So, the number of such pairs is 7203, so the probability is 7203 divided by 9604.But 7203 is 49*147, and 9604 is 98*98 = (49*2)*(49*2) = 49^2 *4. So, 7203/9604 = (49*147)/(49*49*4) )= 147/(49*4) = 3/4. Wait, 147 divided by 49 is 3, so 3/4. Wait, 147/(49*4)=3/4. So, the probability is 3/4? That seems high, but let's check.Wait, 7203 divided by 9604: Let's compute 7203/9604. Let's divide numerator and denominator by 49: 7203 ÷49=147, 9604 ÷49=196. So, 147/196. Then, divide numerator and denominator by 49 again? 147 ÷49=3, 196 ÷49=4. So, 3/4. Yes! So, the probability is 3/4.Wait, that seems a very clean answer. Let me verify again with a smaller example to see if that logic holds.Suppose instead of 98x98, let's say 2x2 grid. Let a and b be 1 or 2. Then, similar logic: if the number of pairs where a is odd or b is odd. For 2x2 grid:Possible a: 1,2 (odd, even)Possible b:1,2 (odd, even)Number of pairs where a is odd or b is odd:(1,1), (1,2), (2,1). So, 3 out of 4. Which is 3/4. So, same as 3/4. So, that seems to hold. So, the general formula is 3/4. Therefore, for the original problem, the probability is 3/4.But wait, but in the problem statement, the starting point is (0, 2004), which is a lattice point, and the endpoint is (a, b) which is also a lattice point. The path is a straight line between them, and the number of lattice points on the path is gcd(a, 2004 - b) + 1. We need this number to be even. So, gcd(a, 2004 - b) + 1 ≡ 0 mod 2, which implies gcd(a, 2004 - b) ≡ 1 mod 2. So, the gcd must be odd. As we saw, this happens if and only if a is odd or b is odd (since 2004 is even, 2004 - b is odd when b is odd). Therefore, the number of such pairs is 3/4 of the total, leading to probability 3/4.But let me check another example to confirm. Suppose starting point is (0, 4) and the square is 2x2. So, possible endpoints (1,1), (1,2), (2,1), (2,2). Then, compute for each endpoint, the number of lattice points on the path from (0,4) to (a, b):For (1,1): delta_x=1, delta_y=3. gcd(1,3)=1. So, number of points=1+1=2. Even. So, counts.For (1,2): delta_x=1, delta_y=2. gcd(1,2)=1. Number of points=2. Even. Counts.For (2,1): delta_x=2, delta_y=3. gcd(2,3)=1. Number of points=2. Even. Counts.For (2,2): delta_x=2, delta_y=2. gcd(2,2)=2. Number of points=3. Odd. Doesn't count.Therefore, 3 out of 4, which is 3/4. So, same result.Another example: starting at (0, 6), endpoint in 3x3 grid. But maybe this is getting too into it. But the pattern seems consistent. Therefore, the answer should be 3/4, so probability 3/4.But let me think again. The key insight here is that the gcd(a, 2004 - b) is odd if and only if a or b is odd. Then, since in each coordinate from 1 to 98, half are odd and half are even (since 98 is even), the number of a odd is 49, and same for b. Then, using inclusion-exclusion, the number of pairs where a is odd or b is odd is 49*98 + 49*98 - 49*49 = 49*(98 + 98 - 49) = 49*147 = 7203. Then, 7203 divided by 98*98 is 7203/9604 = 3/4. Therefore, the probability is 3/4. So, the answer should be boxed{dfrac{3}{4}}.Final AnswerThe probability is boxed{dfrac{3}{4}}.</think>Johann walks from the point ((0, 2004)) to a random lattice point ((a, b)) in the interior of the square with vertices ((0,0)), ((0,99)), ((99,99)), and ((99,0)). We need to find the probability that his path contains an even number of lattice points, including the endpoints.The number of lattice points on a line segment between two points ((x_1, y_1)) and ((x_2, y_2)) is given by (gcd(|x_2 - x_1|, |y_2 - y_1|) + 1). For Johann's path from ((0, 2004)) to ((a, b)), this number is (gcd(a, 2004 - b) + 1). We need this number to be even, which implies (gcd(a, 2004 - b)) must be odd.The greatest common divisor (gcd(a, 2004 - b)) is odd if and only if at least one of (a) or (2004 - b) is odd. Since (2004) is even, (2004 - b) is odd if and only if (b) is odd. Therefore, (gcd(a, 2004 - b)) is odd if and only if (a) is odd or (b) is odd.We need to count the number of pairs ((a, b)) where (1 leq a, b leq 98) and at least one of (a) or (b) is odd. Using the principle of inclusion-exclusion:- The number of odd (a) values is 49, and the number of odd (b) values is 49.- The number of pairs where (a) is odd is (49 times 98).- The number of pairs where (b) is odd is (98 times 49).- The number of pairs where both (a) and (b) are odd is (49 times 49).Thus, the number of favorable pairs is:[49 times 98 + 98 times 49 - 49 times 49 = 49 times (98 + 98 - 49) = 49 times 147 = 7203]The total number of possible pairs ((a, b)) is (98 times 98 = 9604). Therefore, the probability is:[frac{7203}{9604} = frac{3}{4}]The final answer is (boxed{dfrac{3}{4}}).]

🔑:Solution:3 / 4If Johann picks the point (a, b), the path will contain operatorname{gcd}(a, 2004-b)+1 points. There will be an odd number of points in the path if operatorname{gcd}(a, 2004-b) is even, which is true if and only if a and b are both even. Since there are 49^{2} points with a, b both even and 98^{2} total points, the probability that the path contains an even number of points isfrac{98^{2}-49^{2}}{98^{2}}=frac{49^{2}left(2^{2}-1^{2}right)}{49^{2}left(2^{2}right)}=frac{3}{4} .

🧠:[<think>Okay, so I need to find the length of segment AL in triangle ABC where AB to AC ratio is 5:2, and the angle bisector of angle BAC meets BC at L. Also, the magnitude of the vector 2AB + 5AC is 2016. Hmm, let's break this down step by step.First, let me recall some properties. The angle bisector theorem states that the angle bisector divides the opposite side in the ratio of the adjacent sides. So since AB:AC is 5:2, point L divides BC into BL:LC = AB:AC = 5:2. That's straightforward.Now, the main challenge is relating this ratio to the vector given. The vector is 2AB + 5AC, and its magnitude is 2016. I need to connect this information to find AL.Wait, vectors here. Let me think. Let's denote vectors in terms of position vectors maybe. Let me set point A as the origin to simplify the vectors. So, if A is the origin, then vectors AB and AC can be considered as position vectors of points B and C respectively.So, vector AB is from A to B, which would just be the coordinates of B if A is the origin. Similarly, vector AC is the coordinates of C. Then, the vector 2AB + 5AC is just 2B + 5C (since vectors are from the origin). The magnitude of this vector is 2016. So, |2B + 5C| = 2016.But how does this relate to the triangle sides? The sides AB and AC have lengths in the ratio 5:2. So, AB = 5k, AC = 2k for some k. Let me denote AB = 5k and AC = 2k. Then, the actual lengths are 5k and 2k.Since vectors AB and AC have magnitudes 5k and 2k respectively, the magnitude of the vector 2AB + 5AC is given as 2016. So, |2AB + 5AC| = 2016. Let's express this in terms of k.First, let's compute |2AB + 5AC|. To find the magnitude of a vector sum, we can use the formula:|aV + bW|² = a²|V|² + b²|W|² + 2ab(V · W)So, in this case:|2AB + 5AC|² = (2)²|AB|² + (5)²|AC|² + 2*2*5*(AB · AC)Which is 4*(25k²) + 25*(4k²) + 20*(AB · AC)Wait, AB has length 5k, so |AB|² = (5k)^2 = 25k². Similarly, |AC|² = (2k)^2 = 4k². So substituting:4*25k² + 25*4k² + 20*(AB · AC) = 100k² + 100k² + 20*(AB · AC) = 200k² + 20*(AB · AC)So, |2AB + 5AC|² = 200k² + 20*(AB · AC) = (2016)^2.Therefore, 200k² + 20*(AB · AC) = 2016².But I need to find AB · AC. Let's recall that the dot product AB · AC = |AB||AC|cosθ, where θ is the angle at A, which is angle BAC. So, AB · AC = 5k * 2k * cosθ = 10k² cosθ.Therefore, substituting back into the equation:200k² + 20*10k² cosθ = 2016²Simplify:200k² + 200k² cosθ = 2016²Factor out 200k²:200k²(1 + cosθ) = 2016²So, 200k²(1 + cosθ) = (2016)^2. Hmm, that's an equation relating k and cosθ.But we need to find AL. Let me recall that in a triangle, the length of the angle bisector can be found using the formula:AL = (2ab cos (θ/2)) / (a + b)Wait, but this formula might be specific. Let me verify. Alternatively, there's a formula for the length of the angle bisector in terms of the sides. Yes, the angle bisector theorem gives the length of AL.The formula for the length of the angle bisector from angle A is:AL = (2bc / (b + c)) * cos (A/2)But in this case, sides adjacent to angle A are AB and AC, which are of lengths 5k and 2k. Wait, but in the angle bisector formula, the sides adjacent to angle A are usually denoted as b and c in some notations. Wait, perhaps I need to check the exact formula.Alternatively, the formula for the length of the angle bisector can be given by:AL² = AB * AC [1 - (BC² / (AB + AC)^2)]Wait, no, maybe another formula. Let me recall.Yes, the formula for the length of the angle bisector is:AL = (2 * AB * AC * cos (θ/2)) / (AB + AC)Yes, that seems right. Since the angle bisector divides the angle into two equal parts, so θ/2, and using the projection.Alternatively, using the Stewart's theorem. Stewart's theorem relates the length of a cevian (which AL is) to the sides of the triangle. Stewart's theorem states that for a triangle with sides a, b, c, and a cevian of length d that divides the side opposite to the vertex into segments of length m and n, then:b²m + c²n = a(d² + mn)In our case, AL is the cevian from A to BC, which is divided by L into BL:LC = 5:2. So, if BC is divided into BL = 5m and LC = 2m for some m, then BC = 7m. So, in Stewart's theorem:AB² * LC + AC² * BL = BC * (AL² + BL * LC)So, substituting the known ratios:AB = 5k, AC = 2k, BL:LC = 5:2, so let’s let BL = 5m, LC = 2m, so BC = 7m.Then, plugging into Stewart's theorem:(5k)² * 2m + (2k)² * 5m = 7m (AL² + (5m)(2m))Simplify left side:25k² * 2m + 4k² * 5m = 50k²m + 20k²m = 70k²mRight side: 7m (AL² + 10m²)Therefore:70k²m = 7m (AL² + 10m²)Divide both sides by 7m:10k² = AL² + 10m²Thus, AL² = 10k² - 10m² = 10(k² - m²)Hmm, but I don't know the value of m. So, maybe I need another relation to connect m and k.Alternatively, maybe express BC in terms of k using the law of cosines. Since we know sides AB and AC, and angle θ between them, BC can be found as:BC² = AB² + AC² - 2AB * AC cosθSo, BC² = (5k)^2 + (2k)^2 - 2*(5k)*(2k)*cosθ = 25k² + 4k² - 20k² cosθ = 29k² - 20k² cosθBut BC is also equal to 7m (from BL + LC = 5m + 2m = 7m). Therefore:BC² = (7m)^2 = 49m²So, 49m² = 29k² - 20k² cosθTherefore, 20k² cosθ = 29k² - 49m²From earlier, in the equation from the vector magnitude:200k²(1 + cosθ) = 2016²So, 200k² + 200k² cosθ = 2016²But we can replace 200k² cosθ with 200k² * [ (29k² - 49m²) / (20k²) ) ] ?Wait, from 20k² cosθ = 29k² - 49m², so cosθ = (29k² - 49m²)/(20k²)Therefore, substituting back into the equation:200k²(1 + cosθ) = 2016²Which becomes:200k² [1 + (29k² - 49m²)/(20k²)] = 2016²Simplify inside the brackets:1 + (29k² - 49m²)/(20k²) = (20k² + 29k² - 49m²)/20k² = (49k² - 49m²)/20k² = 49(k² - m²)/20k²Therefore, the equation becomes:200k² * [49(k² - m²)/20k²] = 2016²Simplify:200k² * 49(k² - m²) / (20k²) ) = (200 / 20) * 49(k² - m²) = 10 * 49(k² - m²) = 490(k² - m²) = 2016²Therefore, 490(k² - m²) = 2016²But from Stewart's theorem, we had AL² = 10(k² - m²)Thus, AL² = 10(k² - m²) = (2016²)/49Wait, because 490(k² - m²) = 2016² ⇒ (k² - m²) = (2016²)/490 ⇒ AL² = 10*(2016²)/490Simplify 10/490 = 1/49. So, AL² = (2016²)/49 ⇒ AL = 2016/7 = 288Wait, that seems straightforward. Let me check the steps again.From Stewart's theorem, we got AL² = 10(k² - m²)From the vector magnitude, we got 490(k² - m²) = 2016² ⇒ (k² - m²) = (2016²)/490Therefore, AL² = 10*(2016²)/490 = (10/490)*2016² = (1/49)*2016² ⇒ AL = 2016/7 = 288.Yes, that seems correct. Let me verify:2016 divided by 7: 7*288 = 2016, correct. Therefore, AL = 288.So, the length of segment AL is 288.Final Answerboxed{288}</think>Given triangle (ABC) with (AB:AC = 5:2), the angle bisector of (angle BAC) intersects (BC) at point (L). We need to find the length of segment (AL) given that the magnitude of the vector (2 cdot overrightarrow{AB} + 5 cdot overrightarrow{AC}) is 2016.1. Angle Bisector Theorem: By the angle bisector theorem, (BL:LC = AB:AC = 5:2).2. Vector Magnitude: The magnitude of the vector (2 cdot overrightarrow{AB} + 5 cdot overrightarrow{AC}) is given as 2016. Setting (A) as the origin, we denote (overrightarrow{AB}) and (overrightarrow{AC}) as vectors from (A) to (B) and (C) respectively. The magnitude is calculated using the formula for the magnitude of a vector sum: [ |2 cdot overrightarrow{AB} + 5 cdot overrightarrow{AC}|^2 = (2)^2 |overrightarrow{AB}|^2 + (5)^2 |overrightarrow{AC}|^2 + 2 cdot 2 cdot 5 cdot (overrightarrow{AB} cdot overrightarrow{AC}) ] Substituting (|overrightarrow{AB}| = 5k) and (|overrightarrow{AC}| = 2k), we get: [ 4 cdot (25k^2) + 25 cdot (4k^2) + 20 cdot (overrightarrow{AB} cdot overrightarrow{AC}) = 200k^2 + 20 cdot (10k^2 cos theta) = 200k^2(1 + cos theta) ] Given this equals (2016^2), we have: [ 200k^2(1 + cos theta) = 2016^2 ]3. Stewart's Theorem: Using Stewart's theorem for the cevian (AL): [ AB^2 cdot LC + AC^2 cdot BL = BC cdot (AL^2 + BL cdot LC) ] Substituting (AB = 5k), (AC = 2k), (BL = 5m), (LC = 2m), and (BC = 7m), we get: [ (5k)^2 cdot 2m + (2k)^2 cdot 5m = 7m cdot (AL^2 + 10m^2) ] Simplifying, we find: [ 70k^2m = 7m cdot (AL^2 + 10m^2) implies 10k^2 = AL^2 + 10m^2 implies AL^2 = 10(k^2 - m^2) ]4. Relating (k) and (m): Using the law of cosines for (BC): [ BC^2 = 29k^2 - 20k^2 cos theta ] Substituting (BC = 7m), we get: [ 49m^2 = 29k^2 - 20k^2 cos theta ] Solving for (cos theta) and substituting back, we find: [ 200k^2(1 + cos theta) = 2016^2 implies 490(k^2 - m^2) = 2016^2 ] Thus, [ AL^2 = 10 cdot frac{2016^2}{490} = frac{2016^2}{49} implies AL = frac{2016}{7} = 288 ]The length of segment (AL) is (boxed{288}).,<think>Okay, let's see. I need to find the length of segment AL in triangle ABC where AB to AC ratio is 5:2, and the angle bisector of angle BAC intersects BC at point L. Also, the length of the vector 2·AB + 5·AC is 2016. Hmm. Alright, let me break this down step by step.First, the problem mentions triangle ABC with sides AB to AC in the ratio 5:2. So AB is 5 parts and AC is 2 parts. But wait, is the ratio AB:AC = 5:2, meaning AB is longer than AC? Yes, that's how ratios work. So AB is longer. Then, the angle bisector of angle BAC intersects BC at point L. So AL is the angle bisector. There's a theorem about angle bisectors relating the lengths of the sides. The Angle Bisector Theorem, right? It states that the angle bisector divides the opposite side into segments proportional to the adjacent sides. So in this case, BL/LC = AB/AC = 5/2. So BL:LC = 5:2. That's good to note.Next, the vector part: the length of the vector 2·AB + 5·AC is 2016. Hmm. Vectors in triangles. Let me recall that vectors can be used with coordinates or perhaps using the law of cosines. Since we have a vector expression, maybe I need to compute the magnitude of this vector. The length of the vector 2AB + 5AC is 2016. So perhaps I need to express this vector in terms of sides AB and AC, then use the formula for the magnitude of a vector sum. That would involve the magnitudes of AB and AC, the coefficients 2 and 5, and the angle between them, which is angle BAC. Let me write that down.Let’s denote the length of AB as 5k and AC as 2k for some positive real number k. Then, the angle between AB and AC is angle BAC, which we can call θ. The vector 2AB + 5AC would then have a magnitude squared equal to (2AB)^2 + (5AC)^2 + 2*(2AB)*(5AC)*cosθ. Wait, no, hold on. The formula for the magnitude squared of a vector sum a + b is |a|^2 + |b|^2 + 2a·b. The dot product here would be 2AB · 5AC, which is 10|AB||AC|cosθ. So the magnitude squared of 2AB + 5AC is (2|AB|)^2 + (5|AC|)^2 + 2*(2|AB|)*(5|AC|)*cosθ. So that's 4|AB|² + 25|AC|² + 20|AB||AC|cosθ. And the square of the length is 2016². Wait, but the problem says the length is 2016, so this expression equals 2016². Let me confirm: |2AB + 5AC| = 2016, so squared is 2016².But AB and AC are vectors here, right? So AB and AC vectors. So their lengths are |AB| and |AC|. But in the problem, the ratio AB:AC = 5:2. So if we let |AB| = 5k and |AC| = 2k, then we can express everything in terms of k and θ. Then, plugging into the expression:4*(5k)^2 + 25*(2k)^2 + 20*(5k)*(2k)*cosθ = (2016)^2.Let me compute each term:First term: 4*(25k²) = 100k²Second term: 25*(4k²) = 100k²Third term: 20*(10k²)*cosθ = 200k² cosθSo total expression: 100k² + 100k² + 200k² cosθ = 200k² + 200k² cosθ = 200k²(1 + cosθ) = (2016)^2So, 200k²(1 + cosθ) = 2016². Hmm. So that's one equation with variables k and θ.But we need another equation to relate these variables. Since we are dealing with triangle ABC, maybe we can use the Law of Cosines on triangle ABC. In triangle ABC, we have sides AB = 5k, AC = 2k, and BC, which we can call c. The Law of Cosines would relate these sides with angle θ. So:BC² = AB² + AC² - 2*AB*AC*cosθBut wait, angle at A is θ, so Law of Cosines gives BC² = AB² + AC² - 2*AB*AC*cosθ. So:c² = (5k)² + (2k)² - 2*(5k)*(2k)*cosθSimplify:c² = 25k² + 4k² - 20k² cosθc² = 29k² - 20k² cosθBut I don't know c, so maybe that's not helpful yet. Alternatively, maybe we can use the Law of Sines. But we might need another relation.Alternatively, maybe we can find expressions for cosθ in terms of the sides. But since we have two equations:1) 200k²(1 + cosθ) = 2016²2) c² = 29k² - 20k² cosθBut since c is BC, which is divided by L into BL:LC = 5:2. So BL = (5/7)c and LC = (2/7)c. But how does that help us find AL?Wait, we need to find the length of AL. AL is the angle bisector. There is a formula for the length of an angle bisector in a triangle. Let me recall that formula. The length of the angle bisector from angle A is given by:AL = (2ab cos (θ/2)) / (a + b)Wait, is that right? Or another formula. Alternatively, the formula is:AL² = AB * AC - BL * LCWait, Stewart's Theorem. Stewart's Theorem relates the length of a cevian (AL) to the lengths of the sides of the triangle. Stewart's Theorem states that for a triangle with sides a, b, c, and a cevian of length d to the side a, dividing it into m and n, then b²m + c²n = a(d² + mn)In our case, AL is the cevian, which is splitting BC into BL and LC with ratio 5:2. So if BC is of length c, then BL = (5/7)c and LC = (2/7)c. So applying Stewart's Theorem:AB² * LC + AC² * BL = BC * (AL² + BL * LC)Plugging the values:AB²*(2/7)c + AC²*(5/7)c = c*(AL² + (5/7)c*(2/7)c)Simplify:( (25k²)*(2/7)c + (4k²)*(5/7)c ) = c*(AL² + (10/49)c² )Multiply through by 7/c to eliminate denominators:25k²*2 + 4k²*5 = 7*(AL² + (10/49)c² )Which is:50k² + 20k² = 7AL² + (10/7)c²70k² = 7AL² + (10/7)c²Divide both sides by 7:10k² = AL² + (10/49)c²Therefore, AL² = 10k² - (10/49)c²Hmm. So AL² = 10k² - (10/49)c². So if I can express c² in terms of k², then I can find AL.From earlier, we have from the Law of Cosines:c² = 29k² - 20k² cosθSo substitute that into the AL² expression:AL² = 10k² - (10/49)(29k² - 20k² cosθ )= 10k² - (10/49)(29k²) + (10/49)(20k² cosθ )= 10k² - (290/49)k² + (200/49)k² cosθLet me convert 10k² to 490/49 k²:= (490/49)k² - (290/49)k² + (200/49)k² cosθ= (200/49)k² + (200/49)k² cosθFactor out (200/49)k²:= (200/49)k² (1 + cosθ)So AL² = (200/49)k²(1 + cosθ)But from the earlier equation from the vector length:200k²(1 + cosθ) = 2016²So 200k²(1 + cosθ) = 2016² => k²(1 + cosθ) = (2016²)/200Therefore, AL² = (200/49) * (2016²)/200 = (2016²)/49Therefore, AL = 2016 / 7 = 288Wait, so AL is 288? Let me check that again.Because AL² = (200/49)k²(1 + cosθ), and we know from the vector equation that 200k²(1 + cosθ) = (2016)^2. Therefore, k²(1 + cosθ) = (2016)^2 / 200. Then AL² = (200/49) * (2016^2 / 200 ) = 2016² / 49. Taking the square root, AL = 2016 / 7. 2016 divided by 7. Let me compute 2016 ÷ 7. 7*288=2016, yes. 7*280=1960, 2016-1960=56, 56/7=8, so 280+8=288. So AL=288. That seems straightforward. So the length of AL is 288. Therefore, the answer is 288, boxed.Wait, but let me verify the steps again to ensure I didn't skip something important. Let's recap:1. AB:AC = 5:2, so set AB=5k, AC=2k.2. The vector 2AB +5AC has length 2016, so compute its magnitude squared: |2AB +5AC|² = (2*5k)^2 + (5*2k)^2 + 2*(2*5k)*(5*2k)*cosθ. Wait, hold on, no. Wait, hold on. Wait, AB and AC are vectors, so when we take 2AB +5AC, their magnitudes are 2|AB| and 5|AC|, but the angle between AB and AC is angle at A, which is θ. Wait, but hold on. The vector AB is from A to B, and AC is from A to C. So the angle between vectors AB and AC is angle BAC = θ. Therefore, the dot product AB · AC = |AB||AC|cosθ. Therefore, the magnitude squared of 2AB +5AC is (2AB +5AC)·(2AB +5AC) = 4|AB|² + 25|AC|² + 20|AB||AC|cosθ. Which is exactly what I did earlier. So 4*(25k²) + 25*(4k²) + 20*(5k)(2k)cosθ = 100k² + 100k² + 200k² cosθ = 200k²(1 + cosθ). Set equal to 2016². Correct.Then, using Stewart's Theorem to find AL. The key step was applying Stewart's Theorem, which led to AL² = (200/49)k²(1 + cosθ), and since 200k²(1 + cosθ) = 2016², substituting gives AL² = 2016² / 49, so AL = 2016/7 = 288. All steps check out. Therefore, the answer is 288.Final Answerboxed{288}</think>Given triangle (ABC) with (AB:AC = 5:2), the angle bisector of (angle BAC) intersects (BC) at point (L). We need to find the length of segment (AL) given that the length of the vector (2 cdot overrightarrow{AB} + 5 cdot overrightarrow{AC}) is 2016.1. Ratio and Angle Bisector Theorem: - The ratio (AB:AC = 5:2) implies (AB = 5k) and (AC = 2k) for some positive real number (k). - By the Angle Bisector Theorem, (BL:LC = AB:AC = 5:2).2. Vector Magnitude Calculation: - The magnitude of the vector (2 cdot overrightarrow{AB} + 5 cdot overrightarrow{AC}) is 2016. - Using the formula for the magnitude of a vector sum: [ |2 cdot overrightarrow{AB} + 5 cdot overrightarrow{AC}|^2 = (2|AB|)^2 + (5|AC|)^2 + 2 cdot (2|AB|) cdot (5|AC|) cos theta ] - Substituting (AB = 5k) and (AC = 2k): [ 4(25k^2) + 25(4k^2) + 20(10k^2 cos theta) = 2016^2 ] - Simplifying: [ 200k^2(1 + cos theta) = 2016^2 ]3. Stewart's Theorem Application: - Stewart's Theorem for cevian (AL) in triangle (ABC): [ AB^2 cdot LC + AC^2 cdot BL = BC cdot (AL^2 + BL cdot LC) ] - Substituting (BL = frac{5}{7}BC) and (LC = frac{2}{7}BC): [ (25k^2) cdot frac{2}{7}BC + (4k^2) cdot frac{5}{7}BC = BC cdot (AL^2 + frac{10}{49}BC^2) ] - Simplifying and solving for (AL): [ 70k^2 = 7AL^2 + frac{10}{49}BC^2 ] - Using the Law of Cosines for (BC^2): [ BC^2 = 29k^2 - 20k^2 cos theta ] - Substituting back and simplifying: [ AL^2 = frac{200}{49}k^2(1 + cos theta) ] - Using the earlier result (200k^2(1 + cos theta) = 2016^2): [ AL^2 = frac{2016^2}{49} implies AL = frac{2016}{7} = 288 ]Thus, the length of segment (AL) is (boxed{288}).]

🔑:Answer: 288.Solution. Note that by the property of the angle bisector of a triangle B L: L C=B A: A C=5: 2, from which overrightarrow{B L}=5 / 7 cdot overrightarrow{B C}. Now,overrightarrow{A L}=overrightarrow{A B}+overrightarrow{B L}=overrightarrow{A B}+frac{4}{7} cdot overrightarrow{B C}=overrightarrow{A B}+frac{5}{7}(overrightarrow{A C}-overrightarrow{A B})=frac{2}{7} cdot overrightarrow{A B}+frac{5}{7} cdot overrightarrow{A C}=frac{1}{7}(2 cdot overrightarrow{A B}+5 cdot overrightarrow{A C})Therefore, |A L|=1 / 7 cdot|2 cdot overrightarrow{A B}+5 cdot overrightarrow{A C}|=1 / 7 cdot 2016=288.

🧠:[<think>Okay, let's see. The problem says that positive real numbers a and b satisfy the equation a² + 2ab + 4b² = 6. We need to find the maximum value of a + 2b. The options are given as A through D. Hmm, so the question is about maximizing a linear expression a + 2b under a quadratic constraint. This seems like an optimization problem with a constraint. I remember that such problems can sometimes be approached using methods like substitution, completing the square, or maybe even using calculus with Lagrange multipliers. But since it's a multiple-choice question, maybe there's a more straightforward algebraic way. Let me think.First, let me write down the given equation: a² + 2ab + 4b² = 6. And we need to maximize a + 2b. Let's denote S = a + 2b, and we want to find the maximum S such that the equation holds.One approach could be to express a in terms of b (or vice versa) from the equation and substitute into S, then take the derivative to find the maximum. Let's try that. Let me solve the given equation for a in terms of b.Starting with a² + 2ab + 4b² = 6. Let's treat this as a quadratic equation in a:a² + 2b a + (4b² - 6) = 0.Using the quadratic formula, a = [-2b ± sqrt((2b)² - 4*1*(4b² - 6))]/2.Calculating the discriminant:(2b)^2 - 4*1*(4b² - 6) = 4b² - 16b² + 24 = -12b² + 24.So, a = [-2b ± sqrt(-12b² + 24)] / 2.Simplify:a = [-2b ± 2*sqrt(-3b² + 6)] / 2 = -b ± sqrt(6 - 3b²).But since a is a positive real number, we discard the negative square root because sqrt(6 - 3b²) is positive, and -b + sqrt(6 - 3b²) must be positive. So, we have:a = -b + sqrt(6 - 3b²).Therefore, S = a + 2b = (-b + sqrt(6 - 3b²)) + 2b = b + sqrt(6 - 3b²).Now, S is expressed in terms of b: S(b) = b + sqrt(6 - 3b²). To find the maximum of S, we can take the derivative of S with respect to b and set it to zero.Let's compute dS/db:dS/db = 1 + (1/(2*sqrt(6 - 3b²)))*(-6b) = 1 - (3b)/sqrt(6 - 3b²).Set derivative equal to zero:1 - (3b)/sqrt(6 - 3b²) = 0=> (3b)/sqrt(6 - 3b²) = 1Multiply both sides by sqrt(6 - 3b²):3b = sqrt(6 - 3b²)Square both sides:9b² = 6 - 3b²Bring all terms to left side:9b² + 3b² - 6 = 0 => 12b² - 6 = 0 => 12b² = 6 => b² = 6/12 = 1/2 => b = 1/√2 (since b is positive)Then, substitute b = 1/√2 back into expression for a:a = -b + sqrt(6 - 3b²) = -1/√2 + sqrt(6 - 3*(1/2)) = -1/√2 + sqrt(6 - 3/2) = -1/√2 + sqrt(9/2) = -1/√2 + (3)/√2 = ( -1 + 3 )/√2 = 2/√2 = √2.Therefore, a = √2 and b = 1/√2. Then S = a + 2b = √2 + 2*(1/√2) = √2 + √2 = 2√2. Hmm, that's option B. Wait, but the options include 2√5, which is bigger. Did I do something wrong here?Wait, let's check the calculations again. Let's compute S again. If a = √2 and b = 1/√2, then a + 2b = √2 + 2*(1/√2) = √2 + (2/√2) = √2 + √2 = 2√2. That's correct. But why does the answer not match one of the options? Wait, option B is 2√2, which is what we got. But the options include A: 2√5, which is larger. So perhaps 2√2 is the correct answer. Wait, but I need to confirm if this is indeed the maximum.Alternatively, maybe there is a mistake in the substitution. Let me check the steps again.Starting from the original equation: a² + 2ab + 4b² = 6. If a = √2 and b = 1/√2, then:a² = (√2)^2 = 2,2ab = 2*(√2)*(1/√2) = 2*(1) = 2,4b² = 4*(1/2) = 2.Adding them up: 2 + 2 + 2 = 6, which matches. So that's correct.So according to this, the maximum value is 2√2, which is option B. But let me think if there's another way to approach this problem that might give a different answer.Another method could be using the Cauchy-Schwarz inequality. Let me recall that. For vectors u and v, (u ⋅ v)^2 ≤ (u ⋅ u)(v ⋅ v). Maybe we can set up the equation in such a way.Looking at the constraint a² + 2ab + 4b² = 6. Let me try to write this as a quadratic form. Let me see:a² + 2ab + 4b² = (a + b)^2 + 3b² = 6. Hmm, not sure if that helps. Alternatively, maybe group terms:a² + 2ab + 4b² = (a^2 + 2ab + b^2) + 3b² = (a + b)^2 + 3b² = 6. Still, not sure.Alternatively, maybe consider variables substitution. Let me set x = a + 2b, which is the expression we want to maximize. Then, maybe express a in terms of x and b: a = x - 2b. Then substitute into the original equation.Let's try that. Substitute a = x - 2b into a² + 2ab + 4b² = 6:(x - 2b)^2 + 2*(x - 2b)*b + 4b² = 6.Expand each term:First term: (x - 2b)^2 = x² - 4xb + 4b².Second term: 2*(x - 2b)*b = 2xb - 4b².Third term: 4b².Combine all terms:x² - 4xb + 4b² + 2xb - 4b² + 4b² = 6.Simplify:x² - 4xb + 4b² + 2xb - 4b² + 4b² = x² - 2xb + 4b² = 6.So, after substitution, we have x² - 2xb + 4b² = 6. Hmm, but x is the variable we want to maximize. Maybe this isn't helpful. Alternatively, perhaps treat this as a quadratic in b. Let's see:x² - 2xb + 4b² = 6.Rewrite as 4b² - 2xb + (x² - 6) = 0.This is a quadratic equation in terms of b: 4b² - 2xb + (x² - 6) = 0.For real solutions in b, the discriminant must be non-negative. So, discriminant D = ( -2x )² - 4*4*(x² - 6 ) ≥ 0.Calculate D:D = 4x² - 16(x² - 6) = 4x² - 16x² + 96 = -12x² + 96.For real solutions, D ≥ 0:-12x² + 96 ≥ 0 => -12x² ≥ -96 => x² ≤ 8 => x ≤ √8 = 2√2 ≈ 2.828.Hence, the maximum possible value of x = a + 2b is 2√2, which matches our previous result. Therefore, the answer should be B: 2√2.Wait, but let me confirm once again. If we use Lagrange multipliers, would we get the same result?Let me set up the Lagrangian. Let f(a, b) = a + 2b, which we want to maximize. The constraint is g(a, b) = a² + 2ab + 4b² - 6 = 0.The Lagrangian is L(a, b, λ) = a + 2b - λ(a² + 2ab + 4b² - 6).Taking partial derivatives:∂L/∂a = 1 - λ(2a + 2b) = 0,∂L/∂b = 2 - λ(2a + 8b) = 0,∂L/∂λ = -(a² + 2ab + 4b² - 6) = 0.So we have the system:1) 1 - λ(2a + 2b) = 0,2) 2 - λ(2a + 8b) = 0,3) a² + 2ab + 4b² = 6.From equation 1: λ = 1/(2a + 2b).From equation 2: λ = 2/(2a + 8b) = 1/(a + 4b).Therefore, 1/(2a + 2b) = 1/(a + 4b)Cross-multiplying: a + 4b = 2a + 2b => a + 4b - 2a - 2b = 0 => -a + 2b = 0 => a = 2b.So, a = 2b.Substitute a = 2b into the constraint equation: (2b)^2 + 2*(2b)*b + 4b² = 6.Calculate:4b² + 4b² + 4b² = 12b² = 6 => b² = 6/12 = 1/2 => b = 1/√2 (since positive).Thus, a = 2b = 2*(1/√2) = √2.Therefore, a + 2b = √2 + 2*(1/√2) = √2 + √2 = 2√2. Same result.Hence, regardless of the method—whether substitution, Lagrange multipliers, or discriminant approach—the maximum value is 2√2. So the correct answer is B: 2√2.But just to be thorough, let's check if there's a possibility that the maximum could be higher. Suppose we set a very large a and small b, or vice versa. But since both a and b are positive real numbers, let's see.If b approaches 0, then the equation becomes a² = 6 => a = √6. Then a + 2b ≈ √6 ≈ 2.449. But 2√2 ≈ 2.828, which is larger.If a approaches 0, then the equation becomes 4b² = 6 => b = √(6/4) = √(3/2) ≈ 1.2247. Then a + 2b ≈ 0 + 2*1.2247 ≈ 2.449, same as above. So indeed, the maximum seems to be 2√2.Alternatively, another way to think about this is to consider the given quadratic form as a dot product. Let me represent the equation a² + 2ab + 4b² as [a b] * [[1, 1], [1, 4]] * [a; b] = 6. Then, the expression a + 2b is a linear combination. The maximum of a linear function over an ellipse (quadratic form) can be found by the eigenvalue method or by transforming variables. But this might be more complicated. However, since we've already found the answer through multiple methods, I think it's safe to conclude that the maximum is 2√2, which is option B.Final Answerboxed{B}</think>Given that positive real numbers (a) and (b) satisfy (a^2 + 2ab + 4b^2 = 6), we need to find the maximum value of (a + 2b).First, we express (a) in terms of (b) using the quadratic equation (a^2 + 2ab + 4b^2 = 6). Solving for (a) using the quadratic formula, we get:[a = -b + sqrt{6 - 3b^2}]Substituting this into (a + 2b), we get:[S = b + sqrt{6 - 3b^2}]Taking the derivative of (S) with respect to (b) and setting it to zero, we find:[1 - frac{3b}{sqrt{6 - 3b^2}} = 0 implies 3b = sqrt{6 - 3b^2}]Squaring both sides, we solve for (b):[9b^2 = 6 - 3b^2 implies 12b^2 = 6 implies b^2 = frac{1}{2} implies b = frac{1}{sqrt{2}}]Substituting (b = frac{1}{sqrt{2}}) back into the expression for (a), we get:[a = sqrt{2}]Thus, (a + 2b = sqrt{2} + 2 left(frac{1}{sqrt{2}}right) = sqrt{2} + sqrt{2} = 2sqrt{2}).We also verified using the Cauchy-Schwarz inequality and Lagrange multipliers, both leading to the same result. The maximum value of (a + 2b) is (2sqrt{2}).[boxed{B}],<think>Alright, so I have this problem here: Given that positive real numbers (a) and (b) satisfy (a^{2} + 2ab + 4b^{2} = 6), then the maximum value of (a + 2b) is one of the options A through D. Let me try to figure this out step by step.First, the equation given is a quadratic in terms of (a) and (b). The expression we need to maximize is (a + 2b). This seems like an optimization problem with a constraint. I remember from algebra that there are a few methods to tackle such problems. One common method is using substitution to express one variable in terms of the other and then finding the maximum. Another method is using Lagrange multipliers, but since this is likely a high school-level problem, maybe substitution or completing the square would work. There's also the Cauchy-Schwarz inequality, which sometimes helps in these quadratic forms. Let me explore these possibilities.Let me start by trying substitution. Let's take the given equation (a^{2} + 2ab + 4b^{2} = 6). Maybe I can express (a) in terms of (b) or vice versa. Let's try solving for (a). The equation is quadratic in (a):(a^{2} + 2ab + 4b^{2} = 6)This can be written as:(a^{2} + 2ab + b^{2} + 3b^{2} = 6)Which simplifies to:((a + b)^2 + 3b^2 = 6)Hmm, not sure if that helps immediately. Let me see if completing the square is useful here. Alternatively, maybe express (a) in terms of (b):Let me rearrange the original equation:(a^{2} + 2ab = 6 - 4b^{2})Factor the left-hand side:(a(a + 2b) = 6 - 4b^{2})But the expression we need is (a + 2b), which is present here. Let me denote (k = a + 2b). Then, (a = k - 2b). Let's substitute this into the original equation:((k - 2b)^2 + 2(k - 2b)b + 4b^2 = 6)Let me expand each term:First term: ((k - 2b)^2 = k^2 - 4kb + 4b^2)Second term: (2(k - 2b)b = 2kb - 4b^2)Third term: (4b^2)Adding all together:(k^2 - 4kb + 4b^2 + 2kb - 4b^2 + 4b^2 = 6)Simplify term by term:- (k^2) remains.- (-4kb + 2kb = -2kb)- (4b^2 - 4b^2 + 4b^2 = 4b^2)So the equation becomes:(k^2 - 2kb + 4b^2 = 6)Hmm, so now we have (k^2 - 2kb + 4b^2 = 6), and (k = a + 2b). Our goal is to find the maximum value of (k). So we have an equation in terms of (k) and (b), and we need to find the maximum possible (k).Let me rearrange this equation:(k^2 - 2kb + 4b^2 = 6)This is a quadratic equation in terms of (b), but maybe we can treat this as a quadratic in (b) and find for which (k) there exists real solutions for (b). Since (b) is a positive real number, the discriminant must be non-negative. Let me set this up.Consider the equation as a quadratic in (b):(4b^2 - 2kb + k^2 - 6 = 0)For real solutions, discriminant (D) must be ≥ 0.Discriminant (D = (-2k)^2 - 4*4*(k^2 - 6))Calculate:(D = 4k^2 - 16(k^2 - 6))(D = 4k^2 - 16k^2 + 96)(D = -12k^2 + 96)For real solutions, (D ≥ 0):(-12k^2 + 96 ≥ 0)Multiply both sides by -1 (inequality flips):(12k^2 - 96 ≤ 0)Divide both sides by 12:(k^2 - 8 ≤ 0)Thus:(k^2 ≤ 8)Taking square roots:(k ≤ sqrt{8} = 2sqrt{2})But wait, that's option B. But hold on, the original problem says positive real numbers (a) and (b). So even though the discriminant gives (k ≤ 2sqrt{2}), we need to check whether this maximum is achievable when (a) and (b) are positive.Wait, so according to this, the maximum value of (a + 2b) is (2sqrt{2}). However, let me check this with another method to be sure, because sometimes when dealing with quadratics and substitutions, there can be cases where the maximum isn't achieved due to positivity constraints.Alternatively, let's use the Cauchy-Schwarz inequality. The idea is that if we have an expression like (a + 2b), we can relate it to the given quadratic form.Let me recall that Cauchy-Schwarz states that for vectors (mathbf{u}) and (mathbf{v}), we have:[(mathbf{u} cdot mathbf{v})^2 ≤ (mathbf{u} cdot mathbf{u})(mathbf{v} cdot mathbf{v})]So maybe we can represent the given quadratic form as a dot product. Let me see.The given equation is (a^2 + 2ab + 4b^2 = 6). Let me write this as:(a^2 + 2ab + 4b^2 = (a)^2 + (a)(2b) + (2b)^2). Wait, that's not exactly accurate, but maybe there's a way to express this quadratic in terms of a vector.Alternatively, think of the quadratic form as (a^2 + 2ab + 4b^2). Let's see if this can be expressed as the square of a linear combination. For example:Suppose we consider (a + 2b). Let's square it: ((a + 2b)^2 = a^2 + 4ab + 4b^2). Comparing this to the given quadratic form (a^2 + 2ab + 4b^2), the difference is in the middle term. Our squared term has (4ab) instead of (2ab). So perhaps another approach.Alternatively, maybe we can use the method of Lagrange multipliers. Let me set up the Lagrangian function. Let (f(a, b) = a + 2b) be the function to maximize, subject to the constraint (g(a, b) = a^2 + 2ab + 4b^2 - 6 = 0).The Lagrangian is:[mathcal{L}(a, b, lambda) = a + 2b - lambda(a^2 + 2ab + 4b^2 - 6)]Taking partial derivatives:[frac{partial mathcal{L}}{partial a} = 1 - lambda(2a + 2b) = 0][frac{partial mathcal{L}}{partial b} = 2 - lambda(2a + 8b) = 0][frac{partial mathcal{L}}{partial lambda} = -(a^2 + 2ab + 4b^2 - 6) = 0]So from the first equation:(1 = lambda(2a + 2b)) → ( lambda = frac{1}{2(a + b)} )From the second equation:(2 = lambda(2a + 8b)) → ( lambda = frac{2}{2a + 8b} = frac{1}{a + 4b} )Set the two expressions for (lambda) equal:[frac{1}{2(a + b)} = frac{1}{a + 4b}]Cross-multiplying:(a + 4b = 2(a + b))Simplify:(a + 4b = 2a + 2b)Subtract (a + 2b) from both sides:(2b = a)So (a = 2b). Now substitute this back into the constraint equation (a^2 + 2ab + 4b^2 = 6):((2b)^2 + 2*(2b)*b + 4b^2 = 6)Calculate:(4b^2 + 4b^2 + 4b^2 = 6)Total: (12b^2 = 6) → (b^2 = 0.5) → (b = sqrt{0.5} = frac{sqrt{2}}{2}) (since (b) is positive)Then, (a = 2b = 2*(frac{sqrt{2}}{2}) = sqrt{2})Therefore, (a + 2b = sqrt{2} + 2*(frac{sqrt{2}}{2}) = sqrt{2} + sqrt{2} = 2sqrt{2})So according to Lagrange multipliers, the maximum value of (a + 2b) is (2sqrt{2}), which is option B. But wait, earlier when I used substitution and found the discriminant, I got the same answer, but just wanted to confirm via another method.Alternatively, let's check if using Cauchy-Schwarz here is possible. Let's consider the given quadratic form (a^2 + 2ab + 4b^2). Maybe we can write this as a product of vectors. Let me see.Suppose we let vector (mathbf{u} = (a, b)), and the quadratic form can be represented as (mathbf{u} cdot M cdot mathbf{u}^T), where (M) is a matrix. But maybe that's overcomplicating.Alternatively, think of the quadratic form (a^2 + 2ab + 4b^2). Let me try to express this as ((a + b)^2 + 3b^2), which is similar to what I did earlier. Then, since we have ((a + b)^2 + 3b^2 = 6). Then, maybe using Cauchy-Schwarz on the terms (a + b) and (b).But I need to relate this to (a + 2b). Let me denote (x = a + 2b), which is the quantity we want to maximize. Let me express (a) in terms of (x) and (b): (a = x - 2b). Then substitute into the equation:((x - 2b + b)^2 + 3b^2 = 6)Simplify:((x - b)^2 + 3b^2 = 6)Expanding ((x - b)^2 = x^2 - 2xb + b^2), so:(x^2 - 2xb + b^2 + 3b^2 = 6)Which becomes:(x^2 - 2xb + 4b^2 = 6)Wait, this is exactly the same equation I had earlier when substituting (k = x). So again, same result. So either way, this seems to lead to the discriminant method, giving maximum (x = 2sqrt{2}). But since Lagrange multipliers also confirm this, and substitution gives the same, it's probably correct. However, let me check with another approach just to be thorough.Another method is to parametrize the variables. Since the equation (a^2 + 2ab + 4b^2 = 6) is a quadratic, perhaps use trigonometric substitution. For example, considering the equation as an ellipse and parameterizing it. Let me see.First, complete the square for the quadratic equation. Let's see:(a^2 + 2ab + 4b^2 = 6)Earlier, I tried writing this as ((a + b)^2 + 3b^2 = 6). So this is equivalent to ((a + b)^2 + (sqrt{3}b)^2 = 6). Let me set this as an ellipse equation. Let (u = a + b) and (v = sqrt{3}b). Then the equation becomes:(u^2 + v^2 = 6)So this is a circle of radius (sqrt{6}) in the (u-v) plane. But since (a) and (b) are positive, (u = a + b > 0) and (v = sqrt{3}b > 0). So we are looking at the first quadrant portion of this circle.Our target expression is (a + 2b). Express this in terms of (u) and (v):(a + 2b = (a + b) + b = u + b). But (v = sqrt{3}b) implies (b = v/sqrt{3}). So:(a + 2b = u + v/sqrt{3})So we need to maximize (u + v/sqrt{3}) subject to (u^2 + v^2 = 6) with (u > 0) and (v > 0).This is a standard maximization problem on a circle. The maximum of a linear function over a circle occurs where the gradient of the function is parallel to the gradient of the constraint, i.e., using Lagrange multipliers again. But perhaps more straightforwardly, using Cauchy-Schwarz.The expression (u + (1/sqrt{3})v) can be considered as the dot product of the vector ((1, 1/sqrt{3})) and the vector ((u, v)). The maximum value of this dot product is the product of their magnitudes, which is (sqrt{1^2 + (1/sqrt{3})^2} times sqrt{u^2 + v^2}).Compute the first magnitude:(sqrt{1 + 1/3} = sqrt{4/3} = 2/sqrt{3})The second magnitude is (sqrt{6}). Therefore, the maximum value is ( (2/sqrt{3})(sqrt{6}) ) = 2/sqrt{3} * sqrt{6} = 2sqrt{2}). So that's again the same result. Therefore, regardless of the method, the maximum is (2sqrt{2}), which is option B.Wait, but let me check if all steps are valid. For example, in the Cauchy-Schwarz step, the maximum of the dot product is indeed the product of the magnitudes when the vectors are in the same direction. So the maximum occurs when ((u, v)) is a scalar multiple of ((1, 1/sqrt{3})). Let's confirm that.Let ((u, v) = t(1, 1/sqrt{3})). Then, substituting into the constraint (u^2 + v^2 = 6):(t^2(1 + 1/3) = t^2(4/3) = 6) → (t^2 = 6*(3/4) = 18/4 = 9/2) → (t = 3/sqrt{2}). Thus,(u = 3/sqrt{2}), (v = 3/sqrt{2} * 1/sqrt{3} = 3/(sqrt{2}sqrt{3}) = sqrt{3}/sqrt{2})So (a + 2b = u + v/sqrt{3} = 3/sqrt{2} + (sqrt{3}/sqrt{2}) / sqrt{3} = 3/sqrt{2} + 1/sqrt{2} = 4/sqrt{2} = 2sqrt{2}). Perfect, so that checks out.Alternatively, going back to the original variables, when (a = 2b), which we found via Lagrange multipliers, let's verify that. If (a = 2b), then (a = 2b), substituting back into the original equation:( (2b)^2 + 2*(2b)*b + 4b^2 = 4b^2 + 4b^2 + 4b^2 = 12b^2 = 6 Rightarrow b^2 = 0.5 Rightarrow b = sqrt{0.5} = sqrt{2}/2 ), as we found. Then, (a = 2b = sqrt{2}), so (a + 2b = sqrt{2} + 2*(sqrt{2}/2) = sqrt{2} + sqrt{2} = 2sqrt{2}). Yep, all consistent.But just to be absolutely thorough, let me check if there are any other critical points. Suppose we consider the case when either (a) or (b) approaches zero. Since (a) and (b) are positive, they can't be zero, but approaching zero.If (b) approaches zero, the original equation becomes (a^2 = 6) → (a = sqrt{6}). Then (a + 2b ≈ sqrt{6} ≈ 2.449). Since (2sqrt{2} ≈ 2.828), which is larger, so the maximum isn't here.If (a) approaches zero, the equation becomes (4b^2 = 6) → (b = sqrt{6}/2 ≈ 1.2247). Then (a + 2b ≈ 0 + 2*(1.2247) ≈ 2.449), same as before. So in both edge cases, the value is around 2.449, which is less than 2.828. Therefore, the maximum is indeed achieved when both (a) and (b) are positive, not at the boundaries.Hence, all methods converge to the answer (2sqrt{2}), which is option B. Therefore, I think the correct answer is B: (2sqrt{2}).Final Answerboxed{B}</think>Given that positive real numbers (a) and (b) satisfy (a^2 + 2ab + 4b^2 = 6), we need to find the maximum value of (a + 2b).First, we substitute (a = k - 2b) into the given equation, where (k = a + 2b):[(k - 2b)^2 + 2(k - 2b)b + 4b^2 = 6]Expanding and simplifying this equation, we get:[k^2 - 2kb + 4b^2 = 6]This is a quadratic equation in (b):[4b^2 - 2kb + (k^2 - 6) = 0]For (b) to be a real number, the discriminant of this quadratic equation must be non-negative:[(-2k)^2 - 4 cdot 4 cdot (k^2 - 6) geq 0]Simplifying the discriminant:[4k^2 - 16(k^2 - 6) geq 0][4k^2 - 16k^2 + 96 geq 0][-12k^2 + 96 geq 0][12k^2 leq 96][k^2 leq 8][k leq 2sqrt{2}]To verify, we use Lagrange multipliers. Let (f(a, b) = a + 2b) and the constraint (g(a, b) = a^2 + 2ab + 4b^2 - 6 = 0). Solving the system of equations from the partial derivatives, we find (a = 2b). Substituting (a = 2b) into the constraint:[(2b)^2 + 2(2b)b + 4b^2 = 6][12b^2 = 6][b = frac{sqrt{2}}{2}]Then, (a = sqrt{2}) and (a + 2b = sqrt{2} + 2 cdot frac{sqrt{2}}{2} = 2sqrt{2}).Using Cauchy-Schwarz inequality, we parameterize the quadratic form as an ellipse and find the maximum value of (a + 2b) using dot product maximization, confirming the result (2sqrt{2}).Thus, the maximum value of (a + 2b) is (boxed{B}).]

🔑:To find the maximum value of a+2b given that a and b are positive real numbers and satisfy the equation a^{2}+2ab+4b^{2}=6, we proceed as follows:1. Let t=a+2b (where t > 0). We aim to express everything in terms of t to find its maximum value.2. Substituting a=t-2b into the original equation gives us: [ (t-2b)^{2}+2(t-2b)b+4b^{2}=6 ] Simplifying this, we get: [ t^{2}-4tb+4b^{2}+2tb-4b^{2}+4b^{2}=6 ] [ t^{2}-2tb+4b^{2}=6 ] Further simplifying, we find: [ 4b^{2}-2tb+t^{2}-6=0 ]3. For b > 0, the quadratic equation in terms of b must have a positive root. This means the discriminant Delta of the quadratic equation must be non-negative: [ Delta = (-2t)^{2}-4cdot4cdot(t^{2}-6) geq 0 ] Simplifying the discriminant, we get: [ 4t^{2}-16t^{2}+96 geq 0 ] [ -12t^{2}+96 geq 0 ] [ t^{2} leq 8 ] [ t leq 2sqrt{2} ]4. Additionally, since t=a+2b > 0, we also need to ensure that the equation 4b^{2}-2tb+t^{2}-6=0 has a meaningful solution for b. This implies: [ frac{t^{2}-6}{4} > 0 ] [ t^{2} > 6 ] [ t > sqrt{6} ]5. Combining the inequalities, we find that sqrt{6} < t leq 2sqrt{2}. This means the maximum value that t=a+2b can take is 2sqrt{2}.Therefore, the maximum value of a+2b is boxed{2sqrt{2}}, which corresponds to option boxed{B}.