language:en license:mit size_categories:10K<n<100K format:parquet modality:text library:datasets library:pandas library:mlcroissant library:polars region:us
---
language: en
license: mit
tag: curator
---

<a href="https://github.com/bespokelabsai/curator/">
 <img src="https://huggingface.co/datasets/bespokelabs/Bespoke-Stratos-17k/resolve/main/made_with_curator.png" alt="Made with Curator" width=200px>
</a>

## Dataset card for numina_math_1_5_extra_r1

This dataset was made with [Curator](https://github.com/bespokelabsai/curator/).

## Dataset details

A sample from the dataset:

```python
{
    "problem": "A triangle with sides $a, b$, and $c$ is given. Denote by $s$ the semiperimeter, that is $s=(a+b+c) / 2$. Construct a triangle with sides $s-a, s-b$, and $s-c$. This process is repeated until a triangle can no longer be constructed with the sidelengths given.\nFor which original triangles can this process be repeated indefinitely?\nAnswer: Only equilateral triangles.",
    "solution": "The perimeter of each new triangle constructed by the process is $(s-a)+(s-b)+(s-c) = 3s - (a+b+c) = 3s - 2s = s$, that is, it is halved. Consider a new equivalent process in which a similar triangle with side lengths $2(s-a), 2(s-b), 2(s-c)$ is constructed, so the perimeter is kept invariant.\nSuppose without loss of generality that $a \\leq b \\leq c$. Then $2(s-c) \\leq 2(s-b) \\leq 2(s-a)$, and the difference between the largest side and the smallest side changes from $c-a$ to $2(s-a) - 2(s-c) = 2(c-a)$, that is, it doubles. Therefore, if $c-a > 0$ then eventually this difference becomes larger than $a+b+c$, and it's immediate that a triangle cannot be constructed with the side lengths. Hence the only possibility is $c-a = 0 \\Longrightarrow a = b = c$, and it is clear that equilateral triangles can yield an infinite process, because all generated triangles are equilateral.",
    "answer": "a = b = c",
    "problem_type": "Geometry",
    "question_type": "math-word-problem",
    "problem_is_valid": "Yes",
    "solution_is_valid": "Yes",
    "source": "olympiads_ref",
    "synthetic": false,
    "r1_reasoning": "Okay, so I need to figure out for which original triangles the process of constructing a new triangle with sides equal to the semiperimeter minus each original side can be repeated indefinitely. The answer is supposed to be only equilateral triangles, but I need to understand why. Let me start by breaking down the problem.\n\nFirst, let's recall that the semiperimeter \\( s \\) of a triangle with sides \\( a, b, c \\) is \\( s = \\frac{a + b + c}{2} \\). Then, the sides of the next triangle in the process are \\( s - a \\), \\( s - b \\), and \\( s - c \\). The process is repeated until we can't form a triangle anymore. The question is asking for which original triangles this process never stops, i.e., we can keep constructing triangles indefinitely. The answer is that only equilateral triangles have this property. Let me try to verify this.\n\nFirst, let's consider an equilateral triangle. Suppose the original triangle has sides \\( a = b = c \\). Then the semiperimeter \\( s = \\frac{3a}{2} \\). The sides of the next triangle would be \\( s - a = \\frac{3a}{2} - a = \\frac{a}{2} \\), and similarly for the other sides. So the new triangle is also equilateral with sides \\( \\frac{a}{2} \\). Repeating the process, we would get triangles with sides \\( \\frac{a}{4} \\), \\( \\frac{a}{8} \\), etc., which are all valid triangles. So this process can indeed continue indefinitely for an equilateral triangle. That makes sense.\n\nNow, the key part is to show that no other triangles besides equilateral ones can have this indefinite process. Let's suppose we have a non-equilateral triangle. Then, we need to see why the process must eventually terminate. Let's take an example. Suppose we have an isosceles triangle, say with sides \\( 2, 2, 3 \\). Let's compute the semiperimeter: \\( s = \\frac{2 + 2 + 3}{2} = \\frac{7}{2} = 3.5 \\). Then the next sides would be \\( 3.5 - 2 = 1.5 \\), \\( 3.5 - 2 = 1.5 \\), and \\( 3.5 - 3 = 0.5 \\). So the new triangle has sides \\( 1.5, 1.5, 0.5 \\). Let's check if this is a valid triangle. The sum of the two smaller sides must be greater than the largest side. Here, \\( 0.5 + 1.5 = 2.0 > 1.5 \\), so it is valid. Now, let's do the next iteration. The semiperimeter \\( s' = \\frac{1.5 + 1.5 + 0.5}{2} = \\frac{3.5}{2} = 1.75 \\). The new sides are \\( 1.75 - 1.5 = 0.25 \\), \\( 1.75 - 1.5 = 0.25 \\), and \\( 1.75 - 0.5 = 1.25 \\). So the next triangle has sides \\( 0.25, 0.25, 1.25 \\). Checking validity: \\( 0.25 + 0.25 = 0.5 \\), which is not greater than \\( 1.25 \\). Hence, this is not a valid triangle. So the process stops here. So for this isosceles triangle, the process stops after two iterations. Therefore, non-equilateral triangles may terminate quickly.\n\nBut this is just an example. I need to generalize this. Let's consider a general triangle with sides \\( a, b, c \\). Let's denote the sides after one iteration as \\( a' = s - a \\), \\( b' = s - b \\), \\( c' = s - c \\). Let's compute these:\n\n\\( a' = \\frac{a + b + c}{2} - a = \\frac{-a + b + c}{2} \\)\n\nSimilarly,\n\n\\( b' = \\frac{a - b + c}{2} \\)\n\n\\( c' = \\frac{a + b - c}{2} \\)\n\nSo the new sides are linear combinations of the original sides. Let's note that each new side is half the sum of the other two original sides minus half the original side. Hmm.\n\nAlternatively, since \\( a' = \\frac{b + c - a}{2} \\), and similarly for the others. So each new side is the average of the sum of the other two sides minus the original side. Wait, that might not be the clearest way to think about it.\n\nAlternatively, perhaps we can express this transformation in terms of linear algebra. Let me consider the vector of sides \\( (a, b, c) \\) and the transformation \\( T \\) that maps it to \\( (s - a, s - b, s - c) \\). Then, since \\( s = \\frac{a + b + c}{2} \\), we can write:\n\n\\( T(a, b, c) = \\left( \\frac{-a + b + c}{2}, \\frac{a - b + c}{2}, \\frac{a + b - c}{2} \\right) \\)\n\nThis is a linear transformation. Let me represent this as a matrix. Let's see:\n\nThe transformation can be written as:\n\n\\[\n\\begin{pmatrix}\na' \\\\\nb' \\\\\nc'\n\\end{pmatrix}\n= \\frac{1}{2}\n\\begin{pmatrix}\n-1 & 1 & 1 \\\\\n1 & -1 & 1 \\\\\n1 & 1 & -1\n\\end{pmatrix}\n\\begin{pmatrix}\na \\\\\nb \\\\\nc\n\\end{pmatrix}\n\\]\n\nSo the matrix is:\n\n\\[\nM = \\frac{1}{2}\n\\begin{pmatrix}\n-1 & 1 & 1 \\\\\n1 & -1 & 1 \\\\\n1 & 1 & -1\n\\end{pmatrix}\n\\]\n\nInteresting. So each iteration is a multiplication by this matrix \\( M \\). To understand the behavior over multiple iterations, we might need to analyze the eigenvalues or eigenvectors of this matrix. If the original triangle is an eigenvector of this matrix, then the process would scale the sides by the corresponding eigenvalue each time. Let's check if the equilateral triangle is an eigenvector.\n\nSuppose \\( a = b = c \\). Let's denote this common value as \\( x \\). Then, applying the transformation:\n\n\\( a' = \\frac{-x + x + x}{2} = \\frac{x}{2} \\)\n\nSimilarly, \\( b' = \\frac{x}{2} \\), \\( c' = \\frac{x}{2} \\). So the new triangle is also equilateral with sides \\( \\frac{x}{2} \\). Therefore, the vector \\( (x, x, x) \\) is an eigenvector of \\( M \\) with eigenvalue \\( \\frac{1}{2} \\). So each iteration scales the sides by \\( \\frac{1}{2} \\), which is why the process can continue indefinitely, as the sides never become zero, just getting smaller and smaller.\n\nFor non-equilateral triangles, we need to see if repeated application of \\( M \\) leads to a situation where the triangle inequality is violated. Let's consider another example. Take a right-angled triangle, say with sides \\( 3, 4, 5 \\). Let's compute the next sides.\n\nSemiperimeter \\( s = \\frac{3 + 4 + 5}{2} = 6 \\). Then the new sides are \\( 6 - 3 = 3 \\), \\( 6 - 4 = 2 \\), \\( 6 - 5 = 1 \\). So the new triangle has sides \\( 1, 2, 3 \\). Wait, but \\( 1 + 2 = 3 \\), which is not greater than 3. So this is degenerate. Hence, the process stops after one iteration. So even a right-angled triangle stops immediately. Hmm, so maybe only triangles that are \"close enough\" to equilateral can go on for a few steps, but eventually, they fail.\n\nWait, but in the previous isosceles example, it took two iterations. Let me check another triangle. Suppose we take a triangle with sides \\( 2, 3, 4 \\). Let's compute the next sides.\n\nSemiperimeter \\( s = (2 + 3 + 4)/2 = 4.5 \\). New sides: \\( 4.5 - 2 = 2.5 \\), \\( 4.5 - 3 = 1.5 \\), \\( 4.5 - 4 = 0.5 \\). So sides \\( 0.5, 1.5, 2.5 \\). Check triangle inequality: \\( 0.5 + 1.5 = 2.0 \\), which is less than 2.5. So invalid. So process stops here. So even a scalene triangle with sides 2,3,4 stops after one iteration.\n\nWait, but maybe there are some non-equilateral triangles that can go on for a few more steps? Let me try a different triangle. Let's take sides \\( 1, 1, 1 \\) (equilateral). Then, next sides are \\( 0.5, 0.5, 0.5 \\), and so on. So it continues. If I take a triangle that's almost equilateral, say sides \\( 1, 1, 1.1 \\). Let's compute the next sides.\n\nSemiperimeter \\( s = (1 + 1 + 1.1)/2 = 1.55 \\). New sides: \\( 1.55 - 1 = 0.55 \\), \\( 1.55 - 1 = 0.55 \\), \\( 1.55 - 1.1 = 0.45 \\). So sides \\( 0.45, 0.55, 0.55 \\). Check triangle inequality: \\( 0.45 + 0.55 = 1.0 > 0.55 \\), so valid. Next iteration: semiperimeter \\( s' = (0.45 + 0.55 + 0.55)/2 = 0.775 \\). New sides: \\( 0.775 - 0.45 = 0.325 \\), \\( 0.775 - 0.55 = 0.225 \\), \\( 0.775 - 0.55 = 0.225 \\). So sides \\( 0.225, 0.225, 0.325 \\). Check triangle inequality: \\( 0.225 + 0.225 = 0.45 > 0.325 \\). Valid. Next iteration: semiperimeter \\( s'' = (0.225 + 0.225 + 0.325)/2 = 0.3875 \\). New sides: \\( 0.3875 - 0.225 = 0.1625 \\), \\( 0.3875 - 0.225 = 0.1625 \\), \\( 0.3875 - 0.325 = 0.0625 \\). So sides \\( 0.0625, 0.1625, 0.1625 \\). Check triangle inequality: \\( 0.0625 + 0.1625 = 0.225 > 0.1625 \\). Valid. Next iteration: semiperimeter \\( s''' = (0.0625 + 0.1625 + 0.1625)/2 = 0.19375 \\). New sides: \\( 0.19375 - 0.0625 = 0.13125 \\), \\( 0.19375 - 0.1625 = 0.03125 \\), \\( 0.19375 - 0.1625 = 0.03125 \\). So sides \\( 0.03125, 0.03125, 0.13125 \\). Check triangle inequality: \\( 0.03125 + 0.03125 = 0.0625 \\), which is less than 0.13125. Invalid. So the process stops here. So even a triangle very close to equilateral (sides 1,1,1.1) eventually terminates after four iterations. So it seems that unless the triangle is exactly equilateral, the process will terminate after some steps.\n\nTherefore, the key idea is that only equilateral triangles are invariant under this transformation, in the sense that they remain equilateral after each iteration, just scaled down by a factor of 1/2 each time. For any other triangle, the transformation introduces some asymmetry, which eventually leads to a violation of the triangle inequality.\n\nTo formalize this, perhaps we can analyze the transformation matrix \\( M \\). Since the equilateral triangle is an eigenvector with eigenvalue 1/2, other eigenvectors might correspond to other triangles, but if they are not eigenvectors, repeated application of \\( M \\) would lead to a combination of eigenvectors. However, since the other eigenvalues might have magnitudes less than 1/2, causing the sides to decay to zero, but the problem is that even if they decay, the ratios between the sides might change in such a way that the triangle inequality is violated.\n\nAlternatively, perhaps we can consider the ratios of the sides. Let's suppose that the original triangle is not equilateral. Then, the ratios of the sides will change with each iteration. Let's see.\n\nSuppose we have a triangle with sides \\( a, b, c \\), not all equal. Let's compute the ratios of the new sides \\( a', b', c' \\). For simplicity, let's assume \\( a \\geq b \\geq c \\). Then, \\( a' = \\frac{-a + b + c}{2} \\), \\( b' = \\frac{a - b + c}{2} \\), \\( c' = \\frac{a + b - c}{2} \\). Since \\( a \\geq b \\geq c \\), then \\( a' = \\frac{-a + b + c}{2} \\). Since \\( a \\) is the largest, \\( -a + b + c \\) is likely the smallest, so \\( a' \\) might be the smallest new side. Similarly, \\( c' = \\frac{a + b - c}{2} \\), which, since \\( a \\geq b \\geq c \\), \\( a + b - c \\) is the largest, so \\( c' \\) is the largest new side. Wait, let's check:\n\nOriginal sides: \\( a \\geq b \\geq c \\).\n\nCompute \\( a' = (-a + b + c)/2 \\). Since \\( a \\geq b \\geq c \\), then \\( -a + b + c \\leq -a + a + a = a \\), but actually, since \\( b + c \\leq 2a \\) (because \\( a \\geq b, a \\geq c \\)), so \\( -a + b + c \\leq a \\). But depending on the original triangle, this could be positive or not. Wait, but for a valid triangle, \\( b + c > a \\), so \\( -a + b + c > 0 \\). So \\( a' \\) is positive. Similarly, \\( b' = (a - b + c)/2 \\). Since \\( a \\geq b \\), \\( a - b \\geq 0 \\), and \\( c > 0 \\), so \\( b' > 0 \\). \\( c' = (a + b - c)/2 \\). Since \\( a + b > c \\) (triangle inequality), so \\( c' > 0 \\).\n\nBut the ordering of the new sides: Let's see. Suppose \\( a \\geq b \\geq c \\). Then:\n\n\\( c' = (a + b - c)/2 \\). Since \\( a \\geq b \\geq c \\), \\( a + b - c \\geq a \\geq b \\geq c \\). So \\( c' \\) is the largest new side.\n\n\\( b' = (a - b + c)/2 \\). Let's compare \\( b' \\) and \\( a' \\). \\( a' = (-a + b + c)/2 \\), \\( b' = (a - b + c)/2 \\). Let's subtract them: \\( b' - a' = (a - b + c)/2 - (-a + b + c)/2 = (a - b + c + a - b - c)/2 = (2a - 2b)/2 = a - b \\geq 0 \\), since \\( a \\geq b \\). Therefore, \\( b' \\geq a' \\).\n\nSimilarly, compare \\( b' \\) and \\( c' \\). \\( c' = (a + b - c)/2 \\), \\( b' = (a - b + c)/2 \\). Let's subtract: \\( c' - b' = (a + b - c - a + b - c)/2 = (2b - 2c)/2 = b - c \\geq 0 \\), since \\( b \\geq c \\). Therefore, \\( c' \\geq b' \\geq a' \\). So the new sides are ordered \\( c' \\geq b' \\geq a' \\).\n\nTherefore, the largest side in the new triangle is \\( c' = \\frac{a + b - c}{2} \\), and the smallest is \\( a' = \\frac{-a + b + c}{2} \\).\n\nSo, in each iteration, the sides are transformed, and the ordering of the sides changes. The new largest side is \\( \\frac{a + b - c}{2} \\), and the new smallest side is \\( \\frac{-a + b + c}{2} \\).\n\nTo check when the process can be continued indefinitely, we need that after each iteration, the triangle inequalities are satisfied. For the new triangle with sides \\( a', b', c' \\), we need:\n\n1. \\( a' + b' > c' \\)\n2. \\( a' + c' > b' \\)\n3. \\( b' + c' > a' \\)\n\nBut since \\( c' \\geq b' \\geq a' \\), the most restrictive condition is \\( a' + b' > c' \\).\n\nLet's compute \\( a' + b' - c' \\):\n\n\\( a' + b' - c' = \\frac{-a + b + c}{2} + \\frac{a - b + c}{2} - \\frac{a + b - c}{2} \\)\n\nSimplify:\n\nFirst, combine \\( a' + b' \\):\n\n\\( \\frac{(-a + b + c) + (a - b + c)}{2} = \\frac{2c}{2} = c \\)\n\nThen subtract \\( c' \\):\n\n\\( c - \\frac{a + b - c}{2} = \\frac{2c - a - b + c}{2} = \\frac{3c - a - b}{2} \\)\n\nTherefore, \\( a' + b' - c' = \\frac{3c - a - b}{2} \\)\n\nFor the triangle inequality \\( a' + b' > c' \\), we need \\( \\frac{3c - a - b}{2} > 0 \\), which implies \\( 3c > a + b \\).\n\nBut in the original triangle, since \\( a \\geq b \\geq c \\), by triangle inequality, we have \\( a < b + c \\). So \\( a + b < b + c + b = 2b + c \\). But \\( 3c > a + b \\) would require \\( 3c > a + b \\). However, since \\( a \\geq b \\geq c \\), \\( a + b \\geq 2b \\geq 2c \\). Therefore, \\( 3c > a + b \\geq 2c \\), so \\( 3c > 2c \\), which is true, but \\( 3c > a + b \\) is not necessarily true. For example, if \\( a = b = c \\), then \\( 3c = 3a = a + b + c = 3a \\), so \\( 3c = a + b \\), so \\( 3c - a - b = 0 \\), so \\( a' + b' = c' \\). Wait, but in the equilateral case, all sides are equal, so \\( a' = b' = c' \\), so \\( a' + b' = 2c' > c' \\), which is true. Wait, maybe my calculation was wrong.\n\nWait, let's recast. For the equilateral triangle, \\( a = b = c \\). Then \\( a' = b' = c' = \\frac{-a + a + a}{2} = \\frac{a}{2} \\). So \\( a' + b' = \\frac{a}{2} + \\frac{a}{2} = a \\), and \\( c' = \\frac{a}{2} \\). Wait, but \\( a = \\frac{a}{2} \\times 2 \\), so \\( a' + b' = a = 2c' \\), so \\( a' + b' = 2c' > c' \\), which is true. So in the equilateral case, the triangle inequality holds as equality only when sides are scaled down, but since all sides are equal, the sum of any two sides is twice the third, which is greater than the third. Wait, no. If all sides are equal, then the sum of any two sides is twice the third, which is indeed greater than the third. So the triangle inequality holds.\n\nBut in the general case, when we have \\( a \\geq b \\geq c \\), the condition \\( a' + b' > c' \\) becomes \\( \\frac{3c - a - b}{2} > 0 \\), so \\( 3c > a + b \\). However, in the original triangle, we have \\( a < b + c \\). So \\( a + b < b + c + b = 2b + c \\). Therefore, \\( 3c > a + b \\) would require \\( 3c > 2b + c \\), which simplifies to \\( 2c > 2b \\), or \\( c > b \\). But in our ordering, \\( a \\geq b \\geq c \\), so \\( c \\leq b \\). Therefore, \\( 2c \\leq 2b \\), so \\( 3c \\leq 2b + c \\leq 2b + b = 3b \\). But \\( a + b \\geq b + c \\geq 2b \\), so \\( 3c \\leq 3b \\leq a + b + b \\). Wait, this is getting convoluted.\n\nWait, let's take the original triangle with sides \\( a \\geq b \\geq c \\). Then, the triangle inequality tells us \\( a < b + c \\). So \\( a + b < 2b + c \\). For \\( 3c > a + b \\), substituting \\( a < b + c \\), we get \\( 3c > a + b < 3c > (b + c) + b = 2b + c \\), so \\( 3c > 2b + c \\implies 2c > 2b \\implies c > b \\). But since \\( b \\geq c \\), this is impossible unless \\( b = c \\). Therefore, \\( 3c > a + b \\) can only hold if \\( b = c \\) and \\( 3c > a + c \\implies 2c > a \\). But since \\( a \\geq b = c \\), \\( a \\geq c \\), so \\( 2c > a \\geq c \\implies c < a < 2c \\). So if \\( b = c \\) and \\( a \\) is between \\( c \\) and \\( 2c \\), then \\( 3c > a + c \\implies 2c > a \\), which is possible.\n\nFor example, take \\( a = 1.5 \\), \\( b = c = 1 \\). Then, check \\( 3c = 3 \\times 1 = 3 \\), \\( a + b = 1.5 + 1 = 2.5 \\). So \\( 3 > 2.5 \\), so \\( 3c > a + b \\). Therefore, in this case, \\( a' + b' > c' \\). Let's check the next iteration.\n\nOriginal sides: \\( 1.5, 1, 1 \\). Semiperimeter \\( s = (1.5 + 1 + 1)/2 = 1.75 \\). New sides: \\( 1.75 - 1.5 = 0.25 \\), \\( 1.75 - 1 = 0.75 \\), \\( 1.75 - 1 = 0.75 \\). So new sides \\( 0.25, 0.75, 0.75 \\). Check triangle inequality: \\( 0.25 + 0.75 = 1.0 > 0.75 \\), valid. Next iteration: semiperimeter \\( s' = (0.25 + 0.75 + 0.75)/2 = 0.875 \\). New sides: \\( 0.875 - 0.25 = 0.625 \\), \\( 0.875 - 0.75 = 0.125 \\), \\( 0.875 - 0.75 = 0.125 \\). So sides \\( 0.125, 0.125, 0.625 \\). Check triangle inequality: \\( 0.125 + 0.125 = 0.25 \\), which is less than \\( 0.625 \\). Invalid. So even though the first iteration worked, the second iteration fails. So even if \\( 3c > a + b \\), the process might still terminate in the next step.\n\nTherefore, the key point is that unless the triangle is equilateral, each iteration introduces some discrepancy in the side lengths, which eventually leads to a violation of the triangle inequality. In the equilateral case, the sides remain in proportion, so the triangle inequalities continue to hold.\n\nAnother approach might be to consider the angles of the triangle. In an equilateral triangle, all angles are 60 degrees. If we start with a different triangle, the angles change with each iteration. Perhaps the angles tend to 60 degrees if the triangle is approaching equilateral, but due to some instability, any deviation from equilateral leads to a breakdown.\n\nAlternatively, let's consider the sequence of triangles generated by this process. For a non-equilateral triangle, each subsequent triangle becomes more \"flat\" or elongated, until one side becomes too long compared to the sum of the other two. Alternatively, maybe the sides decrease in some pattern that causes the ratios to approach a limit, but if that limit isn't 1:1:1, then eventually the triangle inequality fails.\n\nWait, let's think about the ratios of the sides. Suppose we have a triangle with sides \\( a, b, c \\), not all equal. After one iteration, the sides are \\( a' = \\frac{-a + b + c}{2} \\), \\( b' = \\frac{a - b + c}{2} \\), \\( c' = \\frac{a + b - c}{2} \\). Let's compute the ratios \\( \\frac{a'}{b'} \\), \\( \\frac{b'}{c'} \\), etc., to see if they approach 1.\n\nFor example, take the isosceles triangle \\( 2, 2, 3 \\). After one iteration, sides are \\( 1.5, 1.5, 0.5 \\). The ratios are \\( 1.5/1.5 = 1 \\), \\( 1.5/0.5 = 3 \\). Then next iteration, sides \\( 0.25, 0.25, 1.25 \\). Ratios \\( 1 \\), \\( 5 \\). So the ratios are diverging, not converging to 1. Hence, the process is making the triangle more unequal, leading to a violation.\n\nAlternatively, take a triangle that's almost equilateral, say \\( 1, 1, 1 + \\epsilon \\), where \\( \\epsilon \\) is small. Let's compute the next sides.\n\nSemiperimeter \\( s = \\frac{1 + 1 + 1 + \\epsilon}{2} = \\frac{3 + \\epsilon}{2} \\).\n\nNew sides:\n\n\\( a' = s - a = \\frac{3 + \\epsilon}{2} - 1 = \\frac{1 + \\epsilon}{2} \\)\n\n\\( b' = s - b = \\frac{1 + \\epsilon}{2} \\)\n\n\\( c' = s - c = \\frac{3 + \\epsilon}{2} - (1 + \\epsilon) = \\frac{3 + \\epsilon - 2 - 2\\epsilon}{2} = \\frac{1 - \\epsilon}{2} \\)\n\nSo the new triangle has sides \\( \\frac{1 + \\epsilon}{2}, \\frac{1 + \\epsilon}{2}, \\frac{1 - \\epsilon}{2} \\). The ratio of the sides is \\( 1 + \\epsilon : 1 + \\epsilon : 1 - \\epsilon \\). So the two equal sides are now \\( \\frac{1 + \\epsilon}{2} \\), and the third side is \\( \\frac{1 - \\epsilon}{2} \\). The difference between the equal sides and the third side is \\( \\frac{1 + \\epsilon}{2} - \\frac{1 - \\epsilon}{2} = \\epsilon \\). So the difference remains \\( \\epsilon \\), but scaled down by 1/2. Wait, no. The original difference was \\( (1 + \\epsilon) - 1 = \\epsilon \\). The new difference is \\( \\frac{1 + \\epsilon}{2} - \\frac{1 - \\epsilon}{2} = \\frac{2\\epsilon}{2} = \\epsilon \\). So the absolute difference remains the same, but the sides are scaled down by 1/2. Therefore, the relative difference (i.e., the ratio of the difference to the side lengths) increases. For example, original sides are approximately 1, 1, 1 + \u03b5. After one iteration, sides are approximately 0.5, 0.5, 0.5 - 0.5\u03b5. The difference between the third side and the others is 0.5\u03b5, but the sides themselves are 0.5, so the relative difference is \u03b5, which is the same as before. Wait, but in absolute terms, the difference is halved, but relative to the side lengths, it's the same. Hmm.\n\nWait, let's compute the relative difference. Original triangle: sides 1, 1, 1 + \u03b5. The third side is 1 + \u03b5, which is \u03b5 longer than the others. The relative difference is \u03b5 / 1 = \u03b5.\n\nAfter one iteration: sides \\( \\frac{1 + \\epsilon}{2}, \\frac{1 + \\epsilon}{2}, \\frac{1 - \\epsilon}{2} \\). The third side is \\( \\frac{1 - \\epsilon}{2} \\), which is \\( \\frac{1 + \\epsilon}{2} - \\frac{1 - \\epsilon}{2} = \\epsilon \\) less than the other sides. The relative difference is \\( \\epsilon / \\frac{1 + \\epsilon}{2} \\approx 2\\epsilon \\) (if \u03b5 is small). So the relative difference doubles. Therefore, each iteration approximately doubles the relative difference while scaling down the sides by half. Therefore, even a small \u03b5 will eventually cause the relative difference to become large enough that the triangle inequality is violated.\n\nFor example, starting with \u03b5 = 0.1:\n\nFirst iteration: relative difference \u2248 0.2\n\nSecond iteration: relative difference \u2248 0.4\n\nThird iteration: relative difference \u2248 0.8\n\nFourth iteration: relative difference \u2248 1.6, which would exceed 1, making the triangle inequality fail.\n\nThis suggests that any initial deviation from equilateral will be amplified in relative terms with each iteration, leading to eventual violation of the triangle inequality. Hence, only the equilateral triangle, with \u03b5 = 0, avoids this amplification and can continue indefinitely.\n\nTo formalize this, suppose we have a triangle with sides \\( a, b, c \\), not all equal. Let\u2019s define a measure of deviation from equilateral, such as the ratio of the largest to smallest side, or the difference between the largest and smallest sides. Through each iteration, we can analyze how this measure changes.\n\nAlternatively, consider the transformation in terms of the differences between the sides. Let\u2019s define \\( d_1 = a - b \\), \\( d_2 = b - c \\), assuming \\( a \\geq b \\geq c \\). Then, after one iteration, the new sides are \\( a' = \\frac{-a + b + c}{2} \\), \\( b' = \\frac{a - b + c}{2} \\), \\( c' = \\frac{a + b - c}{2} \\). Let\u2019s compute the new differences.\n\nCompute \\( a' - b' = \\frac{-a + b + c}{2} - \\frac{a - b + c}{2} = \\frac{-2a + 2b}{2} = -a + b = -d_1 \\)\n\nSimilarly, \\( b' - c' = \\frac{a - b + c}{2} - \\frac{a + b - c}{2} = \\frac{-2b + 2c}{2} = -b + c = -d_2 \\)\n\nTherefore, the differences \\( d_1' = |a' - b'| = | -d_1 | = d_1 \\), and \\( d_2' = |b' - c'| = | -d_2 | = d_2 \\). Wait, but the ordering of the sides might change. Wait, in our previous analysis, we saw that the new sides are ordered \\( c' \\geq b' \\geq a' \\). So if originally \\( a \\geq b \\geq c \\), then after transformation, \\( c' \\geq b' \\geq a' \\). Therefore, the new largest side is \\( c' \\), new middle is \\( b' \\), new smallest is \\( a' \\). Therefore, the differences would be \\( c' - b' \\) and \\( b' - a' \\).\n\nCompute \\( c' - b' = \\frac{a + b - c}{2} - \\frac{a - b + c}{2} = \\frac{2b - 2c}{2} = b - c = d_2 \\)\n\nSimilarly, \\( b' - a' = \\frac{a - b + c}{2} - \\frac{-a + b + c}{2} = \\frac{2a - 2b}{2} = a - b = d_1 \\)\n\nTherefore, the new differences are \\( d_2 \\) and \\( d_1 \\), but since the ordering has changed, the new largest difference is \\( d_1 \\), and the next is \\( d_2 \\). Wait, this seems like the differences are swapping places but maintaining their magnitudes. Therefore, the differences themselves are not diminishing; they are persisting or even amplifying when considering the scaling of the sides.\n\nSince each iteration scales all sides by 1/2 (approximately, but not exactly unless equilateral), but the differences between the sides remain the same in absolute terms. Therefore, as the sides get smaller, the relative differences increase, leading to eventual violation of the triangle inequality.\n\nFor example, if we start with differences \\( d_1 \\) and \\( d_2 \\), then after each iteration, the sides are halved, but the differences remain \\( d_1 \\) and \\( d_2 \\). Therefore, the relative differences \\( d_1 / \\text{side length} \\) and \\( d_2 / \\text{side length} \\) double each time. Hence, after \\( n \\) iterations, the relative differences are \\( 2^n \\times \\text{original relative differences} \\), which eventually exceed 1, causing the triangle inequality to fail.\n\nThis analysis shows that any initial difference, no matter how small, will eventually become large enough relative to the sides to violate the triangle inequality. Hence, only when the initial differences are zero (i.e., the triangle is equilateral) can the process continue indefinitely.\n\nAnother angle to consider is invariant properties. For the process to continue indefinitely, each subsequent triangle must satisfy the triangle inequality. The only way this can happen indefinitely is if the ratios of the sides remain fixed, which occurs only in the equilateral case. Any other ratio will change with each iteration, leading towards a degenerate triangle.\n\nAdditionally, we can consider the concept of similarity. For the process to repeat indefinitely, each subsequent triangle must be similar to the previous one. However, unless the original triangle is equilateral, the transformation changes the ratios of the sides, so the triangles are not similar. In the equilateral case, each subsequent triangle is similar with a scaling factor of 1/2.\n\nTo further solidify this, let's assume there exists a non-equilateral triangle for which the process can be repeated indefinitely. Then, the sequence of triangles would have to satisfy the triangle inequality at each step. However, as shown earlier, the relative differences in side lengths grow exponentially with each iteration, which would eventually violate the triangle inequality. Therefore, no such non-equilateral triangle can exist.\n\nIn conclusion, through various examples and analysis of the transformation's effect on side lengths, differences, and ratios, we see that only equilateral triangles maintain the necessary conditions for the process to continue indefinitely. Any deviation from equilateral results in increasing relative differences that ultimately prevent the formation of a valid triangle after a finite number of steps.\n\n**Final Answer**\n\\boxed{\\text{Only equilateral triangles}}",
    "r1_solution": "To determine for which original triangles the process of constructing a new triangle with sides equal to the semiperimeter minus each original side can be repeated indefinitely, we start by analyzing the transformation of the sides of the triangle.\n\nGiven a triangle with sides \\(a, b, c\\) and semiperimeter \\(s = \\frac{a + b + c}{2}\\), the sides of the next triangle are \\(s - a\\), \\(s - b\\), and \\(s - c\\). This transformation can be represented as a linear matrix operation. For an equilateral triangle, this transformation results in another equilateral triangle scaled by a factor of \\(\\frac{1}{2}\\), allowing the process to continue indefinitely.\n\nFor non-equilateral triangles, we analyze the effect of the transformation on the side lengths and the triangle inequality. The key steps are:\n\n1. **Transformation Analysis**: The sides of the new triangle are linear combinations of the original sides. For a non-equilateral triangle, the transformation introduces asymmetry, leading to changes in the side length ratios.\n2. **Triangle Inequality Check**: The most restrictive condition for the new triangle is the sum of the two smaller sides being greater than the largest side. For non-equilateral triangles, this condition eventually fails due to increasing relative differences in side lengths.\n3. **Relative Differences**: Any initial deviation from equilateral results in relative differences that grow exponentially with each iteration. This leads to a violation of the triangle inequality after a finite number of steps.\n4. **Invariance and Similarity**: Only equilateral triangles maintain fixed side ratios under the transformation, ensuring the process can continue indefinitely. Non-equilateral triangles do not maintain similarity, leading to eventual failure.\n\nThrough examples and analysis, we conclude that only equilateral triangles can sustain the process indefinitely. Any non-equilateral triangle will eventually violate the triangle inequality due to increasing relative differences in side lengths.\n\n\\[\n\\boxed{\\text{Only equilateral triangles}}\n\\]"
}
```

## Loading the dataset

You can load this dataset using the following code:

```python
from datasets import load_dataset

dataset = load_dataset("mlfoundations-dev/numina_math_1_5_extra_r1", split="default")
```

