🧠:Okay, so I need to prove that after replacing a lost 2x2 tile with a 1x4 tile, it's impossible to cover the bottom of the rectangular box. Hmm, let me start by understanding the problem thoroughly.First, the original tiling used a combination of 2x2 and 1x4 tiles. Then, one 2x2 tile was removed and replaced with a 1x4 tile. The claim is that now the floor cannot be tiled properly. I need to show why this replacement makes tiling impossible.Maybe I should think about tiling problems in general. Often, these kinds of problems involve parity arguments or coloring. Like, sometimes people use checkerboard colorings to show that certain tilings aren't possible. Let me try that approach here.Let me imagine the rectangular box. Since it was originally covered with 2x2 and 1x4 tiles, its dimensions must be such that they can accommodate both tile types. But maybe the specific dimensions aren't given? Wait, the problem just says it's a rectangular box, so the bottom is a rectangle. Hmm, but we don't know the exact size. Maybe the argument is general regardless of the size, as long as the original tiling was possible with those tiles.Wait, but if the original tiling used both 2x2 and 1x4 tiles, then the area of the rectangle must be a multiple of 4, since each 2x2 tile covers 4 units and each 1x4 covers 4 units as well. So the total area is 4k for some integer k. When we remove a 2x2 tile (area 4) and replace it with a 1x4 tile (area 4), the total area remains the same. So the area isn't the problem here. It must be something else, like the shape or arrangement.Alright, so area is preserved, but tiling becomes impossible. That suggests a more subtle issue, perhaps related to the way the tiles fit together. Let me try coloring the rectangle in a checkerboard pattern. For a standard checkerboard, each 2x2 tile will always cover exactly two black and two white squares. Similarly, a 1x4 tile, if placed horizontally or vertically, will cover either two black and two white squares or... Wait, let's check.In a checkerboard coloring, each 1x4 tile will cover either two black and two white squares or, depending on its position, maybe alternate colors. Let's see: a 1x4 tile placed horizontally on a checkerboard. If it starts on a black square, the colors will be black, white, black, white. So two black and two white. Similarly, if placed vertically, starting on a black square, each row alternates, so over four rows, the colors would be black, white, black, white. Again, two of each. So both the 2x2 and 1x4 tiles cover exactly two black and two white squares each.Therefore, in the original tiling, the entire rectangle must have an equal number of black and white squares. When we remove a 2x2 tile, we are removing two black and two white squares. Then, replacing it with a 1x4 tile, which also covers two black and two white squares. So the total number of black and white squares remains balanced. Therefore, a standard checkerboard coloring doesn't seem to give a contradiction here. Hmm, so maybe this approach isn't sufficient.Alternatively, maybe a different coloring. Sometimes, for 1x4 tiles, people use more complex colorings, like a four-coloring. Let me think. Suppose instead of two colors, we use four colors in a repeating pattern. For example, in each 2x2 block, we can have colors arranged as:1 23 4Then, repeating this pattern across the rectangle. Let me see how the tiles would cover these colors. A 2x2 tile would cover one of each color: 1, 2, 3, 4. A 1x4 tile, if placed horizontally, would cover four consecutive squares. Depending on where it starts, it might cover colors 1, 2, 1, 2 or 3, 4, 3, 4 if aligned vertically. Wait, actually, if the four-coloring is in 2x2 blocks, then horizontally a 1x4 tile would pass through two 2x2 blocks. For example, starting at column 1, it would cover color 1 (from the first 2x2 block), color 2 (still first block), color 1 (next block), color 2 (next block). Similarly, if starting at column 3, same thing. Hmm, so depending on the alignment, maybe a 1x4 tile would cover two 1s and two 2s or two 3s and two 4s?Wait, let me visualize the 4-coloring. If each 2x2 block is colored as:1 23 4Then, the next 2x2 block to the right would be:1 23 4But wait, actually, if it's a repeating 2x2 pattern, then the entire grid would have 1 2 1 2... in the first row, 3 4 3 4... in the second row, then 1 2 1 2... in the third, etc. Wait, no, that's a different coloring. Wait, perhaps I need to tile the entire grid with 2x2 blocks each colored as:1 23 4So in a 4x4 grid, it would be:1 2 1 23 4 3 41 2 1 23 4 3 4Wait, but this is a 2x2 tiling of colors. So each 2x2 tile would exactly cover one of each color. A 1x4 tile placed horizontally would cover two 1s and two 2s, or two 3s and two 4s, depending on the row. Similarly, a vertical 1x4 tile would cover two 1s and two 3s, or two 2s and two 4s. So replacing a 2x2 tile (which covers one of each color) with a 1x4 tile (which covers two of two colors) might change the balance of colors.In the original tiling, each 2x2 tile covers one of each color, so the number of each color is equal. Each 1x4 tile in the original tiling, whether horizontal or vertical, would also cover two pairs of colors, but since there might be multiple 1x4 tiles, maybe the total counts balance out. However, when we remove a 2x2 tile and replace it with a 1x4 tile, perhaps we create an imbalance in the color counts.Let me calculate. Suppose the original tiling had m 2x2 tiles and n 1x4 tiles. Then, the total area is 4m + 4n = 4(m + n), which is the area of the rectangle. Each 2x2 tile contributes 1 of each color (since it covers one 1, 2, 3, 4). Each 1x4 tile, depending on its orientation, contributes either two 1s and two 2s (horizontal in the first row), two 3s and two 4s (horizontal in the second row), two 1s and two 3s (vertical in the first column), or two 2s and two 4s (vertical in the second column). So if all 1x4 tiles are placed in the same orientation, they would contribute two pairs of colors each. However, in the original tiling, the total number of each color (1, 2, 3, 4) must be equal because the rectangle is perfectly covered by these tiles. Therefore, the original configuration must have had a balance where the number of 1s, 2s, 3s, and 4s are equal.But when we remove a 2x2 tile, we remove one of each color. Then, replacing it with a 1x4 tile, which adds two of two colors. For example, if the 1x4 tile is placed horizontally in the first row, it would add two 1s and two 2s. Therefore, the counts would be: color 1 increases by 2 - 1 = +1, color 2 increases by 2 - 1 = +1, color 3 remains -1 (since we removed one and didn't add any), and color 4 remains -1. So now, color 1 and 2 have one more each, and colors 3 and 4 have one less each. Thus, the color counts are no longer equal. Therefore, the tiling is impossible because the new tile introduces an imbalance in the color counts that cannot be compensated for by the remaining tiles.Is this a valid argument? Let me check again. If the original tiling had equal numbers of each color, then after removing a 2x2 tile, each color is decreased by 1. Then, adding a 1x4 tile, which (depending on orientation) adds two of two colors. If we choose horizontal in the top row, adding two 1s and two 2s. Then, compared to the original counts: color 1: original -1 +2 = +1, color 2: original -1 +2 = +1, color 3: original -1 +0 = -1, color 4: original -1 +0 = -1. Therefore, color counts are now (original +1, original +1, original -1, original -1). Thus, the numbers of colors 1 and 2 are now higher by 1 each, and 3 and 4 lower by 1 each. Since all tiles must cover colors in specific amounts, this imbalance would make it impossible to tile the remaining area, as any tile placed would need to maintain the balance, but the remaining tiles can't adjust for this discrepancy.Alternatively, if the replacement 1x4 tile is placed vertically, it would affect two other colors. For example, placed vertically in the first column, it would cover two 1s and two 3s. Then, color 1: -1 +2 = +1, color 3: -1 +2 = +1, color 2: -1 +0 = -1, color 4: -1 +0 = -1. Same kind of imbalance, just different colors.Therefore, regardless of the orientation of the replacement 1x4 tile, it creates an imbalance in the four-coloring, making it impossible to tile the remaining area with the existing tiles, as all remaining tiles (both 2x2 and 1x4) require covering exactly one of each color (for 2x2) or two of two colors (for 1x4). However, with an imbalance, you can't cover the remaining squares without either leaving some colors uncovered or overcovering others.Therefore, this four-coloring argument shows that replacing a 2x2 tile with a 1x4 tile creates a color imbalance that cannot be resolved with the available tiles, hence making it impossible to tile the rectangle.Alternatively, maybe there's another way to see this. Let me think about tiling the rectangle with dominoes or something, but the key is the tile types here. 2x2 and 1x4. The 2x2 is kind of a square block, while 1x4 is a long, skinny tile. Replacing a square with a long tile might disrupt some necessary parity.Wait, but the four-coloring argument seems solid. Let me verify with an example. Suppose the original rectangle is 4x4. Suppose it was tiled with four 2x2 tiles. If we remove one 2x2 tile and replace it with a 1x4 tile. Let's see.Original tiling: four 2x2 tiles. The four-coloring would give four of each color (since each 2x2 tile has one of each color). After removing one 2x2 tile, we have three of each color. Then adding a 1x4 tile. If we place the 1x4 tile horizontally in the first row, covering two 1s and two 2s. Now, color counts: 1: 3 -1 +2 = 4, 2: 3 -1 +2 = 4, 3: 3 -1 +0 = 2, 4: 3 -1 +0 = 2. So colors 1 and 2 have 4 each, 3 and 4 have 2 each. But the total area is 16, which is 4x4. But with the remaining area (16 -4 +4 =16), but the color counts are unequal. To tile the remaining area, which has 4,4,2,2 of colors 1,2,3,4. The existing tiles: three 2x2 tiles and some 1x4 tiles (assuming original was four 2x2 tiles, now three 2x2 and one 1x4). Wait, no, in this case, the original was all 2x2 tiles, and after replacement, it's three 2x2 tiles and one 1x4 tile. But the problem states that they had to replace a lost 2x2 with a 1x4, so now they have the same number of tiles as before? Wait, no: originally, there were some number of 2x2 and 1x4 tiles. Then one 2x2 was removed and a 1x4 was added. So the total number of tiles changes, but the area remains the same. Wait, but in my example, the 4x4 grid originally had four 2x2 tiles. After replacement, it's three 2x2 tiles and one 1x4 tile. The total area is 3*4 +1*4=16, which matches.But in this case, can we tile the 4x4 grid with three 2x2 tiles and one 1x4 tile? Let's see. Suppose we remove a 2x2 tile from the corner, and try to replace it with a 1x4 tile. Let me try to visualize:Imagine the 4x4 grid. If I take out the top-left 2x2 tile, leaving a 2x2 hole. But we are replacing it with a 1x4 tile. Wait, a 1x4 tile can't fill a 2x2 hole. So maybe in the original tiling, there was a mix of tiles. For example, maybe the 4x4 grid was tiled with two 2x2 tiles and two 1x4 tiles. Let me check: two 2x2 tiles cover 8 area, two 1x4 tiles cover 8, total 16. Is that possible?Yes, for example, place two 2x2 tiles in the top-left and bottom-right, and then two horizontal 1x4 tiles in the remaining spaces. But if we remove one 2x2 tile and replace with a 1x4 tile, then we have one 2x2 tile and three 1x4 tiles. Can we tile the 4x4 grid with that? Let me try.Suppose the original tiling was two 2x2 tiles and two 1x4 tiles. Remove one 2x2 tile (area 4), add a 1x4 tile (area 4), so total tiles: one 2x2 and three 1x4 tiles. Total area still 16. But can we arrange them?Let me attempt to place the remaining 2x2 tile somewhere. Suppose we place the 2x2 tile in the top-left corner. Then we have three 1x4 tiles. The remaining area is 12 units, which is three 1x4 tiles. But arranging three 1x4 tiles in the remaining 12 units might not be possible. For example, the remaining area after placing the 2x2 tile is a 4x4 minus 2x2, which is an L-shape of 12 squares. Can an L-shape be tiled with three 1x4 tiles? Probably not, since the L-shape has 6 units along one side and 6 along the other, but 1x4 tiles are straight. Maybe it's possible or not, depending on the exact shape. But perhaps in this case, it's impossible. But this is just an example; the original problem is general.But the four-coloring argument seems more general. In the example above, using the four-coloring, after replacement, we have color counts that are imbalanced, making tiling impossible. So regardless of the original configuration, replacing a 2x2 with a 1x4 will create this imbalance.Alternatively, maybe another approach: consider the original tiling. Each 2x2 tile can be thought of as covering a square block, which is even in both dimensions. The 1x4 tiles are either 1x4 or 4x1, so they cover a line of four squares. When you replace a 2x2 tile with a 1x4 tile, you are introducing a tile that is longer in one dimension. Perhaps this affects some parity in the rows or columns.Suppose we look at the number of tiles crossing certain boundaries. For example, in a tiling with dominoes, sometimes you consider the number of tiles crossing a certain line. Maybe here, we can use a similar approach. If we divide the rectangle into regions and show that the replacement tile changes the parity or makes it impossible to cover.Alternatively, think about the fact that a 2x2 tile is more "balanced" in covering both horizontal and vertical space, while a 1x4 tile is more directional. Maybe the problem arises in the fact that the 1x4 tile forces a certain alignment that can't be matched by the remaining tiles.Wait, going back to the four-coloring. Suppose we use four colors as described, repeating every 2x2 block. Then, in the original tiling, each 2x2 tile covers one of each color, and each 1x4 tile covers two of two colors. So the total number of each color must be equal. After replacement, two colors are overrepresented and two are underrepresented. Since all remaining tiles (both 2x2 and 1x4) can only cover either balanced colors (2x2) or pairs (1x4), you can't compensate for the imbalance. For example, every time you place a 1x4 tile, you must cover two of two colors, but if two colors are already in deficit, you can't use tiles that cover those colors without worsening the deficit.Therefore, the four-coloring argument shows that the replacement makes the color counts unequal, which makes tiling impossible. Therefore, this should be the way to prove it.To formalize this, we can:1. Define a four-coloring of the rectangle as described.2. Show that each original tile (2x2 or 1x4) covers either one of each color or two of two colors.3. Note that the original tiling had an equal number of each color.4. Removing a 2x2 tile removes one of each color.5. Adding a 1x4 tile adds two of two colors, creating an imbalance.6. Argue that no combination of remaining tiles can correct this imbalance, hence tiling is impossible.Therefore, this proves that after the replacement, tiling is impossible.

✅:To prove that replacing a lost (2 times 2) tile with a (1 times 4) tile makes it impossible to cover the box's bottom, we use a four-coloring argument:1. Four-Coloring Scheme: Tile the rectangle with a repeating (2 times 2) color pattern: [ begin{array}{|c|c|} hline 1 & 2 hline 3 & 4 hline end{array} ] Each (2 times 2) tile covers one of each color ((1, 2, 3, 4)), and each (1 times 4) tile (horizontal or vertical) covers two of two colors (e.g., two (1)s and two (2)s if placed horizontally in the first row).2. Original Balance: In the original tiling, the number of each color is equal because (2 times 2) tiles balance all colors, and (1 times 4) tiles uniformly distribute pairs.3. Effect of Replacement: - Removing a (2 times 2) tile decreases each color count by (1). - Adding a (1 times 4) tile increases two colors by (2) each. For example, replacing with a horizontal (1 times 4) tile results in: - Colors (1) and (2): (+1) each (from (-1 + 2)). - Colors (3) and (4): (-1) each (from (-1 + 0)). 4. Imbalance: The new color counts are unequal (two colors exceed the original count, two fall short). All remaining tiles ((2 times 2) and (1 times 4)) strictly preserve or exacerbate this imbalance: - (2 times 2) tiles require one of each color. - (1 times 4) tiles cover two of two colors, unable to address the deficit in the underrepresented colors. Conclusion: The color imbalance caused by replacing a (2 times 2) tile with a (1 times 4) tile cannot be resolved, making it impossible to tile the rectangle. (boxed{text{Replacement creates an irreparable color imbalance, preventing a valid tiling.}})

🔑:1. Consider the Box as a Grid of Unit Squares: Let's assume that the bottom of the rectangular box is divided into unit squares. 2. Coloring the Grid: To simplify the problem, we will use a checkerboard coloring of the grid. Specifically: - Alternate the squares such that each square (i,j) is colored black if i + j is even and white if i + j is odd.3. Tiles and Covered Squares: - Each 2 times 2 tile covers exactly two white squares and two black squares. - Each 1 times 4 tile covers either zero or two squares of the same color and two or four squares of the opposite color (e.g., all white or all black squares if laid horizontally, or alternating black and white squares if laid vertically).4. Even Number of Black and White Squares: Since we are tiling the entire bottom of the box, the number of white squares and the number of black squares must be equal.5. Changes Upon Tile Replacement: Suppose we had an original tiling that completely covered the bottom of the box. When we lose one 2 times 2 tile, we lose exactly one black square and one white square.6. Replacing the Lost Tile: - Replacing it with a 1 times 4 tile instead: - If the 1 times 4 tile is positioned horizontally, it covers exactly two squares of one color (either black or white) and two squares of the other color. - If the 1 times 4 tile is placed vertically, it still does not maintain the balance of colors if it is placed such that it alters the even distribution and overlaps.7. Consequence of the Replacement: After the replacement of the 2 times 2 tile with a 1 times 4 tile, the overall parity and balance of black and white squares covered are altered such that there is no longer an equal number of white and black squares. 8. Conclusion: Therefore, it is impossible to now tile the bottom of the box completely with the remaining tiles as the even distribution balance is broken. Given this logic, we conclude that after the loss and replacement, the tiling proposed cannot fill the box floor effectively: [ boxed{text{It is impossible to re-tile the box floor completely with the remaining tiles.}} ]

🧠:Okay, so I need to simplify this complex algebraic expression. Let me first write it down step by step to make sure I understand the structure. The original expression is:[frac{(a-b)^{2}+ab}{(a+b)^{2}-ab} div frac{a^{5}+b^{5}+a^{2}b^{3}+a^{3}b^{2}}{left(a^{3}+b^{3}+a^{2}b+ab^{2}right) left(a^{3}-b^{3}right)}]Alright, division of two fractions. Remember that dividing by a fraction is the same as multiplying by its reciprocal. So, first, maybe I should rewrite the division as multiplication by flipping the second fraction. Let me do that.So, the expression becomes:[frac{(a-b)^{2}+ab}{(a+b)^{2}-ab} times frac{left(a^{3}+b^{3}+a^{2}b+ab^{2}right) left(a^{3}-b^{3}right)}{a^{5}+b^{5}+a^{2}b^{3}+a^{3}b^{2}}]Now, my task is to simplify this. To do that, I need to factor both numerators and denominators as much as possible and then cancel common factors. Let me start by looking at each part individually.First, let's handle the first fraction: (frac{(a-b)^{2}+ab}{(a+b)^{2}-ab}).Let me compute the numerator and denominator separately.Numerator: ((a - b)^2 + ab). Let's expand ((a - b)^2):[(a - b)^2 = a^2 - 2ab + b^2]So adding ab to that:[a^2 - 2ab + b^2 + ab = a^2 - ab + b^2]Denominator: ((a + b)^2 - ab). Let's expand ((a + b)^2):[(a + b)^2 = a^2 + 2ab + b^2]Subtracting ab:[a^2 + 2ab + b^2 - ab = a^2 + ab + b^2]So the first fraction simplifies to (frac{a^2 - ab + b^2}{a^2 + ab + b^2}). Hmm, interesting. So that part is simplified.Now, moving on to the second fraction's numerator: (left(a^{3}+b^{3}+a^{2}b+ab^{2}right) left(a^{3}-b^{3}right)).Let me first factor each of these terms. Let's look at (a^{3} + b^{3} + a^{2}b + ab^{2}).Wait, (a^3 + b^3) is a sum of cubes, which factors as ((a + b)(a^2 - ab + b^2)). The other part is (a^2b + ab^2 = ab(a + b)). So combining these:[(a^3 + b^3) + (a^2b + ab^2) = (a + b)(a^2 - ab + b^2) + ab(a + b) = (a + b)(a^2 - ab + b^2 + ab) = (a + b)(a^2 + b^2)]Oh, that's a nice simplification! So (a^{3}+b^{3}+a^{2}b+ab^{2} = (a + b)(a^2 + b^2)).Then the other factor in the numerator is (a^{3} - b^{3}), which is a difference of cubes. That factors as ((a - b)(a^2 + ab + b^2)).So putting it together, the numerator of the second fraction is:[(a + b)(a^2 + b^2) times (a - b)(a^2 + ab + b^2)]So multiplying those together:[(a + b)(a - b)(a^2 + b^2)(a^2 + ab + b^2)]Note that ((a + b)(a - b)) is (a^2 - b^2), but maybe we don't need to combine them yet.Now, the denominator of the second fraction is (a^{5}+b^{5}+a^{2}b^{3}+a^{3}b^{2}). Let me see if this can be factored.First, let me arrange the terms:(a^5 + a^3b^2 + a^2b^3 + b^5)Notice that (a^5 + b^5) is a sum of fifth powers, but I don't recall a standard formula for that. However, maybe we can factor by grouping.Looking at the terms:Group as (a^5 + a^3b^2) + (a^2b^3 + b^5)Factor each group:First group: a^3(a^2 + b^2)Second group: b^3(a^2 + b^2)So, factoring out (a^2 + b^2):[(a^3 + b^3)(a^2 + b^2)]Wait, but that's not exactly correct. Wait, let's check:First group: a^3(a^2 + b^2)Second group: b^3(a^2 + b^2)Therefore, factoring out (a^2 + b^2):[(a^3 + b^3)(a^2 + b^2) = (a + b)(a^2 - ab + b^2)(a^2 + b^2)]Ah, right! So the denominator factors as:[(a + b)(a^2 - ab + b^2)(a^2 + b^2)]Therefore, the denominator is ( (a + b)(a^2 - ab + b^2)(a^2 + b^2) ).So now, let's put together the second fraction:Numerator: ((a + b)(a - b)(a^2 + b^2)(a^2 + ab + b^2))Denominator: ((a + b)(a^2 - ab + b^2)(a^2 + b^2))So, when we write the second fraction as numerator over denominator, we have:[frac{(a + b)(a - b)(a^2 + b^2)(a^2 + ab + b^2)}{(a + b)(a^2 - ab + b^2)(a^2 + b^2)}]Now, let's cancel common factors. Both numerator and denominator have (a + b), (a^2 + b^2), so those cancel out.So we are left with:[frac{(a - b)(a^2 + ab + b^2)}{(a^2 - ab + b^2)}]Therefore, the second fraction simplifies to (frac{(a - b)(a^2 + ab + b^2)}{a^2 - ab + b^2}).Now, going back to the entire expression. Remember, the original expression after flipping the second fraction was:First fraction: (frac{a^2 - ab + b^2}{a^2 + ab + b^2})Multiplied by the second simplified fraction: (frac{(a - b)(a^2 + ab + b^2)}{a^2 - ab + b^2})So, multiplying these two fractions together:[frac{a^2 - ab + b^2}{a^2 + ab + b^2} times frac{(a - b)(a^2 + ab + b^2)}{a^2 - ab + b^2}]Looking at this, notice that (a^2 - ab + b^2) in the numerator of the first fraction cancels with the same term in the denominator of the second fraction. Similarly, (a^2 + ab + b^2) in the denominator of the first fraction cancels with the same term in the numerator of the second fraction.So after cancellation, we are left with:[(a - b)]Wait, so does that mean the entire expression simplifies to just (a - b)? Let me verify this step by step to make sure I didn't make a mistake.First, confirm the first fraction's simplification:Original first fraction:Numerator: ((a - b)^2 + ab = a^2 - 2ab + b^2 + ab = a^2 - ab + b^2)Denominator: ((a + b)^2 - ab = a^2 + 2ab + b^2 - ab = a^2 + ab + b^2)So yes, first fraction is (frac{a^2 - ab + b^2}{a^2 + ab + b^2})Then the second fraction after factoring:Numerator became ((a + b)(a - b)(a^2 + b^2)(a^2 + ab + b^2))Denominator became ((a + b)(a^2 - ab + b^2)(a^2 + b^2))Canceling common terms gives (frac{(a - b)(a^2 + ab + b^2)}{a^2 - ab + b^2}), correct.Then multiplying the two fractions:[frac{a^2 - ab + b^2}{a^2 + ab + b^2} times frac{(a - b)(a^2 + ab + b^2)}{a^2 - ab + b^2}]Cancel (a^2 - ab + b^2) top and bottom, and (a^2 + ab + b^2) top and bottom, leaving just (a - b). That seems correct.Therefore, the simplified expression is (a - b).Wait, but let me double-check with specific numbers to ensure I didn't make a mistake.Let me choose a = 2 and b = 1.Compute original expression:First fraction: [(2 - 1)^2 + 2*1] / [(2 + 1)^2 - 2*1] = [1 + 2] / [9 - 2] = 3/7Second fraction: [2^5 +1^5 +2^2*1^3 +2^3*1^2] / [ (2^3 +1^3 +2^2*1 +2*1^2)(2^3 -1^3) ]Compute numerator of second fraction: 32 + 1 + 4*1 + 8*1 = 32 + 1 + 4 + 8 = 45Denominator: (8 + 1 + 4 + 2)(8 - 1) = (15)(7) = 105So the second fraction is 45 / 105 = 3/7Therefore, original expression is (3/7) ÷ (3/7) = 1.But according to our simplification, it should be a - b = 2 - 1 = 1. That matches. So, with a=2, b=1, the original expression equals 1, and our simplified expression also gives 1. So that's correct.Another test case: a=3, b=2.First fraction: [(3-2)^2 + 3*2]/[(3+2)^2 -3*2] = [1 + 6]/[25 -6] =7/19Second fraction numerator: 3^5 +2^5 +3^2*2^3 +3^3*2^2 =243 +32 +9*8 +27*4=243+32+72+108=243+32=275; 275+72=347; 347+108=455Denominator: (3^3 +2^3 +3^2*2 +3*2^2)(3^3 -2^3) = (27 +8 +9*2 +3*4)(27 -8)= (35 +18 +12)(19)= (65)(19)=1235Thus second fraction is 455/1235. Simplify: divide numerator and denominator by 5: 91/247. Wait, 455 ÷5=91, 1235 ÷5=247. Then 91/247. Let's see if it can be reduced. 91=13*7, 247=13*19. So divide by 13: 7/19. So second fraction is 7/19.Thus the original expression is (7/19) ÷ (7/19)=1. According to our simplified expression, a - b =3-2=1. Correct again.So, the simplification seems to hold in these cases. Therefore, the final simplified expression is indeed (a - b).But just to be thorough, let me check a case where a ≠ b +1. Let's take a=5, b=3.Original expression:First fraction: [(5-3)^2 +5*3]/[(5+3)^2 -5*3] = [4 +15]/[64 -15] =19/49Second fraction numerator:5^5 +3^5 +5^2*3^3 +5^3*3^2=3125 +243 +25*27 +125*9=3125+243=3368; 25*27=675; 125*9=1125; 3368+675=4043; 4043+1125=5168Denominator: (5^3 +3^3 +5^2*3 +5*3^2)(5^3 -3^3)=(125 +27 +25*3 +5*9)(125 -27)=(152 +75 +45)(98)=(272)(98)=26656Wait, second fraction numerator is5168 and denominator is26656.Simplify 5168 /26656. Let's divide numerator and denominator by 16: 5168 ÷16=323, 26656 ÷16=1666. 323 and 1666. 323 divides by 17? 17*19=323. 1666 ÷17=98. So 17*19/17*98=19/98. So the second fraction is19/98.Thus original expression is (19/49) ÷ (19/98)= (19/49)*(98/19)=98/49=2. According to a - b=5-3=2. Correct again.So, all test cases are passing. Therefore, the simplified expression is indeed (a - b).But let me just check one more edge case. What if a = b? Wait, but if a = b, then the original expression has denominators like (a + b)^2 -ab. If a = b, then denominator of the first fraction becomes (2a)^2 - a^2 =4a^2 -a^2=3a^2, which is fine. The second fraction's denominator: a^5 +a^5 +a^5 +a^5=4a^5. Numerator: (a^3 +a^3 +a^3 +a^3)(a^3 -a^3)= (4a^3)(0)=0. So the second fraction becomes 0, so division by zero. So when a = b, the expression is undefined. Which is consistent with the simplified expression a - b, which would be zero. But in the original expression, when a = b, the second fraction is 0, so the entire expression is undefined. So the simplification is consistent, as a - b would be zero when a = b, but in the original expression, it's undefined. So all cases where a ≠ b, the expression is valid and simplifies to a - b, but when a = b, both the original expression and simplified expression are undefined or zero respectively. But since division by zero is undefined, we have to state that the expression simplifies to a - b for a ≠ b.But the problem says "simplify the expression", so we can write the simplified form as a - b, noting that a ≠ b to avoid division by zero.Therefore, after thorough analysis and verification, the simplified expression is (a - b).

✅:The simplified expression is (boxed{a - b}).Step-by-Step Explanation:1. Rewrite the Division as Multiplication: [ frac{(a-b)^{2}+ab}{(a+b)^{2}-ab} times frac{left(a^{3}+b^{3}+a^{2}b+ab^{2}right) left(a^{3}-b^{3}right)}{a^{5}+b^{5}+a^{2}b^{3}+a^{3}b^{2}} ]2. Simplify the First Fraction: - Numerator: ((a-b)^2 + ab = a^2 - ab + b^2) - Denominator: ((a+b)^2 - ab = a^2 + ab + b^2) - Result: (frac{a^2 - ab + b^2}{a^2 + ab + b^2})3. Factor the Second Fraction: - Numerator: - (a^3 + b^3 + a^2b + ab^2 = (a + b)(a^2 + b^2)) - (a^3 - b^3 = (a - b)(a^2 + ab + b^2)) - Combined: ((a + b)(a - b)(a^2 + b^2)(a^2 + ab + b^2)) - Denominator: - Factor by grouping: (a^5 + b^5 + a^2b^3 + a^3b^2 = (a + b)(a^2 - ab + b^2)(a^2 + b^2)) - Resulting Fraction: (frac{(a - b)(a^2 + ab + b^2)}{a^2 - ab + b^2})4. Multiply and Cancel Common Factors: [ frac{a^2 - ab + b^2}{a^2 + ab + b^2} times frac{(a - b)(a^2 + ab + b^2)}{a^2 - ab + b^2} ] - Cancel (a^2 - ab + b^2) and (a^2 + ab + b^2) terms.5. Final Simplified Expression: (a - b)Verification: Test cases confirm the simplification is correct for various (a) and (b) (where (a neq b)), ensuring no division by zero occurs in valid cases. The result is (boxed{a - b}).

🔑:Let's start by defining the given expressions:1. Let [A = frac{(a - b)^2 + ab}{(a + b)^2 - ab}]2. Let [B = a^5 + b^5 + a^2 b^3 + a^3 b^2]3. Let [C = left(a^3 + b^3 + a^2 b + ab^2right)left(a^3 - b^3right)]Let's simplify each part of the expression.Simplifying (A):[A = frac{(a - b)^2 + ab}{(a + b)^2 - ab}]Observe that ((a - b)^2 = a^2 - 2ab + b^2) and ((a + b)^2 = a^2 + 2ab + b^2). Therefore,[A = frac{a^2 - 2ab + b^2 + ab}{a^2 + 2ab + b^2 - ab} = frac{a^2 - ab + b^2}{a^2 + ab + b^2}]Simplifying (B):[B = a^5 + a^3 b^2 + b^5 + a^2 b^3]Group terms to factorize:[B = a^3(a^2 + b^2) + b^3(a^2 + b^2) = (a^3 + b^3)(a^2 + b^2)]Simplifying (C):[C = (a^3 + b^3 + a^2 b + ab^2)(a^3 - b^3)]Factorize components:[a^3 + b^3 + a^2 b + ab^2 = (a + b)(a^2 - ab + b^2) + ab(a + b)]Simplifying further, we get:[a^3 + b^3 + a^2 b + ab^2 = (a + b)(a^2 + b^2)]Thus,[C = (a + b)(a^2 + b^2)(a^3 - b^3)]Combining (A), (B), and (C):We now perform the division:[A : frac{B}{C} = frac{frac{a^2 - ab + b^2}{a^2 + ab + b^2}}{frac{(a^3 + b^3)(a^2 + b^2)}{(a + b)(a^2 + b^2)(a^3 - b^3)}}]This can be written as:[= frac{a^2 - ab + b^2}{a^2 + ab + b^2} times frac{(a + b)(a^2 + b^2)(a^3 - b^3)}{(a^3 + b^3)(a^2 + b^2)}]Notice that ((a^2 + b^2)) terms cancel out:[= frac{a^2 - ab + b^2}{a^2 + ab + b^2} times frac{(a + b)(a^3 - b^3)}{(a^3 + b^3)}]Then using factorization (a^3 - b^3 = (a - b)(a^2 + ab + b^2)) and (a^3 + b^3 = (a + b)(a^2 - ab + b^2)), we get:[= frac{a^2 - ab + b^2}{a^2 + ab + b^2} times frac{(a + b)(a - b)(a^2 + ab + b^2)}{(a + b)(a^2 - ab + b^2)} ]Cancel the similar terms:[= frac{a - b}{1}]Conclusion:Thus, the final result is:[boxed{a - b}]

🧠:Okay, so I need to figure out how many non-congruent triangles can be formed using these three segments each of length 2^n for n from 0 to 1009. That's a total of 3030 segments, right? Wait, three segments of each length, so for each n, there are three segments of length 2^n. So, the available lengths are 2^0, 2^1, ..., up to 2^1009, each repeated three times. First, I should recall the triangle inequality theorem. To form a triangle, the sum of any two sides must be greater than the third side. So, given three lengths a, b, c (where a ≤ b ≤ c), the triangle inequality requires that a + b > c. Since all the segments are powers of 2, maybe there's a pattern here.Let me think. The lengths are 1, 2, 4, 8, ..., up to 2^1009. Each length is available three times. So, when choosing three segments to form a triangle, the possible combinations are:1. All three segments of the same length: like (a, a, a). For this to form a triangle, the triangle inequality requires that a + a > a, which simplifies to a > 0. Since all lengths are positive, this is always true. So, each length can form an equilateral triangle. But wait, in this case, since the segments are all the same length, it's an equilateral triangle. But since the problem asks for non-congruent triangles, each equilateral triangle of a different size is non-congruent. So, there are 1010 possible equilateral triangles (since n starts at 0 up to 1009, inclusive, which is 1010 values).2. Two segments of the same length and one different: like (a, a, b), where a ≠ b. Let's assume a < b. Then, the triangle inequality would require a + a > b. But since the lengths are powers of 2, if a is 2^k and b is 2^m where m > k, then 2^k + 2^k = 2^(k+1). So, 2^(k+1) must be greater than 2^m. But since m > k, 2^m is at least 2^(k+1). Therefore, 2^(k+1) > 2^m only if m < k+1, but m > k, so m must equal k+1. Wait, let's check:If a = 2^k, b = 2^(k+1), then a + a = 2*2^k = 2^(k+1) = b. So, 2^(k+1) is not greater than b, it's equal. So, that would not satisfy the triangle inequality. Therefore, even if we take two sides of length a = 2^k and one side of length b = 2^m where m ≥ k+1, the sum a + a = 2^(k+1) ≤ b, which would not satisfy the triangle inequality. So, in this case, triangles with two sides of the same length and the third side longer cannot exist. What if the third side is shorter? Then, if we have (a, a, b) where b < a. Then, the triangle inequality requires that b + a > a, which is trivially true, and a + a > b. Since b < a, then a + a = 2a > b. Since a is at least 1 and b is less than a, but in our case, the lengths are powers of 2. Wait, if b is shorter than a, then b is 2^m where m < k (since a = 2^k). Then, 2a = 2*2^k = 2^(k+1). But b = 2^m, and m < k. So, 2^(k+1) > 2^m for any m < k. So, in this case, triangles with two sides of length a and one side of length b where b < a would satisfy the triangle inequality. So, for example, (2^k, 2^k, 2^m) where m < k. However, such a triangle would be isoceles with two sides longer than the third. But wait, but since the third side is shorter, we need to check if all triangle inequalities are satisfied. The two smaller sides would be 2^m and 2^k. Then, 2^m + 2^k must be greater than 2^k. Which is true since 2^m > 0. So, yes, such a triangle is valid. But how many non-congruent triangles are there of this type?For each a = 2^k, and for each b = 2^m where m < k, we can have a triangle (a, a, b). But since the problem asks for non-congruent triangles, each combination of (a, a, b) where a > b is unique. However, if we have (a, a, b) and (c, c, d), when are they congruent? If a = c and b = d, so each pair (k, m) with m < k gives a unique triangle. So, the number of such triangles would be the number of pairs (k, m) where m < k. But m can range from 0 up to k-1 for each k. Since k can be from 1 to 1009 (since m has to be less than k, and k starts at 1 when m=0). Wait, n starts at 0, so k can be from 0 to 1009. But if k=0, then m would have to be less than 0, which isn't possible. So, k starts at 1. So, for k from 1 to 1009, and m from 0 to k-1. So, the number of such triangles is the sum from k=1 to 1009 of k, since for each k, there are k possible m's. Wait, no. For k=1, m can be 0: 1 value. For k=2, m can be 0,1: 2 values. ... For k=1009, m can be 0,1,...,1008: 1009 values. So, total number is 1 + 2 + ... + 1009. The sum of the first N integers is N(N+1)/2. Here N=1009, so sum is 1009*1010/2. But wait, is that correct?Wait, no. Wait, for each k from 1 to 1009, the number of m's is k (since m goes from 0 to k-1, which is k numbers). So, the total number is sum_{k=1}^{1009} k = (1009)(1010)/2. But this would be a huge number. However, wait, but we have three segments of each length. So, for each triangle (a, a, b), since we need three segments: two a's and one b. But each length has three copies. So, we can form such a triangle as long as we have two a's and one b. Since there are three copies of each, we can do that. But the problem is about non-congruent triangles, so we need to count each distinct triangle (in terms of side lengths) once, regardless of how many times they can be formed. So, the count is purely combinatorial based on the side lengths.But wait, but for each pair (k, m) with k > m, the triangle with sides 2^k, 2^k, 2^m is a valid triangle. So, how many such pairs are there? For each k from 1 to 1009, m from 0 to k-1. So, the total number is 1 + 2 + ... + 1009 = (1009)(1010)/2. But that seems like a lot. However, this would be for triangles with two sides of the same length and a third side shorter. Then, the equilateral triangles are 1010, as earlier. But wait, the equilateral triangles would be counted as a special case where m = k. But in this case, when m = k, we have (a, a, a), which is the equilateral triangle. So, maybe the total triangles with two equal sides are the equilateral ones plus the ones with a different third side. But I need to be careful not to double-count.Wait, in the previous analysis, equilateral triangles are separate. The case (a, a, b) where a ≠ b includes both cases where b < a and b > a. But earlier, we saw that if b > a, then the triangle inequality fails. So, only when b < a, so m < k. So, the triangles with two equal sides and one different are only those with the third side shorter. So, the number is indeed (1009)(1010)/2. Then, the equilateral triangles are 1010. Then, we also have triangles with all sides of different lengths. So, we need to consider that case as well.So, the total number of triangles is the number of equilateral triangles plus the number of isoceles triangles (with two equal sides and one different) plus the number of scalene triangles (all sides different). Let's break it down:1. Equilateral: 1010 (since n ranges from 0 to 1009, inclusive, each gives a unique equilateral triangle).2. Isoceles (two equal sides, one different): (1009)(1010)/2.3. Scalene (all sides different): ?Now, we need to compute the number of scalene triangles. For a scalene triangle, we need three distinct lengths a < b < c such that a + b > c. Again, since all lengths are powers of 2, let's see when this inequality holds.Suppose a = 2^k, b = 2^m, c = 2^n, with k < m < n. Then, the triangle inequality requires that a + b > c. Substituting, 2^k + 2^m > 2^n. Since k < m < n, 2^k + 2^m = 2^k(1 + 2^{m - k}). Let's analyze this expression.Let’s denote d = m - k, so d ≥ 1. Then, 2^k(1 + 2^d) > 2^n. Since n > m, let n = m + e, where e ≥ 1. Then, 2^n = 2^{m + e} = 2^m * 2^e = 2^{k + d} * 2^e. Therefore, the inequality becomes 2^k(1 + 2^d) > 2^{k + d + e}. Dividing both sides by 2^k, we get (1 + 2^d) > 2^{d + e}. Simplify: 1 + 2^d > 2^{d + e}.This can be written as 1 > 2^{d + e} - 2^d = 2^d(2^e - 1). But since d ≥ 1 and e ≥ 1, 2^d(2^e - 1) is at least 2^1(2^1 - 1) = 2*1 = 2. Therefore, 1 > 2 is false. Therefore, the inequality 1 + 2^d > 2^{d + e} is never true. Therefore, scalene triangles with sides 2^k, 2^m, 2^n where k < m < n cannot satisfy the triangle inequality. Wait, that can't be right. Wait, does that mean that no scalene triangles can be formed? That seems surprising.Wait, let's test with some small exponents. Let's take k=0, m=1, n=2. Then, a=1, b=2, c=4. Then, 1 + 2 = 3, which is less than 4, so no. If k=0, m=1, n=1: but n must be greater than m. So, next example: k=0, m=2, n=3. Then, a=1, b=4, c=8. 1 + 4 = 5 < 8. Still no. Another example: k=1, m=2, n=3: 2 + 4 = 6 < 8. Still no. Hmm. What if the two smaller sides add up to more than the largest side? Let's try k= m-1, m, n=m+1. For example, k= m-1, so a=2^{m-1}, b=2^m, c=2^{m+1}. Then, a + b = 2^{m-1} + 2^m = 3*2^{m-1}. Compare to c=2^{m+1}=4*2^{m-1}. So, 3*2^{m-1} < 4*2^{m-1}. Still, a + b < c. So even adjacent exponents don't work. Wait, what if the two smaller exponents are the same? Wait, that would be the isoceles case. So, actually, for scalene triangles, since all sides are different, and each side is a power of 2, the sum of the two smaller sides will always be less than the largest side. Because 2^k + 2^m ≤ 2^{m} + 2^{m} = 2^{m+1}, but if the largest side is 2^n where n ≥ m+1, then 2^{m+1} ≤ 2^n, so 2^k + 2^m ≤ 2^n. Therefore, 2^k + 2^m ≤ 2^n, which means the sum is less than or equal to the largest side. Therefore, the triangle inequality fails. Therefore, there are no scalene triangles. Therefore, the only triangles possible are equilateral and isoceles (with two equal sides and a third side shorter). Therefore, the total number of non-congruent triangles is 1010 (equilateral) plus (1009)(1010)/2 (isoceles). Let me compute that.First, (1009)(1010)/2: 1009 * 1010 = Let's compute 1000*1010 = 1,010,000, and 9*1010 = 9,090. So total is 1,010,000 + 9,090 = 1,019,090. Then divide by 2: 1,019,090 / 2 = 509,545. Then add 1010: 509,545 + 1,010 = 510,555. Wait, but let me check:Wait, 1009*1010/2 is indeed (1009*1010)/2. 1009 is 1010 -1, so 1009*1010 = (1010)^2 - 1010 = 1,020,100 - 1,010 = 1,019,090. Then, divided by 2 is 509,545. Then adding 1010 gives 509,545 + 1,010 = 510,555. So, the total number of non-congruent triangles would be 510,555?But wait, let me double-check. Because when we count the isoceles triangles, we have for each k from 1 to 1009, m from 0 to k-1. So, for k=1, m=0: 1 triangle. For k=2, m=0,1: 2 triangles. ... For k=1009, m=0 to 1008: 1009 triangles. So, the total is the sum from k=1 to 1009 of k. Which is (1009)(1009 + 1)/2 = (1009)(1010)/2 = 509,545. Then add the equilateral triangles: 1010. So, total is 509,545 + 1010 = 510,555.But wait, the problem is about non-congruent triangles. So, we need to ensure that each triangle is counted once, regardless of how it's formed. Since each equilateral triangle is unique in its side length, each isoceles triangle is determined by the two equal sides and the different side. Since the different side is shorter, the triplet (a, a, b) with a > b is unique for each a and b. So, those are all non-congruent. Therefore, the total should indeed be 510,555.But wait, let me confirm once more. Let's take small n to test. Suppose n=0,1. So, available lengths: 1, 2, each with three copies.How many non-congruent triangles?Equilateral: for n=0 (length 1) and n=1 (length 2). So, 2 equilateral triangles.Isoceles triangles:For k=1 (length 2), m=0 (length 1): triangle (2,2,1).For k=2? Wait, in our original problem, n goes up to 1009, but in this test case, n=0,1, so k can only be 1. Then, sum from k=1 to 1 of k = 1. So, 1 isoceles triangle. So total triangles: 2 + 1 = 3.But let's list them:1. (1,1,1)2. (2,2,2)3. (2,2,1)Are there any other triangles? For scalene, we need three different lengths, which would be (1,2,2^something). But here, the only lengths are 1 and 2. So, (1,2,2) is isoceles, already counted. There's no scalene. So, total 3 triangles. Which matches 2 + 1 = 3. So, the formula works here.Another test case: n=0,1,2. Lengths 1,2,4.Equilateral: 3 (1,2,4).Isoceles:For k=1 (2), m=0 (1): (2,2,1)For k=2 (4), m=0 (1): (4,4,1)For k=2 (4), m=1 (2): (4,4,2)So, total isoceles: 1 (for k=1) + 2 (for k=2) = 3. So, total triangles: 3 + 3 = 6.But let's list them:Equilateral: (1,1,1), (2,2,2), (4,4,4)Isoceles: (2,2,1), (4,4,1), (4,4,2)Scalene: (1,2,4) would not work since 1+2=3 < 4. So, no scalene. So total 6 triangles. Which is 3 + 3 = 6. The formula gives 3 (equilateral) + sum_{k=1}^2 k = 3 + 3 = 6. Correct.Thus, the formula seems to hold. Therefore, for the original problem, the total number of non-congruent triangles is 1010 (equilateral) + (1009)(1010)/2 (isoceles) = 510,555.Wait, but let me check again the arithmetic:(1009)(1010)/2 = (1009 * 1010)/2. 1009*1000=1,009,000. 1009*10=10,090. So total is 1,009,000 + 10,090 = 1,019,090. Then divide by 2: 1,019,090 / 2 = 509,545. Then add 1010: 509,545 + 1,010 = 510,555. Yes, that's correct.Therefore, the answer should be 510,555.But wait, let me check once more for n=0,1,2,3. Let's see:Lengths 1,2,4,8. Equilateral: 4.Isoceles:For k=1 (2): m=0 (1) → (2,2,1)For k=2 (4): m=0,1 → (4,4,1), (4,4,2)For k=3 (8): m=0,1,2 → (8,8,1), (8,8,2), (8,8,4)Total isoceles: 1 + 2 + 3 = 6. So total triangles: 4 + 6 = 10.List them:Equilateral: 1,2,4,8.Isoceles: (2,2,1), (4,4,1), (4,4,2), (8,8,1), (8,8,2), (8,8,4).No scalene. So 10 triangles. Formula gives 4 + (3*4)/2 = 4 + 6 = 10. Correct.Therefore, the formula works. So, in the original problem, the answer is 510,555.But wait, the problem statement says "three segments of each length 2^n for n=0,1,⋯,1009". So, n ranges from 0 to 1009 inclusive. That's 1010 values (from 0 to 1009). So, the equilateral triangles are 1010. The isoceles triangles are sum_{k=1}^{1009} k = (1009)(1010)/2 = 509,545. Then total triangles 509,545 + 1010 = 510,555. So, yes.Therefore, the answer is 510555.Final Answerboxed{510555}

✅:To determine the number of non-congruent triangles that can be formed using three segments of each length (2^n) for (n = 0, 1, ldots, 1009), we consider the triangle inequality theorem and the properties of powers of 2.1. Equilateral Triangles: - Each length (2^n) can form an equilateral triangle. There are 1010 such lengths (from (n = 0) to (n = 1009)). - Number of equilateral triangles: (1010).2. Isosceles Triangles (two equal sides and one different): - For each length (2^k) (where (k geq 1)), we can pair it with a shorter length (2^m) (where (m < k)). The number of such pairs is the sum of the first 1009 integers. - The number of isosceles triangles is given by the sum (sum_{k=1}^{1009} k = frac{1009 times 1010}{2} = 509545).3. Scalene Triangles (all sides different): - For three distinct lengths (2^k < 2^m < 2^n), the triangle inequality (2^k + 2^m > 2^n) must hold. However, since (2^k + 2^m leq 2^{m} + 2^{m} = 2^{m+1}), which is always less than (2^n) for (n geq m+1), no scalene triangles can be formed.Summing the equilateral and isosceles triangles:- Total number of non-congruent triangles: (1010 + 509545 = 510555).[boxed{510555}]

🔑:# Step-by-Step Explanation:1. Understanding the Problem: We need to count the number of ways to choose three segments of lengths (2^n) (where (n) ranges from 0 to 1009) that can form a non-degenerate triangle.2. Condition for Triangle Formation: For three positive lengths (a), (b), and (c) to form a triangle, the triangle inequality must be satisfied: [ a + b > c ] [ a + c > b ] [ b + c > a ]3. Segment Lengths Analysis: The segments have lengths (2^0, 2^1, 2^2, ldots, 2^{1009}). We'll examine different possible combinations to see which ones can form a triangle.4. Combination Analysis: Case 1: Three Different Lengths (2^i, 2^j, 2^k) where (i < j < k): [ 2^i + 2^j leq 2^j + 2^j = 2^{j+1} leq 2^k ] This implies (2^i + 2^j leq 2^k), so they cannot form a triangle. Case 2: Two Equal and One Different Length, (2^i, 2^i, 2^j) where (i < j): [ 2^i + 2^i = 2^{i+1} leq 2^j ] This means (2^i + 2^i leq 2^j), so these cannot form a triangle either.5. Valid Combinations: Case 3: Two Lengths Are the Same, (2^i, 2^j, 2^j) where (i < j): [ 2^i + 2^j > 2^j quad text{always true since } 2^i > 0 ] [ 2^j + 2^j > 2^i quad text{true since } 2^j geq 2^i ] Therefore, (2^i, 2^j, 2^j) will always form a triangle when (i < j). Case 4: All Three Lengths Are Equal, (2^k, 2^k, 2^k): This will form an equilateral triangle.6. Counting Valid Combinations: - Combination (2^i, 2^j, 2^j): We have (i < j) and (i, j) ranging from 0 to 1009. The number of such pairs is: [ binom{1010}{2} ] - Combination (2^k, 2^k, 2^k): There are 1010 such combinations (one for each (k) from 0 to 1009).7. Total Valid Combinations: [ binom{1010}{2} + 1010 ] [ = frac{1010 times 1009}{2} + 1010 ] [ = 510555 ] Conclusion:[boxed{510555}]

🧠:Okay, so I have this problem here: Find the number of solutions in natural numbers for the equation (2x + y)(2y + x) = 2017^2017. Hmm, interesting. Let me try to work through this step by step.First, natural numbers usually mean positive integers, right? So x and y have to be positive integers such that when you compute (2x + y) and (2y + x), their product is 2017 raised to the 2017th power. Alright, 2017 is a prime number, right? Let me confirm that. Yeah, 2017 is indeed a prime. So, 2017^2017 is a prime power. That might be useful because when you have a product of two numbers equal to a prime power, the factors themselves must be powers of that prime. So, if (2x + y) and (2y + x) multiply to 2017^2017, then each of them must be of the form 2017^k and 2017^m where k + m = 2017. But wait, hold on. The equation is (2x + y)(2y + x) = 2017^2017. Since 2017 is prime, the only divisors of 2017^2017 are powers of 2017 from 2017^0 up to 2017^2017. But since we're dealing with natural numbers x and y, both (2x + y) and (2y + x) must be at least... let's see. Since x and y are positive integers, 2x + y is at least 2*1 + 1 = 3, and similarly for 2y + x. So both factors are at least 3, meaning that they can't be 1. Therefore, the exponents k and m for 2017 must satisfy k ≥ 1 and m ≥ 1, and k + m = 2017. Therefore, the possible pairs (k, m) are (1, 2016), (2, 2015), ..., (2016, 1). So there are 2016 possible factorizations? Wait, but maybe there's more to it because different factor pairs might lead to the same solution or maybe some factor pairs don't lead to a solution at all. Wait, no. Each factor pair (2017^k, 2017^m) where k + m = 2017 would give a different product, but we need to check if each such pair leads to a valid solution (x, y) in natural numbers. So perhaps not all of these factorizations will result in integer x and y. Therefore, the actual number of solutions might be less than 2016. So we need to find all pairs (a, b) such that a * b = 2017^2017, with a = 2x + y and b = 2y + x, and then solve for x and y in terms of a and b, and check whether x and y are positive integers for each such pair (a, b). So let's formalize that. Let me denote a = 2x + y and b = 2y + x. Then, the product ab = 2017^2017. We need to solve for x and y in terms of a and b. Let's write down the system of equations:2x + y = a x + 2y = bWe can solve this system using linear algebra. Let's write it in matrix form:[2 1][x] = [a] [1 2][y] [b]The determinant of the coefficient matrix is 2*2 - 1*1 = 4 - 1 = 3. Since the determinant is 3, which is invertible modulo 2017^2017 (because 3 and 2017 are coprime, as 2017 is prime and not equal to 3), the system has a unique solution for x and y given any a and b. The solution is:x = (2a - b)/3 y = (2b - a)/3Therefore, for each factor pair (a, b) with a*b = 2017^2017, x and y must be integers. So, (2a - b) and (2b - a) must both be divisible by 3, and also, x and y must be positive integers. So, the key here is that for each divisor a of 2017^2017, we can compute b = 2017^2017 / a, then compute x = (2a - b)/3 and y = (2b - a)/3, and check if x and y are positive integers. But since 2017 is a prime, all divisors a of 2017^2017 are of the form 2017^k where 0 ≤ k ≤ 2017. However, since a and b must be greater than or equal to 3 (as x and y are natural numbers), we have to consider k from 1 up to 2016. Wait, but if k is 0, then a = 1, which would make x = (2*1 - 2017^2017)/3, which is negative, so not a natural number. Similarly, if k = 2017, then a = 2017^2017, and b = 1, leading to y = (2*1 - 2017^2017)/3, which is also negative. So we can restrict k to 1 ≤ k ≤ 2016. But even then, not all of these k will lead to x and y being positive integers. So we need to find all k in 1 ≤ k ≤ 2016 such that (2a - b) and (2b - a) are both divisible by 3 and result in positive integers. Let me try to formalize the divisibility by 3. Let’s substitute a = 2017^k and b = 2017^{2017 - k}. Then x = (2*2017^k - 2017^{2017 - k}) / 3 and y = (2*2017^{2017 - k} - 2017^k) / 3. For x and y to be integers, the numerators must be divisible by 3. So, 2*2017^k ≡ 2017^{2017 - k} mod 3 and 2*2017^{2017 - k} ≡ 2017^k mod 3. But since 2017 mod 3 is equal to 2017 divided by 3. Let's compute 2017 divided by 3. 3*672 = 2016, so 2017 ≡ 1 mod 3. Therefore, 2017 ≡ 1 mod 3. Therefore, 2017^k ≡ 1^k ≡ 1 mod 3 for any integer k. Similarly, 2017^{2017 - k} ≡ 1^{2017 - k} ≡ 1 mod 3. Therefore, substituting back into the equations:2*2017^k - 2017^{2017 - k} ≡ 2*1 - 1 = 1 mod 3 But 1 mod 3 is not congruent to 0, so x would not be an integer. Wait, that can't be right. That would imply that there are no solutions, which seems contradictory. There must be a mistake here. Wait, perhaps I made an error in computing 2017 mod 3. Let's check again. 2017 divided by 3: 3*672 = 2016, so 2017 = 3*672 + 1, so 2017 ≡ 1 mod 3. Correct. So 2017 ≡ 1 mod 3, so any power of 2017 is also ≡ 1 mod 3. Therefore, 2*2017^k ≡ 2*1 ≡ 2 mod 3, and 2017^{2017 - k} ≡ 1 mod 3. Therefore, 2*2017^k - 2017^{2017 - k} ≡ 2 - 1 ≡ 1 mod 3. Similarly, 2*2017^{2017 - k} - 2017^k ≡ 2*1 - 1 ≡ 1 mod 3. So both numerators are congruent to 1 modulo 3, which is not 0. Therefore, x and y would not be integers. But that's a problem. If x and y can't be integers, then the equation has no solutions? But that seems counterintuitive. Maybe I made a mistake in my reasoning. Wait, but the determinant of the system is 3, so in order for x and y to be integers, the right-hand sides (a and b) must satisfy some congruence conditions modulo 3. Since the determinant is 3, the system has integer solutions if and only if the vector (a, b) is in the image of the matrix, which requires that certain congruences hold. Specifically, using Cramer's rule, since the determinant is 3, 3 divides the numerators. But according to the previous calculation, since 2017 ≡ 1 mod 3, so a = 2017^k ≡ 1^k ≡ 1 mod 3, and b = 2017^{2017 - k} ≡ 1^{2017 - k} ≡ 1 mod 3. Therefore, a ≡ 1 mod 3 and b ≡ 1 mod 3. Therefore, 2a - b ≡ 2*1 - 1 = 1 mod 3 2b - a ≡ 2*1 - 1 = 1 mod 3 Therefore, both numerators are congruent to 1 mod 3, which is not 0. Therefore, the system would require dividing by 3, which would not result in integers. So x and y would not be integers. But this suggests that there are no solutions, which can't be right. But the problem states "solutions in natural numbers", so maybe there are no solutions? That seems possible. But maybe I missed something. Let me double-check. Alternatively, maybe 2017 ≡ 1 mod 3, so the equations 2x + y ≡ a ≡ 1 mod 3 and x + 2y ≡ b ≡ 1 mod 3. Let's see: 2x + y ≡ 1 mod 3 x + 2y ≡ 1 mod 3 Let me solve this system of congruences. From the first equation: 2x + y ≡ 1 mod 3 From the second equation: x + 2y ≡ 1 mod 3 Multiply the first equation by 2: 4x + 2y ≡ 2 mod 3 Subtract the second equation: (4x + 2y) - (x + 2y) ≡ 2 - 1 mod 3 3x ≡ 1 mod 3 But 3x ≡ 0 mod 3, so 0 ≡ 1 mod 3, which is a contradiction. Therefore, there are no solutions to the system of congruences. Therefore, there are no natural numbers x and y that satisfy the original equation. Therefore, the number of solutions is zero. Wait, but that seems too straightforward. Let me check with specific numbers. Suppose 2017^2017 is a very large number. Let's take a smaller exponent to test. Let's consider the equation (2x + y)(2y + x) = 2017^1 = 2017. Using the same logic, a and b are factors of 2017, which is prime. So the possible pairs are (1, 2017) and (2017, 1). Let's check both. First pair: a=1, b=2017. Then x=(2*1 - 2017)/3 = (2 - 2017)/3 = (-2015)/3 ≈ -671.666..., which is not a natural number. Similarly, the other pair: a=2017, b=1. Then x=(2*2017 - 1)/3 = (4034 - 1)/3 = 4033/3 ≈ 1344.333..., which is not an integer. So, indeed, even for 2017^1, there are no solutions. Therefore, maybe the conclusion is correct. For each exponent, there are no solutions. Therefore, the answer is 0. But let me check another prime, maybe 3, to see if the reasoning holds. Suppose the equation is (2x + y)(2y + x) = 3^n. Let's take n=1. Then possible factor pairs (1,3) and (3,1). For (a=1, b=3): x=(2*1 - 3)/3 = (-1)/3, invalid. For (a=3, b=1): x=(6 -1)/3=5/3 ≈1.666..., invalid. So no solutions. n=2: 3^2=9. Factor pairs (1,9), (3,3), (9,1). a=1: x=(2 -9)/3=-7/3, invalid. a=3: x=(6 -3)/3=3/3=1; y=(6 -3)/3=1. So x=1, y=1. Wait, that works. So here, with prime 3, when n=2, we have a solution. Wait, so why does 3^2 work? Let's check. If a=3 and b=3, then x=(2*3 - 3)/3=(6-3)/3=1, y=(2*3 -3)/3=1. So (2x + y)=3 and (2y +x)=3, which gives 3*3=9=3^2. So here, even though 3 is congruent to 0 mod 3, but 3 is the prime itself. Wait, but in the case of 2017, 2017 ≡1 mod 3, but 3 is different. Let me analyze this. In the case of prime 3, which is ≡0 mod 3, then 3^k ≡0 mod 3 for any k≥1. So in that case, when we set a=3 and b=3, then 2a - b =6 -3=3, which is divisible by 3, giving x=1. Similarly for y. So in that case, the divisibility works out. But in our original problem, 2017 ≡1 mod 3. So when we set a=2017^k, then a ≡1^k ≡1 mod 3, same for b. Then, 2a - b ≡2*1 -1=1 mod3, which is not 0. Therefore, the numerators are 1 mod3, so division by3 would not result in integers. Therefore, the key difference is that when the prime is congruent to 0 mod3 (like 3 itself), then a and b can be congruent to 0 mod3, allowing the numerators to be divisible by3, leading to integer solutions. But when the prime is congruent to 1 or 2 mod3, then a and b are congruent to1 or something else mod3, making the numerators not divisible by3. Therefore, for the original problem, since 2017 ≡1 mod3, then all a and b=2017^k would be ≡1 mod3, leading to 2a - b ≡1 mod3, which is not 0, so x and y can't be integers. Therefore, there are no solutions. But wait, wait. Let's check another prime, say 7, which is ≡1 mod3. Let’s take 7^2=49. Then factor pairs (1,49), (7,7), (49,1). For a=7, b=7: x=(14 -7)/3=7/3≈2.333..., not integer. For a=1, b=49: x=(2 -49)/3=-47/3≈-15.666..., invalid. For a=49, b=1: x=(98 -1)/3=97/3≈32.333..., invalid. So no solutions here either. Similarly, if we take prime 2, which is ≡2 mod3. Let’s take 2^2=4. Factor pairs (1,4), (2,2), (4,1). a=2, b=2: x=(4 -2)/3=2/3≈0.666..., invalid. a=1: x=(2-4)/3=-2/3 invalid. a=4: x=(8 -1)/3=7/3≈2.333..., invalid. So no solutions. So seems like for primes congruent to1 or 2 mod3, there are no solutions, but for prime=3, which is 0 mod3, there are solutions when the exponent is even? Wait, no. For prime=3, when n=2, we had a solution. Let's check n=3. 3^3=27. Factor pairs (1,27), (3,9), (9,3), (27,1). a=3, b=9: x=(6 -9)/3=-1, invalid. a=9, b=3: x=(18 -3)/3=15/3=5; y=(6 -9)/3=-1, invalid. So only mixed. Wait, maybe there's no solution here. Wait, maybe even for prime=3, higher exponents don't have solutions. Wait, when n=4: 3^4=81. Factor pairs (1,81), (3,27), (9,9), (27,3), (81,1). a=9, b=9: x=(18 -9)/3=9/3=3; y=(18 -9)/3=3. So x=3, y=3. Let's check: (2*3 +3)=9, (2*3 +3)=9, 9*9=81=3^4. So that works. So when exponent is even, for prime=3, there are solutions. So when n is even, setting a=3^{n/2}, b=3^{n/2}, then x=(2a -b)/3=(2*3^{n/2} -3^{n/2})/3=3^{n/2}/3=3^{n/2 -1}. Similarly for y. Therefore, x and y are integers when n is even. But in our original problem, the prime is 2017, which is ≡1 mod3. So, in this case, regardless of the exponent, all factors a and b will be ≡1 mod3, leading to 2a -b ≡1 mod3, so x and y can't be integers. Therefore, there are no solutions. Therefore, the conclusion is that there are no natural number solutions to the equation when the prime is ≡1 mod3, but there can be solutions when the prime is ≡0 mod3 (i.e., the prime is 3). So in our problem, since 2017 ≡1 mod3, there are no solutions. Therefore, the number of solutions is 0. But wait, let me double-check with an example where the prime is 3 and exponent is 2. As we saw, x=1, y=1 is a solution. Let's verify: (2*1 +1)(2*1 +1)=3*3=9=3^2. Correct. Similarly, if we take exponent 4, x=3, y=3: (2*3 +3)(2*3 +3)=9*9=81=3^4. Correct. So for prime 3, even exponents give solutions. But for primes ≡1 mod3, regardless of the exponent, no solutions. Therefore, in our original problem, the answer is 0. But wait, maybe I need to consider that even though 2017 ≡1 mod3, maybe for some exponents k, 2017^k could be ≡ something else mod3? But no, since 2017 ≡1 mod3, any power of it is still ≡1 mod3. So 2017^k ≡1^k=1 mod3. Therefore, a=2017^k ≡1 mod3, and b=2017^{2017 -k}≡1 mod3. Therefore, 2a -b ≡2 -1=1 mod3, same as before. Therefore, regardless of k, 2a -b ≡1 mod3 and 2b -a≡1 mod3. Therefore, x and y can't be integers. Therefore, there are no solutions. Therefore, the final answer is 0. But just to be thorough, let me check if there's another approach. Maybe consider symmetry or substitutions. Let's suppose that x and y are natural numbers. Let's denote u = 2x + y and v = 2y + x. Then, the equation becomes uv = 2017^{2017}. We need to solve for u and v in positive integers such that u =2x + y and v=2y +x. Then, solving for x and y:From u =2x + y and v=2y +x. Let's solve for x and y:From the first equation: y = u -2xSubstitute into the second equation: v =2(u -2x) +x =2u -4x +x =2u -3xTherefore, 2u -3x =vBut we also know that uv =2017^{2017}So substituting v =2u -3x into uv =2017^{2017}, we get u(2u -3x) =2017^{2017}But since y =u -2x must be positive, so u -2x >0 => x < u/2Similarly, v =2u -3x >0 => 2u -3x >0 => x < (2u)/3Therefore, x must be less than the minimum of u/2 and 2u/3, which is u/2. So x < u/2. But x must be a positive integer, so x can be from 1 to floor((u-1)/2). But this approach seems more complicated. Alternatively, maybe express u and v in terms of x and y, but we already did that. Alternatively, note that u and v must satisfy the equation 3x =2u -v and 3y =2v -u. So for x and y to be positive integers, 2u -v must be divisible by3 and positive, and 2v -u must be divisible by3 and positive. Therefore, 2u >v and 2v >u. So combining these, we get u/2 <v <2u. So for each factor pair (u, v) of 2017^{2017}, we need u/2 <v <2u. But since 2017 is prime, u =2017^k and v=2017^{2017 -k} for some k. Then, the inequalities become 2017^k /2 <2017^{2017 -k} <2*2017^kLet me solve these inequalities for k. First inequality: 2017^k /2 <2017^{2017 -k}Multiply both sides by 2: 2017^k <2*2017^{2017 -k}Divide both sides by 2017^{2017 -k}: 2017^{2k -2017} <2Take logarithm base2017: 2k -2017 < log_{2017} 2Since log_{2017} 2 is a very small number (since 2017 is large), approximately 0. Therefore, 2k -2017 <0 => k <2017/2=1008.5Second inequality: 2017^{2017 -k} <2*2017^kDivide both sides by 2017^k: 2017^{2017 -2k} <2Take logarithm base2017: 2017 -2k <log_{2017}2 ≈0Thus, 2017 -2k <0 =>k >2017/2=1008.5Therefore, combining both inequalities: k <1008.5 and k>1008.5, which is impossible. Therefore, there are no integers k satisfying both inequalities. Therefore, there are no such factor pairs (u, v) that satisfy the required inequalities. Therefore, there are no solutions. Therefore, this confirms the previous conclusion that there are no solutions. Hence, the number of solutions is 0. Thus, after considering multiple approaches—modular arithmetic, solving the system of equations, analyzing factor pairs and inequalities—it consistently leads to the conclusion that there are no natural number solutions to the equation. Therefore, the answer should be zero.

✅:The equation ((2x + y)(2y + x) = 2017^{2017}) requires finding natural numbers (x) and (y). Here's the systematic breakdown:1. Prime Factorization Insight: Since (2017) is prime, the factors of (2017^{2017}) are (2017^k) and (2017^{2017 - k}) for (0 leq k leq 2017). However, (x) and (y) being natural numbers implies (2x + y geq 3) and (2y + x geq 3), so (k) ranges from (1) to (2016).2. Solving the System: Let (a = 2x + y) and (b = 2y + x). The system: [ begin{cases} 2x + y = a x + 2y = b end{cases} ] has solutions: [ x = frac{2a - b}{3}, quad y = frac{2b - a}{3}. ] For (x) and (y) to be natural numbers, (2a - b) and (2b - a) must be divisible by 3.3. Modular Analysis: Since (2017 equiv 1 mod 3), all factors (a = 2017^k) and (b = 2017^{2017 - k}) satisfy (a equiv 1 mod 3) and (b equiv 1 mod 3). Substituting: [ 2a - b equiv 2(1) - 1 equiv 1 mod 3, quad 2b - a equiv 2(1) - 1 equiv 1 mod 3. ] Both numerators are congruent to (1 mod 3), not (0), making (x) and (y) non-integers.4. Contradiction in Congruences: The system of congruences: [ begin{cases} 2x + y equiv 1 mod 3 x + 2y equiv 1 mod 3 end{cases} ] leads to (0 equiv 1 mod 3), a contradiction, confirming no solutions exist.5. Inequality Constraints: For valid solutions, (u = 2x + y) and (v = 2y + x) must satisfy (u/2 < v < 2u). However, substituting (u = 2017^k) and (v = 2017^{2017 - k}) leads to conflicting inequalities for (k), proving no such pairs exist.Conclusion: There are no natural number solutions. (boxed{0})

🔑:Given the equation:[(2x + y)(2y + x) = 2017^{2017}]We are asked to determine how many solutions this equation has in natural numbers.# Step 1: Modulo 3 AnalysisFirst, let's consider the properties of the terms (A = 2x + y) and (B = 2y + x) modulo 3.- Suppose (x equiv a pmod{3}) and (y equiv b pmod{3}).- Then: [ A = 2x + y equiv 2a + b pmod{3} ] [ B = 2y + x equiv 2b + a pmod{3} ]# Step 2: Sum of Numbers Modulo 3Next, observe that the sum:[A + B = 2x + y + 2y + x = 3x + 3y equiv 0 pmod{3}]This implies (A) and (B) must satisfy:[(2a + b) + (2b + a) equiv 0 pmod{3}]# Step 3: Implications for (A) and (B)Based on the product being (2017^{2017}), we need to check the properties of (2017 mod 3):[2017 equiv 1 pmod{3}]Thus:[2017^{2017} equiv 1^{2017} equiv 1 pmod{3}]For the product of (A) and (B) to equal (2017^{2017}):[(A times B) equiv 1 pmod{3}]# Step 4: Residues of (A) and (B)Given (A + B equiv 0 pmod{3}), the possible non-zero residues for (A) and (B) under modulo 3 can only be:- (A equiv 1 pmod{3}), (B equiv 2 pmod{3})- or (A equiv 2 pmod{3}), (B equiv 1 pmod{3})In both cases, (A times B equiv 1 times 2 = 2 pmod{3}), which contradicts the necessary condition (1 pmod{3}).# Conclusion:Since no valid (A, B) pairs can satisfy both the sum and the product conditions under modulo 3, there are no solutions for (x) and (y) in natural numbers.[boxed{0}]

🧠:Okay, so I need to find the range of real numbers a such that the modulus of the complex number z = (a + cosθ) + (2a - sinθ)i doesn't exceed 2 for all real θ. Hmm, modulus of a complex number is the square root of the sum of the squares of the real and imaginary parts. So, the modulus squared would be (a + cosθ)^2 + (2a - sinθ)^2 ≤ 4 for all θ. That's because if the modulus is ≤ 2, then the modulus squared is ≤ 4. Maybe working with modulus squared is easier since squaring is a monotonic function for non-negative numbers.Let me write that out:(a + cosθ)^2 + (2a - sinθ)^2 ≤ 4 for all θ ∈ ℝ.Expanding both squares:First term: (a + cosθ)^2 = a² + 2a cosθ + cos²θSecond term: (2a - sinθ)^2 = (2a)^2 - 2*2a*sinθ + sin²θ = 4a² - 4a sinθ + sin²θAdding them together:a² + 2a cosθ + cos²θ + 4a² - 4a sinθ + sin²θ ≤ 4Combine like terms:a² + 4a² = 5a²2a cosθ - 4a sinθ = 2a cosθ - 4a sinθcos²θ + sin²θ = 1So the whole expression becomes:5a² + 2a cosθ - 4a sinθ + 1 ≤ 4Simplify the inequality:5a² + 2a cosθ - 4a sinθ + 1 - 4 ≤ 0Which is:5a² + 2a cosθ - 4a sinθ - 3 ≤ 0So, the problem reduces to ensuring that 5a² + 2a cosθ - 4a sinθ - 3 ≤ 0 for all θ.Hmm. So, the left-hand side (LHS) is a function of θ, and we need this function to be ≤ 0 for all θ. Therefore, the maximum value of the LHS over all θ must be ≤ 0. So, if we can find the maximum of 5a² + 2a cosθ - 4a sinθ - 3 with respect to θ, and then set that maximum ≤ 0, then we can solve for a.The key here is to find the maximum of the expression involving cosθ and sinθ. Let's look at the terms with θ:2a cosθ - 4a sinθ. This can be written as a linear combination of cosθ and sinθ. Remember that expressions like C cosθ + D sinθ can be written as R cos(θ + φ), where R = √(C² + D²). So, the maximum value of such an expression is R, and the minimum is -R.So, let's consider the expression 2a cosθ - 4a sinθ. Let's factor out a:2a cosθ - 4a sinθ = a(2 cosθ - 4 sinθ). So, the coefficient for cosθ is 2, and for sinθ is -4. Therefore, the amplitude R is √(2² + (-4)^2) = √(4 + 16) = √20 = 2√5. Therefore, the maximum value of 2 cosθ - 4 sinθ is 2√5, and the minimum is -2√5. Therefore, the maximum of a(2 cosθ - 4 sinθ) is |a| * 2√5. Wait, but depending on the sign of a. Wait, hold on. If a is positive, then the maximum of a*(2 cosθ - 4 sinθ) would be a*(2√5), and the minimum would be -a*(2√5). Similarly, if a is negative, the maximum and minimum would flip. However, since we are looking for the maximum value of the entire expression 5a² + a(2 cosθ - 4 sinθ) - 3, then regardless of the sign of a, the maximum of the linear term a(2 cosθ - 4 sinθ) would be |a| * 2√5. Because the expression a(2 cosθ - 4 sinθ) can reach up to |a| * 2√5 and down to -|a| * 2√5.Therefore, the maximum value of the entire expression 5a² + 2a cosθ - 4a sinθ - 3 is 5a² + |a| * 2√5 - 3. Therefore, to ensure that this maximum is ≤ 0, we set:5a² + 2√5 |a| - 3 ≤ 0This is an inequality in |a|. Let's denote t = |a|, where t ≥ 0. Then the inequality becomes:5t² + 2√5 t - 3 ≤ 0Now, solve this quadratic inequality for t.First, find the roots of 5t² + 2√5 t - 3 = 0.Using the quadratic formula:t = [-2√5 ± √( (2√5)^2 - 4*5*(-3) ) ] / (2*5)Calculate discriminant:(2√5)^2 - 4*5*(-3) = 4*5 + 60 = 20 + 60 = 80So,t = [ -2√5 ± √80 ] / 10Simplify √80: √80 = √(16*5) = 4√5Therefore,t = [ -2√5 ± 4√5 ] / 10So, two solutions:1. t = [ -2√5 + 4√5 ] / 10 = (2√5)/10 = √5/52. t = [ -2√5 - 4√5 ] / 10 = (-6√5)/10 = -3√5/5But since t = |a| ≥ 0, the negative solution is invalid.Thus, the quadratic is ≤ 0 between t = -3√5/5 and t = √5/5. But since t ≥ 0, the valid interval is t ∈ [0, √5/5]. Therefore, |a| ≤ √5/5, which implies that a ∈ [ -√5/5, √5/5 ]Wait, but hold on. Let me check again. The quadratic 5t² + 2√5 t - 3 is a parabola opening upwards (since the coefficient of t² is positive). So, it is below zero between its two roots. But since one root is negative and the other is positive, the inequality 5t² + 2√5 t - 3 ≤ 0 is satisfied for t between -3√5/5 and √5/5. However, since t = |a| ≥ 0, the solution is t ∈ [0, √5/5]. Therefore, |a| ≤ √5/5, so a ∈ [ -√5/5, √5/5 ].But wait, the problem states that the modulus does not exceed 2 for all θ ∈ ℝ. But is this the correct range for a? Let me verify.Wait, but earlier, when we said that the maximum of 5a² + 2a cosθ - 4a sinθ - 3 is 5a² + 2√5 |a| - 3. But is this correct?Wait, the expression is 5a² + [2a cosθ - 4a sinθ] - 3. The term [2a cosθ - 4a sinθ] can be written as a*(2 cosθ - 4 sinθ). The maximum of this term is |a| * √(2² + (-4)^2) = |a| * √(4 + 16) = |a| * √20 = |a| * 2√5. So, the maximum of the entire expression 5a² + a*(2 cosθ - 4 sinθ) - 3 is 5a² + |a|*2√5 - 3. Since the coefficient of the quadratic term in t is positive, this parabola in t opens upwards. Therefore, the inequality 5a² + 2√5 |a| - 3 ≤ 0 holds when |a| is between the roots. But since one root is negative and the other is positive, only the positive root is relevant here. So, indeed |a| must be ≤ √5/5. So, the range is [ -√5/5, √5/5 ].But wait, let me test a = √5/5. Let's compute the modulus squared:z = (a + cosθ) + (2a - sinθ)iSo modulus squared is (a + cosθ)^2 + (2a - sinθ)^2.Plugging a = √5/5:First term: (√5/5 + cosθ)^2 = (cosθ + √5/5)^2Second term: (2*√5/5 - sinθ)^2 = ( (2√5)/5 - sinθ )^2Compute the sum:= (cosθ + √5/5)^2 + ( (2√5)/5 - sinθ )^2Expand both:First term: cos²θ + 2*(√5/5)*cosθ + (√5/5)^2Second term: sin²θ - 2*(2√5/5)*sinθ + (2√5/5)^2Adding them:cos²θ + sin²θ + 2*(√5/5) cosθ - 4*(√5/5) sinθ + (√5/5)^2 + (2√5/5)^2Simplify:1 + (2√5/5 cosθ - 4√5/5 sinθ) + (5/25 + 20/25)Calculate constants:5/25 + 20/25 = 25/25 = 1So total expression becomes:1 + (2√5/5 cosθ - 4√5/5 sinθ) + 1 = 2 + (2√5/5 cosθ - 4√5/5 sinθ)Now, the term (2√5/5 cosθ - 4√5/5 sinθ) can be written as (2√5/5) cosθ - (4√5/5) sinθ. Let's factor out √5/5:√5/5 [2 cosθ - 4 sinθ]. The maximum of 2 cosθ - 4 sinθ is √(2² + (-4)^2) = √(4 + 16) = √20 = 2√5. Therefore, the maximum of the term is √5/5 * 2√5 = (√5 * 2√5)/5 = (2*5)/5 = 2.Therefore, the maximum value of modulus squared is 2 + 2 = 4, which is equal to 4. So the modulus is 2. Therefore, when a = √5/5, the modulus does reach 2 for some θ, which is allowed since the problem says the modulus does not exceed 2. Similarly, when a = -√5/5, let's check:a = -√5/5Compute modulus squared:( -√5/5 + cosθ )^2 + ( -2√5/5 - sinθ )^2First term: (cosθ - √5/5)^2 = cos²θ - 2*(√5/5) cosθ + (√5/5)^2Second term: (-2√5/5 - sinθ)^2 = (sinθ + 2√5/5)^2 = sin²θ + 4√5/5 sinθ + (2√5/5)^2Adding them:cos²θ + sin²θ - 2*(√5/5) cosθ + 4√5/5 sinθ + (√5/5)^2 + (2√5/5)^2Simplify:1 + (-2√5/5 cosθ + 4√5/5 sinθ) + (5/25 + 20/25)Again, constants sum to 1, so total expression:1 + (-2√5/5 cosθ + 4√5/5 sinθ) + 1 = 2 + (-2√5/5 cosθ + 4√5/5 sinθ)Factor out terms:= 2 + (√5/5)( -2 cosθ + 4 sinθ )The term (-2 cosθ + 4 sinθ) has a maximum amplitude of √((-2)^2 + 4^2) = √(4 + 16) = √20 = 2√5. Therefore, the maximum of (√5/5)( -2 cosθ + 4 sinθ ) is √5/5 * 2√5 = 2*5/5 = 2. So the maximum modulus squared is 2 + 2 = 4, again modulus 2. So even when a is negative, the maximum modulus is 2, which is allowed.Therefore, according to this, the range of a is [ -√5/5, √5/5 ].Wait, but let me check with a different value. Suppose a is 0. Then z = cosθ + (-sinθ)i, modulus squared is cos²θ + sin²θ = 1 ≤ 4. So modulus is 1 ≤ 2, which is okay. So a=0 is included in the interval, which is correct. What if a=1? Then modulus squared is (1 + cosθ)^2 + (2 - sinθ)^2. Let's compute:(1 + 2 cosθ + cos²θ) + (4 - 4 sinθ + sin²θ) = 1 + 4 + 2 cosθ - 4 sinθ + cos²θ + sin²θ = 5 + 2 cosθ -4 sinθ. The maximum of 2 cosθ -4 sinθ is 2√5, so maximum modulus squared is 5 + 2√5. If a=1, 5*(1)^2 + 2√5*1 -3 = 5 + 2√5 -3 = 2 + 2√5 ≈ 2 + 4.472 = 6.472 > 4. Therefore, the modulus squared would exceed 4, which is not allowed. So a=1 is not in the range, which is consistent with our solution.Another test: a=√5/5 ≈ 0.447. Let's compute modulus squared:5a² + 2√5 |a| -3. Wait, no. Wait, modulus squared is (a + cosθ)^2 + (2a - sinθ)^2. For a=√5/5, as we computed earlier, the maximum modulus squared is 4, so modulus is 2. So that's acceptable. For a=√5/5 + ε, where ε is a small positive number, the modulus squared would exceed 4 for some θ. Therefore, the upper bound is indeed √5/5.Similarly, for a negative, say a=-√5/5 ≈ -0.447, modulus squared is 4, as we saw. If a is less than -√5/5, like a=-1, then modulus squared would be (-1 + cosθ)^2 + (-2 - sinθ)^2. Let's compute:(1 - 2 cosθ + cos²θ) + (4 +4 sinθ + sin²θ) = 1 +4 -2 cosθ +4 sinθ +1 = 6 -2 cosθ +4 sinθ. The maximum of -2 cosθ +4 sinθ is √(4 +16)=√20=2√5. So maximum modulus squared is 6 +2√5 ≈6 +4.472≈10.472>4. So modulus exceeds 2, which is not allowed. Hence a must be between -√5/5 and √5/5.Therefore, the range is [ -√5/5, √5/5 ].But let me think again: when we derived the inequality 5a² + 2√5 |a| -3 ≤0, leading to |a| ≤√5/5, but in reality, when a is negative, the term 2a cosθ -4a sinθ becomes negative times cosθ and positive times sinθ. But regardless, when we take the absolute value, it's |a| multiplied by the amplitude. So the analysis is the same whether a is positive or negative. Hence, the result is symmetric around 0, hence the interval is [ -√5/5, √5/5 ].Alternatively, maybe we can parametrize the expression 2a cosθ -4a sinθ as a single sinusoidal function. Let me check.Expressed as 2a cosθ -4a sinθ = R cos(θ + φ), where R = √( (2a)^2 + (-4a)^2 ) = √(4a² +16a²)=√20a²=2√5 |a|.But since cos(θ + φ) ranges between -1 and 1, the maximum of 2a cosθ -4a sinθ is 2√5 |a|, as we found earlier.Therefore, modulus squared expression:5a² + 2a cosθ -4a sinθ +1 ≤4 → 5a² + (2a cosθ -4a sinθ) ≤3 → 5a² + 2√5 |a| cos(θ + φ) ≤3.Wait, but since cos(θ + φ) can be 1 or -1, the maximum of 5a² + 2√5 |a| cos(θ + φ) is 5a² + 2√5 |a|. Therefore, the maximum of modulus squared is 5a² + 2√5 |a|. Therefore, to have 5a² + 2√5 |a| ≤4? Wait, no. Wait, earlier we thought modulus squared was 5a² + 2a cosθ -4a sinθ +1. Wait, perhaps I made a miscalculation earlier.Wait, let's redo the expansion step carefully.Original modulus squared:(a + cosθ)^2 + (2a - sinθ)^2.Expand:(a² + 2a cosθ + cos²θ) + (4a² -4a sinθ + sin²θ)Combine terms:a² +4a²=5a²2a cosθ -4a sinθcos²θ + sin²θ=1So total modulus squared =5a² + 2a cosθ -4a sinθ +1.Therefore, the inequality is:5a² + 2a cosθ -4a sinθ +1 ≤4Therefore:5a² + 2a cosθ -4a sinθ ≤3Then, the maximum of 2a cosθ -4a sinθ is 2√5 |a| as before. So the maximum of the left-hand side is 5a² + 2√5 |a|. Therefore, we need 5a² + 2√5 |a| ≤3.Wait, wait, this contradicts the previous conclusion where I thought modulus squared was 5a² + 2√5 |a| -3 ≤0. Wait, no. Wait, the modulus squared is 5a² + 2a cosθ -4a sinθ +1 ≤4. Therefore, subtract 4: 5a² + 2a cosθ -4a sinθ +1 -4 ≤0 → 5a² + 2a cosθ -4a sinθ -3 ≤0. Then, the maximum of this expression is 5a² + 2√5 |a| -3. So the inequality is 5a² + 2√5 |a| -3 ≤0. So that is correct.Wait, but earlier, when I considered modulus squared for a=√5/5, I got 4, which is exactly 4. So modulus squared ≤4 is acceptable. Therefore, the maximum modulus squared is 4. Therefore, when 5a² + 2√5 |a| -3 =0, the maximum modulus squared is 4. So solving 5a² + 2√5 |a| -3 =0 gives |a|=√5/5. Therefore, the range is |a| ≤√5/5. Therefore, the answer is a ∈ [-√5/5, √5/5].Therefore, I think this is the correct answer. Let me check once more with a=√5/5. Then modulus squared is 5*(5/25) + 2√5*(√5/5) -3 = 5*(1/5) + 2*5/5 -3 = 1 + 2 -3=0. Wait, no. Wait, perhaps confusion. Wait, modulus squared is 5a² + 2a cosθ -4a sinθ +1. For a=√5/5, modulus squared is 5*(5/25) + 2*(√5/5)cosθ -4*(√5/5) sinθ +1 = (25/25) + (2√5/5 cosθ -4√5/5 sinθ) +1 = 1 + (2√5/5 cosθ -4√5/5 sinθ) +1 = 2 + (2√5/5 cosθ -4√5/5 sinθ). The maximum of (2√5/5 cosθ -4√5/5 sinθ) is √( (2√5/5)^2 + (-4√5/5)^2 )=√( (20/25 + 80/25 ))=√(100/25)=√4=2. Therefore, maximum modulus squared is 2 + 2=4. Therefore, modulus is 2. Which is acceptable. So indeed, when |a|=√5/5, the modulus reaches 2, which is the maximum allowed. For |a|>√5/5, the maximum modulus would exceed 2. Therefore, the range is correct.Therefore, the final answer is a ∈ [ -√5/5, √5/5 ]. In boxed form, that's boxed{left[ -dfrac{sqrt{5}}{5}, dfrac{sqrt{5}}{5} right]}.Final AnswerThe range of the real number ( a ) is boxed{left[ -dfrac{sqrt{5}}{5}, dfrac{sqrt{5}}{5} right]}.

✅:To find the range of the real number ( a ) such that the modulus of the complex number ( z = (a + cos theta) + (2a - sin theta)i ) does not exceed 2 for all ( theta in mathbf{R} ), we start by considering the modulus squared of ( z ):[|z|^2 = (a + cos theta)^2 + (2a - sin theta)^2]Expanding and simplifying the expression:[(a + cos theta)^2 + (2a - sin theta)^2 = a^2 + 2a cos theta + cos^2 theta + 4a^2 - 4a sin theta + sin^2 theta]Combining like terms and using the Pythagorean identity ( cos^2 theta + sin^2 theta = 1 ):[5a^2 + 2a cos theta - 4a sin theta + 1]We need this expression to be less than or equal to 4 for all ( theta ):[5a^2 + 2a cos theta - 4a sin theta + 1 leq 4]Subtracting 4 from both sides:[5a^2 + 2a cos theta - 4a sin theta - 3 leq 0]The term ( 2a cos theta - 4a sin theta ) can be written as ( a(2 cos theta - 4 sin theta) ). The maximum value of ( 2 cos theta - 4 sin theta ) is ( sqrt{2^2 + (-4)^2} = 2sqrt{5} ). Therefore, the maximum value of ( a(2 cos theta - 4 sin theta) ) is ( |a| cdot 2sqrt{5} ).Thus, the inequality becomes:[5a^2 + 2sqrt{5} |a| - 3 leq 0]Let ( t = |a| ). The quadratic inequality in ( t ) is:[5t^2 + 2sqrt{5} t - 3 leq 0]Solving this quadratic inequality using the quadratic formula:[t = frac{-2sqrt{5} pm sqrt{(2sqrt{5})^2 + 4 cdot 5 cdot 3}}{2 cdot 5} = frac{-2sqrt{5} pm sqrt{20 + 60}}{10} = frac{-2sqrt{5} pm 4sqrt{5}}{10}]This gives the roots:[t = frac{sqrt{5}}{5} quad text{and} quad t = -frac{3sqrt{5}}{5}]Since ( t geq 0 ), the valid solution is ( t = frac{sqrt{5}}{5} ). Therefore, ( |a| leq frac{sqrt{5}}{5} ), leading to the range:[a in left[ -frac{sqrt{5}}{5}, frac{sqrt{5}}{5} right]]Thus, the range of the real number ( a ) is boxed{left[ -dfrac{sqrt{5}}{5}, dfrac{sqrt{5}}{5} right]}.

🔑:1. Given the complex number: [ z = (a + cos theta) + (2a - sin theta),mathrm{i} ] we need to find the range of the real number (a) such that the modulus (|z|) does not exceed 2. 2. The modulus of (z) is given by: [ |z| = sqrt{(a + cos theta)^2 + (2a - sin theta)^2} ] We set up the inequality: [ |z| leq 2 ] which translates to: [ sqrt{(a + cos theta)^2 + (2a - sin theta)^2} leq 2 ]3. Squaring both sides of the inequality, we get: [ (a + cos theta)^2 + (2a - sin theta)^2 leq 4 ]4. Expanding the terms inside the square, we obtain: [ (a^2 + 2a cos theta + cos^2 theta) + ((2a)^2 - 4a sin theta + sin^2 theta) leq 4 ] Simplifying further: [ a^2 + 2a cos theta + cos^2 theta + 4a^2 - 4a sin theta + sin^2 theta leq 4 ]5. Collecting similar terms: [ 5a^2 + 2a (cos theta - 2 sin theta) + cos^2 theta + sin^2 theta leq 4 ]6. Using the Pythagorean identity (cos^2 theta + sin^2 theta = 1), we have: [ 5a^2 + 2a (cos theta - 2 sin theta) + 1 leq 4 ]7. Isolating the expression involving (a): [ 5a^2 + 2a (cos theta - 2 sin theta) leq 3 ]8. To simplify the term (2a (cos theta - 2 sin theta)), let (R = sqrt{1^2 + (-2)^2} = sqrt{5}), and define (varphi) so that (cos varphi = frac{1}{sqrt{5}}) and (sin varphi = frac{2}{sqrt{5}}). Thus: [ cos theta - 2 sin theta = sqrt{5} cdot sin (theta - varphi) ]9. Rewriting the inequality: [ 5a^2 + 2a (-sqrt{5} sin (theta - varphi)) leq 3 ]10. Since (|sin (theta - varphi)| leq 1), the inequality becomes: [ 5a^2 - 2 sqrt{5} |a| leq 3 ]11. Rearranging terms to isolate (|a|): [ 5a^2 - 2 sqrt{5}|a| - 3 leq 0 ]12. Solving the quadratic inequality for (|a|) using the quadratic formula (a = -frac{-2sqrt{5}}{2 cdot 5}) gives: [ |a| leq frac{sqrt{5}}{5} ]13. Therefore, the value of (a) must lie within the range: [ a in left[-frac{sqrt{5}}{5}, frac{sqrt{5}}{5}right] ] Conclusion[boxed{left[-frac{sqrt{5}}{5}, frac{sqrt{5}}{5}right]}]

🧠:Okay, let's tackle this geometry problem step by step. First, I need to make sure I understand the problem correctly. We have a rectangle ABCD with an area of 1 square unit. The center of the incircle of triangle ABC is point O. Then, we need to consider another rectangle that has O and D as opposite corners, with the other two vertices on the perimeter of the original rectangle. The task is to find the area of this new rectangle.Let me start by visualizing the original rectangle ABCD. Since it's a rectangle, opposite sides are equal and all angles are 90 degrees. Let's denote the vertices in order: A, B, C, D. So, AB and CD are one pair of opposite sides, and BC and DA are the other pair. The area is 1, so if AB has length 'a' and BC has length 'b', then a*b = 1.Next, we need to consider triangle ABC. Wait, triangle ABC is formed by vertices A, B, and C. But in a rectangle, points A, B, and C form a right triangle with right angle at B. Because from A to B to C, angle at B is 90 degrees. So triangle ABC is a right-angled triangle with legs AB and BC, and hypotenuse AC.Now, the problem mentions the incircle of triangle ABC. The center of the incircle (incenter) of a triangle is the intersection point of the angle bisectors. For a right-angled triangle, the inradius can be calculated, and the incenter is located at a specific position relative to the sides.Let me recall the formula for the inradius of a right-angled triangle. The inradius 'r' is given by r = (a + b - c)/2, where 'a' and 'b' are the legs, and 'c' is the hypotenuse. Here, in triangle ABC, legs are AB = a and BC = b, so hypotenuse AC = sqrt(a² + b²). Therefore, the inradius r = (a + b - sqrt(a² + b²))/2.The coordinates of the incenter can also be determined. In a right-angled triangle, the inradius is at a distance 'r' from each of the sides. Since the right angle is at B, the incenter should be 'r' units away from each of the legs AB and BC, and also 'r' units away from the hypotenuse AC. So, if we place the triangle in a coordinate system with point B at the origin (0,0), point A at (a,0), and point C at (0,b), then the incenter O would be at (r, r). Because it's 'r' units away from both the x-axis and y-axis, which are the legs of the triangle.Wait, but in the problem statement, the original figure is a rectangle ABCD, so the coordinates might be different. Let me clarify the positions. Let's assign coordinates to the rectangle to make it easier. Let's set point A at (0,0), B at (a,0), C at (a,b), and D at (0,b). Then, triangle ABC has vertices at A(0,0), B(a,0), and C(a,b). Wait, but in this case, triangle ABC is not a right-angled triangle. Wait, no. Let me check.If we have rectangle ABCD with coordinates A(0,0), B(a,0), C(a,b), D(0,b), then the triangle ABC is formed by points A(0,0), B(a,0), and C(a,b). Is this a right-angled triangle? Let's see. The sides AB is from (0,0) to (a,0), BC is from (a,0) to (a,b), and AC is from (0,0) to (a,b). So, angle at B is between AB and BC. AB is horizontal, BC is vertical, so angle at B is 90 degrees. Yes, triangle ABC is right-angled at B. So the inradius formula applies here.So, in this coordinate system, the inradius r is (AB + BC - AC)/2 = (a + b - sqrt(a² + b²))/2, as I mentioned earlier. The incenter O is located at (a - r, r). Wait, why? Because in a right-angled triangle, the inradius is located at a distance 'r' from each leg. Since the legs here are AB (along the x-axis from (0,0) to (a,0)) and BC (along the y-axis from (a,0) to (a,b)), the incenter should be 'r' units away from both AB and BC. So from AB (the x-axis), moving up 'r' units, and from BC (the vertical line x=a), moving left 'r' units. Therefore, the coordinates of O would be (a - r, r). Wait, that makes sense. Because if we start at point B(a,0), moving left 'r' along AB would take us to x = a - r, and moving up 'r' from AB (y=0) would take us to y = r. Therefore, O is at (a - r, r).Alternatively, if we model the triangle ABC in a coordinate system where B is at (a,0), then maybe the inradius is at (a - r, 0 + r). Yes, that seems right.Now, the problem mentions a new rectangle with opposite corners at O and D, where D is (0,b) in our coordinate system. So, the new rectangle has vertices O(a - r, r) and D(0,b), and the other two vertices are on the perimeter of the original rectangle. We need to find the area of this rectangle.Let me visualize this. The rectangle connects O and D, with the other two vertices on the sides of the original rectangle ABCD. So, the sides of the new rectangle must be parallel to the sides of the original rectangle, or not necessarily? Wait, since it's a rectangle, opposite sides are equal and all angles are 90 degrees. The other two vertices must lie on the perimeter (i.e., the edges) of the original rectangle.So, the new rectangle is formed by O, D, and two other points on the sides of ABCD. Let's denote the other two vertices as E and F, such that OEFD is a rectangle. Wait, actually, the rectangle has O and D as opposite corners, so the rectangle could be O, some point on the perimeter, D, and another point on the perimeter.Alternatively, perhaps the rectangle is such that one side is from O to a point on a side of ABCD, and another side from D to another point on another side of ABCD, forming a rectangle. Hmm, this is a bit abstract. Maybe assigning coordinates will help.Let me denote the coordinates:Original rectangle:A(0,0), B(a,0), C(a,b), D(0,b).Incenter O of triangle ABC is at (a - r, r), where r = (a + b - sqrt(a² + b²))/2.Point D is at (0,b).We need to find the rectangle with O(a - r, r) and D(0,b) as opposite corners, with the other two vertices on the perimeter of ABCD.So, the rectangle must have sides that are either horizontal/vertical or at some angle. But since the original rectangle has sides aligned with the axes, and the new rectangle's other vertices lie on the perimeter (edges) of ABCD, perhaps the sides of the new rectangle are also aligned with the axes? Or maybe not necessarily.Wait, in a rectangle, opposite sides are equal and parallel, and all angles are 90 degrees. So, if O and D are opposite corners, then the sides of the new rectangle would be along the lines connecting O to one of the intermediate vertices and D to the other. Since the other two vertices are on the perimeter of ABCD, they must lie on one of the four sides: AB, BC, CD, or DA.Let me try to parametrize the problem. Let's denote the other two vertices as E and F, lying on the perimeter of ABCD. Then, the rectangle is OEFD, with O, E, F, D as vertices. Alternatively, it could be O, F, D, E or some other order, but the key is that O and D are opposite corners.To form a rectangle with O and D as opposite corners, the midpoint of the diagonal OD should be the same for both diagonals. Also, the other two vertices E and F must satisfy that OE and DF are sides, or some other configuration.Alternatively, perhaps the sides of the new rectangle are parallel to the sides of the original rectangle. If that's the case, then the coordinates of E and F can be found by moving horizontally and vertically from O and D.Wait, let's suppose that the new rectangle has sides parallel to the original rectangle. Then, from point O(a - r, r), moving horizontally to the left until we reach the side AD (which is x=0), but point D is already at (0,b). Alternatively, moving vertically up or down.Wait, this might not be straightforward. Let me sketch mentally.If the rectangle has O and D as opposite corners, and sides parallel to the original rectangle, then the other two vertices would be at (0, r) and (a - r, b). Let's check if these points lie on the perimeter of ABCD.Point (0, r): This is on side AD (from (0,0) to (0,b)), since 0 ≤ r ≤ b. Similarly, point (a - r, b): This is on side DC (from (0,b) to (a,b)), but x-coordinate a - r. However, side DC is from (a,b) to (0,b), but actually, side DC is from D(0,b) to C(a,b). Wait, no: in the original rectangle, the sides are AB(a,0)-B(a,0), BC(a,0)-C(a,b), CD(a,b)-D(0,b), and DA(0,b)-A(0,0). Wait, no. Actually, in the rectangle ABCD, the sides are AB (from A(0,0) to B(a,0)), BC (from B(a,0) to C(a,b)), CD (from C(a,b) to D(0,b)), and DA (from D(0,b) to A(0,0)).So, side CD is from C(a,b) to D(0,b), which is the top side. So, moving from (a,b) to (0,b). Therefore, any point on CD has y-coordinate b and x-coordinate between 0 and a. Similarly, side DA is from D(0,b) to A(0,0), which is the left side, x=0, y from b to 0.Therefore, if we consider the other two vertices of the new rectangle, they must lie on these sides. Let's assume that the rectangle has sides parallel to the original rectangle. Then, the new rectangle would have sides that are horizontal and vertical. So, from O(a - r, r), moving horizontally left to x=0, which would be point (0, r) on DA, and moving vertically up to y=b, which would be point (a - r, b) on CD. Then, connecting these points: O(a - r, r) to (0, r) to D(0,b) to (a - r, b) and back to O. Wait, but this is not a rectangle. Because from (0, r) to D(0,b) is vertical, and from D(0,b) to (a - r, b) is horizontal. But then from (a - r, b) back to O(a - r, r) is vertical. Wait, no, that would form a rectangle. Wait, let me check:Points O(a - r, r), (0, r), D(0,b), and (a - r, b). Connecting these in order: O to (0, r) is horizontal left, (0, r) to D(0,b) is vertical up, D(0,b) to (a - r, b) is horizontal right, and (a - r, b) back to O(a - r, r) is vertical down. Yes, this forms a rectangle with sides parallel to the original rectangle. The length from O to (0, r) is (a - r) - 0 = a - r. The height from (0, r) to D(0,b) is b - r. Similarly, the sides of the rectangle are (a - r) and (b - r). Therefore, the area would be (a - r)(b - r). But wait, is this the rectangle intended in the problem?But the problem states that the other two vertices lie on the perimeter of the original rectangle. In this case, (0, r) is on DA (from D(0,b) to A(0,0)), and (a - r, b) is on CD (from C(a,b) to D(0,b)). So yes, both points are on the perimeter. Therefore, this would be a valid rectangle. However, the problem says "a rectangle with vertices O and D as opposite corners, and the other two vertices lying on the perimeter of the original rectangle." So, this seems to fit. Then, the area would be (a - r)(b - r). But we need to confirm if this is indeed the case.Alternatively, maybe the rectangle is not aligned with the axes. Let's consider that possibility. Suppose the rectangle is not aligned, so the sides are not horizontal or vertical. Then, the other two vertices would lie on different sides of the original rectangle. For example, one could be on AB and another on BC, but given the positions of O and D, it's unclear. But perhaps the rectangle must have its sides aligned because otherwise, it's complicated to have two vertices on the perimeter. Let me check.Alternatively, perhaps there's another configuration. Let's consider that the two other vertices are on sides AB and BC. But O is located near the center of triangle ABC, so maybe that's not possible. Alternatively, one on AD and one on DC, which is the case we considered before.Given that the problem says "the other two vertices lying on the perimeter of the original rectangle," it's possible that they can be on any of the sides. However, given the positions of O and D, the most straightforward rectangle would have the other two vertices on DA and DC, as we considered.But let's verify this. Suppose we have O(a - r, r) and D(0,b). If we connect O to D via two other points on the perimeter, then depending on the path, the rectangle could be formed. Let me think.Another approach: The rectangle with O and D as opposite vertices must have its sides such that the vectors from O to one vertex and from O to another vertex are perpendicular. Wait, perhaps parametrizing the coordinates.Let me denote the two other vertices as E and F. Since E and F are on the perimeter of ABCD, their coordinates must lie on the sides AB, BC, CD, or DA.Let's suppose E is on side DA (x=0, y between 0 and b) and F is on side DC (y=b, x between 0 and a). Then, the rectangle would be O, E, D, F. Let's check if this can form a rectangle.If E is on DA, say E(0, e), and F is on DC, say F(f, b), then the rectangle OEFD should satisfy that OE is perpendicular to ED, and so on. Wait, no. In a rectangle, adjacent sides are perpendicular. So, the vectors OE and OF should be perpendicular.Wait, actually, in the rectangle OEFD, the sides OE and ED should be adjacent and perpendicular. Wait, maybe I need to think in terms of coordinates.Alternatively, since O and D are opposite corners, the other two vertices E and F must satisfy that E is adjacent to O and D, and F is adjacent to O and D. Wait, this is getting confusing.Alternatively, using vectors: the vector from O to E and the vector from E to D should be perpendicular. Similarly, the vector from D to F and from F to O should be perpendicular. Hmm, this might not be the simplest approach.Wait, perhaps it's better to use coordinate geometry. Let me assign coordinates as before:A(0,0), B(a,0), C(a,b), D(0,b). O is at (a - r, r).We need to find points E and F on the perimeter of ABCD such that OEFD is a rectangle.Since O and D are opposite corners, the midpoint of the diagonal OD should be the same as the midpoint of the diagonal EF. The midpoint of OD is ((a - r + 0)/2, (r + b)/2) = ((a - r)/2, (r + b)/2). Therefore, the midpoint of EF must also be this point. So, if E is (x1, y1) and F is (x2, y2), then (x1 + x2)/2 = (a - r)/2 and (y1 + y2)/2 = (r + b)/2. Therefore, x1 + x2 = a - r and y1 + y2 = r + b.Additionally, since E and F are on the perimeter of ABCD, their coordinates must satisfy the conditions of being on one of the sides:- Side AB: y=0, 0 ≤ x ≤ a- Side BC: x=a, 0 ≤ y ≤ b- Side CD: y=b, 0 ≤ x ≤ a- Side DA: x=0, 0 ≤ y ≤ bSo, possible positions for E and F:Case 1: E is on DA (x=0) and F is on CD (y=b).Then E is (0, y1) and F is (x2, b). Then from the midpoint equations:0 + x2 = a - r ⇒ x2 = a - ry1 + b = r + b ⇒ y1 = rTherefore, E is (0, r) and F is (a - r, b). Then, checking if these points form a rectangle with O and D.Points: O(a - r, r), E(0, r), D(0, b), F(a - r, b). Connecting these points: O to E is a horizontal line left to (0, r), E to D is vertical up to (0, b), D to F is horizontal right to (a - r, b), and F to O is vertical down to (a - r, r). This is indeed a rectangle with sides parallel to the original rectangle. The length is a - r (from x=0 to x=a - r) and the height is b - r (from y=r to y=b). Therefore, the area is (a - r)(b - r).Case 2: E and F on other sides.Suppose E is on AB (y=0) and F is on BC (x=a).Then E is (x1, 0) and F is (a, y2).From midpoint equations:x1 + a = a - r ⇒ x1 = -r (which is not possible since x1 must be between 0 and a). Therefore, this case is invalid.Case 3: E on AB (y=0) and F on CD (y=b).E is (x1, 0), F is (x2, b).Midpoint equations:x1 + x2 = a - r0 + b = r + b ⇒ 0 = r, which is not possible unless r=0, which is not the case. Hence, invalid.Case 4: E on BC (x=a) and F on DA (x=0).E is (a, y1), F is (0, y2).Midpoint equations:a + 0 = a - r ⇒ 0 = -r ⇒ invalid.Similarly, other combinations can be checked, but likely only Case 1 is valid. Therefore, the rectangle is the one with vertices at O(a - r, r), E(0, r), D(0, b), and F(a - r, b). Therefore, area is (a - r)(b - r).Now, we need to compute this area. Recall that a*b = 1, and r = (a + b - sqrt(a² + b²))/2.Let me express (a - r)(b - r) in terms of a and b.First, expand the product:(a - r)(b - r) = ab - a r - b r + r²We know that ab = 1.Compute the other terms:a r + b r = r(a + b)r² = [ (a + b - sqrt(a² + b²)) / 2 ]²Therefore,(a - r)(b - r) = 1 - r(a + b) + r²But we can substitute r = (a + b - c)/2, where c = sqrt(a² + b²).Let me denote c = sqrt(a² + b²).Then,r = (a + b - c)/2So,(a - r)(b - r) = 1 - [ (a + b - c)/2 ]*(a + b) + [ (a + b - c)/2 ]²Let me compute each term step by step.First term: 1Second term: [ (a + b - c)/2 ]*(a + b) = (a + b)(a + b - c)/2Third term: [ (a + b - c)/2 ]² = (a + b - c)² / 4So,(a - r)(b - r) = 1 - ( (a + b)(a + b - c)/2 ) + ( (a + b - c)^2 ) / 4Let me combine these terms:Let me denote S = a + b, and c = sqrt(a² + b²). Then,(a - r)(b - r) = 1 - ( S(S - c)/2 ) + ( (S - c)^2 ) / 4Expand the terms:First, compute S(S - c)/2:= (S² - S c)/2Second, compute (S - c)^2 / 4:= (S² - 2 S c + c²)/4Therefore,(a - r)(b - r) = 1 - (S² - S c)/2 + (S² - 2 S c + c²)/4Combine the terms:Let's express all terms with denominator 4:= 1 - 2(S² - S c)/4 + (S² - 2 S c + c²)/4= 1 + [ -2(S² - S c) + (S² - 2 S c + c²) ] / 4Simplify the numerator:-2(S² - S c) + (S² - 2 S c + c²)= -2 S² + 2 S c + S² - 2 S c + c²= (-2 S² + S²) + (2 S c - 2 S c) + c²= (-S²) + 0 + c²= c² - S²Therefore,(a - r)(b - r) = 1 + (c² - S²)/4But c² = a² + b², and S = a + b, so S² = a² + 2ab + b²Therefore,c² - S² = (a² + b²) - (a² + 2ab + b²) = -2abThus,(a - r)(b - r) = 1 + (-2ab)/4 = 1 - (ab)/2But ab = 1, so:= 1 - 1/2 = 1/2Wait, this simplifies to 1/2. Therefore, the area of the new rectangle is 1/2.But wait, this is fascinating. Regardless of the values of a and b (as long as ab = 1), the area of the new rectangle is always 1/2. That's interesting. So the area is 1/2 square units.But let me verify this conclusion step by step to ensure there are no mistakes.We started with ab = 1.Expressed (a - r)(b - r) in terms of a and b, substituted r, expanded, and through algebraic manipulation found that the area simplifies to 1/2.Alternatively, is there a simpler way to see this?Alternatively, consider that in any rectangle with area 1, regardless of its aspect ratio, the area of this particular rectangle formed by the incenter and the opposite vertex is always 1/2. That seems counter-intuitive, but the mathematics shows it. Let me test with a specific case.Let me take a square, which is a special case of a rectangle where a = b. Since ab = 1, a = b = 1. Then, the original rectangle is a square with side length 1.In this case, triangle ABC is a right-angled triangle with legs of 1 and 1. The hypotenuse is sqrt(2). The inradius r = (1 + 1 - sqrt(2))/2 = (2 - sqrt(2))/2 ≈ (2 - 1.414)/2 ≈ 0.293.The coordinates of O would be (a - r, r) = (1 - (2 - sqrt(2))/2, (2 - sqrt(2))/2) = ( (2 - (2 - sqrt(2)))/2, (2 - sqrt(2))/2 ) = (sqrt(2)/2, (2 - sqrt(2))/2 )Point D is at (0,1).The other vertices of the new rectangle would be (0, r) and (a - r, 1) = (sqrt(2)/2, 1).So, the new rectangle has length sqrt(2)/2 - 0 = sqrt(2)/2 and height 1 - r = 1 - (2 - sqrt(2))/2 = (2 - (2 - sqrt(2)))/2 = sqrt(2)/2.Therefore, the area is (sqrt(2)/2)*(sqrt(2)/2) = (2)/4 = 1/2. Which matches the previous result.So in the case of a square, the area is indeed 1/2. Therefore, this provides confirmation.Another test case: let a = 2, b = 1/2, so that ab = 1.Then, triangle ABC has legs a=2, b=1/2, hypotenuse c = sqrt(4 + 0.25) = sqrt(4.25) = sqrt(17)/2 ≈ 2.0616.Inradius r = (2 + 0.5 - sqrt(17)/2)/2 = (2.5 - 2.0616/2)/2. Wait, let's compute it properly:r = (a + b - c)/2 = (2 + 0.5 - sqrt(17)/2)/2 = (2.5 - (sqrt(17)/2))/2 = (5/2 - sqrt(17)/2)/2 = (5 - sqrt(17))/4 ≈ (5 - 4.123)/4 ≈ 0.877/4 ≈ 0.219.Then, (a - r)(b - r) = (2 - 0.219)(0.5 - 0.219) ≈ (1.781)(0.281) ≈ 0.502, which is approximately 0.5. Given rounding errors, this is close to 1/2. Therefore, this also supports the conclusion.Therefore, regardless of the dimensions of the original rectangle (as long as ab=1), the area of the new rectangle is 1/2. Therefore, the answer is 1/2.Final AnswerThe area of the new rectangle is boxed{dfrac{1}{2}}.

✅:Given a rectangle (ABCD) with an area of 1 square unit, we need to find the area of a new rectangle with vertices (O) (the incenter of triangle (ABC)) and (D) as opposite corners, and the other two vertices lying on the perimeter of the original rectangle.1. Coordinates and Dimensions: - Assign coordinates: (A(0,0)), (B(a,0)), (C(a,b)), (D(0,b)). - The area of the rectangle is (ab = 1).2. Incenter Calculation: - Triangle (ABC) is a right-angled triangle with legs (AB = a) and (BC = b). - The hypotenuse (AC = sqrt{a^2 + b^2}). - The inradius (r) of triangle (ABC) is given by: [ r = frac{a + b - sqrt{a^2 + b^2}}{2} ] - The coordinates of the incenter (O) are ((a - r, r)).3. New Rectangle Coordinates: - The new rectangle has vertices (O(a - r, r)) and (D(0, b)) with the other two vertices on the perimeter of the original rectangle. - The other two vertices are found to be ((0, r)) and ((a - r, b)) on the sides (DA) and (CD) respectively.4. Area Calculation: - The length and height of the new rectangle are ((a - r)) and ((b - r)). - The area of the new rectangle is: [ (a - r)(b - r) ] - Substituting (r = frac{a + b - sqrt{a^2 + b^2}}{2}) and simplifying using algebraic manipulation and properties of the rectangle, the area simplifies to: [ frac{1}{2} ]Thus, the area of the new rectangle is (boxed{dfrac{1}{2}}).

🔑:To solve the given problem, we need to determine the area of a new rectangle where O and D are opposite vertices. Here’s the step-by-step solution:1. Understanding the Elements: - The given rectangle ABCD has an area of 1. - A, B, and C are vertices of ABC which form a triangle whose incircle’s center is O. - We need to find the area of the rectangle formed by O and D as opposite vertices, with the other two vertices on the perimeter of the original rectangle.2. Analyzing the Incircle: The incircle of triangle ABC is tangent to sides AB, BC, and CA at D_{1}, D_{2}, and D_{3} respectively. Let’s assume the sides of rectangle ABCD are a and b such that: [ ab = 1 ]3. Area of triangle ABC: The area K of triangle ABC will also be: [ K = frac{1}{2} cdot a cdot b = frac{1}{2} cdot 1 = frac{1}{2} ]4. Using Properties of Right Triangles and Incircle: The incenter O of triangle ABC is also the centroid, splitting the original triangle into smaller triangles of equal area summing up to K = frac{1}{2}.5. Forming the New Rectangle: Consider the new rectangle ODEF with O (in the triangle, it’s the incenter), D as diagonal vertices: - By symmetry and area distribution, cutting through line AC, it breaks down the required region symmetry into a similar structure.6. Area of the New Rectangle: As the newly formed rectangle ODEF will be half of the initial rectangle, its area text{Area}_{new} will be: [ text{Area}_{new} = frac{1}{2} cdot text{Area}_{ABCD} = frac{1}{2} cdot 1 = frac{1}{2} ]7. Conclusion: The area of the new rectangle formed by the incenter O and vertex D as opposite vertices within the original unit rectangle ABCD is: [ boxed{frac{1}{2}} ]