💡Alright, so I've got this problem about friendships among students at an Olympiad. It says that no matter how you split the students into two contest halls, there are always two friends in different halls. That sounds like it's related to graph theory, right? Maybe something about connectedness or something else.Let me try to model this. If I think of each student as a vertex in a graph, and an edge between two vertices if they're friends. So, the problem is saying that no matter how you partition the graph into two sets, there's always at least one edge between the two sets. That means the graph isn't disconnected, right? Because if it were disconnected, you could put each connected component in a separate hall, and there wouldn't be any friends across halls. So, the graph must be connected.But wait, it's not just connected; it's also not complete. Because the problem mentions that there are students who aren't friends. So, the graph is connected but not complete. That means there are some pairs of students who aren't friends, but you can still get from any student to any other student through a series of friendships.Now, the problem is asking me to show that for a fixed student A, there exist two other students B and C such that in the group {A, B, C}, there are exactly two friendships. So, in other words, among A, B, and C, exactly two of the three possible friendships exist.Let me think about how to approach this. Maybe I can use some properties of connected graphs. Since the graph is connected, there's a path between any two students. But it's not complete, so there must be some pairs that aren't directly connected.Suppose I pick a student A. Since the graph is connected, A is friends with at least one other student. Let's say A is friends with B. Now, if A is friends with everyone, then the graph would be complete, which it's not. So, A must not be friends with at least one student, say C.But wait, if A isn't friends with C, then since the graph is connected, there must be a path from A to C through other students. Let's say the shortest path from A to C is A-B-D-C. So, A is friends with B, B is friends with D, D is friends with C.Now, in this case, looking at the set {A, B, D}, how many friendships are there? A is friends with B, B is friends with D, but A isn't friends with D. So, there are exactly two friendships in this set. That seems to satisfy the condition.But wait, the problem says that no matter how you partition the students into two halls, there are always two friends in different halls. So, does that affect how we choose B and C? Maybe not directly, because we're just looking for the existence of such B and C.Let me try another angle. Suppose for contradiction that for every pair of students B and C, the set {A, B, C} has either one or three friendships. So, either A is friends with both B and C, but B and C aren't friends, or A is friends with both B and C, and B and C are friends.But if A is friends with both B and C, and B and C are friends, then {A, B, C} has three friendships. If A is friends with both B and C, but B and C aren't friends, then {A, B, C} has two friendships. Wait, that contradicts the assumption. So, maybe my initial thought was wrong.Alternatively, maybe if the graph is connected and not complete, there must be a path of length two from A to some other student, which would give us exactly two friendships in the set {A, B, C}.Wait, let's think about it more carefully. If the graph is connected, there's a path from A to every other student. If the graph isn't complete, there's at least one student who isn't directly connected to A. Let's call that student C. Since the graph is connected, there must be a path from A to C, say A-B-C. So, A is friends with B, B is friends with C, but A isn't friends with C. Therefore, in the set {A, B, C}, there are exactly two friendships: A-B and B-C.That seems to work. So, in this case, B is a friend of A, and C is a friend of B but not a friend of A. Therefore, the set {A, B, C} has exactly two friendships.But wait, the problem says "there exist students B and C" such that in {A, B, C}, there are exactly two friendships. So, in this case, B and C are such students. Therefore, we've found them.But let me make sure I'm not missing something. The problem also mentions that no matter how you partition the students into two halls, there are always two friends in different halls. Does that affect the argument? Maybe not directly, because we're just looking for the existence of such B and C, regardless of the partitioning.Alternatively, maybe the condition about partitioning implies that the graph is connected, which we already used. So, the key point is that the graph is connected but not complete, so there must be a path of length two from A to some other student, giving us exactly two friendships in the set {A, B, C}.I think that makes sense. So, to summarize, since the graph is connected, there's a path from A to any other student. Since the graph isn't complete, there's at least one student not directly connected to A, so the path from A to that student must be at least length two. Therefore, there must be a student B who is a friend of A and a friend of C, where C is not a friend of A. Thus, in the set {A, B, C}, there are exactly two friendships.Yeah, that seems to cover it. I don't see any flaws in this reasoning. So, I think that's the solution.

💡Alright, so I need to show that if a fraction a/b lies between two consecutive Fibonacci ratios, then the denominator b must be at least as big as the next Fibonacci number. Hmm, okay, let's break this down.First, the Fibonacci sequence is defined as f₁ = 1, f₂ = 1, and each subsequent term is the sum of the two previous ones. So, f₃ = 2, f₄ = 3, f₅ = 5, and so on. The problem says that a/b is between fₙ/fₙ₋₁ and fₙ₊₁/fₙ. I need to show that b is at least fₙ₊₁.Let me think about the properties of Fibonacci numbers and fractions. I remember something about Farey sequences and mediants. The mediant of two fractions a/b and c/d is (a+c)/(b+d). It always lies between the two fractions. Also, if two fractions are consecutive in a Farey sequence, their determinant |ad - bc| is 1. Maybe that's useful here.In the case of Fibonacci numbers, I recall that consecutive Fibonacci ratios are actually consecutive in some Farey sequence. So, fₙ/fₙ₋₁ and fₙ₊₁/fₙ are consecutive in the Farey sequence of some order. That means their determinant should be 1. Let me check that.Calculating the determinant: |fₙ * fₙ - fₙ₋₁ * fₙ₊₁|. Using the identity for Fibonacci numbers, fₙ₊₁ = fₙ + fₙ₋₁. So, fₙ₊₁ = fₙ + fₙ₋₁. Let's substitute that in:|fₙ * fₙ - fₙ₋₁ * (fₙ + fₙ₋₁)| = |fₙ² - fₙ₋₁ fₙ - fₙ₋₁²| = |fₙ² - fₙ fₙ₋₁ - fₙ₋₁²|.Hmm, I think there's a Fibonacci identity that says fₙ² - fₙ fₙ₋₁ - fₙ₋₁² = (-1)ⁿ⁻¹. So, the absolute value would be 1. Yes, that's right! So, |fₙ * fₙ - fₙ₋₁ * fₙ₊₁| = 1. That means fₙ/fₙ₋₁ and fₙ₊₁/fₙ are consecutive in some Farey sequence.Now, if a/b lies between these two fractions, then by the property of Farey sequences, the denominator b must be greater than or equal to the sum of the denominators of the two consecutive terms. Wait, is that the case?Actually, I think the property is that if you have two consecutive fractions a/b and c/d in a Farey sequence, then any fraction between them must have a denominator greater than or equal to the sum of b and d. But in this case, the two fractions are fₙ/fₙ₋₁ and fₙ₊₁/fₙ. Their denominators are fₙ₋₁ and fₙ. So, the sum of the denominators is fₙ₋₁ + fₙ, which is fₙ₊₁. So, does that mean that b must be at least fₙ₊₁?Wait, let me make sure. The property says that if you have two consecutive terms in a Farey sequence, then any fraction between them has a denominator greater than or equal to the sum of the denominators of the two terms. So, in this case, since fₙ/fₙ₋₁ and fₙ₊₁/fₙ are consecutive in some Farey sequence, any fraction a/b between them must satisfy b ≥ fₙ₋₁ + fₙ = fₙ₊₁.That seems to be the case. So, is that the proof? It feels a bit quick, but maybe it's correct. Let me try to formalize it.Suppose that a/b lies between fₙ/fₙ₋₁ and fₙ₊₁/fₙ. Since fₙ/fₙ₋₁ and fₙ₊₁/fₙ are consecutive in a Farey sequence, their determinant is 1, as we saw earlier. Then, by the property of Farey sequences, any fraction between them must have a denominator at least equal to the sum of the denominators of the two terms. Since the denominators are fₙ₋₁ and fₙ, their sum is fₙ₊₁. Therefore, b must be at least fₙ₊₁.Wait, but is this property always true? Let me think about an example. Take n=3, so f₃=2, f₂=1, f₄=3. So, the fractions are 2/1 and 3/2. Any fraction between 2 and 1.5 must have denominator at least 3. Let's see, for example, 5/3 is approximately 1.666, which is between 1.5 and 2. The denominator is 3, which is equal to f₄=3. Another example, 7/4 is 1.75, which is also between 1.5 and 2, and 4 is greater than 3. So, it seems to hold.Another example, n=4: f₄=3, f₃=2, f₅=5. So, fractions are 3/2 and 5/3. Any fraction between 1.5 and approximately 1.666 must have denominator at least 5. Let's see, 8/5 is 1.6, which is between 1.5 and 1.666, and 5 is equal to f₅=5. Another fraction, 11/7 ≈ 1.571, which is between 1.5 and 1.666, and 7 is greater than 5. So, again, it holds.Okay, so the property seems to hold in these examples. Therefore, I think the argument is correct. The key points are:1. Fibonacci ratios fₙ/fₙ₋₁ and fₙ₊₁/fₙ are consecutive in a Farey sequence because their determinant is 1.2. Any fraction between two consecutive terms in a Farey sequence must have a denominator at least equal to the sum of the denominators of the two terms.3. The sum of fₙ₋₁ and fₙ is fₙ₊₁, so b must be at least fₙ₊₁.Therefore, the conclusion is that b ≥ fₙ₊₁.**Final Answer**boxed{b geq f_{n+1}}

💡Okay, so I have this problem about a rectangular solid and a cube being removed from it. I need to find the difference in surface area between the original solid and the new solid after the cube is removed. The options are given, and I need to figure out if the surface area is less or more by 12, 8, or if it's the same.First, let me visualize the original solid. It's a rectangular prism with dimensions 4 feet by 5 feet by 6 feet. That means it has six faces: two with dimensions 4x5, two with 4x6, and two with 5x6. The surface area of a rectangular prism is calculated by the formula 2(lw + lh + wh), where l is length, w is width, and h is height.So, plugging in the numbers: 2*(4*5 + 4*6 + 5*6) = 2*(20 + 24 + 30) = 2*74 = 148 square feet. Okay, so the original surface area is 148 square feet.Now, a cube with sides of 2 feet is removed from the middle of one of the 4x5 faces. Let me think about what this does to the surface area. When you remove a cube from a solid, you're taking away some volume, but you're also creating new surfaces where the cube was removed.The cube has sides of 2 feet, so each face of the cube is 2x2 feet. Since the cube is removed from the middle of a face, one face of the cube was originally part of the 4x5 face of the solid. When we remove the cube, we're taking away that 2x2 area from the original face, but we're also exposing the other five faces of the cube.Wait, no. Actually, when you remove a cube from the middle of a face, you're removing a part of that face, but you're also creating a hole. The hole has its own inner surfaces. So, the original face loses a 2x2 area, but the inner surfaces of the hole add new area.Let me clarify: the original 4x5 face had an area of 20 square feet. Removing a 2x2 square from it reduces the area of that face by 4 square feet. But now, the hole has four new inner faces, each of which is 2x2 feet. So, each inner face is 4 square feet, and there are four of them, so that's 16 square feet added.Wait, but the cube has six faces, right? So, if we remove a cube from the middle of a face, we're removing one face (the one that was part of the original solid) and exposing the other five faces of the cube. But actually, no, because the cube is removed from the middle, so the hole is open on one side, meaning only five faces are exposed? Or is it four?Hmm, I think it's four because the hole is on one face, so the inner surfaces are four sides of the cube. Let me think: if you have a cube and you remove it from the middle of a face, the hole will have four inner walls, each of which is 2x2 feet. So, that's four faces, each 4 square feet, totaling 16 square feet.But wait, the original face loses 4 square feet, so the net change is the loss of 4 square feet minus the gain of 16 square feet, which is a net gain of 12 square feet. So, the surface area increases by 12 square feet.But let me double-check. The original surface area is 148. After removing the cube, the surface area becomes 148 - 4 + 16 = 160. So, the difference is 160 - 148 = 12 more.Wait, but I'm not sure if it's four or five faces. If the cube is removed from the middle, does it expose five faces or four? Because the cube has six faces, but one face was part of the original solid, so when you remove it, you're left with five faces exposed. But in reality, when you remove a cube from the middle of a face, you're creating a hole that has four inner walls, not five, because the fifth face would be the back of the cube, which is not exposed.Wait, no, actually, when you remove a cube from the middle of a face, you're creating a hole that has four inner walls, each corresponding to the sides of the cube. The fifth face would be the back of the cube, but since it's removed from the middle, the back face is not exposed; it's just open space. So, actually, only four inner faces are exposed.Therefore, the calculation would be: original surface area minus the area removed (4 square feet) plus the area of the four inner faces (16 square feet). So, 148 - 4 + 16 = 160. The difference is 12 more.But wait, another way to think about it is that removing the cube removes one face of the cube (4 square feet) from the original solid, but adds five new faces (each 4 square feet) because the cube had six faces, one of which was part of the original solid. So, 5*4 = 20 added, minus the 4 removed, so net change is +16. But that would make the surface area 148 + 16 = 164, which is a difference of 16 more, but that's not one of the options.Hmm, I'm confused now. Let me try to clarify.When you remove a cube from the middle of a face, you're taking away a part of that face, which reduces the surface area by the area of the cube's face (4 square feet). However, you're also exposing the inner surfaces of the hole, which are the other five faces of the cube. Each of these faces is 4 square feet, so that's 5*4 = 20 square feet added.Therefore, the net change is -4 + 20 = +16. So, the surface area increases by 16 square feet. But 16 isn't one of the options. The options are 12 less, 12 more, 8 less, 8 more, or the same.Wait, maybe I'm overcomplicating it. Let's think differently. The original surface area is 148. When you remove the cube, you're removing a 2x2x2 cube. The surface area change is the area removed from the original face minus the area added by the inner surfaces.The area removed from the original face is 2x2 = 4 square feet. The inner surfaces added are four sides of the cube, each 2x2, so 4*4 = 16 square feet. So, the net change is -4 + 16 = +12. Therefore, the surface area increases by 12 square feet.Yes, that makes sense. So, the difference is 12 more. So, the answer should be B) 12 more.Wait, but earlier I thought it was four inner faces, but now I'm considering five. I think the confusion is whether the back face of the cube is exposed or not. Since the cube is removed from the middle of a face, the back face of the cube is not exposed; it's just open space. Therefore, only four inner faces are exposed, each 2x2, so 4*4=16. The area removed is 4, so net change is +12.Yes, that seems correct. So, the surface area increases by 12 square feet.

💡Okay, so I need to figure out what number Emily adds to 50 squared to get 51 squared, and what number she subtracts from 50 squared to get 49 squared. Hmm, let's start by recalling what 50 squared is. I know that 50 times 50 is 2500. So, 50² = 2500.Now, for 51 squared, Emily is adding some number to 2500. I think she's using a method where she breaks down 51 into 50 plus 1. So, 51² would be (50 + 1)². I remember there's a formula for expanding squares of binomials, like (a + b)² = a² + 2ab + b². Let me apply that here.So, (50 + 1)² = 50² + 2*50*1 + 1². That would be 2500 + 100 + 1. Adding those together, 2500 + 100 is 2600, and then plus 1 is 2601. So, 51² is 2601.Since Emily started with 50², which is 2500, she must have added 101 to get to 2601. Wait, let me check that. 2500 + 101 is indeed 2601, so that makes sense. So, the number she adds is 101.Now, for 49 squared, Emily is subtracting a number from 50². I think she's using a similar approach by breaking down 49 into 50 minus 1. So, 49² would be (50 - 1)². Again, using the binomial expansion formula, (a - b)² = a² - 2ab + b².Applying that here, (50 - 1)² = 50² - 2*50*1 + 1². That would be 2500 - 100 + 1. Subtracting 100 from 2500 gives 2400, and then adding 1 gives 2401. So, 49² is 2401.Since Emily started with 50², which is 2500, she must have subtracted 99 to get to 2401. Let me verify that. 2500 - 99 is 2401, which matches. So, the number she subtracts is 99.Wait, let me make sure I didn't make a mistake in the calculations. For 51², expanding (50 + 1)² gives 50² + 100 + 1, which is 2500 + 100 + 1 = 2601. That seems correct. And for 49², expanding (50 - 1)² gives 50² - 100 + 1, which is 2500 - 100 + 1 = 2401. That also checks out.So, Emily adds 101 to 50² to get 51² and subtracts 99 from 50² to get 49². I think that's the answer.

💡Okay, so I have this parametric curve given by (x, y) = (3 cos t - 2 sin t, 3 sin t). I need to find the coefficients a, b, c such that the curve can be expressed in the form ax² + bxy + cy² = 1. Hmm, that sounds like converting parametric equations into a Cartesian equation. I remember that usually, you can eliminate the parameter t by expressing cos t and sin t in terms of x and y, and then using the Pythagorean identity cos² t + sin² t = 1. Let me write down the given parametric equations:x = 3 cos t - 2 sin t y = 3 sin tSo, from the second equation, y = 3 sin t, I can solve for sin t:sin t = y / 3That's straightforward. Now, I need to express cos t in terms of x and y. Let's rearrange the first equation:x = 3 cos t - 2 sin tI can substitute sin t from the second equation into this:x = 3 cos t - 2*(y / 3) x = 3 cos t - (2y)/3Let me solve for cos t:3 cos t = x + (2y)/3 cos t = (x + (2y)/3) / 3 cos t = x/3 + (2y)/9Okay, so now I have expressions for both cos t and sin t in terms of x and y:cos t = x/3 + (2y)/9 sin t = y/3Now, since cos² t + sin² t = 1, I can substitute these expressions into that identity:[ (x/3 + (2y)/9 ) ]² + [ (y/3) ]² = 1Let me compute each term step by step.First, let's compute (x/3 + (2y)/9 )²:Let me write it as [ (x/3) + (2y/9) ]². To expand this, I can use the formula (a + b)² = a² + 2ab + b².So,(x/3)² + 2*(x/3)*(2y/9) + (2y/9)² = x²/9 + (4xy)/27 + 4y²/81Okay, that's the first term. Now, the second term is (y/3)², which is y²/9.So, putting it all together:x²/9 + (4xy)/27 + 4y²/81 + y²/9 = 1Now, let's combine like terms. First, I'll write all terms with denominators:x²/9 + (4xy)/27 + (4y²)/81 + y²/9I notice that the y² terms have denominators 81 and 9. Let me convert y²/9 to 9y²/81 so that they have the same denominator:x²/9 + (4xy)/27 + (4y²)/81 + (9y²)/81Now, combine the y² terms:4y²/81 + 9y²/81 = 13y²/81So, now the equation becomes:x²/9 + (4xy)/27 + 13y²/81 = 1Now, I need to express this in the form ax² + bxy + cy² = 1. So, let's write each term with coefficients:a x² + b x y + c y² = 1Comparing term by term:x² term: a = 1/9 xy term: b = 4/27 y² term: c = 13/81Wait, but hold on. In the equation I have, the coefficients are 1/9, 4/27, and 13/81. So, is that the final answer? Let me double-check my calculations.First, expanding (x/3 + 2y/9)^2:(x/3)^2 = x²/9 2*(x/3)*(2y/9) = 4xy/27 (2y/9)^2 = 4y²/81Then, adding (y/3)^2 = y²/9.So, total equation:x²/9 + 4xy/27 + 4y²/81 + y²/9 = 1Convert y²/9 to 9y²/81:x²/9 + 4xy/27 + 4y²/81 + 9y²/81 = 1Combine y² terms: 4 + 9 = 13, so 13y²/81.So, equation is:x²/9 + 4xy/27 + 13y²/81 = 1Therefore, coefficients are a = 1/9, b = 4/27, c = 13/81.Wait, but in the initial problem statement, the user had a different answer. They had a = 1/9, b = 4/27, c = 5/27. Hmm, that's conflicting with my result here. Did I make a mistake somewhere?Let me go through the steps again.Starting from:x = 3 cos t - 2 sin t y = 3 sin tExpress sin t = y / 3Then, x = 3 cos t - 2*(y / 3) x = 3 cos t - (2y)/3 So, 3 cos t = x + (2y)/3 cos t = (x + (2y)/3) / 3 cos t = x/3 + (2y)/9So, cos t = x/3 + 2y/9Then, cos² t + sin² t = 1:(x/3 + 2y/9)^2 + (y/3)^2 = 1Expanding (x/3 + 2y/9)^2:= (x/3)^2 + 2*(x/3)*(2y/9) + (2y/9)^2 = x²/9 + (4xy)/27 + 4y²/81Adding (y/3)^2 = y²/9:Total equation:x²/9 + 4xy/27 + 4y²/81 + y²/9 = 1Convert y²/9 to 9y²/81:x²/9 + 4xy/27 + 4y²/81 + 9y²/81 = 1Combine y² terms: 4 + 9 = 13, so 13y²/81.Thus, equation is:x²/9 + 4xy/27 + 13y²/81 = 1So, coefficients are a = 1/9, b = 4/27, c = 13/81.Wait, so why does the initial problem have c = 5/27? Maybe I made a mistake in the initial steps.Alternatively, perhaps the user made a mistake in their own solution? Let me check their steps.In the given solution, they expanded the quadratic form:a(3 cos t - 2 sin t)^2 + b(3 cos t - 2 sin t)(3 sin t) + c(3 sin t)^2Which expands to:a(9 cos² t - 12 cos t sin t + 4 sin² t) + b(9 cos t sin t - 6 sin² t) + c(9 sin² t)Then, grouping terms:9a cos² t + (-12a + 9b) cos t sin t + (4a - 6b + 9c) sin² t = 1Then, setting coefficients equal:9a = 1 -12a + 9b = 0 4a - 6b + 9c = 1Solving:a = 1/9 -12*(1/9) + 9b = 0 => -4/3 + 9b = 0 => 9b = 4/3 => b = 4/27 4*(1/9) - 6*(4/27) + 9c = 1 4/9 - 24/27 + 9c = 1 Convert to common denominator 27: 12/27 - 24/27 + 9c = 1 -12/27 + 9c = 1 -4/9 + 9c = 1 9c = 1 + 4/9 = 13/9 c = 13/81Wait, so according to this, c = 13/81, which is the same as my result. But in the given solution, they have c = 5/27. That must be a mistake in their solution.Wait, let me check their third equation:4a - 6b + 9c = 1Plugging a = 1/9, b = 4/27:4*(1/9) = 4/9 -6*(4/27) = -24/27 = -8/9 So, 4/9 - 8/9 + 9c = 1 (4 - 8)/9 + 9c = 1 -4/9 + 9c = 1 9c = 1 + 4/9 = 13/9 c = 13/81Yes, that's correct. So in their solution, they wrote c = 5/27, which is incorrect. It should be 13/81.Therefore, my initial calculation is correct, and the given solution had an error in the value of c.So, the correct ordered triple is (1/9, 4/27, 13/81).But let me double-check my own work to make sure.Starting from parametric equations:x = 3 cos t - 2 sin t y = 3 sin tExpress sin t = y / 3Then, x = 3 cos t - 2*(y / 3) x = 3 cos t - (2y)/3 So, 3 cos t = x + (2y)/3 cos t = (x + (2y)/3) / 3 = x/3 + 2y/9Then, cos² t + sin² t = 1:(x/3 + 2y/9)^2 + (y/3)^2 = 1Expanding:(x²)/9 + (4xy)/27 + (4y²)/81 + y²/9 = 1Convert y²/9 to 9y²/81:x²/9 + 4xy/27 + 4y²/81 + 9y²/81 = 1 x²/9 + 4xy/27 + 13y²/81 = 1So, coefficients:a = 1/9 b = 4/27 c = 13/81Yes, that's consistent. So, the correct answer should be (1/9, 4/27, 13/81). The initial solution provided had c = 5/27, which is incorrect.Therefore, I think my answer is correct.

💡Okay, so I have this problem where I need to find the sum of m and b, where the line y = mx + b is the line of reflection that maps the point (2,3) to its image (10,7). Hmm, reflections... I remember that reflecting a point over a line means that the line is the perpendicular bisector of the segment connecting the original point and its image. So, first, I should probably find the midpoint between (2,3) and (10,7) because that midpoint should lie on the line y = mx + b.Let me calculate the midpoint. The midpoint formula is ((x1 + x2)/2, (y1 + y2)/2). Plugging in the values, that would be ((2 + 10)/2, (3 + 7)/2). So, that's (12/2, 10/2), which simplifies to (6,5). Okay, so the midpoint is (6,5). That means the line y = mx + b passes through (6,5). Got that.Now, I also remember that the line of reflection is perpendicular to the segment connecting the original point and its image. So, I need to find the slope of the segment connecting (2,3) and (10,7) first. The slope formula is (y2 - y1)/(x2 - x1). Plugging in the values, that's (7 - 3)/(10 - 2) = 4/8 = 1/2. So, the slope of the segment is 1/2.Since the line of reflection is perpendicular to this segment, its slope should be the negative reciprocal of 1/2. The negative reciprocal of 1/2 is -2. So, the slope m of the line y = mx + b is -2. That seems right because perpendicular lines have slopes that multiply to -1, and 1/2 * (-2) = -1, which checks out.Now that I have the slope m = -2 and a point (6,5) that lies on the line, I can use the point-slope form to find the equation of the line. The point-slope form is y - y1 = m(x - x1). Plugging in the values, that's y - 5 = -2(x - 6). Let me simplify that. Distribute the -2 on the right side: y - 5 = -2x + 12. Then, add 5 to both sides to get y by itself: y = -2x + 17. So, the equation of the line is y = -2x + 17.Therefore, m is -2 and b is 17. The problem asks for m + b, so I just add them together: -2 + 17 = 15. So, the sum of m and b is 15.Wait, let me double-check my steps to make sure I didn't make a mistake. First, midpoint: (6,5) seems correct. Slope between the two points: 1/2, that's right. Negative reciprocal is -2, that makes sense. Then, using point-slope: y - 5 = -2(x - 6). Distribute correctly: y - 5 = -2x + 12. Add 5: y = -2x + 17. Yeah, that looks good. So, m + b is indeed 15. I think that's the correct answer.**Final Answer**The value of ( m + b ) is boxed{15}.