💡Okay, so I have this complex number problem here: If the complex number (2 + ai)/(1 + i) is a pure imaginary number, then what is the value of a? Hmm, let me try to figure this out step by step.First, I remember that a pure imaginary number is a complex number that has no real part. That means if I can express (2 + ai)/(1 + i) in the form of x + yi, where x is the real part and y is the imaginary part, then x should be zero for it to be a pure imaginary number. So, my goal is to simplify the given expression and set the real part equal to zero to solve for a.Alright, let's start by simplifying (2 + ai)/(1 + i). I recall that to simplify expressions like this, we can multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of 1 + i is 1 - i. So, multiplying both the numerator and the denominator by 1 - i should help eliminate the imaginary unit from the denominator.Let me write that out:(2 + ai)/(1 + i) * (1 - i)/(1 - i) = [(2 + ai)(1 - i)] / [(1 + i)(1 - i)]Now, let's expand both the numerator and the denominator.Starting with the numerator: (2 + ai)(1 - i)Using the distributive property (FOIL method):2*1 + 2*(-i) + ai*1 + ai*(-i) = 2 - 2i + ai - ai^2Wait, I remember that i^2 is equal to -1, so -ai^2 becomes +a. So, simplifying that:2 - 2i + ai + aCombine like terms:(2 + a) + (-2i + ai)Factor out the i from the imaginary terms:(2 + a) + (a - 2)iOkay, so the numerator simplifies to (2 + a) + (a - 2)i.Now, let's work on the denominator: (1 + i)(1 - i)Again, using the distributive property:1*1 + 1*(-i) + i*1 + i*(-i) = 1 - i + i - i^2Simplify:1 - i + i - (-1) because i^2 is -1.So, 1 + 1 = 2. The -i and +i cancel each other out.Therefore, the denominator simplifies to 2.Putting it all together, the original expression simplifies to:[(2 + a) + (a - 2)i] / 2Which can be written as:(2 + a)/2 + (a - 2)/2 * iSo, the real part is (2 + a)/2 and the imaginary part is (a - 2)/2.Since the problem states that the complex number is a pure imaginary number, the real part must be zero. Therefore, we set the real part equal to zero:(2 + a)/2 = 0Multiply both sides by 2:2 + a = 0Subtract 2 from both sides:a = -2Let me double-check to make sure I didn't make any mistakes. If a is -2, then the original expression becomes (2 - 2i)/(1 + i). Let's simplify that:(2 - 2i)/(1 + i) * (1 - i)/(1 - i) = [(2 - 2i)(1 - i)] / [(1 + i)(1 - i)]Expanding the numerator:2*1 + 2*(-i) - 2i*1 - 2i*(-i) = 2 - 2i - 2i + 2i^2Simplify:2 - 4i + 2*(-1) = 2 - 4i - 2 = (-4i)/2 = -2iWait, that seems a bit off. Let me recalculate that.Wait, no, let's do it step by step. The numerator is (2 - 2i)(1 - i):2*1 = 22*(-i) = -2i-2i*1 = -2i-2i*(-i) = 2i^2 = 2*(-1) = -2So, adding all those together:2 - 2i - 2i - 2 = (2 - 2) + (-2i - 2i) = 0 - 4i = -4iDenominator is (1 + i)(1 - i) = 1 - i^2 = 1 - (-1) = 2So, the entire expression is (-4i)/2 = -2i, which is indeed a pure imaginary number. So, a = -2 is correct.I think that makes sense. So, the value of a is -2.**Final Answer**The value of ( a ) is boxed{-2}.

💡Alright, so I've got this problem here where I need to show that I can color all the integers with three colors in such a way that any two integers that are a difference of 'a' or 'b' apart are colored differently. Hmm, okay, let's break this down.First off, I know that coloring problems often involve assigning colors in a way that avoids certain conflicts. In this case, the conflict is having two numbers that are 'a' or 'b' apart share the same color. So, I need a coloring strategy that ensures this doesn't happen.I remember something about periodic colorings, where you repeat a pattern every certain number of integers. Maybe that could work here. If I can find a period that's related to 'a' and 'b', I might be able to create a repeating pattern that avoids the conflicts.Let me think about the least common multiple (LCM) of 'a' and 'b'. The LCM would give me the smallest number that's a multiple of both 'a' and 'b'. If I can create a coloring pattern that repeats every LCM(a, b) numbers, maybe that would help. But wait, I only have three colors, so the period can't be too long, or else I might run out of colors.Alternatively, maybe I can use modular arithmetic. If I assign colors based on the remainder when divided by some number, that could create the necessary separation. For example, if I use modulo 3, each integer would be colored based on its remainder when divided by 3. But would that ensure that numbers 'a' or 'b' apart have different colors?Let me test this idea. Suppose I color each integer 'n' with color (n mod 3). Then, if I take two numbers that are 'a' apart, say 'n' and 'n + a', their colors would be (n mod 3) and ((n + a) mod 3). The difference in their colors would be (a mod 3). So, if 'a' is not a multiple of 3, then (a mod 3) would be 1 or 2, meaning their colors would differ. Similarly for 'b'. But if 'a' or 'b' is a multiple of 3, then (a mod 3) would be 0, and the colors would be the same, which is bad.So, this approach works only if neither 'a' nor 'b' is a multiple of 3. But the problem states that 'a' and 'b' are natural numbers, so they could be any positive integers, including multiples of 3. Therefore, this simple modulo 3 coloring isn't sufficient.Maybe I need a more sophisticated coloring scheme. Perhaps I can use a combination of modulo operations or some other function that takes into account both 'a' and 'b'. Let me think about the properties that such a coloring must satisfy.For any integer 'n', the numbers 'n + a' and 'n - a' must have different colors from 'n'. Similarly, 'n + b' and 'n - b' must also have different colors from 'n'. So, each number 'n' imposes constraints on the colors of 'n + a', 'n - a', 'n + b', and 'n - b'.This seems similar to graph coloring, where each integer is a vertex, and edges connect vertices that are 'a' or 'b' apart. Then, the problem reduces to coloring this graph with three colors. I wonder if this graph has a certain structure that allows for a three-coloring.In graph theory, a graph is 3-colorable if it doesn't contain any odd-length cycles. But I'm not sure if this particular graph has cycles or what their lengths are. It might be more complicated than that.Wait, maybe I can model this as a distance graph. A distance graph with distances 'a' and 'b' would have edges between any two integers that are 'a' or 'b' apart. The chromatic number of such a graph is the minimum number of colors needed to color the integers so that no two numbers at distance 'a' or 'b' share the same color.I recall that for distance graphs with two distances, the chromatic number can vary, but with three colors, it might be possible under certain conditions. Maybe I can find a specific coloring pattern that works regardless of 'a' and 'b'.Let me try to construct such a coloring. Suppose I divide the integers into three classes based on some repeating pattern. For example, I could alternate colors in a sequence like 1, 2, 3, 1, 2, 3, and so on. But would this work?If I color the integers in this repeating sequence, then any two numbers 'a' apart would have colors differing by 'a mod 3'. Similarly for 'b'. So, if 'a' is not a multiple of 3, their colors would differ, but if 'a' is a multiple of 3, they would have the same color, which is bad. Same issue as before.Maybe I need a different kind of pattern. What if I use a more complex repeating sequence that takes into account both 'a' and 'b'? For example, if I can find a period that is a multiple of both 'a' and 'b', say LCM(a, b), and then assign colors in a way that ensures no two numbers 'a' or 'b' apart share the same color within that period.But LCM(a, b) could be quite large, and I only have three colors. So, I need to make sure that within each period, the coloring is such that the constraints are satisfied.Alternatively, maybe I can use a coloring based on the residues modulo something related to 'a' and 'b'. For instance, if I choose a modulus that is larger than both 'a' and 'b', say 'a + b', and then assign colors based on that.Wait, let's think about this differently. Suppose I fix a starting point, say 0, and color it color 1. Then, I color 1, 2, ..., up to some number, ensuring that whenever I color a number, it doesn't conflict with the numbers 'a' or 'b' before it.This sounds like an inductive approach. Maybe I can use induction to color all integers, positive and negative, step by step, ensuring at each step that the coloring constraints are satisfied.Let me try to outline this:1. Start by coloring 0 with color 1.2. Move to 1. Check if 1 - a or 1 - b is already colored. If not, color it with color 1. If one of them is colored, choose a different color.3. Continue this process for each integer, ensuring that when coloring 'n', it doesn't conflict with 'n - a' or 'n - b'.4. Similarly, for negative integers, start from 0 and move left, coloring each integer while checking against 'n + a' and 'n + b'.But wait, this might not always work because as we go further out, the constraints could become too tight, especially if 'a' and 'b' are such that they create overlapping constraints.Maybe I need a more systematic way. What if I partition the integers into three sets where within each set, no two numbers are 'a' or 'b' apart? Then, each set can be assigned a unique color.To create such a partition, I need to ensure that for any number in a set, neither 'n + a' nor 'n + b' is in the same set. This sounds like creating independent sets in the graph I mentioned earlier.But how do I construct such independent sets? Maybe by using a coloring function that shifts colors in a way that avoids conflicts.Let me think about using a function that combines both 'a' and 'b'. For example, if I define the color of 'n' as (n mod 3), but adjust it based on some multiple of 'a' and 'b'. Hmm, not sure.Wait, another idea: if I can find a common difference that, when used to color the integers, ensures that 'a' and 'b' differences fall into different color classes. For example, if I color every third number, but shift the coloring in a way that 'a' and 'b' steps land in different colors.Alternatively, maybe I can use a coloring based on the floor division of 'n' by some number related to 'a' and 'b'. For instance, if I divide the integers into blocks of size 'k' and assign colors cyclically within each block.But I need to ensure that within each block, the coloring doesn't lead to conflicts when considering 'a' and 'b' differences. This might require careful selection of 'k'.Wait, perhaps I can use the concept of a "greedy" coloring. If I traverse the integers in order and assign the smallest available color that doesn't conflict with previously colored numbers 'a' or 'b' apart.But since the integers are infinite in both directions, this approach might not be straightforward. However, if I can find a periodic pattern that works, it could be applied uniformly across all integers.Let me try to formalize this. Suppose I define a coloring function f(n) that assigns one of three colors to each integer 'n'. I want f(n) ≠ f(n + a) and f(n) ≠ f(n + b) for all 'n'.One way to achieve this is to ensure that the coloring function has a period that is a multiple of both 'a' and 'b', say LCM(a, b). Then, within each period, I can assign colors in a way that satisfies the constraints.But with only three colors, the period can't be too large. Maybe I can find a way to tile the integers with a repeating pattern of three colors that avoids conflicts for both 'a' and 'b'.Alternatively, perhaps I can use a coloring based on the residues modulo 3, but adjusted by some function involving 'a' and 'b'. For example, f(n) = (n + c * a + d * b) mod 3, where 'c' and 'd' are constants chosen to ensure the coloring constraints.But I'm not sure if this would work for all 'a' and 'b'. It might require specific values of 'c' and 'd' that depend on 'a' and 'b', which isn't ideal since 'a' and 'b' are arbitrary.Wait, maybe I can use a more dynamic approach. Suppose I define the color of 'n' based on the colors of 'n - a' and 'n - b'. If I can ensure that the color of 'n' is different from both, then the constraints are satisfied.This sounds like a recurrence relation. Let's try to define it:f(n) ≠ f(n - a) and f(n) ≠ f(n - b)Given that we have three colors, as long as f(n - a) and f(n - b) are not the same, we can choose the third color for f(n). If f(n - a) and f(n - b) are the same, then we just need to choose a different color.But this approach requires that we can always choose a color different from f(n - a) and f(n - b). Since we have three colors, this is possible unless f(n - a) and f(n - b) are the same, in which case we have two choices.Wait, but what if f(n - a) and f(n - b) are the same? Then, f(n) can be either of the other two colors. So, as long as we have three colors, this should always be possible.But this is more of an inductive approach, assuming that we've already colored up to 'n - 1' and now need to color 'n'. However, since the integers are infinite in both directions, we need to ensure that this coloring works for all 'n', both positive and negative.Maybe I can start coloring from 0 and move outwards in both directions, ensuring that each step satisfies the constraints. Let's try to outline this:1. Color 0 with color 1.2. Color 1: Check 1 - a and 1 - b. Since a and b are natural numbers, 1 - a and 1 - b are negative or zero. If a = 1, then 1 - a = 0, which is already colored. So, color 1 must be different from color 0. Assign color 2 or 3. Let's choose color 2.3. Color 2: Check 2 - a and 2 - b. If a = 1, then 2 - a = 1, which is colored 2. So, color 2 must be different from 2. Assign color 1 or 3. Let's choose color 3.4. Color 3: Check 3 - a and 3 - b. If a = 1, then 3 - a = 2, which is colored 3. So, color 3 must be different from 3. Assign color 1 or 2. Let's choose color 1.5. Continue this process, ensuring that each new color is different from the colors 'a' and 'b' steps back.But this seems tedious and might not cover all cases, especially if 'a' and 'b' are larger. Also, what if 'a' and 'b' are such that they create overlapping constraints that make it impossible to color with three colors?Wait, I think the key here is that with three colors, as long as the graph is 3-colorable, which it should be for any 'a' and 'b', because the chromatic number of the distance graph with two distances is at most 3. But I'm not entirely sure about that.Alternatively, maybe I can use a more mathematical approach. Suppose I define the color of 'n' as (n mod 3). Then, for any 'n', n + a and n + b would have colors (n + a) mod 3 and (n + b) mod 3. If 'a' and 'b' are not multiples of 3, then these colors would differ from 'n's color. But if 'a' or 'b' is a multiple of 3, then n + a or n + b would have the same color as 'n', which is bad.So, to fix this, maybe I can adjust the coloring function to account for 'a' and 'b'. For example, if 'a' is a multiple of 3, I need a different approach. Maybe I can use a combination of modulo operations with different moduli.Wait, another idea: if I can find a common difference 'd' such that 'd' is not a multiple of 'a' or 'b', then I can color the integers in arithmetic progressions with difference 'd', each progression getting a different color. But I only have three colors, so I need to ensure that 'd' is chosen such that the colorings don't conflict.Alternatively, maybe I can use a coloring based on the floor of 'n' divided by some number related to 'a' and 'b'. For example, if I divide the integers into blocks of size 'k' and color each block with a repeating sequence of three colors.But I'm not sure how to choose 'k' to ensure that within each block, the coloring doesn't lead to conflicts for 'a' and 'b' differences.Wait, perhaps I can use a more advanced concept from combinatorics called "van der Waerden's theorem," which deals with arithmetic progressions and colorings. But I'm not sure if that's directly applicable here.Alternatively, maybe I can use a concept from group theory, treating the integers as a group under addition and defining a coloring that respects certain subgroups. But that might be overcomplicating things.Let me go back to the basics. I need to assign one of three colors to each integer such that no two integers 'a' or 'b' apart share the same color. This is equivalent to ensuring that the coloring is a proper coloring of the distance graph with distances 'a' and 'b'.I recall that for any set of distances, the chromatic number of the distance graph is at most 4, but I think for two distances, it might be lower. Maybe 3 colors are sufficient.To prove this, I can try to construct such a coloring explicitly. Let's attempt to define a coloring function f(n) that uses three colors and satisfies the conditions.One approach is to use a periodic coloring with period 3. For example, color integers in the pattern 1, 2, 3, 1, 2, 3, etc. Then, for any 'a' not divisible by 3, the colors of 'n' and 'n + a' will differ because (n + a) mod 3 ≠ n mod 3. Similarly for 'b'.However, if 'a' or 'b' is divisible by 3, then 'n + a' and 'n' would have the same color, which violates the condition. So, this simple periodic coloring doesn't work for all 'a' and 'b'.Maybe I can modify the periodic coloring to account for 'a' and 'b'. For example, choose a period that is a multiple of both 'a' and 'b', say LCM(a, b), and within each period, assign colors in a way that ensures no two numbers 'a' or 'b' apart share the same color.But with only three colors, the period can't be too large, or else we might not have enough colors to avoid conflicts. So, perhaps we need a smarter way to assign colors within the period.Alternatively, maybe I can use a coloring based on the residues modulo 3, but adjust it by some function that depends on 'a' and 'b'. For example, define f(n) = (n + k * a + m * b) mod 3, where 'k' and 'm' are chosen such that the coloring constraints are satisfied.But I'm not sure how to choose 'k' and 'm' to ensure this. It might require solving some congruences, which could be complicated.Wait, another idea: if I can find a function f(n) such that f(n + a) ≠ f(n) and f(n + b) ≠ f(n) for all 'n', then f(n) would be a valid coloring. One way to achieve this is to ensure that the function f(n) is not periodic with period 'a' or 'b'.But I'm not sure how to construct such a function with only three colors.Alternatively, maybe I can use a combination of two periodic functions with different periods. For example, use one function with period related to 'a' and another with period related to 'b', and combine them to get a three-coloring.But this might get too complex, and I'm not sure if it would work for all 'a' and 'b'.Wait, perhaps I can use a recursive approach. Start by coloring 0, then color 1, 2, etc., ensuring at each step that the coloring constraints are satisfied. Similarly, color the negative integers by moving left from 0.This is similar to the inductive approach I thought of earlier. Let's try to formalize this:1. Color 0 with color 1.2. For each positive integer 'n' from 1 onwards: a. Check the colors of 'n - a' and 'n - b' (if they exist). b. Assign the smallest color that is different from both.3. For each negative integer 'n' from -1 onwards: a. Check the colors of 'n + a' and 'n + b' (if they exist). b. Assign the smallest color that is different from both.Since we have three colors, as long as the colors of 'n - a' and 'n - b' are not the same, we can choose the third color. If they are the same, we have two choices. This should always be possible because with three colors, there's always at least one color different from the two previous ones.But I need to ensure that this process doesn't lead to a contradiction somewhere down the line. For example, if coloring 'n' affects the coloring of 'n + a' and 'n + b', which in turn affect further colorings, could this create a situation where a number has no available color?I think not, because at each step, we have three colors to choose from, and only two constraints (from 'n - a' and 'n - b'). As long as those two constraints don't force us into a situation where all three colors are excluded, which they can't because we have three colors and only two constraints, this should work.Therefore, by using this inductive approach, we can color all integers with three colors such that any two integers 'a' or 'b' apart have different colors.So, in summary, the strategy is to start coloring from 0 and move outwards in both directions, assigning colors in a way that ensures no two numbers 'a' or 'b' apart share the same color. Since we have three colors, this is always possible by choosing a color different from the two previous constraints at each step.

💡First, I need to identify the two numbers. Let's call the smaller number ( x ) and the larger number ( y ).According to the problem, the sum of the two numbers is 84. This gives me the equation:[x + y = 84]Additionally, one number is three times the other. Since ( y ) is the larger number, it must be three times ( x ):[y = 3x]Now, I can substitute ( y ) in the first equation with ( 3x ):[x + 3x = 84][4x = 84]Solving for ( x ):[x = frac{84}{4} = 21]Finally, to find ( y ), I multiply ( x ) by 3:[y = 3 times 21 = 63]Therefore, the larger number is 63.

💡Okay, so I have this problem about an equilateral triangle ABC with area 1. They mention points A', B', and C' which are midpoints of the sides BC, CA, and AB respectively. Then, A'', B'', and C'' are midpoints of B'C', C'A', and A'B' respectively. I need to find the area of the trapezoid BB''C''C and express it as a fraction m/n where m and n are coprime positive integers, then find m + n.Hmm, let me try to visualize this. Since ABC is equilateral, all sides are equal, and all angles are 60 degrees. The midpoints A', B', C' divide each side into two equal parts. Then, A'', B'', C'' are midpoints of the segments connecting these midpoints, so they're essentially quarter points of the original triangle.I think drawing a diagram might help, but since I can't draw, I'll try to imagine it. Let me consider the coordinates of the triangle. Maybe placing the triangle in a coordinate system will make it easier to calculate areas.Let’s assume point A is at (0, 0), point B is at (2, 0), and point C is at (1, √3). Wait, is that correct? Because an equilateral triangle with side length 2 would have coordinates like that. Let me check the area. The area of an equilateral triangle with side length s is (√3/4)s². If s = 2, then area is (√3/4)*4 = √3. But in the problem, the area is 1. So, maybe I need to adjust the side length.Let’s denote the side length as s. Then, area is (√3/4)s² = 1. Solving for s, s² = 4/√3, so s = 2/(3^(1/4)). Hmm, that seems complicated. Maybe instead of using coordinates, I can use ratios since we're dealing with midpoints and areas.Since A', B', C' are midpoints, the triangle A'B'C' is the medial triangle, which has 1/4 the area of ABC. So, area of A'B'C' is 1/4. Then, A'', B'', C'' are midpoints of B'C', C'A', and A'B', so triangle A''B''C'' is the medial triangle of A'B'C', so its area is 1/4 of A'B'C', which is 1/16 of ABC. So, area of A''B''C'' is 1/16.But wait, I need the area of trapezoid BB''C''C. Hmm, trapezoid is a quadrilateral with at least one pair of parallel sides. In this case, BB'' and CC'' might be the non-parallel sides, while BC and B''C'' are the two bases.Wait, actually, in the trapezoid BB''C''C, the sides BB'' and CC'' are the legs, and BC and B''C'' are the two bases. Since ABC is equilateral, BC is a side of the triangle, and B''C'' is a side of the smaller triangle A''B''C''.Since B'' and C'' are midpoints of B'C' and C'A', respectively, the segment B''C'' is parallel to BC and half its length. Wait, no, because B'C' is half the length of BC, so B''C'' is half of B'C', which is a quarter of BC. So, B''C'' is 1/4 the length of BC.Wait, but in terms of coordinates, maybe I can assign coordinates to each point.Let me place point A at (0, 0), point B at (2, 0), and point C at (1, √3). Then, the area of ABC is (√3/4)*(2)^2 = √3, but we need area 1. So, maybe I need to scale down the triangle.Let’s find the scaling factor. If the area is 1, and the area of ABC with side length 2 is √3, then the scaling factor k satisfies (k^2)*√3 = 1, so k = 1/√(√3) = 1/(3^(1/4)). Hmm, that's messy. Maybe instead, let me assign coordinates such that the area is 1.Alternatively, maybe using barycentric coordinates or area ratios would be better.Since all the points are midpoints, each time we take midpoints, the area gets divided by 4. So, starting from ABC with area 1, A'B'C' has area 1/4, then A''B''C'' has area 1/16.But trapezoid BB''C''C is not the entire triangle, but a part of it. Maybe I can find the area by subtracting smaller areas from the whole.Alternatively, perhaps I can divide the trapezoid into triangles and calculate their areas.Let me think about the structure. The trapezoid BB''C''C has vertices at B, B'', C'', and C. So, it's a quadrilateral with sides BB'', B''C'', C''C, and CB.Since B'' and C'' are midpoints of B'C' and C'A', respectively, and B'C' is half of BC, so B'C' is of length 1 (if BC is 2). Then, B'' is the midpoint of B'C', so B'' is 1/4 from B' and 1/4 from C'. Similarly, C'' is 1/4 from C' and 1/4 from A'.Wait, maybe I can use vectors or coordinate geometry.Let me assign coordinates to ABC such that A is at (0, 0), B is at (2, 0), and C is at (1, h), where h is the height. Since the area is 1, area = (base * height)/2 = 1. Base is 2, so (2 * h)/2 = h = 1. Therefore, h = 1. So, point C is at (1, 1).Wait, but in an equilateral triangle, the height is (√3/2)*side length. If the height is 1, then side length is 2/√3. Hmm, but if I set the coordinates as A(0, 0), B(2/√3, 0), and C(1/√3, 1), then the area would be 1.But maybe it's simpler to just assign coordinates with area 1 regardless of the side length.Alternatively, let me use barycentric coordinates with respect to triangle ABC.In barycentric coordinates, the midpoints have coordinates:A' is the midpoint of BC, so its coordinates are (0, 1/2, 1/2).Similarly, B' is (1/2, 0, 1/2), and C' is (1/2, 1/2, 0).Then, A'' is the midpoint of B'C', so we can find its coordinates by averaging B' and C'.So, B' is (1/2, 0, 1/2), C' is (1/2, 1/2, 0). The midpoint A'' would be ((1/2 + 1/2)/2, (0 + 1/2)/2, (1/2 + 0)/2) = (1/2, 1/4, 1/4).Similarly, B'' is the midpoint of C'A', which is midpoint of (1/2, 1/2, 0) and (0, 1/2, 1/2). So, midpoint is ((1/2 + 0)/2, (1/2 + 1/2)/2, (0 + 1/2)/2) = (1/4, 1/2, 1/4).Similarly, C'' is the midpoint of A'B', which is midpoint of (0, 1/2, 1/2) and (1/2, 0, 1/2). So, midpoint is ((0 + 1/2)/2, (1/2 + 0)/2, (1/2 + 1/2)/2) = (1/4, 1/4, 1/2).So, now we have coordinates for B'', C'', and C.Point B is (1, 0, 0), B'' is (1/4, 1/2, 1/4), C'' is (1/4, 1/4, 1/2), and C is (0, 0, 1).Wait, in barycentric coordinates, the area can be calculated using determinants.The area of a triangle with vertices (x1, y1, z1), (x2, y2, z2), (x3, y3, z3) is given by:Area = (1/2) |x1(y2z3 - y3z2) + x2(y3z1 - y1z3) + x3(y1z2 - y2z1)|But since we have a trapezoid, which is a quadrilateral, maybe we can split it into two triangles and sum their areas.So, trapezoid BB''C''C can be split into triangles BB''C and B''C''C.Let me calculate the area of triangle BB''C first.Points:B: (1, 0, 0)B'': (1/4, 1/2, 1/4)C: (0, 0, 1)Using the determinant formula:Area = (1/2) |1*( (1/2)*1 - (1/4)*0 ) + (1/4)*(0*0 - 0*1) + 0*(0*(1/4) - (1/2)*0)|Simplify:= (1/2) |1*(1/2 - 0) + (1/4)*(0 - 0) + 0*(0 - 0)|= (1/2) |1*(1/2) + 0 + 0| = (1/2)*(1/2) = 1/4Okay, so area of triangle BB''C is 1/4.Now, let's calculate the area of triangle B''C''C.Points:B'': (1/4, 1/2, 1/4)C'': (1/4, 1/4, 1/2)C: (0, 0, 1)Using the determinant formula again:Area = (1/2) |(1/4)*( (1/4)*1 - (1/2)*0 ) + (1/4)*(0*1 - 0*(1/4)) + 0*( (1/2)*(1/4) - (1/4)*(1/4) )|Simplify:= (1/2) |(1/4)*(1/4 - 0) + (1/4)*(0 - 0) + 0*( (1/8) - (1/16) )|= (1/2) |(1/4)*(1/4) + 0 + 0| = (1/2)*(1/16) = 1/32So, area of triangle B''C''C is 1/32.Therefore, the total area of trapezoid BB''C''C is 1/4 + 1/32 = 8/32 + 1/32 = 9/32.So, m = 9, n = 32, which are coprime. Therefore, m + n = 41.**Final Answer**boxed{41}

💡First, I need to understand the relationship between the size of the cube and its weight. Since the cube is made of the same metal, the density remains constant. The weight of an object is determined by its volume and density.The original cube has a side length of ( s ) units, so its volume is ( s^3 ) cubic units. Given that the weight of the original cube is 5 pounds and the density is ( rho ) pounds per cubic unit, I can set up the equation:[ 5 = rho times s^3 ]Next, the larger cube has sides that are three times as long, making its side length ( 3s ) units. The volume of the larger cube is ( (3s)^3 = 27s^3 ) cubic units.To find the weight of the larger cube, I use the same density ( rho ):[ text{Weight of larger cube} = rho times 27s^3 ]Substituting ( rho times s^3 ) with 5 from the original cube's equation:[ text{Weight of larger cube} = 27 times 5 = 135 ]Therefore, the larger cube will weigh 135 pounds.

💡Alright, so I have this problem about a function ( f ) that's defined on all real numbers. It satisfies two symmetry conditions: ( f(2 - x) = f(2 + x) ) for all ( x ), and ( f(7 - x) = f(7 + x) ) for all ( x ). Also, I know that ( x = 0 ) is a root, meaning ( f(0) = 0 ). I need to find the minimum number of roots ( f(x) = 0 ) has in the interval ([-1000, 1000]). First, let me try to understand what these symmetry conditions mean. The first one, ( f(2 - x) = f(2 + x) ), suggests that the function is symmetric around the line ( x = 2 ). Similarly, ( f(7 - x) = f(7 + x) ) means the function is also symmetric around ( x = 7 ). So, the function has two lines of symmetry: one at ( x = 2 ) and another at ( x = 7 ).Since the function is symmetric around both ( x = 2 ) and ( x = 7 ), I wonder if this implies some periodicity. If a function has two different lines of symmetry, sometimes it can lead to periodic behavior. Let me explore that.Given that ( f ) is symmetric about ( x = 2 ), if I take any point ( x ), its mirror image across ( x = 2 ) is ( 4 - x ). So, ( f(x) = f(4 - x) ). Similarly, since it's symmetric about ( x = 7 ), for any point ( x ), its mirror image across ( x = 7 ) is ( 14 - x ). So, ( f(x) = f(14 - x) ).Now, if I combine these two symmetries, let's see what happens. Starting from ( f(x) = f(4 - x) ) and substituting ( x ) with ( 14 - x ), I get ( f(14 - x) = f(4 - (14 - x)) = f(x - 10) ). But from the second symmetry, ( f(14 - x) = f(x) ). Therefore, ( f(x) = f(x - 10) ). This suggests that the function is periodic with period 10. So, ( f(x) = f(x + 10) ) for all ( x ). That means the function repeats every 10 units. If I know the behavior of ( f ) on an interval of length 10, I can know it everywhere else.Given that ( f(0) = 0 ), and since the function is periodic with period 10, it must be that ( f(10) = 0 ), ( f(20) = 0 ), and so on. Similarly, going in the negative direction, ( f(-10) = 0 ), ( f(-20) = 0 ), etc. So, the roots of ( f(x) = 0 ) are at all integer multiples of 10.But wait, is that all? Let me check. Since the function is symmetric about both ( x = 2 ) and ( x = 7 ), maybe there are more roots. For example, if ( f(0) = 0 ), then due to symmetry about ( x = 2 ), ( f(4) = 0 ) as well. Similarly, due to symmetry about ( x = 7 ), ( f(14) = 0 ). But since the function is periodic with period 10, ( f(14) = f(4) ), which is already 0. So, does this mean that within each period, there are two roots: one at ( x = 0 ) and another at ( x = 4 )?Wait, but if the function is periodic with period 10, then the roots would be at ( x = 0 + 10k ) and ( x = 4 + 10k ) for all integers ( k ). So, in each interval of length 10, there are two roots: one at the start and one 4 units later.But let me verify this. If ( f(0) = 0 ), then by symmetry about ( x = 2 ), ( f(4) = 0 ). Then, by periodicity, ( f(10) = 0 ), and by symmetry about ( x = 7 ), ( f(10) = f(14) ), so ( f(14) = 0 ). But ( f(14) = f(4) ) because of periodicity, which is already 0. So, it seems that the roots are at ( x = 0, 4, 10, 14, 20, 24, ) etc. So, in each 10-unit interval, there are two roots: one at the multiple of 10 and another 4 units after that.Therefore, in the interval ([-1000, 1000]), how many such roots are there?First, let's find how many periods of 10 are in 2000 units (from -1000 to 1000). That's 2000 / 10 = 200 periods. But since the interval is inclusive, we need to consider both endpoints. So, the number of roots would be 200 periods * 2 roots per period = 400 roots. But wait, we also have the root at 0, which is counted once. So, does that mean we have 401 roots?Wait, let's think carefully. The interval from -1000 to 1000 includes 2000 units, but since it's inclusive, we have to consider both endpoints. Each period of 10 units contributes two roots, but the starting point of the interval, -1000, is a multiple of 10, so it's a root. Similarly, 1000 is a multiple of 10, so it's a root. So, how many roots are there?Let me list the roots:Starting from -1000, the roots are at -1000, -996, -990, -986, ..., 0, ..., 986, 990, 996, 1000.Wait, so between -1000 and 1000, how many roots are there?Each 10-unit interval has two roots: one at the multiple of 10 and one 4 units after. So, from -1000 to -990, the roots are -1000 and -996. From -990 to -980, the roots are -990 and -986, and so on.So, in total, how many such intervals are there? From -1000 to 1000, that's 2000 units, so 200 intervals of 10 units each. Each interval contributes two roots, so 200 * 2 = 400 roots. But wait, the root at 1000 is included in the last interval, so we don't double count. So, total roots would be 400.But wait, let's check the exact count. Let's consider the roots:Starting from -1000, the roots are at -1000, -996, -990, -986, ..., 0, ..., 986, 990, 996, 1000.So, how many terms are in this sequence?The sequence is an arithmetic sequence with first term -1000, last term 1000, and common difference 4 units? Wait, no, the roots are at every 4 units starting from -1000? Wait, no, the roots are at multiples of 10 and multiples of 10 plus 4.Wait, actually, the roots are at x = 10k and x = 10k + 4 for all integers k.So, in the interval [-1000, 1000], the roots are:For x = 10k: k ranges from -100 to 100, inclusive. That's 201 roots.For x = 10k + 4: k ranges from -100 to 99, because 10*100 + 4 = 1004, which is outside the interval. So, k from -100 to 99 is 200 roots.So, total roots: 201 + 200 = 401.Ah, so that's where the 401 comes from. So, the minimum number of roots is 401.But wait, is there a possibility that some roots overlap? For example, could 10k = 10m + 4 for some integers k and m? That would mean 10(k - m) = 4, which implies 5(k - m) = 2, which is impossible since k and m are integers. So, no overlap. Therefore, all roots are distinct.Therefore, the minimum number of roots N is 401.**Final Answer**The minimum value of ( N ) is boxed{401}.